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Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Chemical_Thermodynamics_(Supplement_to_Shepherd_et_al.)/04%3A_Fundamental_2_-_Counting_Configurations/4.04%3A_Applying_the_Laws_of_Probability | The laws of probability apply to events that are independent. If the result of one trial depends on the result of another trial, we may still be able to use the laws of probability. However, to do so, we must know the nature of the interdependence. If the activity associated with event C precedes the activity associated with event D , the probability of D may depend on whether C occurs. Suppose that the first activity is tossing a coin and that the second activity is drawing a card from a deck; however, the deck we use depends on whether the coin comes up heads or tails. If the coin is heads, we draw a card from an ordinary deck; if the coin is tails, we draw a coin from a deck with the face cards removed. Now we ask about the probability of drawing an ace. If the coin is heads, the probability of drawing an ace is \({4}/{52}={1}/{13}\). If the coin is tails, the probability of drawing an ace is \({4}/{40}={1}/{10}\). The combination coin is heads and card is ace has probability: \(\left({1}/{2}\right)\left({1}/{13}\right)={1}/{26}\). The combination coin is tails and card is ace has probability \(\left({1}/{2}\right)\left({1}/{10}\right)={1}/{20}\). In this case, the probability of drawing an ace depends on the modification we make to the deck based on the outcome of the coin toss. Applying the laws of probability is straightforward. An example that illustrates the application of these laws in a transparent way is provided by villages First, Second, Third, and Fourth, which are separated by rivers. (See Figure 1.) Bridges \(1\), \(2\), and \(3\) span the river between First and Second. Bridges \(a\) and \(b\) span the river between Second and Third. Bridges \(A\), \(B\), \(C\), and \(D\) span the river between Third and Fourth. A traveler from First to Fourth who is free to take any route he pleases has a choice from among \(3\times 2\times 4=24\) possible combinations. Let us consider the probabilities associated with various events: There are 24 possible routes. If a traveler chooses his route at random, the probability that he will take any particular route is \({1}/{24}\). This illustrates our assumption that each event in a set of \(N\) exhaustive and mutually exclusive events occurs with probability \({1}/{N}\). If he chooses a route at random, the probability that he goes from First to Second by either bridge \(1\) or bridge \(2\) is \(P\left(1\right)+P\left(2\right)=\ {1}/{3}+{1}/{3}={2}/{3}\). This illustrates the calculation of the probability of alternative events. The probability of the particular route \(2\to a\to C\) is \(P\left(2\right)\times P\left(a\right)\times P\left(C\right)=\left({1}/{3}\right)\left({1}/{2}\right)\left({1}/{4}\right)={1}/{24}\), and we calculate the same probability for any other route from First to Fourth. This illustrates the calculation of the probability of a compound event. If he crosses bridge \(1\), the probability that his route will be \(2\to a\to C\) is zero, of course. The probability of an event that has already occurred is 1, and the probability of any alternative is zero. If he crosses bridge \(1,\) \(P\left(1\right)=1\), and \(P\left(2\right)=P\left(3\right)=0\). Given that a traveler has used bridge \(1\), the probability of the route \(1\to a\to C\) becomes the probability of path \(a\to C\), which is \(P\left(a\right)\times P\left(C\right)=\left({1}/{2}\right)\left({1}/{4}\right)={1}/{8}\). Since \(P\left(1\right)=1\), the probability of the compound event \(1\to a\to C\) is the probability of the compound event \(a\to C\). The outcomes of rolling dice, rolling provide more illustrations. If we roll two dice, we can classify the possible outcomes according to the sums of the outcomes for the individual dice. There are thirty-six possible outcomes. They are displayed in Table 1. Table 1: Outcomes from tossing two dice 0 1 2 3 4 5 6 7 NaN Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for first die Outcome for second die NaN 1 2 3 4 5 6 Outcome for second die 1 2 3 4 5 6 7 Outcome for second die 2 3 4 5 6 7 8 Outcome for second die 3 4 5 6 7 8 9 Outcome for second die 4 5 6 7 8 9 10 Outcome for second die 5 6 7 8 9 10 11 Outcome for second die 6 7 8 9 10 11 12 Let us consider the probabilities associated with various dice-throwing events: The probability of any given outcome, say the first die shows \(2\) and the second die shows \(3\), is \({1}/{36}\). Since the probability that the first die shows \(3\) while the second die shows \(2\) is also \({1}/{36}\), the probability that one die shows \(2\) and the other shows \(3\) is \[P\left(3\right)\times P\left(2\right)+P\left(2\right)\times P\left(3\right) =\left({1}/{36}\right)+\left({1}/{36}\right) ={1}/{18}. \nonumber \] Four different outcomes correspond to the event that the score is \(5\). Therefore, the probability of rolling \(5\) is \[P\left(1\right)\times P\left(4\right)+P\left(2\right)\times P\left(3\right) +P\left(3\right)\times P\left(2\right)+P\left(4\right)\times P\left(1\right) ={1}/{9} \nonumber \] The probability of rolling a score of three or less is the probability of rolling \(2\), plus the probability of rolling \(3\) which is \(\left({1}/{36}\right)+\left({2}/{36}\right)={3}/{36}={1}/{12}\) Suppose we roll the dice one at a time and that the first die shows \(2\). The probability of rolling \(7\) when the second die is thrown is now \({1}/{6}\), because only rolling a \(5\) can make the score 7, and there is a probability of \({1}/{6}\) that a \(5\) will come up when the second die is thrown. Suppose the first die is red and the second die is green. The probability that the red die comes up \(2\) and the green die comes up \(3\) is \(\left({1}/{6}\right)\left({1}/{6}\right)={1}/{36}\). Above we looked at the number of outcomes associated with a score of \(3\) to find that the probability of this event is \({1}/{18}\). We can use another argument to get this result. The probability that two dice roll a score of three is equal to the probability that the first die shows \(1\) or \(2\) times the probability that the second die shows whatever score is necessary to make the total equal to three. This is: \[\begin{align*} P\left(first\ die\ shows\ 1\ or\ 2\right)\times \left({1}/{6}\right) &= \left[\left({1}/{6}\right)+\left({1}/{6}\right)\right]\times {1}/{6} \\[4pt] &={2}/{36} \\[4pt]& ={1}/{18} \end{align*} \nonumber \] Application of the laws of probability is frequently made easier by recognizing a simple restatement of the requirement that events be mutually exclusive. In a given trial, either an event occurs or it does not. Let the probability that an event A occurs be \(P\left(A\right)\). Let the probability that event A does not occur be \(P\left(\sim A\right)\). Since in any given trial, the outcome must belong either to event A or to event \(\sim A\), we have \[P\left(A\right)+P\left(\sim A\right)=1 \nonumber \] For example, if the probability of success in a single trial is \({2}/{3}\), the probability of failure is \({1}/{3}\). If we consider the outcomes of two successive trials, we can group them into four events. Event SS: First trial is a success; second trial is a success. Event SF: First trial is a success; second trial is a failure. Event FS: First trial is a failure; second trial is a success. Event FF: First trial is a failure; second trial is a failure. Using the laws of probability, we have \[ \begin{align*} 1 &=P\left(Event\ SS\right)+P\left(Event\ SF\right)+P\left(Event\ FS\right)+\ P(Event\ FF) \\[4pt] &=P_1\left(S\right)\times P_2\left(S\right)+P_1\left(S\right)\times P_2\left(F\right) +P_1(F)\times P_2(S)+P_1(F)\times P_2(F) \end{align*} \nonumber \] where \(P_1\left(X\right)\) and \(P_2\left(X\right)\) are the probability of event \(X\) in the first and second trials, respectively. This situation can be mapped onto a simple diagram. We represent the possible outcomes of the first trial by line segments on one side of a unit square \(P_1\left(S\right)+P_1\left(F\right)=1\). We represent the outcomes of the second trial by line segments along an adjoining side of the unit square. The four possible events are now represented by the areas of four mutually exclusive and exhaustive portions of the unit square as shown in Figure 2. |
Courses/University_of_Georgia/CHEM_3212%3A_Physical_Chemistry_II/07%3A_Entropy_Part_II/7.03%3A_The_Third_Law_of_Thermodynamics | One important consequence of Botlzmann’s proposal is that a perfectly ordered crystal (i.e. one that has only one energetic arrangement in its lowest energy state) will have an entropy of 0. This makes entropy qualitatively different than other thermodynamic functions. For example, in the case of enthalpy, it is impossible have a zero to the scale without setting an arbitrary reference (which is that the enthalpy of formation of elements in their standard states is zero.) But entropy has a natural zero! It is the state at which a system has perfect order. This also has another important consequence, in that it suggests that there must also be a zero to the temperature scale. These consequences are summed up in the Third Law of Thermodynamics . The entropy of a perfectly ordered crystal at 0 K is zero. This also suggests that absolute molar entropies can be calculated by \[S = \int_o^{T} \dfrac{C}{T} dT \nonumber \] where \(C\) is the heat capacity. An entropy value determined in this manner is called a Third Law Entropy . Naturally, the heat capacity will have some temperature dependence. It will also change abruptly if the substance undergoes a phase change. Unfortunately, it is exceedingly difficult to measure heat capacities very near zero K. Fortunately, many substances follow the Debye Extrapolation in that at very low temperatures, their heat capacities are proportional to T 3 . Using this assumption, we have a temperature dependence model that allows us to extrapolate absolute zero based on the heat capacity measured at as low a temperature as can be found. Example \(\PageIndex{1}\) SiO 2 is found to have a molar heat capacity of 0.777 J mol -1 K -1 at 15 K (Yamashita, et al., 2001). Calculate the molar entropy of SiO 2 at 15 K. Solution Using the Debye model, the heat capacity is given by The value of a can be determined by The entropy is then calculated by Calculating a third Law Entropy Start at 0 K, and go from there! |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/05%3A_Thermochemistry/5.07%3A_Exercises | 1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not? 2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile. 3. Explain the difference between heat capacity and specific heat of a substance. 4. Calculate the heat capacity, in joules and in calories per degree, of the following: 28.4 g of water 1.00 oz of lead 5. Calculate the heat capacity, in joules and in calories per degree, of the following: 45.8 g of nitrogen gas 1.00 pound of aluminum metal 6. How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? 7. How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C? 8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? 9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase? 10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 11. A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. What is the specific heat of the substance? If it is one of the substances found in Table 5.1, what is its likely identity? 12. An aluminum kettle weighs 1.05 kg. What is the heat capacity of the kettle? How much heat is required to increase the temperature of this kettle from 23.0 °C to 99.0 °C? How much heat is required to heat this kettle from 23.0 °C to 99.0 °C if it contains 1.25 L of water (density of 0.997 g/mL and a specific heat of 4.184 J/g °C)? 13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 10 6 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)? 14. A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers. 15. Would the amount of heat measured for the reaction in Example 5.5 be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 16. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. 17. Would the amount of heat absorbed by the dissolution in Example 5.6 appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer. 18. How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat. 19. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water. 20. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer. 21. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C. 22. A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure 5.12. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal? 23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in Figure 5.12, what is the resulting temperature of the water? 24. A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure 5.12). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 25. Dissolving 3.0 g of CaCl 2 ( s ) in 150.0 g of water in a calorimeter (Figure 5.12) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 26. When 50.0 g of 0.200 M NaCl( aq ) at 24.1 °C is added to 100.0 g of 0.100 M AgNO 3 ( aq ) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl( s ) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced. 27. The addition of 3.15 g of Ba(OH) 2 ·8H 2 O to a solution of 1.52 g of NH 4 SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: Ba(OH) 2 ·8H 2 O( s ) + 2NH 4 SCN( aq ) ⟶ Ba(SCN) 2 ( aq ) + 2NH 3 ( aq ) + 10H 2 O( l ) 28. The reaction of 50 mL of acid and 50 mL of base described in Example 5.5 increased the temperature of the solution by 6.9 ºC. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 ºC? Explain your answer. 29. If the 3.21 g of NH 4 NO 3 in Example 5.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer. 30. When 1.0 g of fructose, C 6 H 12 O 6 ( s ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is q for this combustion? 31. When a 0.740-g sample of trinitrotoluene (TNT), C 7 H 5 N 2 O 6 , is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample? 32. One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter (Figure 5.17), the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 10 3 pounds). 33. The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g? 34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? 35. What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g? 36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 10 3 Calories if the average number of Calories for fat is 9.1 Calories/g? 37. A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g? 38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. 39. Explain how the heat measured in Example 5.5 differs from the enthalpy change for the exothermic reaction described by the following equation: 40. Using the data in the check your learning section of Example 5.5, calculate Δ H in kJ/mol of AgNO 3 ( aq ) for the reaction: 41. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of NH 4 NO 3 under the conditions described in Example 5.6. 42. Calculate Δ H for the reaction described by the equation. ( Hint : Use the value for the approximate amount of heat absorbed by the reaction that you calculated in a previous exercise.) 43. Calculate the enthalpy of solution (Δ H for the dissolution) per mole of CaCl 2 (refer to Exercise 5.25). 44. Although the gas used in an oxyacetylene torch (Figure 5.7) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 5.2. Considering the conditions for which the tabulated data are reported, suggest an explanation. 45. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions? 46. How much heat is produced by combustion of 125 g of methanol under standard state conditions? 47. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions? 48. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions? 49. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions? 50. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? If both solutions are at the same temperature and the specific heat of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation? 51. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents? 52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO 2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl 2 F 2 (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO 2 and CCl 2 F 2 are and respectively. 53. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C. 54. Which of the enthalpies of combustion in Table 5.2 the table are also standard enthalpies of formation? 55. Does the standard enthalpy of formation of H 2 O( g ) differ from Δ H ° for the reaction 56. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO( s ) to Hg( l ) and O 2 ( g ) under standard conditions? 57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn 3 O 4 ( s ) at standard state conditions? 58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe 2 O 3 ( s ) at standard state conditions? 59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid: Determine the total energy change for the production of one mole of aqueous nitric acid by this process. 60. Both graphite and diamond burn. For the conversion of graphite to diamond: Which produces more heat, the combustion of graphite or the combustion of diamond? 61. From the molar heats of formation in Appendix G, determine how much heat is required to evaporate one mole of water: 62. Which produces more heat? or for the phase change 63. Calculate for the process from the following information: 64. Calculate for the process from the following information: 65. Calculate Δ H for the process from the following information: 66. Calculate for the process from the following information: 67. Calculate the standard molar enthalpy of formation of NO( g ) from the following data: 68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions: 70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. 71. The decomposition of hydrogen peroxide, H 2 O 2 , has been used to provide thrust in the control jets of various space vehicles. Using the data in Appendix G, determine how much heat is produced by the decomposition of exactly 1 mole of H 2 O 2 under standard conditions. 72. Calculate the enthalpy of combustion of propane, C 3 H 8 ( g ), for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of propane is −104 kJ/mol. 73. Calculate the enthalpy of combustion of butane, C 4 H 10 ( g ) for the formation of H 2 O( g ) and CO 2 ( g ). The enthalpy of formation of butane is −126 kJ/mol. 74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned? 75. The white pigment TiO 2 is prepared by the reaction of titanium tetrachloride, TiCl 4 , with water vapor in the gas phase: How much heat is evolved in the production of exactly 1 mole of TiO 2 ( s ) under standard state conditions? 76. Water gas, a mixture of H 2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: Under the conditions of the reaction, methanol forms as a gas. Calculate for this reaction and for the condensation of gaseous methanol to liquid methanol. (c) Calculate the heat of combustion of 1 mole of liquid methanol to H 2 O( g ) and CO 2 ( g ). 77. In the early days of automobiles, illumination at night was provided by burning acetylene, C 2 H 2 . Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC 2 : Calculate the standard enthalpy of the reaction. The of CaC 2 is −15.14 kcal/mol. 78. From the data in Table 5.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO( g ), CH 4 ( g ), or C 2 H 2 ( g ). 79. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 10 5 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane). 80. Ethanol, C 2 H 5 OH, is used as a fuel for motor vehicles, particularly in Brazil. Write the balanced equation for the combustion of ethanol to CO 2 ( g ) and H 2 O( g ), and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol. The density of ethanol is 0.7893 g/mL. Calculate the enthalpy of combustion of exactly 1 L of ethanol. Assuming that an automobile’s mileage is directly proportional to the heat of combustion of the fuel, calculate how much farther an automobile could be expected to travel on 1 L of gasoline than on 1 L of ethanol. Assume that gasoline has the heat of combustion and the density of n–octane, C 8 H 18 density = 0.7025 g/mL). 81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B 2 H 6 , produces B 2 O 3 ( s ) and H 2 O( g )], methane [CH 4 , produces CO 2 ( g ) and H 2 O( g )], and hydrazine [N 2 H 4 , produces N 2 ( g ) and H 2 O( g )]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The 82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions? 83. Ethylene, C 2 H 4 , a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. Using the data in the table in Appendix G, calculate Δ H ° for the reaction. 84. The oxidation of the sugar glucose, C 6 H 12 O 6 , is described by the following equation: The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose? How many Calories can be produced by the metabolism of 1.0 g of glucose? 85. Propane, C 3 H 8 , is a hydrocarbon that is commonly used as a fuel. Write a balanced equation for the complete combustion of propane gas. Calculate the volume of air at 25 °C and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent O 2 by volume. (Hint: We will see how to do this calculation in a later chapter on gases—for now use the information that 1.00 L of air at 25 °C and 1.00 atm contains 0.275 g of O 2 .) The heat of combustion of propane is −2,219.2 kJ/mol. Calculate the heat of formation, of propane given that of H 2 O( l ) = −285.8 kJ/mol and of CO 2 ( g ) = −393.5 kJ/mol. Assuming that all of the heat released in burning 25.0 grams of propane is transferred to 4.00 kilograms of water, calculate the increase in temperature of the water. 86. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house). (f) How many kilowatt–hours (1 kWh = 3.6 10 6 J) of electricity would be required to provide the heat necessary to heat the house? Note electricity is 100% efficient in producing heat inside a house. (g) Although electricity is 100% efficient in producing heat inside a house, production and distribution of electricity is not 100% efficient. The efficiency of production and distribution of electricity produced in a coal-fired power plant is about 40%. A certain type of coal provides 2.26 kWh per pound upon combustion. What mass of this coal in kilograms will be required to produce the electrical energy necessary to heat the house if the efficiency of generation and distribution is 40%? |
Courses/Chandler_Gilbert_Community_College/Fundamental_Organic_ala_Mech/05%3A_Unit_2/5.04%3A_Apply_to_Acids_and_Bases/5.4.05%3A_Acids_and_Bases_-_The_Lewis_Definition | The Lewis definition of acids and bases is more encompassing than the Brønsted–Lowry definition because it’s not limited to substances that donate or accept just protons. A Lewis acid is a substance that accepts an electron pair, and a Lewis base is a substance that donates an electron pair. The donated electron pair is shared between the acid and the base in a covalent bond. Lewis Acids and the Curved Arrow Formalism The fact that a Lewis acid is able to accept an electron pair means that it must have either a vacant, low-energy orbital or a polar bond to hydrogen so that it can donate H + (which has an empty 1 s orbital). Thus, the Lewis definition of acidity includes many species in addition to H + . For example, various metal cations, such as Mg 2 + , are Lewis acids because they accept a pair of electrons when they form a bond to a base. We’ll also see in later chapters that certain metabolic reactions begin with an acid–base reaction between Mg 2 + as a Lewis acid and an organic diphosphate or triphosphate ion as the Lewis base. In the same way, compounds of group 3A elements, such as BF 3 and AlCl 3 , are Lewis acids because they have unfilled valence orbitals and can accept electron pairs from Lewis bases, as shown in Figure \(\PageIndex{1}\). Similarly, many transition-metal compounds, such as TiCl 4 , FeCl 3 , ZnCl 2 , and SnCl 4 , are Lewis acids. Look closely at the acid–base reaction in Figure \(\PageIndex{1}\), and notice how it's shown. Dimethyl ether, the Lewis base, donates an electron pair to a vacant valence orbital of the boron atom in BF 3 , a Lewis acid. The direction of electron-pair flow from base to acid is shown using a curved arrow, just as the direction of electron flow from one resonance structure to another was shown using curved arrows in Section 2.6. We’ll use this curved-arrow notation throughout the remainder of this text to indicate electron flow during reactions, so get used to seeing it. Some further examples of Lewis acids follow: Lewis Bases The Lewis definition of a base—a compound with a pair of nonbonding electrons that it can use to bond to a Lewis acid—is similar to the Brønsted–Lowry definition. Thus, H 2 O, with its two pairs of nonbonding electrons on oxygen, acts as a Lewis base by donating an electron pair to an H + in forming the hydronium ion, H 3 O + . In a more general sense, most oxygen- and nitrogen-containing organic compounds can act as Lewis bases because they have lone pairs of electrons. A divalent oxygen compound has two lone pairs of electrons, and a trivalent nitrogen compound has one lone pair. Note in the following examples that some compounds can act as both acids and bases, just as water can. Alcohols and carboxylic acids, for instance, act as acids when they donate an H + but as bases when their oxygen atom accepts an H + . Note also that some Lewis bases, such as carboxylic acids, esters, and amides, have more than one atom with a lone pair of electrons and can therefore react at more than one site. Acetic acid, for example, can be protonated either on the doubly bonded oxygen atom or on the singly bonded oxygen atom. The reaction normally occurs only once in such instances, and the more stable of the two possible protonation products are formed. For acetic acid, protonation by reaction with sulfuric acid occurs on the doubly bonded oxygen because that product is stabilized by two resonance forms. Worked Example \(\PageIndex{1}\): Using Curved Arrows to Show Electron Flow Using curved arrows, show how acetaldehyde, CH 3 CHO, can act as a Lewis base. Strategy A Lewis base donates an electron pair to a Lewis acid. We therefore need to locate the electron lone pairs on acetaldehyde and use a curved arrow to show the movement of a pair toward the H atom of the acid. Solution Exercise \(\PageIndex{1}\) Using curved arrows, show how the species in part (a) can act as Lewis bases in their reactions with HCl, and show how the species in part (b) can act as Lewis acids in their reaction with OH – . CH 3 CH 2 OH, HN(CH 3 ) 2 , P(CH 3 ) 3 H 3 C + , B(CH 3 ) 3 , MgBr 2 Answer Exercise \(\PageIndex{2}\) Imidazole, which forms part of amino acid histidine, can act as both an acid and a base. Look at the electrostatic potential map of imidazole, and identify the most acidic hydrogen atom and the most basic nitrogen atom. Draw structures for the resonance forms of the products that result when imidazole is protonated by an acid and deprotonated by a base. Answer |
Courses/Portland_Community_College/CH106%3A_Allied_Health_Chemistry_III/04%3A_Aldehydes_and_Ketones/4.02%3A_Aldehydes_and_Ketones-_Structure_and_Names | Learning Objectives Identify the general structure for an aldehyde and a ketone. Use common names to name aldehydes and ketones. Use the IUPAC system to name aldehydes and ketones. The next functional group we consider, the carbonyl group, has a carbon-to-oxygen double bond. Carbonyl groups define two related families of organic compounds: the aldehydes and the ketones. The carbonyl group is ubiquitous in biological compounds. It is found in carbohydrates, fats, proteins, nucleic acids, hormones, and vitamins—organic compounds critical to living systems. In a ketone, two carbon groups are attached to the carbonyl carbon atom. The following general formulas, in which R represents an alkyl group and Ar stands for an aryl group, represent ketones. In an aldehyde, at least one of the attached groups must be a hydrogen atom. The following compounds are aldehydes: In condensed formulas, we use CHO to identify an aldehyde rather than CO H , which might be confused with an alcohol. This follows the general rule that in condensed structural formulas H comes after the atom it is attached to (usually C, N, or O). The carbon-to-oxygen double bond is not shown but understood to be present. Because they contain the same functional group, aldehydes and ketones share many common properties, but they still differ enough to warrant their classification into two families. Naming Aldehydes and Ketones Both common and International Union of Pure and Applied Chemistry (IUPAC) names are frequently used for aldehydes and ketones, with common names predominating for the lower homologs. The common names of aldehydes are taken from the names of the acids into which the aldehydes can be converted by oxidation . The stems for the common names of the first four aldehydes are as follows: 1 carbon atom: form - 2 carbon atoms: acet - 3 carbon atoms: propion - 4 carbon atoms: butyr - Because the carbonyl group in a ketone must be attached to two carbon groups, the simplest ketone has three carbon atoms. It is widely known as acetone , a unique name unrelated to other common names for ketones. Generally, the common names of ketones consist of the names of the groups attached to the carbonyl group, followed by the word ketone . (Note the similarity to the naming of ethers.) Another name for acetone, then, is dimethyl ketone. The ketone with four carbon atoms is ethyl methyl ketone. Example \(\PageIndex{1}\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Solution This compound has the carbonyl group on an end carbon atom, so it is an aldehyde. This compound has the carbonyl group on an interior carbon atom, so it is a ketone. Both alkyl groups are propyl groups. The name is therefore dipropyl ketone. This compound has the carbonyl group between two alkyl groups, so it is a ketone. One alkyl group has three carbon atoms and is attached by the middle carbon atom; it is an isopropyl group. A group with one carbon atom is a methyl group. The name is therefore isopropyl methyl ketone. Exercise \(\PageIndex{1}\) Classify each compound as an aldehyde or a ketone. Give the common name for each ketone. Here are some simple IUPAC rules for naming aldehydes and ketones: The stem names of aldehydes and ketones are derived from those of the parent alkanes, defined by the longest continuous chain (LCC) of carbon atoms that contains the functional group. For an aldehyde, drop the - e from the alkane name and add the ending - al . Methanal is the IUPAC name for formaldehyde, and ethanal is the name for acetaldehyde. For a ketone, drop the - e from the alkane name and add the ending - one . Propanone is the IUPAC name for acetone, and butanone is the name for ethyl methyl ketone. To indicate the position of a substituent on an aldehyde, the carbonyl carbon atom is always considered to be C1; it is unnecessary to designate this group by number. To indicate the position of a substituent on a ketone, number the chain in the manner that gives the carbonyl carbon atom the lowest possible number. In cyclic ketones, it is understood that the carbonyl carbon atom is C1. Example \(\PageIndex{2}\) Give the IUPAC name for each compound. Solution There are five carbon atoms in the LCC . The methyl group (CH 3 ) is a substituent on the second carbon atom of the chain; the aldehyde carbon atom is always C1. The name is derived from pentane. Dropping the - e and adding the ending - al gives pentanal. The methyl group on the second carbon atom makes the name 2-methylpentanal. There are five carbon atoms in the LCC. The carbonyl carbon atom is C3, and there are methyl groups on C2 and C4. The IUPAC name is 2,4-dimethyl-3-pentanone. There are six carbon atoms in the ring. The compound is cyclohexanone. No number is needed to indicate the position of the carbonyl group because all six carbon atoms are equivalent. Exercise Give the IUPAC name for each compound. Example \(\PageIndex{3}\) Draw the structure for each compound. 7-chlorooctanal 4-methyl–3-hexanone Solution The octan - part of the name tells us that the LCC has eight carbon atoms. There is a chlorine (Cl) atom on the seventh carbon atom; numbering from the carbonyl group and counting the carbonyl carbon atom as C1, we place the Cl atom on the seventh carbon atom. The hexan - part of the name tells us that the LCC has six carbon atoms. The 3 means that the carbonyl carbon atom is C3 in this chain, and the 4 tells us that there is a methyl (CH 3 ) group at C4: Exercise \(\PageIndex{3}\) Draw the structure for each compound. 5-bromo-3-iodoheptanal 5-bromo-4-ethyl-2-heptanone Summary The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone . Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an - al ending for an aldehydes and an - one ending for a ketone. |
Courses/Taft_College/CHEM_1510%3A_Introductory_College_Chemistry/16%3A_Acids_and_Bases/16.08%3A_The_pH_and_pOH_Scales_-_Ways_to_Express_Acidity_and_Basicity | Learning Objectives Define pH and pOH. Determine the pH of acidic and basic solutions. Determine the hydronium ion concentration and pOH from pH. As we have seen, \([H_3O^+]\) and \([OH^−]\) values can be markedly different from one aqueous solution to another. So chemists defined a new scale that succinctly indicates the concentrations of either of these two ions. \(pH\) is a logarithmic function of \([H_3O^+]\): \[pH = −\log[H_3O^+] \label{pH} \] \(pH\) is usually (but not always) between 0 and 14. Knowing the dependence of \(pH\) on \([H_3O^+]\), we can summarize as follows: If pH < 7, then the solution is acidic. If pH = 7, then the solution is neutral. If pH > 7, then the solution is basic. This is known as the \(pH\) scale. The pH scale is the range of value s from 0 to 14 that describes the acidity or basicity of a solution. You can use \(pH\) to make a quick determination whether a given aqueous solution is acidic, basic, or neutral. Figure \(\PageIndex{1}\) illus trates this relationship, along with some examples of various solutions. Because hydrogen ion concentrations are generally less than one (for example \(1.3 \times 10^{-3}\,M\)), th e log of the number will be a negative number. To make pH even easier to work with, pH is defined as the negative log of \([H_3O^+]\) , which will give a positive value for pH. Example \(\PageIndex{1}\) Label each solution as acidic, basic, or neutral based only on the stated \(pH\). milk of magnesia, pH = 10.5 pure water, pH = 7 wine, pH = 3.0 Answer With a pH greater than 7, milk of magnesia is basic. (Milk of magnesia is largely Mg(OH) 2 .) Pure water, with a pH of 7, is neutral. With a pH of less than 7, wine is acidic. Exercise \(\PageIndex{1}\) Identify each substance as acidic, basic, or neutral based only on the stated \(pH\). human blood with \(pH\) = 7.4 household ammonia with \(pH\) = 11.0 cherries with \(pH\) = 3.6 Answer a basic Answer b basic Answer c acidic Calculating pH from Hydronium Concentration The pH of solutions can be determined by using logarithms as illustrated in the next example for stomach acid. Stomach acid is a solution of \(HCl\) with a hydronium ion concentration of \(1.2 \times 10^{−3}\; M\), what is the \(pH\) of the solution? \[ \begin{align} \mathrm{pH} &= \mathrm{-\log [H_3O^+]} \nonumber \\ &=-\log(1.2 \times 10^{−3}) \nonumber \\ &=−(−2.92)=2.92 \nonumber \end{align} \nonumber \] Logarithms To get the log value on your calculator, enter the number (in this case, the hydronium ion concentration) first, then press the LOG key. If the number is 1.0 x 10 -5 (for [H 3 O + ] = 1.0 x 10 -5 M) you should get an answer of "-5". If you get a different answer, or an error, try pressing the LOG key before you enter the number. Example \(\PageIndex{2}\): Converting Ph to Hydronium Concentration Find the pH, given the \([H_3O^+]\) of the following: 1 ×10 -3 M 2.5 ×10 -11 M 4.7 ×10 -9 M Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: [H3O+] =1 × 10−3 M [H3O+] =2.5 ×10-11 M [H3O+] = 4.7 ×10-9 M Find: ? pH Plan the problem. Need to use the expression for pH (Equation \ref{pH}). pH = - log [H3O+] Calculate. Now substitute the known quantity into the equation and solve. pH = - log [1 × 10−3 ] = 3.0 (1 decimal places since 1 has 1 significant figure) pH = - log [2.5 ×10-11] = 10.60 (2 decimal places since 2.5 has 2 significant figures) pH = - log [4.7 ×10-9] = 8.30 (2 decimal places since 4.7 has 2 significant figures) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise \(\PageIndex{2}\) Find the pH, given [H 3 O + ] of the following: 5.8 ×10 -4 M 1.0×10 -7 Answer a 3.22 Answer b 7.00 Calculating Hydronium Concentration from pH Sometimes you need to work "backwards"—you know the pH of a solution and need to find \([H_3O^+]\), or even the concentration of the acid solution. How do you do that? To convert pH into \([H_3O^+]\) we solve Equation \ref{pH} for \([H_3O^+]\). This involves taking the antilog (or inverse log) of the negative value of pH . \[[\ce{H3O^{+}}] = \text{antilog} (-pH) \nonumber \] or \[[\ce{H_3O^+}] = 10^{-pH} \label{ph1} \] As mentioned above, different calculators work slightly differently—make sure you can do the following calculations using your calculator. Calculator Skills We have a solution with a pH = 8.3. What is [H 3 O + ] ? With some calculators you will do things in the following order: Enter 8.3 as a negative number (use the key with both the +/- signs, not the subtraction key). Use your calculator's 2nd or Shift or INV (inverse) key to type in the symbol found above the LOG key. The shifted function should be 10 x . You should get the answer 5.0 × 10 -9 . Other calculators require you to enter keys in the order they appear in the equation. Use the Shift or second function to key in the 10 x function. Use the +/- key to type in a negative number, then type in 8.3. You should get the answer 5.0 × 10 -9 . If neither of these methods work, try rearranging the order in which you type in the keys. Don't give up—you must master your calculator! Example \(\PageIndex{3}\): Calculating Hydronium Concentration from pH Find the hydronium ion concentration in a solution with a pH of 12.6. Is this solution an acid or a base? How do you know? Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: pH = 12.6 Find: [H3O+] = ? M Plan the problem. Need to use the expression for [H3O+] (Equation \ref{ph1}). [H3O+] = antilog (-pH) or [H3O+] = 10-pH Calculate. Now substitute the known quantity into the equation and solve. [H3O+] = antilog (12.60) = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) or [H3O+] = 10-12.60 = 2.5 x 10-13 M (2 significant figures since 4.7 has 12.60 2 decimal places) The other issue that concerns us here is significant figures. Because the number(s) before the decimal point in a logarithm relate to the power on 10, the number of digits after the decimal point is what determines the number of significant figures in the final answer: Exercise \(\PageIndex{3}\) If moist soil has a pH of 7.84, what is [H 3 O + ] of the soil solution? Answer 1.5 x 10 -8 M The pOH scale As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. \[\text{pOH} = -\text{log} \left[ \ce{OH^-} \right] \nonumber \] The relation between the hydronium and hydroxide ion concentrations expressed as p-functions is easily derived from the \(K_w\) expression: \[K_\ce{w}=\ce{[H_3O^+][OH^- ]} \label{\(\PageIndex{6}\)} \] \[-\log K_\ce{w}=\mathrm{-\log([H_3O^+][OH^−])=-\log[H_3O^+] + -\log[OH^-]}\label{\(\PageIndex{7}\)} \] \[\mathrm{p\mathit{K}_w=pH + pOH} \label{\(\PageIndex{8}\)} \] At 25 °C, the value of \(K_w\) is \(1.0 \times 10^{−14}\), and so: \[\mathrm{14.00=pH + pOH} \label{\(\PageIndex{9}\)} \] The hydronium ion molarity in pure water (or any neutral solution) is \( 1.0 \times 10^{-7}\; M\) at 25 °C. The pH and pOH of a neutral solution at this temperature are therefore: \[\mathrm{pH=-\log[H_3O^+]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\PageIndex{1}\)0} \] \[\mathrm{pOH=-\log[OH^−]=-\log(1.0\times 10^{−7}) = 7.00} \label{\(\PageIndex{1}\)1} \] And so, at this temperature, acidic solutions are those with hydronium ion molarities greater than \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities less than \( 1.0 \times 10^{-7}\; M\) (corresponding to pH values less than 7.00 and pOH values greater than 7.00). Basic solutions are those with hydronium ion molarities less than \( 1.0 \times 10^{-7}\; M\) and hydroxide ion molarities greater than \( 1.0 \times 10^{-7}\; M\) (corresponding to pH values greater than 7.00 and pOH values less than 7.00). Example \(\PageIndex{4}\): Find the pOH of a solution with a pH of 4.42. Solution Steps for Problem Solving Unnamed: 1 Identify the "given" information and what the problem is asking you to "find." Given: pH =4.42 Find: ? pOH Plan the problem. Need to use the expression pOH = 14 - pH Calculate. Now substitute the known quantity into the equation and solve. pOH=14−4.42=9.58 Think about your result. The pH is that of an acidic solution, and the resulting pOH is the difference after subtracting from 14. The answer has two significant figures because the given pH has two decimal places. Exercise \(\PageIndex{4}\) The pH of a solution is 8.22. What is the pOH? Answer 5.78 The diagram below shows all of the interrelationships between [H3O+][H3O+], [OH−][OH−], pH, and pOH. Contributions & Attributions Peggy Lawson (Oxbow Prairie Heights School). Funded by Saskatchewan Educational Technology Consortium. Template:OpenStax |
Bookshelves/General_Chemistry/Chemistry_2e_(OpenStax)/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.06%3A_Occurrence_Preparation_and_Properties_of_Carbonates | Learning Objectives By the end of this section, you will be able to: Describe the preparation, properties, and uses of some representative metal carbonates The chemistry of carbon is extensive; however, most of this chemistry is not relevant to this chapter. The other aspects of the chemistry of carbon will appear in the chapter covering organic chemistry. In this chapter, we will focus on the carbonate ion and related substances. The metals of groups 1 and 2, as well as zinc, cadmium, mercury, and lead(II), form ionic carbonates —compounds that contain the carbonate anions, The metals of group 1, magnesium, calcium, strontium, and barium also form hydrogen carbonates —compounds that contain the hydrogen carbonate anion, \(\ce{HCO3^{−}}\), also known as the bicarbonate anion . With the exception of magnesium carbonate, it is possible to prepare carbonates of the metals of groups 1 and 2 by the reaction of carbon dioxide with the respective oxide or hydroxide. Examples of such reactions include: \[\begin{align*} \ce{Na_2 O(s) + CO_2(g) &-> Na_2CO_3(s) } \\[4pt][4pt] \ce{Ca( OH )_2(s) + CO_2(g) &-> CaCO_3(s) + H2O(l)} \end{align*} \] The carbonates of the alkaline earth metals of group 12 and lead(II) are not soluble. These carbonates precipitate upon mixing a solution of soluble alkali metal carbonate with a solution of soluble salts of these metals. Examples of net ionic equations for the reactions are: \[\begin{align*} \ce{Ca^{2+}(aq) + CO_3^{2-}(aq) &-> CaCO_3(s)} \\[4pt][4pt] \ce{Pb^{2+}(aq) + CO_3^{2-}(aq) &-> PbCO_3(s)} \end{align*} \] Pearls and the shells of most mollusks are calcium carbonate. Tin(II) or one of the trivalent or tetravalent ions such as Al 3 + or Sn 4 + behave differently in this reaction as carbon dioxide and the corresponding oxide form instead of the carbonate. Alkali metal hydrogen carbonates such as NaHCO 3 and CsHCO 3 form by saturating a solution of the hydroxides with carbon dioxide. The net ionic reaction involves hydroxide ion and carbon dioxide: \[\ce{OH^{-}(aq) + CO_2(aq) -> HCO_3^{-}(aq)} \nonumber \] It is possible to isolate the solids by evaporation of the water from the solution. Although they are insoluble in pure water, alkaline earth carbonates dissolve readily in water containing carbon dioxide because hydrogen carbonate salts form. For example, caves and sinkholes form in limestone when CaCO 3 dissolves in water containing dissolved carbon dioxide: \[\ce{CaCO_3(s) + CO_2(aq) + H2O(l) -> Ca^{2+}(aq) + 2 HCO_3^{-}(aq)} \nonumber \] Hydrogen carbonates of the alkaline earth metals remain stable only in solution; evaporation of the solution produces the carbonate. Stalactites and stalagmites, like those shown in Figure \(\PageIndex{1}\): , form in caves when drops of water containing dissolved calcium hydrogen carbonate evaporate to leave a deposit of calcium carbonate. The two carbonates used commercially in the largest quantities are sodium carbonate and calcium carbonate. In the United States, sodium carbonate is extracted from the mineral trona, Na 3 (CO 3 )(HCO 3 )(H 2 O) 2 . Following recrystallization to remove clay and other impurities, heating the recrystallized trona produces Na 2 CO 3 : \[\ce{2 Na_3(CO3)(HCO3)(H2O)2(s) -> 3 Na_2CO_3(s) + 5 H2O(l) + CO_2(g)} \nonumber \] Carbonates are moderately strong bases. Aqueous solutions are basic because the carbonate ion accepts hydrogen ion from water in this reversible reaction: \[\ce{CO_3^{2-}(aq) + H2O(l) <=> HCO_3^{-}(aq) + OH^{-}(aq)} \nonumber \] Carbonates react with acids to form salts of the metal, gaseous carbon dioxide, and water. The reaction of calcium carbonate, the active ingredient of the antacid Tums, with hydrochloric acid(stomach acid), as shown in Figure \(\PageIndex{2}\), illustrates the reaction: \[\ce{CaCO_3(s) + 2 HCl(aq) -> CaCl_2(aq) + CO_2(g) + H2O(l)} \nonumber \] Other applications of carbonates include glass making—where carbonate ions serve as a source of oxide ions—and synthesis of oxides. Hydrogen carbonates are amphoteric because they act as both weak acids and weak bases. Hydrogen carbonate ions act as acids and react with solutions of soluble hydroxides to form a carbonate and water: \[\ce{KHCO_3(aq) + KOH(aq) -> K_2 CO_3(aq) + H2O(l)} \nonumber \] With acids, hydrogen carbonates form a salt, carbon dioxide, and water. Baking soda(bicarbonate of soda or sodium bicarbonate) is sodium hydrogen carbonate. Baking powder contains baking soda and a solid acid such as potassium hydrogen tartrate(cream of tartar), KHC 4 H 4 O 6 . As long as the powder is dry, no reaction occurs; immediately after the addition of water, the acid reacts with the hydrogen carbonate ions to form carbon dioxide: \[\ce{HC_4 H_4 O_6^{-}(aq) + HCO_3^{-}(aq) -> C_4 H_4 O_6^{2-}(aq) + CO_2(g) + H2O(l)} \nonumber \] Dough will trap the carbon dioxide, causing it to expand during baking, producing the characteristic texture of baked goods. |
Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/02%3A_Molecular_Structure/2.02%3A_Lewis_Structures_and_Molecular_Shape/2.2.03%3A_Resonance | Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structures are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds. Sometimes One Lewis Structure Is Not Enough Sometimes, even when formal charges are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (\(\ce{O3}\)), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Ozone (\(O_3\)) 1. We know that ozone has a V-shaped structure, so one O atom is central: 2. Each O atom has 6 valence electrons, for a total of 18 valence electrons. 3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives with 14 electrons left over. 4. If we place three lone pairs of electrons on each terminal oxygen, we obtain and have 2 electrons left over. 5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom: 6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O 2 (120.7 pm). Equivalent Lewis dot structures, such as those of ozone, are called resonance structures . The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound: The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures. When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures. The Carbonate (\(CO_3^{2−} \)) Ion Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O 3 , though, the actual structure of CO 3 2− is an average of three resonance structures. 1. Because carbon is the least electronegative element, we place it in the central position: 2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons. 3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon: 4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge: 5. No electrons are left for the central atom. 6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices: As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion: The actual structure is an average of these three resonance structures. The Nitrate (\(NO_3^-\)) ion 1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons 2. Draw the bond connectivities: 3. Add octet electrons to the atoms bonded to the center atom: 4. Place any leftover electrons (24-24 = 0 ) on the center atom: 5. Does the central atom have an octet? NO , it has 6 electrons Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet: 6. Does the central atom have an octet? YES Are there possible resonance structures? YES Note: We would expect that the bond lengths in the \(\ce{NO_3^{-}}\) ion to be somewhat shorter than a single bond. Warning For second period elements you cannot exced the octet, even if additional double or triple bonds would reduce the formal charges in the structure. Example \(\PageIndex{1}\): Benzene Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (\(\ce{C6H6}\)) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene. Given: molecular formula and molecular geometry Asked for: resonance structures Strategy: Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet. Draw the resonance structures for benzene. Solution: A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following: Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons. B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following: Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds. C There are, however, two ways to do this: Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon: Exercise \(\PageIndex{1}\): Nitrate Ion The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO 2 − ). Answer Resonance structures are particularly common in oxoanions of the p -block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene. How do resonance structures represent the true structure? If several reasonable resonance forms for a molecule exists, the "actual electronic structure" of the molecule will probably be intermediate between all the forms that you can draw. The classic example is benzene in Example \(\PageIndex{1}\). One would expect the double bonds to be shorter than the single bonds, but if once overlays the two structures, you see that one structure has a single bond where the other structure has a double bond. The best measurements that we can make of benzene do not show two bond lengths - instead, they show that the bond length is intermediate between the two resonance structures. Resonance structures is a mechanism that allows us to use all of the possible resonance structures to try to predict what the actual form of the molecule would be. Single bonds, double bonds, triple bonds, +1 charges, -1 charges, these are our limitations in explaining the structures, and the true forms can be in between - a carbon-carbon bond could be mostly single bond with a little bit of double bond character and a partial negative charge, for example. Summary Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. These structures are written with a double-headed arrow between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others. |
Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/07%3A_Energy_and_Chemical_Processes/7.1%3A_Energy_and_Its_Units | Skills to Develop To define energy and he a t. Energy is the ability to do work. You can understand what this means by thinking about yourself when you feel “energetic.” You feel ready to go—to jump up and get something done. When you have a lot of energy, you can perform a lot of work. By contrast, if you do not feel energetic, you have very little desire to do much of anything. This description is not only applicable to you but also to all physical and chemical processes. The quantity of work that can be done is related to the quantity of energy available to do it. Energy can be transferred from one object to another if the objects have different temperatures. The transfer of energy due to temperature differences is called heat. For example, if you hold an ice cube in your hand, the ice cube slowly melts as energy in the form of heat is transferred from your hand to the ice. As your hand loses energy, it starts to feel cold. Because of their interrelationships, energy, work, and heat have the same units. The SI unit of energy, work, and heat is the joule (J). A joule is a tiny amount of energy. For example, it takes about 4 J to warm 1 mL of H 2 O by 1°C. Many processes occur with energy changes in thousands of joules, so the kilojoule (kJ) is also common. Another unit of energy, used widely in the health professions and everyday life, is the calorie (cal). The calorie was initially defined as the amount of energy needed to warm 1 g of H 2 O by 1°C, but in modern times, the calorie is related directly to the joule, as follows: \[1\; cal = 4.184\; J\] We can use this relationship to convert quantities of energy, work, or heat from one unit to another. Although the joule is the proper SI unit for energy, we will use the calorie or the kilocalorie (or Calorie) in this chapter because they are widely used by health professionals. The calorie is used in nutrition to express the energy content of foods. However, because a calorie is a rather small quantity, nutritional energies are usually expressed in kilocalories (kcal), also called Calories (capitalized; Cal). For example, a candy bar may provide 120 Cal (nutritional calories) of energy, which is equal to 120,000 cal. Figure \(\PageIndex{1}\) shows an example. Proteins and carbohydrates supply 4 kcal/g, while fat supplies 9 kcal/g. Figure \(\PageIndex{1}\): Nutritional Energy. A sample nutrition facts label, with instructions from the U.S. Food and Drug Administration. Image used with permission from Wikipedia. Example \(\PageIndex{1}\) The energy content of a single serving of bread is 70.0 Cal. What is the energy content in calories? In joules? SOLUTION This is a simple conversion-factor problem. Using the relationship 1 Cal = 1,000 cal, we can answer the first question with a one-step conversion: \(\mathrm{70.0 \: Cal\times\dfrac{1,000\: cal}{1\: Cal}= 70,000\: cal}\) Then we convert calories into joules \(\mathrm{70,000 \: cal \times \dfrac{4.184\: J}{1\: cal}=293,000\: J}\) and then kilojoules \(\mathrm{293,000\: J\times\dfrac{1\: kJ}{1,000\: J}=293\: kJ}\) The energy content of bread comes mostly from carbohydrates. Exercise \(\PageIndex{1}\) The energy content of one cup of honey is 1,030 Cal. What is its energy content in calories and joules? To Your Health: Energy Expenditures Most health professionals agree that exercise is a valuable component of a healthy lifestyle. Exercise not only strengthens the body and develops muscle tone but also expends energy. After obtaining energy from the foods we eat, we need to expend that energy somehow, or our bodies will store it in unhealthy ways (e.g., fat). Like the energy content in food, the energy expenditures of exercise are also reported in kilocalories, usually kilocalories per hour of exercise. These expenditures vary widely, from about 440 kcal/h for walking at a speed of 4 mph to 1,870 kcal/h for mountain biking at 20 mph. Table \(\PageIndex{1}\) lists the energy expenditure for a variety of exercises. Exercise Energy Expended (kcal/h) aerobics, low-level 325 basketball 940 bike riding, 20 mph 830 golfing, with cart 220 golfing, carrying clubs 425 jogging, 7.5 mph 950 racquetball 740 skiing, downhill 520 soccer 680 walking upstairs 1200 yoga 280 Because some forms of exercise use more energy than others, anyone considering a specific exercise regimen should consult with his or her physician first. Summary Energy is the ability to do work. Heat is the transfer of energy due to temperature differences. Energy and heat are expressed in units of joules. Concept Review Exercises What is the relationship between energy and heat? What units are used to express energy and heat? Answers Heat is the exchange of energy from one part of the universe to another. Heat and energy have the same units. Joules and calories are the units of energy and heat. Exercises Define energy . What is heat? What is the relationship between a calorie and a joule? Which unit is larger? What is the relationship between a calorie and a kilocalorie? Which unit is larger? Express 1,265 cal in kilocalories and in joules. Express 9,043.3 J in calories and in kilocalories. One kilocalorie equals how many kilojoules? One kilojoule equals how many kilocalories? Many nutrition experts say that an average person needs 2,000 Cal per day from his or her diet. How many joules is this? Baby formula typically has 20.0 Cal per ounce. How many ounces of formula should a baby drink per day if the RDI is 850 Cal? Answers Energy is the ability to do work. 1 cal = 4.184 J; the calorie is larger. 1.265 kcal; 5,293 J 1 kcal = 4.184 kJ 8.4 × 10 6 J |
Courses/San_Diego_Miramar_College/Chem_103%3A_Fundamentals_of_GOB_Chemistry_(Garces)/08%3A_Gases/8.02%3A_Kinetic_Molecular_Theory-_A_Model_for_Gases | Learning Objectives State the major concepts behind the kinetic theory of gases. Relate the general properties of gases to the kinetic theory. Gases were among the first substances studied in terms of the modern scientific method, which was developed in the 1600s. It did not take long to recognize that gases all shared certain physical behaviors, suggesting that all gases could be described by one all-encompassing theory. Today, that theory is the kinetic theory of gases . It is based on the following statements: Gases consist of tiny particles of matter that are in constant motion. Gas particles are constantly colliding with each other and the walls of a container. These collisions are elastic; that is, there is no net loss of energy from the collisions. Gas particles are separated by large distances, with the size of a gas particle tiny compared to the distances that separate them. There are no interactive forces (i.e., attraction or repulsion) between the particles of a gas. The average speed of gas particles is dependent on the temperature of the gas. Figure \(\PageIndex{1}\) shows a representation of how we mentally picture the gas phase. This model of gases explains some of the physical properties of gases. Because most of a gas is empty space, a gas has a low density and can expand or contract under the appropriate influence. The fact that gas particles are in constant motion means that two or more gases will always mix, as the particles from the individual gases move and collide with each other. An ideal gas is a gas that exactly follows the statements of the kinetic theory. Unfortunately, real gases are not ideal. Many gases deviate slightly from agreeing perfectly with the kinetic theory of gases. However, most gases adhere to the statements so well that the kinetic theory of gases is well accepted by the scientific community. The physical behavior of gases is explained by the kinetic theory of gases. An ideal gas adheres exactly to the kinetic theory of gases. |
Ancillary_Materials/Worksheets/Worksheets%3A_Inorganic_Chemistry/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry/05%3A_Stereochemistry/5.11%3A_Carbohydrates_in_Cyclic_Form | Carbohydrates are complicated molecules. In solution, they slowly change into different isomers. In water, most carbohydrates can change from one form to another quickly; they are described as being in equilibrium with different structures. That means they can change back and forth. The most prevalent form for most carbohydrates is a ring. One of the oxygens farther along the chain can reach around and bond to the carbon in the C=O at the head of the chain. When that happens, there are two possible orientations of the oxygen at the head of the chain. The two stereoisomers that result are diastereomers. Some of the chiral centers are the same, but one is different. Figure SC11.1. D-threose and its two possible five-membered-ring forms. For example, D-threose can form two different five-membered rings (that is, rings made from a circle of five atoms). In carbohydrates, a five-membered ring with an oxygen in it is called a furanose (another common form, a six-membered ring, is called a pyranose). There are two diastereomers formed. they are sometimes called anomers. That means they differ in 3D space at the anomeric center; the anomeric center is the C=O carbon to which an oxygen binds to form the ring. These drawings, by the way, are called Haworth projections. They are commonly used in biochemistry to depict the cyclic forms of sugars. They seem to suggest a funny see-saw shape for the carbon atoms, but the idea here is not to convey the shape of the molecule perfectly. Just like in Fischer projections, the main concern is to quickly convey stereochemical relationships. In a Haworth projection, you can easily see whether two groups are cis to each other or trans to each other. Note that a Haworth projection of an odd-numbered ring is always drawn with the ring oxygen in the back. We are looking at the ring from the edge and slightly above, so the "back" is the upper edge of the polygon. It's just like if you were looking at a swimming pool, and the far corner of the pool appeared in your field of vision above the near corner. For an even-numbered ring, the same idea applies, but the ring oxygen is always drawn on the top edge and to the right. If you want to get an idea of the true shape of the molecule, you can see it in the ball-and-stick models below. Figure SC11.2. Ball-and-stick model of β-D-threofuranose. Figure SC11.3. Ball-and-stick model of α-D-threofuranose. Go to Animation SC11.1. A three-dimensional model of β-D-threofuranose. Go to Animation SC11.2. Another three-dimensional model of α-D-threofuranose. Carbohydrates are normally found in their cyclic forms, not their chain forms. For every ring a carbohydrate can form, there are always two diastereomeric forms called anomers. The two terms, α and β, that described the two different stereoisomers of the cyclic form of threose refer to the position of the OH group on the newly formed chiral center (the old C=O was trigonal planar, not tetrahedral, so it wasn't chiral). The terms are defined by looking at the Haworth projection of the molecule. If the new OH is in the upper position, it is termed β (think butterfly); it is is in the lower position, it is termed α (think ant). Carbohydrates do not always make five-membered rings. Sometimes they make six-membered rings instead. A five-membered ring with an oxygen in it is called a furanose. A six-membered ring with an oxygen in it is called a pyranose. Ribose is a five-carbon sugar that could make either a five-membered or a six-membered ring, although we usually see it in its five-membered ring. Figure SC11.4. Ribose in equilibrium between its chain form and some of its ring forms. The presence of different diastereomers that can change back and forth complicates things. If you wanted to measure the optical activity of a sugar, you would probably open a bottle of the solid carbohydrate, dissolve some up in water and put the sample in a polarimeter. However, the optical rotation would change over time as the pure solid slowly turned into other isomers. This process is called mutarotation (which means changing optical rotation). For this reason, the time at which a measurement was performed is often reported with optical rotation values of carbohydrates. For example, the optical rotation value for D-threose may be listed as [α] D -12.3 (20 min, c = 4, water). Carbohydrates are important partly because they are structural units in biology. Carbohydrates are incorporated into DNA and RNA as well as important enzymatic cofactors such as ATP, NADH and Acetyl Coenzyme A. Very frequently, the unique pieces attached to the carbohydrates to make these different molecules are attached via the anomeric carbon. Specific OH groups on carbohydrates are frequently substituted with other groups to make more complex molecules. One more piece of terminology may be useful to review here. There are terms used in ring structures to describe whether two groups are attached to the same face or on opposite faces. For example, in β-D-threofuranose, the second hydroxy group is cis to the one on the anomeric center, but in α-D-threofuranose, the second hydroxy group is trans to the one on the anomeric center. Two groups on the same face of a ring are described as being cis to each other. Two groups on the opposite faces of a ring are described as being trans to each other. The terms cis and trans can be applied to any rings, not just carbohydrates. Problem SC11.1. In the following carbohydrate, glucose, show the cyclic structures that would form if the indicated oxygen bonded to the carbonyl carbon (the C=O in the CHO group at the top). You should move a proton from one oxygen to another so that all the atoms have the usual number of bonds. Problem SC11.2 . β-deoxyribofuranose (below) is a building block of DNA. a) What is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring? b) In the other possible 5-membered ring form of deoxyribose, what is the relationship between the two hydroxy (OH) groups that are directly attached to the 5-membered ring? |
Bookshelves/Analytical_Chemistry/Instrumental_Analysis_(LibreTexts)/25%3A_Voltammetry/25.06%3A_Stripping_Methods | Another important voltammetric technique is stripping voltammetry, which consists of three related techniques: anodic stripping voltammetry, cathodic stripping voltammetry, and adsorptive stripping voltammetry. Because anodic stripping voltammetry is the more widely used of these techniques, we will consider it in greatest detail. Anodic stripping voltammetry consists of two steps (Figure \(\PageIndex{1}\)). The first step is a controlled potential electrolysis in which we hold the working electrode—usually a hanging mercury drop or a mercury film electrode—at a cathodic potential sufficient to deposit the metal ion on the electrode. For example, when analyzing Cu 2 + the deposition reaction is \[\mathrm{Cu}^{2+}+2 e^{-} \rightleftharpoons \mathrm{Cu}(\mathrm{Hg}) \label{sv1} \] where Cu(Hg) indicates that the copper is amalgamated with the mercury. This step serves as a means of concentrating the analyte by transferring it from the larger volume of the solution to the smaller volume of the electrode. During most of the electrolysis we stir the solution to increase the rate of deposition. Near the end of the deposition time we stop the stirring—eliminating convection as a mode of mass transport—and allow the solution to become quiescent. Typical deposition times of 1–30 min are common, with analytes at lower concentrations requiring longer times. In the second step, we scan the potential anodically—that is, toward a more positive potential. When the working electrode’s potential is sufficiently positive, the analyte is stripped from the electrode, returning to solution in its oxidized form. \[\mathrm{Cu}(\mathrm{Hg})\rightleftharpoons \text{ Cu}^{2+}+2 e^{-} \label{sv2} \] Monitoring the current during the stripping step gives a peak-shaped voltammogram, as shown in Figure \(\PageIndex{1}\). The peak current is proportional to the analyte’s concentration in the solution. Because we are concentrating the analyte in the electrode, detection limits are much smaller than other electrochemical techniques. An improvement of three orders of magnitude—the equivalent of parts per billion instead of parts per million—is routine. Applications Anodic stripping voltammetry is very sensitive to experimental conditions, which we must carefully control to obtain results that are accurate and precise. Key variables include the area of the mercury film or the size of the hanging Hg drop, the deposition time, the rest time, the rate of stirring, and the scan rate during the stripping step. Anodic stripping voltammetry is particularly useful for metals that form amalgams with mercury, several examples of which are listed in Table \(\PageIndex{1}\). anodic stripping voltammetry cathodic stripping voltammetry adsorptive stripping voltammetry Bi3+ Br– bilirubin Cd2+ Cl– codeine Cu2+ I– cocaine Ga3+ mercaptans (RSH) digitoxin In3+ S2– dopamine Pb2+ SCN– heme Tl+ NaN monesin Sn2+ NaN testosterone Zn2+ NaN NaN Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. Source: Compiled from Peterson, W. M.; Wong, R. V. Am. Lab. November 1981, 116–128; Wang, J. Am. Lab. May 1985, 41–50. The experimental design for cathodic stripping voltammetry is similar to anodic stripping voltammetry with two exceptions. First, the deposition step involves the oxidation of the Hg electrode to \(\text{Hg}_2^{2+}\), which then reacts with the analyte to form an insoluble film at the surface of the electrode. For example, when Cl – is the analyte the deposition step is \[2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \text{ Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-} \label{sv3} \] Second, stripping is accomplished by scanning cathodically toward a more negative potential, reducing \(\text{Hg}_2^{2+}\) back to Hg and returning the analyte to solution. \[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons 2 \mathrm{Hg}( l)+2 \mathrm{Cl}^{-}(a q) \label{sv4} \] Table \(\PageIndex{1}\) lists several analytes analyzed successfully by cathodic stripping voltammetry. In adsorptive stripping voltammetry, the deposition step occurs without electrolysis. Instead, the analyte adsorbs to the electrode’s surface. During deposition we maintain the electrode at a potential that enhances adsorption. For example, we can adsorb a neutral molecule on a Hg drop if we apply a potential of –0.4 V versus the SCE, a potential where the surface charge of mercury is approximately zero. When deposition is complete, we scan the potential in an anodic or a cathodic direction, depending on whether we are oxidizing or reducing the analyte. Examples of compounds that have been analyzed by absorptive stripping voltammetry also are listed in Table \(\PageIndex{1}\). |
Courses/Lumen_Learning/Book%3A_Western_Civilization_(Lumen)/Ch._12_The_Rise_of_Nation-States/13.16%3A_The_Sun-King_and_Authoritarianism | Learning Objective Describe Louis XIV’s views on royal power and how he expanded his own authority Key Points At the time of King Louis XIII’s death in 1643, Louis XIV was only five years old. His mother, Anne of Austria, was named regent, but she entrusted the government to the chief minister, Cardinal Mazarin. Mazarin’s policies paved the way for the authoritarian reign of Louis XIV. Louis began his personal reign with administrative and fiscal reforms. National debt was quickly reduced through more efficient taxation, although reforms imposing taxes on the aristocracy were late and of limited outcome. Louis and his administration also bolstered French commerce and trade by establishing new industries in France and instituted reforms in military administration that curbed the independent spirit of the nobility by imposing order at court and in the army. Louis also attempted uniform regulation of civil procedure throughout legally irregular France by issuing a comprehensive legal code, the “Grande Ordonnance de Procédure Civile” of 1667, also known as the Code Louis. One of his most infamous decrees was the Code Noir, which sanctioned slavery in French colonies. Louis also attached nobles to his court at Versailles and thus achieved increased control over the French aristocracy. An elaborate court ritual by which the king observed the aristocracy and distributed his favors was created to ensure the aristocracy remained under his scrutiny. Following consistent efforts to limit religious tolerance, Louis XIV issued the Edict of Fontainebleau, which revoked the Edict of Nantes and repealed all the privileges that arose therefrom. Terms cuius regio, eius religio A Latin phrase that literally means “Whose realm, his religion,” meaning that the religion of the ruler was to dictate the religion of those ruled. At the Peace of Augsburg of 1555, which ended a period of armed conflict between Roman Catholic and Protestant forces within the Holy Roman Empire, the rulers of the German-speaking states and Charles V, the emperor, agreed to accept this principle. Declaration of the Clergy of France A four-article document of the 1681 Assembly of the French clergy promulgated in 1682, which codified the principles of Gallicanism into a system for the first time in an official and definitive formula. Gallicanism The belief that popular civil authority—often represented by the monarchs’ authority or the state’s authority—over the Catholic Church is comparable to the pope’s. Edict of Fontainebleau A 1685 edict issued by Louis XIV of France, also known as the Revocation of the Edict of Nantes. The Edict of Nantes (1598) had granted the Huguenots the right to practice their religion without persecution from the state. Code Noir A decree originally passed by France’s King Louis XIV in 1685 that defined the conditions of slavery in the French colonial empire, restricted the activities of free black persons, forbade the exercise of any religion other than Roman Catholicism, and ordered all Jews out of France’s colonies. The Sun King At the time of King Louis XIII’s death in 1643, Louis XIV was only five years old. His mother, Anne of Austria, was named regent in spite of her late husband’s wishes. Anne assumed the regency but entrusted the government to the chief minister, Cardinal Mazarin, who helped her expand the limited power her husband had left her. He functioned essentially as the co-ruler of France alongside Queen Anne during her regency, and until his death effectively directed French policy alongside the monarch. In 1651, when Louis XIV officially came of age, Anne’s regency legally ended. However, she kept much power and influence over her son until the death of Mazarin. On the death of Mazarin in 1661, Louis assumed personal control of the reins of government and astonished his court by declaring that he would rule without a chief minister. Louis XIV, King of France, in 1661 by Charles Le Brun. After Mazarin’s death in 1661, Louis assumed personal control of the reins of government and astonished his court by declaring that he would rule without a chief minister: “Up to this moment I have been pleased to entrust the government of my affairs to the late Cardinal. It is now time that I govern them myself. You [he was talking to the secretaries and ministers of state] will assist me with your counsels when I ask for them. I request and order you to seal no orders except by my command … I order you not to sign anything, not even a passport … without my command; to render account to me personally each day and to favor no one.” Reforms Louis began his personal reign with administrative and fiscal reforms. In 1661, the treasury verged on bankruptcy. To rectify the situation, Louis chose Jean-Baptiste Colbert as Controller-General of Finances in 1665. Colbert reduced the national debt through more efficient taxation. Excellent results were achieved, and the deficit of 1661 turned into a surplus in 1666. However, to support the reorganized and enlarged army, the panoply of Versailles, and the growing civil administration, the king needed a good deal of money, but methods of collecting taxes were costly and inefficient. The main weakness of the existing system arose from an old bargain between the French crown and nobility: the king might raise without consent if only he refrained from taxing the nobles. Only towards the close of his reign, under extreme stress of war, was Louis able, for the first time in French history, to impose direct taxes on the aristocracy. This was a step toward equality before the law and toward sound public finance, but so many concessions and exemptions were won by nobles and bourgeois that the reform lost much of its value. Louis and Colbert also had wide-ranging plans to bolster French commerce and trade. Colbert’s administration established new industries, encouraged domestic manufacturers and inventors, and invited manufacturers and artisans from all over Europe to France. This aimed to decrease foreign imports while increasing French exports, hence reducing the net outflow of precious metals from France. Louis also instituted reforms in military administration and, with the help of his trusted experts, curbed the independent spirit of the nobility by imposing order at court and in the army. Gone were the days when generals protracted war at the frontiers while bickering over precedence and ignoring orders from the capital and the larger politico-diplomatic picture. The old military aristocracy ceased to have a monopoly over senior military positions and rank. Louis also attempted uniform regulation of civil procedure throughout legally irregular France by issuing a comprehensive legal code, the “Grande Ordonnance de Procédure Civile” of 1667, also known as the Code Louis. Among other things, it prescribed baptismal, marriage, and death records in the state’s registers, not the church’s, and also strictly regulated the right of the Parlements to remonstrate. The Code Louis played an important part in French legal history as the basis for the Napoleonic code, itself the origin of many modern legal codes.One of Louis’s most infamous decrees was the Grande Ordonnance sur les Colonies of 1685, also known as the Code Noir (“black code”). It sanctioned slavery and limited the ownership of slaves in the colonies to Roman Catholics only. It also required slaves to be baptized. Centralization of Power Louis initially supported traditional Gallicanism, which limited papal authority in France, and convened an Assembly of the French clergy in November 1681. Before its dissolution eight months later, the assembly had accepted the Declaration of the Clergy of France, which increased royal authority at the expense of papal power. Without royal approval, bishops could not leave France and appeals could not be made to the pope. Additionally, government officials could not be excommunicated for acts committed in pursuance of their duties. Although the king could not make ecclesiastical law, all papal regulations without royal assent were invalid in France. Unsurprisingly, the pope repudiated the declaration. Louis also attached nobles to his court at Versailles and thus achieved increased control over the French aristocracy. Apartments were built to house those willing to pay court to the king. However, the pensions and privileges necessary to live in a style appropriate to their rank were only possible by waiting constantly on Louis. For this purpose, an elaborate court ritual was created where the king became the center of attention and was observed throughout the day by the public. With his excellent memory, Louis could see who attended him at court and who was absent, facilitating the subsequent distribution of favors and positions. Another tool Louis used to control his nobility was censorship, which often involved opening letters to discern their author’s opinion of the government and king. Moreover, by entertaining, impressing, and domesticating nobles with extravagant luxury and other distractions, Louis not only cultivated public opinion of himself, but also ensured the aristocracy remained under his scrutiny. This, along with the prohibition of private armies, prevented the aristocracy from passing time on their own estates and in their regional power-bases, from which they historically had waged local wars and plotted resistance to royal authority. Louis thus compelled and seduced the old military aristocracy (the “nobility of the sword”) into becoming his ceremonial courtiers, further weakening their power. In their place, he raised commoners or the more recently ennobled bureaucratic aristocracy as presumably easier to control. Religion Finally, Louis dramatically limited religious tolerance in France, as he saw the persistence of Protestantism as a disgraceful reminder of royal powerlessness. Responding to petitions, Louis initially excluded Protestants from office, constrained the meeting of synods, closed churches outside Edict of Nantes-stipulated areas, banned Protestant outdoor preachers, and prohibited domestic Protestant migration. He also disallowed Protestant-Catholic intermarriages where third parties objected, encouraged missions to the Protestants, and rewarded converts to Catholicism. In 1681, Louis dramatically increased his persecution of Protestants. The principle of cuius regio, eius religio generally had also meant that subjects who refused to convert could emigrate, but Louis banned emigration and effectively insisted that all Protestants must be converted. In 1685, he issued the Edict of Fontainebleau, which cited the redundancy of privileges for Protestants given their scarcity after the extensive conversions. It revoked the Edict of Nantes and repealed all the privileges that arose therefrom. By his edict, Louis no longer tolerated Protestant groups, pastors, or churches to exist in France. No further churches were to be constructed, and those already existing were to be demolished. Pastors could choose either exile or a secular life. Those Protestants who had resisted conversion were now to be baptized forcibly into the established church. Sources CC licensed content, Shared previously Boundless World History. Authored by : Boundless. Located at : https://www.boundless.com/world-history/textbooks/boundless-world-history-textbook/ . License : CC BY-SA: Attribution-ShareAlike |
Courses/Indiana_Tech/EWC%3A_CHEM_1020_-_General_Chemistry_I_(Budhi)/01%3A_Matter_Measurement_and_Problem_Solving/1.9%3A_Matter_Measurement_and_Problem_Solving_(Exercises) | Template:HideTOC These are homework exercises to accompany the Textmap created for Chemistry: A Molecular Approach by Nivaldo Tro. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here . In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission. Q1.35 A household receives a $125 electricity bill. The cost of electricity is $0.150 /kWh. How much energy, in joules (J), did the household use? Strategy Step 1 : You must first multiply the total cost of the bill ($125) by 1 kWh and then divide that answer by the cost of each kW ($0.150 kWh). You should end up with an equation that looks like this: \[($125)\times \dfrac{1\;kWh}{0.150\;kWh}\nonumber \] Step 2 : Once you do this math, then you should receive the answer of 833.33 kWh. This is the number of kWh used during the billing cycle. Step 3 : Next, you will then take your solution found during step 1 and multiply it by \(\dfrac{1}{2.78\times 10^{-7}}\) in order to convert kWh to kWh/J. You should have the equation: \[(833.33)\times \dfrac{1}{2.78\cdot 10^{-7}}\nonumber \] Step 4 : After completing step 3, you should receive an answer of \[3.0 \times 10^{9} J\nonumber \] Step 5 : YOU ARE FINISHED! Q1.36 Determine whether the mixtures are Homogeneous or Heterogeneous. State whether it is a compound or pure substance? Mud and Water Salt and water Chicken noodle soup Gold \(H_{2}O\) Strategy Homogeneous: mixtures are uniformly distributed, easily dissolved, of a single phase, also they are easily dispersed through the membrane. Heterogeneous: mixtures are unequally distributed, do not dissolve in water, and could be of different phases. A pure substance: has a definite and constant composition, it can be a compound or and element. A element: is composed of a single atom. A compound is composed of two or more elements. For example, \(H_{2}O\) is a ratio of two hydrogen atoms and one oxygen atom to form \(H_{2}O\). So with this strategy you can solve the following question. Mud and Water: Heterogeneous mixture Salt and water: Homogeneous mixture Chicken noodle soup: Heterogeneous mixture Gold: (element) \(H_{2}O\): Compound Q1.36 Classify each substance as a pure substance or a mixture. If it is a pure substance, classify it as an element or a compound. If it is a mixture, classify it as homogeneous or heterogeneous. Plain yogurt Chicken noodle soup Titanium Table salt What we know: A pure substance is made up of only one type of particle. A mixture is a substance composed of two or more particles. An element is a substance that cannot be chemically broken down into simpler substances. A compound is a substance composed of two or more elements. A heterogeneous mixture is one in which the composition varies from one region of the mixture to another. A homogeneous mixture is one with the same composition throughout. Solution mixture, homogeneous mixture, heterogeneous pure substance, element pure substance, compound Q1.41 Classify Each statement as an Observation, Theory, or Law. All matter that exists is made up of Submicroscopic particles known as atoms. If an iron rod is placed in a closed container and it rusts, the mass of the container and its contents do not change. Matter is neither created nor destroyed When a candle is burned, heat is released Strategy You need to understand what an Observation, Theory or law is before you classify it in the question. Observation : An outcome of events that is being viewed Theory : Preserved explanation for observations and laws Law : Summarizes past observations and predicts future ones Now that we have defined them, we need to apply it to the question (AnswerS IN BOLD) This statement is proposing all matter is made up of atoms so it would become a theory because it is an idea that studies propose is true. This statement is saying, without matter being added to the container, As you See it rust, the mass doesn't change so this is assumed by Observation. This statement is called the Law of conservation of mass, so because this is true, the statement is a Law. If you are looking at a candle and see the flame, then you should understand that heat is being released. You tell this from Observation . Q1.45 When 12 g of sodium reacts with 23.5 g of chlorine to form sodium chloride, how many grams of sodium chloride is formed? (Assume that sodium chloride is the only product). Solution: What we know: We have a chemical reaction between sodium and chlorine to form only sodium chloride. We are also given the masses of the reactants, 12 g of sodium and 23.5 g of chlorine. We also know that the Law of Conservation of Mass tells us that matter is neither created nor destroyed in a chemical reaction. What we're asked for: The mass, in grams, of sodium chloride formed from the reaction. Strategy: A. Find the sum of the masses of the reactants from the chemical equation. Solution 12 g + 23.5 g = 35.5 g 35.5 g of sodium chloride is formed from the reaction. Q1.55 According to Dalton's Atomic Theory, what are compounds made up of? Strategy Find the summary of Dalton's Atomic Theory Read through it Locate where it talks about what compounds are made of Write your answer Solution Compounds are made up of two or more different atoms. Q1.64 Isotopes are atoms that have the same number of protons but differ in the number of neutrons. Write the chemical symbols for each of the following isotopes: the chlorine isotope with 18 neutrons the chlorine isotope with 20 neutrons the calcium isotope with 20 neutrons the neon isotope with 11 neutrons Strategy : Determine the given element's atomic number from the periodic table. Add the number of neutrons given in the question to the atomic number to calculate the mass number of the isotope. The mass number will be represented in the upper left corner of the chemical symbol, while the atomic number will be represented in the lower left corner. Helpful Hints: A chemical symbol is written as \[_{Z}^{A}\textrm{X}\nonumber \] . The A stands for the number of protons and neutrons, also known as the mass number. The Z stands for the number of protons, also known as the atomic number. Subtracting Z from A will give you the number of neutrons. Example: \[_{19}^{39}\textrm{K}\nonumber \] --> 39-19= 20 neutrons Therefore, adding the given number of neutrons to the atomic number, Z , gives you the mass number. Solution : \[_{17}^{35}\textrm{Cl}\nonumber \] \[_{17}^{37}\textrm{Cl}\nonumber \] \[_{20}^{40}\textrm{Ca}\nonumber \] \[_{10}^{21}\textrm{Ne}\nonumber \] Q1.65 Give the number of protons and neutrons in each isotope listed \[_{19}^{41}\textrm{K}\nonumber \] \[_{4}^{11}\textrm{Be}\nonumber \] \[_{22}^{45}\textrm{Ti}\nonumber \] \[_{8}^{16}\textrm{O}\nonumber \] Strategy Take the larger number and subtract the smaller number from it. The smaller number is the atomic number (z), which indicates the number of protons the element has. The larger number indicates the number of protons and neutron present in the element. 41 K has an atomic mass of 41. The atomic mass is the number of protons and neutrons present in the element. To get the number of neutrons, subtract the atomic number (19) from the atomic mass (41). When done so you find that K 41 has 19 protons and 22 neutrons. Keep in mind the atomic number is the number of protons present in the element, so the only thing you are trying to find through subtraction is the number of neutrons. For 11 Be, the same thing occurs. Take the atomic number and subtract it from the atomic mass. The atomic mass is 11 and the atomic number is 4, so, 11 minus for equals 7. The answer for part b. is 4 protons and 7 neutrons. For 45 Ti, you subtract 22 from 45, which gives you 22 protons and 23 neutrons. For 16 O, the same method is applied. Subtract 8 from 16 and you get 8 protons and 8 electrons. Q1.66 Question: Find the amount of protons and neutrons in the following atoms: \[_{6}^{14}\textrm{C}\nonumber \] \[_{8}^{18}\textrm{O}\nonumber \] \[_{25}^{55}\textrm{Mn}\nonumber \] \[_{30}^{64}\textrm{Zn}\nonumber \] Strategy First identify the subscript, the bottom number on the left hand side of the element's symbol. This is the atomic number and indicates how many protons the atom's nucleus contains. All isotopes of the same element contain the same number of protons. Next, identify the superscript, the top number left hand side of the element symbol. This is the atomic mass number, the total combined number of protons and neutrons. Different isotopes of a given element differ in the number of neutrons contained in their nuclei. The number of protons can be found simply by identifying the atomic number. To find the number of neutrons in the atom, the atomic number is subtracted from the atomic mass number. Solution A The atomic number of carbon is 6, the atomic number of oxygen is 8, the atomic number of manganese is 25, and the atomic number of zinc is 30. Therefore, we now know the number of protons contained within each atom since this number is represented by the atomic number. B The atomic mass number of the carbon isotope is 14, the mass number of the oxygen isotope is 18, the mass number of the manganese isotope is 55, and the mass number of the zinc isotope is 64. This is the combined number of protons and neutrons in each respective isotope. C Carbon isotope: \[14-6=8\, neutrons\nonumber \] Oxygen isotope: \[18-8=10\, neutrons\nonumber \] Manganese isotope: \[55-25=30\, neutrons\nonumber \] Zinc isotope: \[64-30=34\, neutrons\nonumber \] Q1.64 Write the isotopic symbols in the fo rm \(\ce{^A_ZX}\) for each isotope. The oxygen isotope with 9 neutrons. The neon isotope with 10 neutrons. The chlorine isotope with 18 neutrons. The carbon isotope with 8 neutrons. Strategy Identify each element’s atomic number (number of protons). Find the atomic mass of each element by adding the number of protons and the number of neutrons. Properly write the isotopic symbol of each element by placing the atomic mass as the superscript and the atomic number (number of protons) as the subscript. Solution Atomic Numbers (Number of Protons): Oxygen: 8 protons Neon: 10 protons Chlorine: 17 protons Carbon: 6 protons Atomic Mass: Oxygen: 8+9=17 Neon: 10+10=20 Chlorine: 17+18=35 Carbon: 6+8=14 Isotopic Symbols: Oxygen: 17 8 O Neon: 20 10 Ne Chlorine: 35 17 Cl Carbon: 14 6 C Q1.69 Question: Determine the number of protons and electrons in each of the following ions. Strategy First locate each element on the periodic table. Identify the atomic number associated with each element, or the number in between the element's symbol and scientific name. The atomic number is the number of protons as well as the number of electrons an element possesses in it's neutral state. Next, identify the superscript of each element, or the number to the top right of the element's symbol. This number should include a "+" or "-" sign next to it, indicating the element's charge. A "+" indicates a positively charged ion and a "-" indicates a negatively charged ion. An element becomes positively charged when it loses electrons, and negative when it gains electrons. To solve for each element's total number of electrons, take that element's atomic number, and inversely add or subtract the number in the superscript according to the element's charge. If an element has a "+" sign, it is losing the amount of electrons indicated by the superscript number, so subtract this number from the element's atomic number. The same goes if an element has a "-" sign, instead adding the number in the superscript to the element's atomic number. Solution (1) N 3- A. Nitrogen's atomic number is 7, so in a neutral state, Nitrogen has 7 protons and electrons. B. Nitrogen's superscript is "3-" indicating that it is gaining 3 electrons C. Because it gains 3 electrons, N 3- now has a total of 10 electrons. (2) Be 2 + A. Beryllium's atomic number is 4, so Beryllium possesses 4 protons and electrons B. The superscript "2+" indicates beryllium is losing 2 electrons C. 2 electrons are subtracted from Beryllium's total 4, so it now has 2 electrons. (3) K + A. Potassium as a neutral element possesses 19 protons and electrons, indicated by it's atomic number of 19. B. The superscript "+" indicates Potassium will lose 1 electron from it's total. C. 19 total electrons minus 1 makes K + number of electrons to 18. (4) Se 2 - A. Selenium's atomic number is 34, indicating 34 protons and electrons at ground state. B. It' superscript "2-" indicates that it will be gaining 2 electrons to it's total. C. 34 electrons plus 2 makes selenium's total number of electrons 34. Q 1.73 The atomic mass for Helium is 4.003 amu and the mass spectrum for Helium shows a spike at this mass. Meanwhile, the atomic mass for Bromine is 79.904 amu; however, the mass spectrum for Bromine does not show a peak at that mass. What causes the difference? Solution The best way to start this problem is to recall what the atomic mass means in the context of the problem. This is the standard atomic mass, which is the weighted average of all the isotopes. "Weighted" in this situation means that the abundance of each isotope factors in to how it factors in to the overall the standard atomic mass for that particular element. Next step is to look at Helium. Helium has two isotopes and is stable at either; however, in atmospheric values, Helium is found most abundantly as 4 He (99.999863%) and much less commonly found as 3 He (0.000137%). Consequently, 3 He does not have a strong influence on the standard atomic mass of Helium or much expression on the mass spectrum which measures relative abundance. Examining Bromine reveals two major isotopes for that element as well. Both 79 Br as well as 81 Br are rather prevalent ( 79 Br with 50.69% and 81 Br with 49.31%). This means that their relative abundance is high enough that they would make a significant impact on both the mass spectrum as well as the standard atomic weight of Bromine. To connect the dots, the dichotomy between the two elements comes from the abundance of their isotopes. Bromine's two isotopes almost equally comprise the atomic mass so the atomic mass they produce does not match up with an actual relatively abundant isotope of Br so therefore the line for that particular atomic mass is absent on the mass spectrum for Br. Helium, meanwhile, is primarily influenced by the 3 He because of its incredibly high abundance and therefore the mass spectrum and the standard atomic mass are very similar. Q1.74 The atomic mass of chlorine is 35.453 amu. There are two stable isotopes for chlorine. Would either of these isotopes have a mass of 35.453 amu? Why or why not? Important things to consider The atomic mass is the sum of the protons and neutrons. Isotopes vary in the amount of neutrons of the same element. Each isotope will have a different mass number depending upon how many neutrons are present. amu = atomic mass unit unit used for indicating mass on an atomic scale Answer No, because isotopes are variants of an element that differ in the number of neutrons present, but they have the same number of protons (atomic number the same). |
Courses/Furman_University/CHM101%3A_Chemistry_and_Global_Awareness_(Gordon)/01%3A_Introduction_to_Chemistry | What is chemistry? Simply put, chemistry is the study of the interactions of matter with other matter and with energy. This seems straightforward enough. However, the definition of chemistry includes a wide range of topics that must be understood to gain a mastery of the topic or even take additional courses in chemistry. Get ready for a fantastic journey through a world of wonder, delight, and knowledge. One of the themes of this book is “chemistry is everywhere,” and indeed it is; you would not be alive if it weren’t for chemistry because your body is a big chemical machine. If you don’t believe it, don’t worry. Every chapter in this book contains examples that will show you how chemistry is, in fact, everywhere. So enjoy the ride—and enjoy chemistry. 1.1: What is Chemistry? Chemistry is the study of matter—what it consists of, what its properties are, and how it changes. Being able to describe the ingredients in a cake and how they change when the cake is baked is called chemistry. Matter is anything that has mass and takes up space—that is, anything that is physically real. Some things are easily identified as matter—this book, for example. 1.2: Using the Scientific Method Scientists search for answers to questions and solutions to problems by using a procedure called the scientific method. This procedure consists of making observations, formulating hypotheses, and designing experiments, which in turn lead to additional observations, hypotheses, and experiments in repeated cycles 1.3: Pure and Applied Research In science, we usually talk about two types of research: pure and applied. Pure research focuses on answering basic questions such as, "how do gases behave?" Applied research would be involved in the process of developing specific preparation for a gas in order for it to be produced and delivered efficiently and economically. This division sounds like it would be easy to make, but sometimes we cannot draw a clear line between what is "pure" and what is "applied". 1.4: Classification and Properties of Matter The properties that chemists use to describe matter fall into two general categories. Physical properties are characteristics that describe matter. They include characteristics such as size, shape, color, and mass. Chemical properties are characteristics that describe how matter changes its chemical structure or composition. An example of a chemical property is flammability—a material’s ability to burn—because burning (also known as combustion) changes the chemical composition of a material. 1.5: Applying Properties - MSDS/SDS A Material Safety Data Sheet (MSDS) is designed to provide both workers and emergency personnel with the proper procedures for handling or working with a particular substance. MSDS's include information such as physical data (melting point, boiling point, flash point etc.), toxicity, health effects, first aid, reactivity, storage, disposal, protective equipment, and spill/leak procedures. 1.6: All About the Elements We will delve more into the initial construction and modifications of today's periodic table. For now, you need to be aware that the periodic table has columns (known as families/groups). Horizontal rows are called periods. On the table below, the element symbols are either one of two letters. Previously, some of the newly discovered elements used three letter symbols. These three letter symbols corresponded to a Latin numbering system until the elements were given official names. |
Bookshelves/Environmental_Chemistry/Green_Chemistry_and_the_Ten_Commandments_of_Sustainability_(Manahan)/07%3A_Chemistry_of_Life_and_Green_Chemistry/7.01%3A_New_Page | Biochemistry is the science of chemical processes that occur in living organisms. 1 By its nature biochemistry is a green chemical and biological science. This is because over eons of evolution organisms have evolved that carry out biochemical processes sustainably. Because the enzymes that carry out biochemical processes can only function under mild conditions, particularly of temperature, biochemical processes take place under safe conditions, avoiding the high temperatures, high pressures, and corrosive and reactive chemicals that often characterize synthetic chemical operations. Therefore, it is appropriate to refer to green biochemistry . The ability of organisms to carry out chemical processes is truly amazing, even more so when one considers that many of them occur in single-celled organisms. Photosynthetic cyanobacteria consisting of individual cells less than a micrometer (μm) in size can make all the complex biochemicals they need to exist and reproduce using sunlight for energy and simple inorganic substances such as CO 2 , K + ion, NO 3 - ion and HPO 4 2- ion for raw materials. Beginning soon after conditions on Earth became hospitable to life, these photosynthetic bacteria produced the oxygen that now composes about 20% of Earth’s atmosphere. Fossilized stromatolites (bodies of sedimentary materials bound together by films produced by microorganisms) produced by cyanobacteria have been demonstrated dating back 2.8 billion years, and this remarkable microorganism that converts atmospheric carbon dioxide to biomass and atmospheric N 2 to chemically fixed N may have been on Earth as long as 3.5 billion years ago. It is fascinating to view single live cells of animal-like protozoa through an optical microscope. An ameba appears as a body of cellular protoplasm and moves by oozing about like a living blob of jelly. Examination of Euglena protozoa may show a cell several μm in size with many features including a cell nucleus that serves to direct metabolism and reproduction, green chloroplasts for photosynthetic production of biomass, a red eye-spot sensitive to light, a mouth-like contractile vacuole by which the cell expels excess water, and a thin tail-like structure(flagella) that moves rapidly and propels the cell. More detailed examination by electron microscope of such cells and those that make up more complex organisms reveals many more cell parts that are involved with biochemical function. At least a rudimentary knowledge of biochemistry is needed to understand green chemistry, environmental chemistry, and sustainability science and technology. One reason is the ability of organisms to synthesize a vast variety of substances. The most obvious of these is biomass made by the photosynthetic fixation of carbon dioxide and that forms the basis of nature’s food webs.Organisms make many of the materials upon which humans rely. In addition to food, one such material is the lignocellulose that composes most of plant biomass such as wood used for construction, paper-making, and fuel. Very complex molecules are made by organisms, for example, human insulin produced by genetically engineered organisms. Organisms make materials under very mild conditions compared to those used in the anthrosphere. An important example is chemically fixed nitrogen from the atmosphere which is produced synthetically in the anthrosphere as ammonia (NH 3 ) at high temperatures and pressures whereas Rhizobium bacteria attached to the roots of soybeans and other legume plants fix nitrogen in the mild conditions of the soil environment. Increasingly as supplies of petroleum and other non-renewable raw materials become more scarce, humans are turning to microorganisms and plants to make essential materials. Another major reason for considering biochemistry as part of green chemistry and sustainability is the protection of organisms from products and processes in the anthrosphere. It is essential to know the potential toxic effects of various materials, a subject addressed by toxicological chemistry . 2 One of the fundamental goals of green chemistry is to minimize the production and use of products that may have adverse environmental effects. Sustainability of the entire planet requires that humans not disperse into the environment substances that may undergo bioaccumulation and be toxic to humans and other organisms. Biochemical processes not only are profoundly influenced by chemical species in the environment, they largely determine the nature of these species, their degradation, and even their syntheses, particularly in the aquatic and soil environments. The study of such phenomena forms the basis of environmental biochemistry. This chapter is designed to give an overview of biochemistry and how it relates to green chemistry and sustainability science and technology. A glance at the structural formulas of some of the biochemicals shown in this chapter gives a hint of the complexity of biochemistry. This chapter is designed to provide a basic understanding of this complex science with enough detail for it to be meaningful but to avoid overwhelming the reader. It begins with an overview of the four major classes of biochemicals— proteins, carbohydrates, lipids, and nucleic acids. Many of the compounds in these classes are polymers with molecular masses of the order of a million or even larger. Proteins and nucleic acids consist of macromolecules, lipids are usually relatively small molecules, carbohydrates range from small sugar molecules to high-molar-mass macromolecules such as those in cellulose. The behavior of a substance in a biological system depends to a large extent upon whether the substance is hydrophilic (“water-loving”) or hydrophobic (“water-hating”). Some important toxic substances are hydrophobic, a characteristic that enables them to traverse cell membranes readily and to bioaccumulate in lipid (fat) tissue. Many hydrocarbons and organohalide compounds synthesized from hydrocarbons are hydrophobic. Part of the detoxification process carried on by living organisms is to render such molecules hydrophilic, therefore water-soluble and readily eliminated from the body. |
Courses/Riverland_Community_College/CHEM_1000_-_Introduction_to_Chemistry_(Riverland)/16%3A_Solutions | Template:HideTOC Solutions play a very important role in many biological, laboratory, and industrial applications of chemistry. Of particular importance are solutions involving substances dissolved in water, or aqueous solutions. Solutions represent equilibrium systems, and the lessons learned in the last chapter will be of particular importance again. Quantitative measurements of solutions are another key component of this chapter. Solutions can involve all physical states—gases dissolved in gases (the air around us), solids dissolved in solids (metal alloys), and liquids dissolved in solids (amalgams—liquid mercury dissolved in another metal such as silver, tin or copper). This chapter is almost exclusively concerned with aqueous solutions, substances dissolved in water. 16.1: Water - Some Unique Properties Earth is the only known body in our solar system that has liquid water existing freely on its surface. That is a good thing because life on Earth would not be possible without the presence of liquid water. 16.2: Solutions - Homogeneous Mixtures There are two types of mixtures: mixtures in which the substances are evenly mixed together (called a homogenous mixture, or solution) and a mixture in which the substances are not evenly mixed (called a heterogeneous mixture). When a solution, or homogenous mixture, is said to have uniform properties throughout, the definition is referring to properties at the particle level. 16.3: The Dissolution Process When a solute dissolves, its individual particles are surrounded by solvent molecules and are separated from each other. 16.4: Solutions of Solids Dissolved in Water- How to Make Rock Candy Solutions can be formed in a variety of combinations using solids, liquids, and gases. We also know that solutions have constant composition and we can also vary this composition up to a point to maintain the homogeneous nature of the solution. Reasons for why solutions form will be explored in this section, along with a discussion of why water is used most frequently to dissolve substances of various types. 16.5: Solutions of Gases in Water Several factors affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases. This is one of the major impacts resulting from the thermal pollution of natural bodies of water. 16.6: Solution Concentration- Mass Percent To define a solution precisely, we need to state its concentration: how much solute is dissolved in a certain amount of solvent. Words such as "dilute" or "concentrated" are used to describe solutions that have a little or a lot of dissolved solute, respectively. However "dilute" and "concentrated" are relative terms, and have meanings dependent on various factors. The mass/mass percent (% m/m) is defined as the mass of a solute divided by the mass of a solution times 100. 16.7: Solution Concentration- Molarity Another way of expressing concentration is to give the number of moles of solute per unit volume of solution. Of all the quantitative measures of concentration, molarity is the one used most frequently by chemists. Molarity is defined as the number of moles of solute per liter of solution. The symbol for molarity is MM or moles/liter. Chemists also use square brackets to indicate a reference to the molarity of a substance. 16.8: Solution Dilution We are often concerned with how much solute is dissolved in a given amount of solution. We will begin our discussion of solution concentration with two related and relative terms—dilute and concentrated. 16.9: Freezing Point Depression and Boiling Point Elevation Freezing point depression and boiling point elevation are "colligative properties" that depend on the concentration of solute in a solvent, but not on the type of solute. What this means for the example above is that people in colder climates don't necessarily need salt to get the same effect on the roads—any solute will work. The higher the concentration of solute, the more these colligative properties will change. 16.10: Osmosis and Diffusion Fish cells, like all cells, have semipermeable membranes. Eventually, the concentration of "stuff" on either side of them will even out. A fish that lives in salt water will have somewhat salty water inside itself. Put it in freshwater, and the freshwater will, through osmosis, enter the fish, causing its cells to swell, and the fish will die. What will happen to a freshwater fish in the ocean? Template:HideTOC |
Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/10%3A_Organic_Functional_Groups_-_Introduction_to_Acid-Base_Chemistry/10.04%3A_Physical_Properties_of_Esters | Learning Objectives Compare the boiling points of esters with alcohols of similar molar mass. Compare the solubilities of esters in water with the solubilities of comparable alkanes and alcohols in water. Ester molecules are polar but have no hydrogen atom attached directly to an oxygen atom. They are therefore incapable of engaging in intermolecular hydrogen bonding with one another and thus have considerably lower boiling points than their isomeric carboxylic acids counterparts. Because ester molecules can engage in hydrogen bonding with water molecules, however, esters of low molar mass are somewhat soluble in water. Borderline solubility occurs in those molecules that have three to five carbon atoms. Table \(\PageIndex{1}\) lists the physical properties of some common esters. Esters are common solvents. Ethyl acetate is used to extract organic solutes from aqueous solutions—for example, to remove caffeine from coffee. It also is used to remove nail polish and paint. Cellulose nitrate is dissolved in ethyl acetate and butyl acetate to form lacquers. The solvent evaporates as the lacquer “dries,” leaving a thin film on the surface. High boiling esters are used as softeners (plasticizers) for brittle plastics. Condensed Structural Formula Name Molar Mass Melting Point (°C) Boiling Point (°C) Aroma HCOOCH3 methyl formate 60 −99 32 NaN HCOOCH2CH3 ethyl formate 74 −80 54 rum CH3COOCH3 methyl acetate 74 −98 57 NaN CH3COOCH2CH3 ethyl acetate 88 −84 77 NaN CH3CH2CH2COOCH3 methyl butyrate 102 −85 102 apple CH3CH2CH2COOCH2CH3 ethyl butyrate 116 −101 121 pineapple CH3COO(CH2)4CH3 pentyl acetate 130 −71 148 pear CH3COOCH2CH2CH(CH3)2 isopentyl acetate 130 −79 142 banana CH3COOCH2C6H5 benzyl acetate 150 −51 215 jasmine CH3CH2CH2COO(CH2)4CH3 pentyl butyrate 158 −73 185 apricot CH3COO(CH2)7CH3 octyl acetate 172 −39 210 orange Summary Esters have polar bonds but do not engage in hydrogen bonding and are therefore intermediate in boiling points between the nonpolar alkanes and the alcohols, which engage in hydrogen bonding. Ester molecules can engage in hydrogen bonding with water, so esters of low molar mass are therefore somewhat soluble in water. |
Courses/Brevard_College/CHE_104%3A_Principles_of_Chemistry_II/04%3A_Thermochemistry_and_Thermodynamics/4.18%3A_Spontaneous_and_Nonspontaneous_Reactions | Nitroglycerin is a tricky substance. An active ingredient in dynamite (where it is stabilized), "raw" nitroglycerin is very unstable. Physical shock will cause the material to explode. The reaction is shown below. \[4 \ce{C_3H_5(ONO_2)_3} \rightarrow 12 \ce{CO_2} + 10 \ce{H_2O} + 6 \ce{N_2} + \ce{O_2}\nonumber \] The explosion of nitroglycerin releases large volumes of gases and is very exothermic. Spontaneous Reactions Reactions are favorable when they result in a decrease in enthalpy and an increase in entropy of the system. When both of these conditions are met, the reaction occurs naturally. A spontaneous reaction is a reaction that favors the formation of products at the conditions under which the reaction is occurring. A roaring bonfire is an example of a spontaneous reaction, since it is exothermic (there is a decrease in the energy of the system as energy is released to the surroundings as heat). The products of a fire are composed partly of gases such as carbon dioxide and water vapor. The entropy of the system increases during a combustion reaction. The combination of energy decrease and entropy increase dictates that combustion reactions are spontaneous reactions. A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, it must be endothermic, accompanied by a decrease in entropy, or both. Our atmosphere is composed primarily of a mixture of nitrogen and oxygen gases. One could write an equation showing these gases undergoing a chemical reaction to form nitrogen monoxide: \[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO} \left( g \right)\nonumber \] Fortunately, this reaction is nonspontaneous at normal temperatures and pressures. It is a highly endothermic reaction with a slightly positive entropy change \(\left( \Delta S \right)\). Nitrogen monoxide is capable of being produced at very high temperatures and has been observed to form as a result of lightning strikes. One must be careful not to confuse the term spontaneous with the notion that a reaction occurs rapidly. A spontaneous reaction is one in which product formation is favored, even if the reaction is extremely slow. A piece of paper will not suddenly burst into flames, although its combustion is a spontaneous reaction. What is missing is the required activation energy to get the reaction started. If the paper were to be heated to a high enough temperature, it would begin to burn, at which point the reaction would proceed spontaneously until completion. In a reversible reaction, one reaction direction may be favored over the other. Carbonic acid is present in carbonated beverages. It decomposes spontaneously to carbon dioxide and water, according to the following reaction. \[\ce{H_2CO_3} \left( aq \right) \rightleftharpoons \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber \] If you were to start with pure carbonic acid in water and allow the system to come to equilibrium, more than \(99\%\) of the carbonic acid would be converted into carbon dioxide and water. The forward reaction is spontaneous because the products of the forward reaction are favored at equilibrium. In the reverse reaction, carbon dioxide and water are the reactants, and carbonic acid is the product. When carbon dioxide is bubbled into water, less than \(1\%\) is converted to carbonic acid when the reaction reaches equilibrium. The reverse reaction, as written above, is not spontaneous. Summary Spontaneous and nonspontaneous reactions are defined. Examples of both types of reactions are given. |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/18%3A_Reactions_of_Aromatic_Compounds/18.11%3A__NAS_Reactions_-_the_Elimination-Addition_(Benzyne)_Mechanism | Elimination-Addition Mechanism of Nucleophilic Aromatic Substitution via Benzyne (Arynes) The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around \(340^\text{o}\) and this reaction was once an important commercial process for the production of benzenol (phenol): In addition, aryl chlorides, bromides, and iodides can be converted to areneamines \(\ce{ArNH_2}\) by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as \(-33^\text{o}\) with liquid ammonia as solvent: However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, the entering group does not always occupy the same position on the ring as that vacated by the halogen substituent . For example, the hydrolysis of 4-chloromethylbenzene at \(340^\text{o}\) gives an equimolar mixture of 3- and 4-methylbenzenols: Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change (Section 1-1H): The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed elimination of hydrogen halide \(\left( \ce{HX} \right)\) from the aryl halide - as illustrated below for the amination of bromobenzene: Elimination The product of the elimination reaction is a highly reactive intermediate \(9\) called benzyne , or dehydrobenzene , which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product: Addition The rearrangements in these reactions result from the reaction of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, \(10\), from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols: In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in Section 14-10C and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as \(11\), in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without: When \(11\) decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself: Exercise 25. When p -chlorotoluene is reacted with NaOH, two products are seen. While when m -chlorotoluene is reacted with NaOH, three products are seen. Explain this. Answer 25. You need to look at the benzyne intermediates. The para substituted only allows for two products, while the para produces two different alkynes which give three different products. |
Bookshelves/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.2%3A_Evolution_of_Atomic_Theory | Learning Objectives Outline milestones in the development of modern atomic theory Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford Describe the three subatomic particles that compose atoms Introduce the term isotopes In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work. Atomic Theory after the Nineteenth Century If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure \(\PageIndex{1}\)). Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron , a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “ electr ic i on .” In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure \(\PageIndex{2}\)). Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 \(\times\) 10 −19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 \(\times\) 10 −19 C), two electrons (2 times 1.6 \(\times\) 10 −19 C), three electrons (3 times 1.6 \(\times\) 10 −19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 \(\times\) 10 11 C/kg), it only required a simple calculation to determine the mass of the electron as well. \[\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1} \] Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka , who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure \(\PageIndex{3}\)). The next major development in understanding the atom came from Ernest Rutherford , a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle. What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure \(\PageIndex{4}\)). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you” 1 (p. 68). Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions: The volume occupied by an atom must consist of a large amount of empty space. A small, relatively heavy, positively charged body, the nucleus , must be at the center of each atom. This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure \(\PageIndex{5}\)). After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton , the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today. Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes —atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery. One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons , uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this chapter. Summary Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes. Footnotes Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science , eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf . Glossary alpha particle (α particle) positively charged particle consisting of two protons and two neutrons electron negatively charged, subatomic particle of relatively low mass located outside the nucleus isotopes atoms that contain the same number of protons but different numbers of neutrons neutron uncharged, subatomic particle located in the nucleus proton positively charged, subatomic particle located in the nucleus nucleus massive, positively charged center of an atom made up of protons and neutrons |
Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/01%3A_Orientation_to_this_course | 1.1: Pre-lab orientation assignment 1.2: Introductory Details Experiments have been set up for students working in small groups. Be sure to have a work plan that makes efficient use of your time. Lengthy sample preparation procedures should be started early in the six hours assigned to a given experiment. 1.3: Rubrics The points available for the semester and that will be used to calculate your final grade are described here. TAs will use rubrics similar to the ones given here to score your assignments. Rubrics are subject to minor changes during the semester. 1.4: Statements |
Courses/SUNY_Oneonta/Chem_322_Lecture_Content/06%3A_Nucleophilic_Acyl_Substitution_Reactions/6.10%3A_Nucleophilic_Acyl_Substitution_Reactions_in_the_Laboratory | All of the biological nucleophilic acyl substitution reactions we have seen so far have counterparts in laboratory organic synthesis. Mechanistically, one of the biggest differences between the biological and the lab versions is that the lab reactions usually are run with a strong acid or base as a catalyst, whereas biological reactions are of course taking place at physiological \(pH.\) When proposing mechanisms, then, care must be taken to draw intermediates in their reasonable protonation states: for example, a hydronium ion (\(H_3O^+\)) intermediate is reasonable to propose in an acidic reaction, but a hydroxide (\(OH^-\)) intermediate is not. Ester reactions - bananas, soap and biodiesel Acid-catalyzed synthesis of flavor compounds such as isopentyl acetate (an ester with the flavor of banana) is simple to carry out in the lab. In this esterification reaction, acetic acid is combined with isopentyl alcohol along with a catalytic amount of sulfuric acid. Acid-catalyzed esterification (laboratory reaction): Mechanism: The carbonyl oxygen of acetic acid is first protonated (step 1), which draws electron density away from the carbon and increases its electrophilicity. In step 2, the alcohol nucleophile attacks: notice that under acidic conditions, the nucleophile is not deprotonated simultaneously as it attacks (as we would show in a biochemical mechanism), and the tetrahedral intermediate is a cation rather than an anion. In step 3, a proton is transferred from one oxygen atom to another, creating a good leaving group (water) which is expelled in step 4. Finally (step 5), the carbonyl oxygen on the ester is deprotonated, regenerating the catalytic acid. This reaction is highly reversible, because carboxylic acids are approximately as reactive as esters. In order to obtain good yields of the ester, an excess of acetic acid can be used, which by Le Chatelier's principle (see your General Chemistry textbook for a review) shifts the equilibrium toward the ester product. Saponification is a common term for base-induced hydrolysis of an ester. For example, methyl benzoate will hydrolyze to benzoate and methanol when added to water with a catalytic amount of sodium hydroxide. Mechanism of base-catalyzed ester hydrolysis (saponification): Addition of the base provides hydroxide ion to act as a nucleophile (hydroxide is of course a better nucleophile than water) in step 1. The tetrahedral intermediate (anionic in this case, because the reaction conditions are basic) then collapses in step 2, and the alkoxide (\(CH_3O^-\)) leaves. We are not used to seeing alkoxides or hydroxides as leaving groups in biochemical reactions, because they are strong bases - but in a basic solution, this is a reasonable chemical step. Step 3 is simply an acid-base reaction between the carboxylic acid and the alkoxide. Note that this is referred to as base-induced rather than base-catalyzed because hydroxide is not regenerated, and thus a full molar equivalent of base must be used. The saponification process derives its name from the ancient craft of soap-making, in which the ester groups of triacylglycerols in animal fats are hydrolized under basic conditions to glycerol and fatty acyl anions (see section 2.5A for a reminder of how fatty acyl anions work as soap). We learned earlier about transesterification reactions in the context of the chemical mechanism of aspirin. Transesterification also plays a key role in a technology that is already an important component in the overall effort to develop environmentally friendly, renewable energy sources: biodeisel. You may have heard stories about people running their cars on biodeisel from used french fry oil. To make biodeisel, triacylglycerols in fats and oils can be transesterified with methanol or ethanol under basic conditions. The fatty acyl methyl and ethyl ester products are viable motor fuels. Exercise \(\PageIndex{1}\) Draw structures of the carboxylic acid and alcohol starting materials that could be used to synthesize the fragrant fruit esters shown in section 6.2. Exercise \(\PageIndex{2}\) What would happen if you tried to synthesize isopentyl acetate (banana oil) with basic rather than acidic conditions? Would this work? Exercise \(\PageIndex{3}\) Consider the reverse direction of the acid-catalyzed esterification reaction. What would you call this reaction in organic chemistry terms? Exercise \(\PageIndex{4}\) An alternative way to synthesize esters is to start with a carboxylate and an alkyl halide. Draw a mechanism for such a synthesis of methyl benzoate - what type of reaction mechanism is this? Acid chlorides and acid anhydrides In the cell, acyl phosphates and thioesters are the most reactive of the carboxylic acid derivatives. In the organic synthesis lab, their counterparts are acid chlorides and acid anhydrides, respectively. Of the two, acid chlorides are the more reactive, because the chloride ion is a weaker base and better leaving group than the carboxylate ion (the \(pK_a\) of \(HCl\) is -7, while that of carboxylic acids is about 4.5: remember, a stronger conjugate acid means a weaker conjugate base). Acid chlorides can be prepared from carboxylic acids using \(SOCl_2\): Acid anhydrides can be prepared from carboxylic acids and an acid chloride under basic conditions: Acetic anhydride is often used to prepare acetate esters and amides from alcohols and amines, respectively. The synthesis of aspirin and acetaminophen are examples: A carboxylic acid cannot be directly converted into an amide because the amine nucleophile would simply act as a base and deprotonate the carboxylic acid: Instead, the carboxylic acid is first converted to an acid chloride (in other words, the carboxylic acid is activated), then the acid chloride is combined with an amine to make the amide. This sequence of reactions is a direct parallel to the biochemical glutamine and asparagine synthase reactions we saw earlier in the chapter ( section 11.5), except that the activated form of carboxylic acid is an acid chloride instead of an acyl phosphate or acyl-AMP. Exercise \(\PageIndex{5}\) For the preparation of the amide below, sh ow a starting carboxylate and amine and the intermediate acid chloride species. Polyesters and polyamides If you have ever had the misfortune of undergoing surgery or having to be stitched up after a bad cut, it is likely that you benefited from our increasing understanding of polymers and carboxylic ester chemistry. Polyglycolic acid is a material commonly used to make dissolving sutures. It is a polyester - a polymer linked together by ester groups - and is formed from successive acyl substitution reactions between the alcohol group on one end of a glycolic acid monomer and the carboxylic acid group on a second: The resulting polymer - in which each strand is generally several hundred to a few thousand monomers long - is strong, flexible, and not irritating to body tissues. It is not, however, permanent: the ester groups are reactive to gradual, spontaneous hydrolysis at physiological \(pH\), which means that the threads will dissolve naturally over several weeks, eliminating the need for them to be cut out by a doctor. Exercise \(\PageIndex{6}\) Dacron, a polyester used in clothing fiber, is made of alternating dimethyl terephthalate and ethylene glycol monomers. Draw the structure of a Dacron tetramer (in other words, four monomers linked together). Water is a side product of glycolic acid polymerization. What is the equivalent side product in Dacron production? Exercise \(\PageIndex{7}\) Nylon 6,6 is a widely used polyamide composed of alternating monomers. Nylon 6,6 has the structure shown below -the region within the parentheses is the repeating unit, with 'n' indicating a large number of repeats. Identify the two monomeric compounds used to make the polymer. The Gabriel synthesis of primary amines The Gabriel synthesis, named after the 19 th -century German chemist Siegmund Gabriel, is a useful way to convert alkyl halides to amines and another example of \(S_N2\) and acyl substitution steps in the laboratory. The nitrogen in the newly introduced amine group comes from phthalimide. In the first step of the reaction, phthalimide is deprotonated by hydroxide, then in step 2 it acts as a nucleophile to displace a halide in an \(S_N2\) reaction (phthalimide is not a very powerful nucleophile, so this reaction works only with unhindered primary or methyl halides). Step 3 is simply a pair of hydrolytic acyl substitution steps to release the primary amine, with an aromatic dicarboxylate by-product. Exercise \(\PageIndex{8}\) Phthalimide contains an ' imide ' functional group, and has a \(pK_a\) of approximately 10. What makes the imide group so much more acidic than an amide, which has a \(pK_a\) of approximately 17? As an alternative procedure, release of the amine in step 3 can be carried out with hydrazine (\(H_2NNH_2\)) instead of hydroxide. Again, this occurs through two nucleophilic acyl substitution reactions. In 2000, chemists at MIT synthesizing a porphyrin-containing molecule introduced two amine groups using the Gabriel synthesis with hydrazine. Porphyrins, which include the 'heme' in our red blood cells, are an important family of biomolecules with a variety of biochemical function (J. Org. Chem. 2000, 65, 5298). |
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.23%3A_Synthetic_Macromolecules-_Some_Applied_Organic_Chemistry | Other important classes of substances containing very large molecules are the plastics and artificial fibers, which are such a conspicuous, though not always a positive, feature of modern life. Most of these materials are made in the same basic way. The starting materials or monomers are relatively simple molecules—usually carbon compounds derived from petroleum—which can be persuaded to link up with each other in order to form a long chain of repeating units called a polymer . If we think of the monomer as a bead, then the polymer corresponds to a string of beads. Polymers are usually classified into two types: addition polymers and condensation polymers , according to the kind of reaction by which they are made. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.05%3A_Filtering_Methods | There are many methods used to separate a mixture containing a solid and liquid. If the solid settles well, the liquid can sometimes be poured off (decanted). If the solid has very small sized particles or forms a cloudy mixture, the mixture can sometimes be centrifuged or passed through a filter pipette (on the microscale, < 5 mL). 1.5A: Overview of Methods The most common methods of solid-liquid separation in the organic lab are gravity and suction filtration. Gravity filtration refers to pouring a solid-liquid mixture through a funnel containing a filter paper, allowing the liquid to seep through while trapping the solid on the paper. Suction filtration is a similar process with the difference being the application of a vacuum beneath the funnel in order to pull liquid through the filter paper with suction. 1.5B: Decanting When there is a need to separate a solid-liquid mixture, on occasion it is possible to pour off the liquid while leaving the solid behind. This process is called decanting, and is the simplest separation method. 1.5C: Gravity Filtration Gravity filtration is generally used when the filtrate (liquid that has passed through the filter paper) will be retained, while the solid on the filter paper will be discarded. 1.5D: Suction Filtration Suction filtration (vacuum filtration) is the standard technique used for separating a solid-liquid mixture when the goal is to retain the solid (for example in crystallization). Similar to gravity filtration, a solid-liquid mixture is poured onto a filter paper, with the main difference being that the process is aided by suction beneath the funnel. 1.5E: Hot Filtration A hot filtration is generally used in some crystallization, when a solid contains impurities that are insoluble in the crystallization solvent. It is also necessary in crystallization when charcoal is used to remove highly colored impurities from a solid, as charcoal is so fine that it cannot be removed by decanting. 1.5F: Pipette Filtration (Microscale) For the separation of small volumes (< 10mL ) of solid-liquid mixtures, pipette filters are ideal as filter papers absorb a significant amount of material. Pipette filtration may also be used if small amounts of solid are noticed in NMR or GC samples, as both instruments require analysis of liquids without suspended solids. 1.5G: Centrifugation Centrifugation is used for the separation of solid-liquid mixtures that are stubborn to settle or difficult to otherwise filter. It uses centrifugal force by rapidly spinning samples so that the solid is forced to the bottom of the tube. In this section is shown centrifugation of a suspension of yellow lead(II) iodide in water (Figure 1.90b). |
Courses/SUNY_Potsdam/Book%3A_Organic_Chemistry_I_(Walker)/05%3A_Physical_Organic_Chemistry/5.06%3A_Reactive_intermediates | In chemistry, a reactive intermediate or an intermediate is a short-lived, high-energy, highly reactive molecule. When generated in a chemical reaction, it will quickly convert into a more stable molecule. Only in exceptional cases can these compounds be isolated and stored, e.g. low temperatures, matrix isolation. When their existence is indicated, reactive intermediates can help explain how a chemical reaction takes place. Most chemical reactions take more than one elementary step to complete, and a reactive intermediate is a high-energy, yet stable, product that exists only in one of the intermediate steps. The series of steps together make a reaction mechanism. A reactive intermediate differs from a reactant or product or a simple reaction intermediate only in that it cannot usually be isolated but is sometimes observable only through fast spectroscopic methods. It is stable in the sense that an elementary reaction forms the reactive intermediate and the elementary reaction in the next step is needed to destroy it. When a reactive intermediate is not observable, its existence must be inferred through experimentation. This usually involves changing reaction conditions such as temperature or concentration and applying the techniques of chemical kinetics, chemical thermodynamics, or spectroscopy. We will often refer to certain reactive intermediates based on carbon, viz., carbocations, radicals, carbanions and carbenes. Common features Reactive intermediates have several features in common: low concentration with respect to reaction substrate and final reaction product often generated on chemical decomposition of a chemical compound it is often possible to prove the existence of this species by spectroscopic means cage effects have to be taken into account often stabilization by conjugation or resonance often difficult to distinguish from a transition state prove existence by means of chemical trapping References ^ Carey, Francis A.; Sundberg, Richard J.; (1984). Advanced Organic Chemistry Part A Structure and Mechanisms (2nd ed.). New York N.Y.: Plenum Press. ISBN 0-306-41198-9 . ^ March Jerry; (1885). Advanced Organic Chemistry reactions, mechanisms and structure (3rd ed.). New York: John Wiley & Sons, inc. ISBN 0-471-85472-7 ^ Gilchrist, T. L. (1966). Carbenes nitrenes and arynes . Springer US. ISBN 9780306500268 . ^ Moss, Robert A.; Platz, Matthew S.; Jones, Jr., Maitland (2004). Reactive intermediate chemistry . Hoboken, N.J.: Wiley-Interscience. ISBN 9780471721499 . Carbocations (R+) A carbocation is an ion with a positively-charged carbon atom . Among the simplest examples are methenium CH 3 + , methanium CH 5 + , and ethanium C 2 H 7 + . Until the early 1970s, all carbocations were called carbonium ions.[1] In present-day chemistry, a carbocation is any positively charged carbon atom, classified in two main categories according to the valence of the charged carbon: +3 in carbenium ions (protonated carbenes), +5 or +6 in the carbonium ions (protonated alkanes, named by analogy to ammonium). These are much less common. Structure and properties The charged carbon atom in a carbocation is a “sextet”, i.e. it has only six electrons in its outer valence shell instead of the eight valence electrons that ensures maximum stability (octet rule). Therefore, carbocations are often reactive, seeking to fill the octet of valence electrons as well as regain a neutral charge. One could reasonably assume a carbocation to have $$sp^3$$ hybridization with an empty $$sp_3$$ orbital giving positive charge. However, the reactivity of a carbocation more closely resembles $$sp^2$$ hybridization with atrigonal planar molecular geometry. An example is the methyl cation, $$CH_3^+$$. Order of stability of examples of tertiary (3 o ), secondary (2 o ), and primary (1 o ) alkyl carbenium ions , as well as the methyl cation (far right). The methyl group is so unstable it is only observed in the gas phase. Carbocations are often the target of nucleophilic attack by nucleophiles such as water or halide ions. Carbocations typically undergo rearrangement reactions from less stable structures to equally stable or more stable ones with rate constants in excess of 10 9 /sec. This fact complicates synthetic pathways to many compounds. For example, when 3-pentanol is heated with aqueous HCl, the initially formed 3-pentyl carbocation rearranges to a statistical mixture of the 3-pentyl and 2-pentyl. These cations react with chloride ion to produce about 1/3 3-chloropentane and 2/3 2-chloropentane. A carbocation may be stabilized by resonance by a carbon-carbon double bond next to the ionized carbon. Such cations as allyl cation CH 2 =CH–CH 2 + and benzyl cation C 6 H 5 –CH 2 + are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the π bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge – and, therefore, stabilizes the carbocation. References Gold Book definition for carbocation Radicals A radicals is a seven electron intermediate that adopts a flat, sp 2 structure despite the fact that it has four electron groups; the lone electron resides in a half-filled p-orbital. This sp 2 structure allows radicals to delocalize the single electron through resonance. We will study radical reactions in detail in the second semester. Being short of the octet, radicals are electrophilic, and therefore they are stabilized by alkyl groups. Thus the order for stability is the same as for carbocations, namely tertiary > secondary > primary > methyl . Carbanions A carbanion is an eight electron intermediate with an sp 3 structure as shown in A. Despite its full octet, it is very reactive due to the fact that carbon is not very electronegative. Although it is sp 3 , it can participate in resonance because it can easily re-hybridize to an sp 2 structure (see B), which allows overlap. Carbanions are electron-rich and nucleophilic, so in fact they are destabilized by alkyl groups. This means that the order for stability is the opposite of that for carbocations, namely methyl > primary > secondary > tertiary . Carbenes Carbenes are the least obvious of the four common intermediates; in most cases they have a six-electron sp 2 structure that has a lone pair but no overall charge. Although they are short of a full octet, they also have a reactive lone pair, so (depending on structure) carbenes can be either electrophilic or nucleophilic, or sometimes both – they just like to react with almost anything! We will learn about carbene reactions in section 10.7. CC licensed content, Shared previously Reactive_intermediate. Authored by : Wikipedia contributors. Provided by : Wikimedia Foundation. Located at : https://en.wikipedia.org/wiki/Reactive_intermediate . Project : Wikipedia. License : CC BY-SA: Attribution-ShareAlike Carbocations. Authored by : Libretexts contributors. Provided by : UC Davis. Located at : https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbocations . Project : Libretexts. License : CC BY-NC: Attribution-NonCommercial |
Courses/Palomar_College/PC%3A_CHEM100_-_Fundamentals_of_Chemistry/15%3A_Chemical_Bonding/15.4%3A_Lewis_Structures%3A_Counting_Valence_Electrons | Learning Objectives Draw Lewis structures for covalent compounds. The following procedure can be used to construct Lewis electron structures for more complex molecules and ions: How-to: Constructing Lewis electron structures 1. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO 3 2− , for example, we add two electrons to the total because of the −2 charge. 2. Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl 4 and CO 3 2− , which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H 2 O, for example, there is a bonding pair of electrons between oxygen and each hydrogen. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs. 5. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms. 7. Final check Always make sure all valence electrons are accounted for and each atom has an octet of electrons except for hydrogen (with two electrons). The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal. Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed. Example \(\PageIndex{1}\): Water Write the Lewis Structure for H 2 O. Solution Steps for Writing Lewis Structures Example \(\PageIndex{1}\) 1. Determine the total number of valence electrons in the molecule or ion. Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. 2. Arrange the atoms to show specific connections. H O H Because H atoms are almost always terminal, the arrangement within the molecule must be HOH. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Placing one bonding pair of electrons between the O atom and each H atom gives H -O- H with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. 5. If any electrons are left over, place them on the central atom. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure: 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. Not necessary 7. Final check The Lewis structure gives oxygen an octet and each hydrogen two electrons, Example \(\PageIndex{2}\) Write the Lewis structure for the \(CH_2O\) molecule Solution Steps for Writing Lewis Structures Example \(\PageIndex{2}\) 1. Determine the total number of valence electrons in the molecule or ion. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. 2. Arrange the atoms to show specific connections. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. 3. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. Placing a bonding pair of electrons between each pair of bonded atoms gives the following: Six electrons are used, and 6 are left over. 4. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. 5. If any electrons are left over, place them on the central atom. Not necessary There are no electrons left to place on the central atom. 6. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: 7. Final check Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. Exercise \(\PageIndex{1}\) Write Lewis electron structures for CO 2 and SCl 2 , a vile-smelling, unstable red liquid that is used in the manufacture of rubber. Answer CO 2 . Answer SCl 2 . The United States Supreme Court has the unenviable task of deciding what the law is. This responsibility can be a major challenge when there is no clear principle involved or where there is a new situation not encountered before. Chemistry faces the same challenge in extending basic concepts to fit a new situation. Drawing of Lewis structures for polyatomic ions uses the same approach, but tweaks the process a little to fit a somewhat different set of circumstances. Writing Lewis Structures for Polyatomic Ions Recall that a polyatomic ion is a group of atoms that are covalently bonded together and which carry an overall electrical charge. The ammonium ion, \(\ce{NH_4^+}\), is formed when a hydrogen ion \(\left( \ce{H^+} \right)\) attaches to the lone pair of an ammonia \(\left( \ce{NH_3} \right)\) molecule in a coordinate covalent bond. When drawing the Lewis structure of a polyatomic ion, the charge of the ion is reflected in the number of total valence electrons in the structure. In the case of the ammonium ion: \(1 \: \ce{N}\) atom \(= 5\) valence electrons \(4 \: \ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons subtract 1 electron for the \(1+\)charge of the ion total of 8 valence electrons in the ion It is customary to put the Lewis structure of a polyatomic ion into a large set of brackets, with the charge of the ion as a superscript outside the brackets. Exercise \(\PageIndex{2}\) Draw the Lewis electron dot structure for the sulfate ion. Answer Exceptions to the Octet Rule As important and useful as the octet rule is in chemical bonding, there are some well-known violations. This does not mean that the octet rule is useless—quite the contrary. As with many rules, there are exceptions, or violations. There are three violations to the octet rule. Odd-electron molecules represent the first violation to the octet rule. Although they are few, some stable compounds have an odd number of electrons in their valence shells. With an odd number of electrons, at least one atom in the molecule will have to violate the octet rule. Examples of stable odd-electron molecules are NO, NO 2 , and ClO 2 . The Lewis electron dot diagram for NO is as follows: Although the O atom has an octet of electrons, the N atom has only seven electrons in its valence shell. Although NO is a stable compound, it is very chemically reactive, as are most other odd-electron compounds. Electron-deficient molecules represent the second violation to the octet rule. These stable compounds have less than eight electrons around an atom in the molecule. The most common examples are the covalent compounds of beryllium and boron. For example, beryllium can form two covalent bonds, resulting in only four electrons in its valence shell: Boron commonly makes only three covalent bonds, resulting in only six valence electrons around the B atom. A well-known example is BF 3 : The third violation to the octet rule is found in those compounds with more than eight electrons assigned to their valence shell. These are called expanded valence shell molecules. Such compounds are formed only by central atoms in the third row of the periodic table or beyond that have empty d orbitals in their valence shells that can participate in covalent bonding. One such compound is PF 5 . The only reasonable Lewis electron dot diagram for this compound has the P atom making five covalent bonds: Formally, the P atom has 10 electrons in its valence shell. Example \(\PageIndex{3}\): Octet Violations Identify each violation to the octet rule by drawing a Lewis electron dot diagram. ClO SF 6 Solution a. With one Cl atom and one O atom, this molecule has 6 + 7 = 13 valence electrons, so it is an odd-electron molecule. A Lewis electron dot diagram for this molecule is as follows: b. In SF 6 , the central S atom makes six covalent bonds to the six surrounding F atoms, so it is an expanded valence shell molecule. Its Lewis electron dot diagram is as follows: Exercise \(\PageIndex{3}\): Xenon Difluoride Identify the violation to the octet rule in XeF 2 by drawing a Lewis electron dot diagram. Answer: The Xe atom has an expanded valence shell with more than eight electrons around it. Summary Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter bonding pairs , which are shared by two atoms, and lone pairs , which are not shared between atoms. Lewis structures for polyatomic ions follow the same rules as those for other covalent compounds. There are three violations to the octet rule: odd-electron molecules, electron-deficient molecules, and expanded valence shell molecules |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Chemistry_(Blinder)/01%3A_Chapters/1.03%3A_Quantum_Mechanics_of_Some_Simple_Systems | The simple quantum-mechanical problem we have just solved can provide an instructive application to chemistry: the free-electron model (FEM) for delocalized \(\pi\)-electrons. The simplest case is the 1,3-butadiene molecule \[\rho =2\psi _{1}^2+2\psi _{2}^2\label{28}\] A chemical interpretation of this picture might be that, since the \(\pi\)-electron density is concentrated between carbon atoms 1 and 2, and between 3 and 4, the predominant structure of butadiene has double bonds between these two pairs of atoms. Each double bond consists of a \(\pi\)-bond, in addition to the underlying \(\sigma\)-bond. However, this is not the complete story, because we must also take account of the residual \(\pi\)-electron density between carbons 2 and 3. In the terminology of valence-bond theory, butadiene would be described as a resonance hybrid with the contributing structures CH 2 =CH-CH=CH 2 (the predominant structure) and ºCH 2 - CH=CH-CH 2 º (a secondary contribution). The reality of the latter structure is suggested by the ability of butadiene to undergo 1,4-addition reactions. The free-electron model can also be applied to the electronic spectrum of b utadiene and other linear polyenes . The lowest unoccupied molecular orbital ( LUMO ) in butadiene corresponds to the n=3 particle-in-a-box state. Neglecting electron-electron interaction, the longest-wavelength (lowest-energy) electronic transition should occur from n=2, the highest occupied molecular orbital (HOMO). The energy difference is given by \[\Delta E=E_{3}-E_{2}=(3^2-2^2)\dfrac{h^2}{8mL^2}\label{29}\] Here m represents the mass of an electron (not a butadiene molecule!), 9.1x10 -31 Kg, and L is the effective length of the box, 4x1.40x10 -10 m. By the Bohr frequency condition \[\Delta E=h\upsilon =\dfrac{hc}{\lambda }\label{30}\] The wavelength is predicted to be 207 nm. This compares well with the experimental maximum of the first electronic absorption band, \(\lambda_{max} \approx\) 210 nm, in the ultraviolet region. We might therefore be emboldened to apply the model to predict absorption spectra in higher polyenes CH 2 =(CH-CH=) n- 1 CH 2 . For the molecule with 2 n carbon atoms ( n double bonds), the HOMO → LUMO transition corresponds to n → n + 1, thus \[\dfrac{hc}{\lambda} \approx \begin{bmatrix}(n+1)^2-n^2\end{bmatrix}\dfrac{h^2}{8m(2nL_{CC})^2}\label{31}\] A useful constant in this computation is the Compton wavelength \[\dfrac{h}{mc}= 2.426 \times 10^{-12} m.\] For n =3, hexatriene, the predicted wavelength is 332 nm, while experiment gives \(\lambda _{max}\approx \) 250 nm. For n =4, octatetraene, FEM predicts 460 nm, while \(\lambda _{max}\approx\) 300 nm. Clearly the model has been pushed beyond range of quantitate validity, although the trend of increasing absorption band wavelength with increasing n is correctly predicted. Incidentally, a compound should be colored if its absorption includes any part of the visible range 400-700 nm. Retinol (vitamin A), which contains a polyene chain with n =5, has a pale yellow color. This is its structure: |
Courses/Saint_Francis_University/Chem_114%3A_Human_Chemistry_II_(Hargittai)/18%3A_Amino_Acids_and_Proteins/18.05%3A_Handedness | Learning Objectives Explain how a peptide is formed from individual amino acids. Explain why the sequence of amino acids in a protein is important. Two or more amino acids can join together into chains called peptides. In an earlier chapter, we discussed the reaction between ammonia and a carboxylic acid to form an amide. In a similar reaction, the amino group on one amino acid molecule reacts with the carboxyl group on another, releasing a molecule of water and forming an amide linkage: An amide bond joining two amino acid units is called a peptide bond . Note that the product molecule still has a reactive amino group on the left and a reactive carboxyl group on the right. These can react with additional amino acids to lengthen the peptide. The process can continue until thousands of units have joined, resulting in large proteins. A chain consisting of only two amino acid units is called a dipeptide ; a chain consisting of three is a tripeptide . By convention, peptide and protein structures are depicted with the amino acid whose amino group is free (the amino-terminal or N-terminal end) on the left and the amino acid with a free carboxyl group (the carboxyl-terminal or C-terminal end) to the right. Individual amino acids joined in a chain are called residues . The general term peptide refers to an amino acid chain of unspecified length. However, chains of about 50 amino acids or more are usually called proteins or polypeptides. In its physiologically active form, a protein may be composed of one or more polypeptide chains. Note: Order is Important For peptides and proteins to be physiologically active, it is not enough that they incorporate certain amounts of specific amino acids. The order, or sequence , in which the amino acids are connected is also of critical importance. Bradykinin is a nine-amino acid peptide (Figure \(\PageIndex{1}\)) produced in the blood that has the following amino acid sequence: arg-pro-pro-gly-phe-ser-pro-phe-arg This peptide lowers blood pressure, stimulates smooth muscle tissue, increases capillary permeability, and causes pain. When the order of amino acids in bradykinin is reversed, arg-phe-pro-ser-phe-gly-pro-pro-arg the peptide resulting from this synthesis shows none of the activity of bradykinin. Just as millions of different words are spelled with our 26-letter English alphabet, millions of different proteins are made with the 20 common amino acids. However, just as the English alphabet can be used to write gibberish, amino acids can be put together in the wrong sequence to produce nonfunctional proteins. Although the correct sequence is ordinarily of utmost importance, it is not always absolutely required. Just as you can sometimes make sense of incorrectly spelled English words, a protein with a small percentage of “incorrect” amino acids may continue to function. However, it rarely functions as well as a protein having the correct sequence. There are also instances in which seemingly minor errors of sequence have disastrous effects. For example, in some people, every molecule of hemoglobin (a protein in the blood that transports oxygen) has a single incorrect amino acid unit out of about 300 (a single valine replaces a glutamic acid). That “minor” error is responsible for sickle cell anemia, an inherited condition that usually is fatal. Summary The amino group of one amino acid can react with the carboxyl group on another amino acid to form a peptide bond that links the two amino acids together. Additional amino acids can be added on through the formation of addition peptide (amide) bonds. A sequence of amino acids in a peptide or protein is written with the N-terminal amino acid first and the C-terminal amino acid at the end (writing left to right). Concept Review Exercises Distinguish between the N-terminal amino acid and the C-terminal amino acid of a peptide or protein. Describe the difference between an amino acid and a peptide. Amino acid units in a protein are connected by peptide bonds. What is another name for the functional group linking the amino acids? Answers The N-terminal end is the end of a peptide or protein whose amino group is free (not involved in the formation of a peptide bond), while the C-terminal end has a free carboxyl group. A peptide is composed of two or more amino acids. Amino acids are the building blocks of peptides. amide bond Exercises Draw the structure for each peptide. gly-val val-gly Draw the structure for cys-val-ala. Identify the C- and N-terminal amino acids for the peptide lys-val-phe-gly-arg-cys. Identify the C- and N-terminal amino acids for the peptide asp-arg-val-tyr-ile-his-pro-phe. Answers C-terminal amino acid: cys; N-terminal amino acid: lys |
Courses/Widener_University/Widener_University%3A_Chem_135/10%3A_Liquids_and_Solids | The great distances between atoms and molecules in a gaseous phase, and the corresponding absence of any significant interactions between them, allows for simple descriptions of many physical properties that are the same for all gases, regardless of their chemical identities. As described in the final module of the chapter on gases, this situation changes at high pressures and low temperatures—conditions that permit the atoms and molecules to interact to a much greater extent. In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. 10.1: Prelude to Liquids and Solids In the liquid and solid states, these interactions are of considerable strength and play an important role in determining a number of physical properties that do depend on the chemical identity of the substance. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined. 10.2: Intermolecular Forces The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient kinetic energy to move past each other. Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. 10.3: Properties of Liquids The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension. Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for surface wetting and capillary rise. 10.4: Phase Transitions Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. 10.E: Liquids and Solids (Exercises) End of chapter homework problems for Chapter \(\PageIndex{1}\). |
Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.01%3A_Prelude_to_Electronic_Structure | Once scientists had accepted the idea that electrons were constituents of all matter, theories attempting to explain just how electrons were incorporated in the structure of the atom began to develop. This was especially true after Rutherford had discovered that most of the volume of an atom was occupied by electrons. Both chemists and physicists became interested in the electronic structure of atoms—the chemists because they wanted to explain valence and bonding, and the physicists because they wanted to explain the spectra of atoms,the light emitted when gaseous atoms were raised to a high temperature or bombarded by electrons. We'll explore their work in a later chapter . Electrons and Light Once we realize that there would be no light, no colors, no vision, without the pattern of behavior exhibited by electrons constrained to the nuclei of atoms, understanding electronic structure becomes much more interesting. Later we'll see how physicists and physical chemists developed theories to explain the spectra that were observed when atoms convert electrical, heat, and chemical energy into heat. Light from the sun, when passed through a prism (or through the water droplets of a rainbow), produces a continuous visible spectrum, where all wavelengths are represented: Figure \(\PageIndex{1}\) A rainbow displays the whole of the visible spectrum, from purple to red. But when collections individual atoms (in the gas phase) absorb energy, they emit energy as light with only certain wavelengths, in a line spectrum, like those displayed by electrically excited hydrogen or neon gas, or by potassium or barium salts in a flame: Figure \(\PageIndex{2}\) Simulation of a hydrogen emission spectrum Figure \(\PageIndex{3}\) Simulation of a neon emission spectrum Figure \(\PageIndex{4}\) Simulation of a potassium flame spectrum Figure \(\PageIndex{5}\) Simulation of a barium flame spectrum These emission spectra presented a puzzle: why did electrons, which had absorbed energy to increase their separation from the nucleus, emit energy as light of only certain wavelengths? The Danish physicist Niels Bohr (1885 to 1962) proposed the first theory able to explain this phenomenon, in terms of electrons with wave-like properties. Electrons and Chemical Properties Surprisingly, the same theory of electronic structure that explained emission spectra also explained chemical properties of the elements and bonding. The chief contributors to this development, which occurred mainly during the 15 years between 1910 and 1925, was the U.S. chemist Gilbert Newton Lewis (1875 to 1946). Ideas about the electronic structures of atoms developed during the first half of the twentieth century. The periodic repetition of chemical properties discovered by Mendeleev led G. N. Lewis to the conclusion that atoms must have a shell structure. This was confirmed by wave mechanics . Only certain specific wave patterns are possible for an electron in an atom, and these electron clouds are arranged in concentric shells. The energy of each electron in an atom depends on how strongly the electron is attracted by the positive charge on the nucleus and on how much it is repelled by other electrons. Although each electron cannot be assigned a precise trajectory or orbit in an atom, its wave pattern allows us to determine the probability that it will be at a certain location. From this the energy of each electron and the order of filling orbitals can be obtained. Thus we can determine the electron configuration for an atom of any element. Such electron configurations correlate with the periodic table . Because electrons in inner orbitals screen outer electrons from nuclear charge, the fourth and higher shells begin to fill before d (and sometimes f ) subshells in previous shells are occupied. This overlap in energies of shells explains why Lewis’ ideas are less useful for elements in the fourth and subsequent rows of the periodic table. It also accounts for the steady variation in properties of transition metals across the table, and for the nearly identical characteristics of inner transition elements as opposed to the large differences from one group of representative elements to the next. Although some added complication arises from the wave-mechanical picture, it does confirm Lewis’ basic postulate that valence electrons determine chemical properties and influence the bonding of one atom to another. In other pages you will see how rearrangement of valence electrons can hold atoms together, and how different kinds of bonds result in different macroscopic properties. |
Courses/Smith_College/CHM_223_Chemistry_III%3A_Organic_Chemistry_(2024)/06%3A_Carbonyl_Alpha-Substitution_Reactions/6.01%3A_Chapter_Objectives_and_Introduction_to_Carbonyl_Alpha-Substitution_Reactions | Objectives After completing this section, you should be able to write a general mechanism for an alpha substitution reaction of a carbonyl compound. Key Terms Make certain that you can define, and use in context, the key terms below. alpha ( α ) position alpha substitution reaction Study Notes An “alpha substitution reaction” of a carbonyl compound is a reaction in which one of the hydrogen atoms on the carbon adjacent to the carbonyl group is substituted by some other atom or group. Attack by the electrophile (E + ) can occur on the enol or enolate intermediate. There are four common types of reactions involving compounds containing a carbonyl bond. The first two, nucleophilic addition and nucleophilic acyl substitution, have been discussed in CHM 222. Nucleophilic addition occurs due to the electrophilic nature of the carbonyl carbon. After addition of a nucleophile, the carbonyl becomes a tetrahedral alkoxide intermediate which is usually protonated to become an -OH group. Nucleophilic Addition to a Carbonyl Nucleophilic acyl substitution is similar in that a tetrahedral alkoxide intermediate is formed after nucleophilic addition to the carbonyl. However, subsequent removal of the leaving group allows for the C=O (carbonyl) bond to reform. Overall, there is a substitution of the leaving group with the incoming nucleophile. Nucleophilic Acyl Substitution Involving a Carbonyl Reactions at The Alpha Carbon The remaining common carbonyl reaction types are α-substitutions and carbonyl condensations. Both utilize the special properties of carbons directly adjacent to carbonyls which are called α-carbons . These reactions, which can be regarded as the backbone of much synthetic organic chemistry, usually result in the replacement of a hydrogen attached to an α-carbon with some type of electrophile. These reactions involve two new nucleophilic species called the enol and the enolate. This chapter will focus on α-substitutions reactions. Although there are many carbonyl containing functional groups, the initial investigation in this chapter will focus on α-substitutions reactions using aldehydes and ketones. Important examples considered in this chapter include α-halogenation and α-alkylation. |
Courses/University_of_Alberta_Augustana_Campus/AUCHE_110_-_General_Chemistry_I_(Elizabeth_McGinitie)/01%3A__Review/1.15%3A_Stoichiometric_Calculations-_Amounts_of_Reactants_and_Products | Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions. Balancing In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients. Reactants to Products A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation: \[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber \] In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the Law of Conservation of Mass , which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium (\(Na\)), hydrogen (\(H\)), and chloride (\(Cl\)) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements. Stoichiometric Coefficients In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation: \[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber \] we can determine that 2 moles of \(HCl\) will react with 2 moles of \(Na_{(s)}\) to form 2 moles of \(NaCl_{(aq)}\) and 1 mole of \(H_{2(g)}\). If we know how many moles of \(Na\) reacted, we can use the ratio of 2 moles of \(NaCl\) to 2 moles of Na to determine how many moles of \(NaCl\) were produced or we can use the ratio of 1 mole of \(H_2\) to 2 moles of \(Na\) to convert to \(NaCl\). This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems. Example 1 Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction. \[\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber \] Solution Start by counting the number of atoms of each element. UNBALANCED Element Reactant (# of atoms) Product (# of atoms) Pb 1 1 O 8 9 H 6 2 S 1 2 The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of \(H_2SO_4\) to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of \(H_2O\) where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced. \[\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber \] BALANCED Element Reactant (# of atoms) Product (# of atoms) Pb 1 1 O 12 12 H 8 8 S 2 2 Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last. A balanced equation ultimately has to satisfy two conditions. The numbers of each element on the left and right side of the equation must be equal. The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions . Stoichiometry and Balanced Equations In stoichiometry, balanced equations make it possible to compare different elements through the stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful. Example 2 There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent? Solution The equation for this can be written as \[\ce{I + 2S \rightarrow IS2}\nonumber \] where \(I\) represents invitations, \(S\) represents stamps, and \(IS_2\) represents the sent party invitations consisting of one invitation and two stamps. Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation. Invitations Stamps Party Invitations Sent In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed. Example 3 What is the limiting reagent in this example? Solution Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using stoichiometric factors. 12 I x ( 1IS 2 /1I) = 12 IS 2 possible 20 S x ( 1IS 2 /2S) = 10 IS 2 possible When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as stoichiometric proportions . Types of Reactions There are 6 basic types of reactions. Combustion : Combustion is the formation of CO 2 and H 2 O from the reaction of a chemical and O 2 Combination (synthesis) : Combination is the addition of 2 or more simple reactants to form a complex product. Decomposition: Decomposition is when complex reactants are broken down into simpler products. Single Displacement : Single displacement is when an element from on reactant switches with an element of the other to form two new reactants. Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants. Acid-Base: Acid- base reactions are when two reactants form salts and water. Molar Mass Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass Example 4 What is the molar mass of H 2 O? Solution \[\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol \nonumber \] Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa. Example 5: Combustion of Propane Propane (\(\ce{C_3H_8}\)) burns in this reaction: \[\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber \] If 200 g of propane is burned, how many g of \(H_2O\) is produced? Solution Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of \(C_3H_8\) to moles of \(C_3H_8\) then from moles of \(C_3H_8\) to moles of \(H_2O\). Then convert from moles of \(H_2O\) to grams of \(H_2O\). Step 1: 200 g \(C_3H_8\) is equal to 4.54 mol \(C_3H_8\). Step 2: Since there is a ratio of 4:1 \(H_2O\) to \(C_3H_8\), for every 4.54 mol \(C_3H_8\) there are 18.18 mol \(H_2O\). Step 3: Convert 18.18 mol \(H_2O\) to g \(H_2O\). 18.18 mol \(H_2O\) is equal to 327.27 g \(H_2O\). Variation in Stoichiometric Equations Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis. Density Density (\(\rho\)) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used. Volume x (Mass/Volume) = Mass Mass x (Volume/Mass) = Volume Percent Mass Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule. Example 6 A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there? Solution 10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon 0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon Molarity Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions. Example 7 How much 5 M stock solution is needed to prepare 100 mL of 2 M solution? Solution 100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution. These ratios of molarity, density, and mass percent are useful in complex examples ahead. Determining Empirical Formulas An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present. Example 8: Combustion of Organic Molecules 1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO 2 and 0.599 g of H 2 O. What is the empirical formula of the organic molecule? Solution This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here. \[ \ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber \] Since all the moles of C and H in CO 2 and H 2 O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample. 0.0333mol CO 2 ( 1mol C/ 1mol CO 2 ) = 0.0333mol C in unknown 0.599g H 2 O ( 1mol H 2 O/ 18.01528g H 2 O)( 2mol H/ 1mol H 2 O) = 0.0665 mol H in unknown Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO 2 and H 2 O. This will give you the number of moles from both the unknown organic molecule and the O 2 so you must subtract the moles of oxygen transferred from the O 2 . Moles of oxygen in CO 2 : 0.0333mol CO 2 ( 2mol O/ 1mol CO 2 ) = 0.0666 mol O Moles of oxygen in H 2 O: 0.599g H 2 O ( 1mol H 2 O/18.01528 g H 2 O)( 1mol O/ 1mol H 2 O) = 0.0332 mol O Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O 2 reactant. 0.333mol CO 2 (44.0098g CO 2 / 1mol CO 2 ) = 1.466g CO 2 1.466g CO 2 + 0.599g H 2 O - 1.000g unknown organic = 1.065g O 2 Moles of oxygen in O 2 1.065g O 2 ( 1mol O 2 / 31.9988g O 2 )( 2mol O/ 1mol O 2 ) = 0.0666mol O Moles of oxygen in unknown (0.0666mol O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number. (1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O From this ratio, the empirical formula is calculated to be CH 2 O. Determining Molecular Formulas To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula. Example 9 In the example above, it was determined that the unknown molecule had an empirical formula of CH 2 O. 1. Find the molar mass of the empircal formula CH 2 O. 12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH 2 O 2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol. 3. Divide the experimentally determined molecular mass by the mass of the empirical formula. (120.056 g/mol) / (30.026 g/mol) = 3.9984 4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass. 5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula. CH 2 O * 4 = ? C: 1 * 4 = 4 H: 2 * 4 = 8 O 1 * 4 = 4 CH 2 O * 4 = C 4 H 8 O 4 6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass. molar mass of C 4 H 8 O 4 = 120.104 g/mol experimentally determined mass = 120.056 g/mol % error = | theoretical - experimental | / theoretical * 100% % error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100% % error = 0.040 % Example 10: Complex Stoichiometry Problem An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm 3 . He accidentally breaks off a 1.203 cm 3 piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with the copper, how many grams of H 2 (g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.) Solution Step 1 : Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H + (aq) and Fe(s). The given product is H2(g) and based on knowledge of redox reactions, the other product must be Fe 2 + (aq). \[\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber \] Step 2: Write down all the given information Alloy density = (3.15g alloy/ 1L alloy) x grams of alloy = 45% copper = ( 45g Cu(s)/ 100g alloy) x grams of alloy = 55% iron(II) = ( 55g Fe(s)/ 100g alloy) 1 liter alloy = 1000cm 3 alloy alloy sample = 1.203cm 3 alloy Step 3: Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed. Step 4: Start with the compound you know the most about and use given ratios to convert it to the desired compound. Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted. 1.203cm 3 alloy( 1liter alloy/ 1000cm 3 alloy)(3.15g alloy/ 1liter alloy)( 55g Fe(s)/ 100g alloy)( 1mol Fe(s)/55.8g Fe(s))=3.74 x 10 -5 mol Fe(s) Make sure all the units cancel out to give you moles of \(\ce{Fe(s)}\). The above conversion involves using multiple stoichiometric relationships from density, percent mass, and molar mass. The balanced equation must now be used to convert moles of Fe(s) to moles of H 2 (g). Remember that the balanced equation's coefficients state the stoichiometric factor or mole ratio of reactants and products. 3.74 x 10 -5 mol Fe (s) ( 1mol H 2 (g)/ 1mol Fe(s)) = 3.74 x 10 -5 mol H 2 (g) Step 5: Check units The question asks for how many grams of H 2 (g) were released so the moles of H 2 (g) must still be converted to grams using the molar mass of H 2 (g). Since there are two H in each H 2 , its molar mass is twice that of a single H atom. molar mass = 2(1.00794g/ mol ) = 2.01588g/ mol 3.74 x 10 -5 mol H 2 (g) (2.01588g H 2 (g)/ 1mol H 2 (g)) = 7.53 x 10 -5 g H 2 (g) released Problems Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems. 1) Why are the following equations not considered balanced? \(H_2O_{(l)} \rightarrow H_{2(g)} + O_{2(g)}\) \(Zn_{(s)} + Au^+_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Ag_{(s)}\) 2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction. 3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution? 4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below. \[\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber \] 5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO 2 and 0.388g H 2 O. Knowing that all the carbon and hydrogen atoms in CO 2 and H 2 O came from the 0.777g sample, what is the empirical formula of the organic compound? Weblinks for further reference 1. Refer to http://chemistry.about.com/cs/stoich.../aa042903a.htm as an outside resource on how to balance chemical reactions. 2. Refer to http://www.learnchem.net/tutorials/stoich.shtml as an outside resource on stoichiometry. References T. E. Brown, H.E LeMay, B. Bursten, C. Murphy. Chemistry: The Central Science. Prentice Hall, January 8, 2008. J. C. Kotz P.M. Treichel, J. Townsend. Chemistry and Chemical Reactivity. Brooks Cole, February 7, 2008. Petrucci, harwood, Herring, Madura. General Chemistry Principles & Modern Applications. Prentice Hall. New Jersey, 2007. |
Courses/SUNY_Oneonta/Organic_Chemistry_with_a_Biological_Emphasis_(SUNY_Oneonta)/12%3A_Reactions_at_the_-Carbon_Part_I/12.03%3A_Isomerization_at_the_-Carbon | Enolate ions, as well as enols and enamines (section 7.6) are the key reactive intermediates in many biochemical isomerization reactions. Isomerizations can involve either the interconversion of constitutional isomers, in which bond connectivity is altered, or of stereoisomers, where the stereochemical configuration is changed. Enzymes that interconvert constitutional isomers are usually called isomerases, while those that interconvert the configuration of a chiral carbon are usually referred to as racemases or epimerases. Carbonyl isomerization One very important family of isomerase enzymes catalyzes the shifting of a carbonyl group in sugar molecules, often converting between a ketose and an aldose (recall that the terms ketose and aldolse refer to sugar molecules containing ketone and aldehyde groups, respectively). Carbonyl isomerization: Mechanism: The ketose species is first converted to its enol tautomer in step 1 (actually, this particular intermediate is known as an 'ene-diol' rather than an enol, because there are hydroxyl groups on both sides of the carbon-carbon double bond). Step 2 leads to the aldose, and is simply another tautomerization step. However, because there is a hydroxyl group on the adjacent (blue) carbon, a carbonyl can form there as well as at the red carbon. An example is the glycolysis pathway reaction catalyzed by the enzyme triose phosphate isomerase (EC 5.3.1.1). Here, dihydroxyacetone phosphate (DHAP) is reversibly converted to glyceraldehyde phosphate (GAP). Notice that DHAP is achiral while GAP is chiral, and that a new chiral center is introduced at the middle (red) carbon of GAP. As you should expect, the enzyme is stereoselective: in step 2 a proton is delivered to the red carbon, from behind the plane of the page, to yield the R enantiomer. Also in the glycolysis pathway, glucose-6-phosphate (an aldose) and fructose-6-phosphate (a ketose) are interconverted through an ene-diol intermediate (EC 5.3.1.9) by an enzyme that is closely related to triose-phosphate isomerase. D raw the ene-diol intermediate in the phosphoglucose isomerase reaction. Stereoisomerization at the \(\alpha\)-carbon Enolates are a common intermediate in reactions where the stereochemical configuration of a chiral \(\alpha\)-carbon is interconverted. These are commonly referred to as racemization or epimerization reactions, depending on whether the interconverted isomers are enantiomers or epimers (recall that the term 'epimer' refers to a pair of diastereomers that differ by a single chiral center). Racemization/ epimerization: Mechanism: These reactions proceed though a deprotonation-reprotonation mechanism, illustrated above. In step 1, the chiral a-carbon is deprotonated, leading to a planar, achiral enolate. In step 2, a proton is delivered back to the a-carbon, but from the opposite side from which the proton was taken in step 1, resulting in the opposite stereochemistry at this carbon. Two acid-base groups, positioned at opposing sides of the enzyme's active site, work in tandem to accomplish this feat. The proteins and peptides in all known living things are constructed almost exclusively of L-amino acids, but in rare cases scientists have identified peptides which incorporate D-amino acids, which have the opposite stereochemistry at the a-carbon. Amino acid racemase enzymes catalyze the interconversion of L and D amino acids. As you may recall from the introductory section to this chapter, the venom of the male platypus contains a neurotoxic peptide in which an L-leucine amino acid has been converted by a racemase enzyme to D-leucine. In another example, the cell walls of bacteria are constructed in part of peptides containing D-glutamate, converted from L-glutamate by the enzyme glutamate racemase. (EC 5.1.1.3) (Biochemistry 2001, 40, 6199). A reaction (EC 5.1.3.1) in sugar metabolism involves the interconversion of the epimers ribulose-5-phosphate and xylulose-5-phosphate. The enzyme that catalyzes this reaction is called an 'epimerase'. (J. Mol. Biol. 2003, 326, 127). Draw a reasonable mechanism for the ribulose-5-phosphate epimerization reaction above. Your mechanism should show an enolate intermediate and specify stereochemistry throughout. Predict the products of epimerization reactions starting with each of the substrates shown. Hint Carbons next to imine groups can also be considered \(\alpha\)-carbons! Recall from chapter 3 that a major issue with the drug thalidomide is the fact that the R enantiomer, which is an effective sedative, rapidly isomerizes in the body to the terotogenic (mutation-causing) S enantiomer. Note that the chiral center in thalidomide is an \(\alpha\)-carbon! Recently chemists reported the synthesis of a thalidomide derivative in which the carbonyl group is replaced by an 'oxetane' ring, with the aim of making an isotopically stable form of the drug (because the carbonyl group has been removed, racemization is no longer possible - there is no a-carbon!) (Org. Lett. 2013, 15, 4312.) Alkene regioisomerization The position of an alkene group can also be changed through a reaction in which the first step is abstraction of an \(\alpha\)-proton and formation of an enolate intermediate. degradation pathway for unsaturated fatty acids (fatty acids whose hydrocarbon chains contain one or more double bonds) involves the 'shuffling' of the position of a carbon-carbon double bond, from a cis bond between carbon #3 and carbon #4 to a trans bond between carbon #2 and carbon #3. This is accomplished by the enzyme enoyl \(CoA\) isomerase (EC 5.3.3.8). (J. Biol Chem 2001, 276, 13622). Alkene isomerization: Mechanism: Consider the structures of the substrate and product of the isomerization reaction above. What two factors contribute to the thermodynamic 'driving force' for the transformation? The reaction below is part of the biosynthetic pathway for menthol. Suggest a mechanism that includes an enolate intermediate. |
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Chemistry_143_-_Introductory_Chemistry_(Bunag)/04%3A_Atomic_Structure/4.17%3A_Isotopes | Are all the members of the football team shown above identical? They are on the same team and are all known by the same team name, but there are individual differences among the players. We do not expect the kicker to be as big as the quarterback. The tight end is very likely to weigh less than the defensive tackle on the other side of the ball. They play as a unit, but they have different weights and heights. What are Isotopes? The history of the atom is full of some of these differences. Although John Dalton stated in his atomic theory of 1804 that all atoms of an element are identical, the discovery of the neutron began to show that this assumption was not correct. The study of radioactive materials (elements that spontaneously give off particles to form new elements) by Frederick Soddy (1877-1956) gave important clues about the internal structure of atoms. His work showed that some substances with different radioactive properties and different atomic masses were in fact the same element. He coined the term isotope from the Greek roots isos (íσος “equal”) and topos (τóπος “place”). He described isotopes as, “Put colloquially, their atoms have identical outsides but different insides.” Soddy won the Nobel Prize in Chemistry in 1921 for his work. As stated earlier, not all atoms of a given element are identical. Specifically, the number of neutrons can be variable for many elements. As an example, naturally occurring carbon exists in three forms. Each carbon atom has the same number of protons (6), which is its atomic number. Each carbon atom also contains six electrons in order to maintain electrical neutrality. However the number of neutrons varies as six, seven, or eight. Isotopes are atoms that have the same number atomic number, but different mass numbers due to a change in the number of neutrons. The three isotopes of carbon can be referred to as carbon-12 (\(^{12} _6C\)), carbon-13 (\(^{13} _6C\)), and carbon-14 (\(^{14} _6C\)) refers to the nucleus of a given isotope of an element. A carbon atom is one of three different nuclides. Most elements naturally consist of mixtures of isotopes. Carbon has three natural isotopes, while some heavier elements can have many more. Tin has ten stable isotopes, the most of any element. While the presence of isotopes affects the mass of an atom, it does not affect its chemical reactivity. Chemical behavior is governed by the number of electrons and the number of protons. Carbon-13 behaves chemically in exactly the same way as the more plentiful carbon-12. Summary Isotopes are atoms that have the same atomic number, but different mass numbers due to a change in the number of neutrons. The term nuclide refers to the nucleus of a given isotope of an element. The atomic mass of an atom equals the sum of the protons and the neutrons. Review What are isotopes? Why do different isotopes of an element generally have the same physical and chemical properties? How would the nucleus of the hydrogen-1 and hydrogen-2 differ? Relate the concepts of isotope and mass number. All oxygen atoms have eight protons, and most have eight neutrons as well. What is the mass number of an oxygen isotope that has nine neutrons? What is the name of this isotope? An isotope of yttrium has 39 protons and 59 neutrons. What is the mass number of that isotope? An isotope with a mass number of 193 has 116 neutrons. What is the atomic number of this isotope? An isotope of barium (atomic number 56) has an mass of 138. How many neutrons are in the nucleus of this isotope? Relate the concepts of isotope and mass number. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Valence_Bond_Theory_and_Hybrid_Atomic_Orbitals | However, the application of VSEPR theory can be expanded to complicated molecules such as 0 H H H O | | | // H-C-C=C=C-C=C-C-C | | \ H N O-H / \ H H By applying the VSEPR theory, one deduces the following results: \(\ce{H-C-C}\) bond angle = 109 o \(\ce{H-C=C}\) bond angle = 120 o , geometry around \(\ce{C}\) trigonal planar \(\ce{C=C=C}\) bond angle = 180 o , in other words linear \(\ce{H-N-C}\) bond angle = 109 o , tetrahedral around \(\ce{N}\) \(\ce{C-O-H}\) bond angle = 105 or 109 o , 2 lone electron pairs around \(\ce{O}\) Confidence Building Questions In terms of valence bond theory, how is a chemical bond formed? Hint: A chemical bond is due to the overlap of atomic orbitals. Discussion - Molecular orbital theory considers the energy states of the molecule. When one s and two p atomic orbitals are used to generate hybrid orbitals, how many hybrid orbitals will be generated? Hint: Using three atomic orbitals generates three hybrid orbitals. Discussion - Number of orbitals does not change in hybridization of atomic orbitals. In the structures of \(\ce{SO2}\) and \(\ce{NO2}\) , what are the values of the bond angles? Hint: The bond angles are expected to be less than 120 degrees. Discussion - Since the lone electron pair in \(\ce{:SO2}\) and lone electron in \(\ce{.NO2}\) take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds. What is the geometrical shape of the molecule \(\ce{CH4}\) , methane? Hint: Methane molecules are tetrahedral. Discussion - The 4 \(\ce{H}\) atoms form a tetrahedron, and methane has a tetrahedral shape. What do you expect the bond angles to be in the \(\ce{NH4+}\) ion? Hint: All bond angles are 109.5 degrees, the ideal value for a symmetric tetrahedral structure. Discussion - The structure of this ion is very similar to that of \(\ce{CH4}\). What hybrid orbitals does the \(\ce{C}\) atom use in the compound \(\ce{H-C\equiv C-H}\) , in which the molecule is linear? Hint: The sp hybrid orbitals are used by the \(\ce{C}\) atom. Discussion - Sigma ( s ) bonds are due to sp hybrid orbitals, and 2 p orbitals are used for pi ( p ) bonds. The two sigma bonds for each \(\ce{C}\) are due to overlap of sp hybrid orbitals of each \(\ce{C}\) atom. What hybrid orbitals does \(\ce{C}\) use in the molecule: H O=C< H This is a trigonal planar molecule. It is called formaldehyde, a solvent for preserving biological samples. The compound has an unpleasant smell. Hint: The \(\ce{C}\) atom uses sp 2 hybrid orbitals. Skill - The \(\ce{C}\) atom has 3 sigma ( s ) bonds by using three sp 2 hybrid orbitals and a pi ( p ) bond, due to one 2 p orbital. What is the shape of the molecule \(\ce{SF6}\) ? Hint: Its shape is octahedral. Discussion - Since the \(\ce{S}\) atom uses d 2 sp 3 hybrid orbitals, you expect the shape to be octahedral. The \(\ce{F}\) atoms form an octahedron around the sulfur. Phosphorus often forms a five coordinated compound \(\ce{PX5}\) . What hybrid orbitals does \(\ce{P}\) use in these compounds? Hint: The \(\ce{P}\) atom uses dsp 3 hybrid orbitals. Discussion - A total of 5 atomic orbitals are used in the hybridization: one 3 d , one 3 s and three 3 p orbitals. The dsp 3 hybrid orbitals of \(\ce{P}\) give rise to a trigonal bipyramidal coordination around the \(\ce{P}\) atom. The energy of d orbitals in \(\ce{N}\) is not compatible with 2 s and 2 p orbitals for hybridization. Thus, you seldom encounter a compound with formula \(\ce{NX5}\) with \(\ce{N}\) as the central atom. |
Courses/El_Paso_Community_College/CHEM1306%3A_Health_Chemistry_I_(Rodriguez)/07%3A_Energy_and_Chemical_Reactions/7.01%3A_Energy | Learning Objectives Define heat and work. Distinguish between kinetic energy and potential energy. State the law of conservation of matter and energy. Just like matter, energy is a term that we are all familiar with and use on a daily basis. Before you go on a long hike, you eat an energy bar; every month, the energy bill is paid; on TV , politicians argue about the energy crisis. But what is energy? If you stop to think about it, energy is very complicated. When you plug a lamp into an electric socket, you see energy in the form of light, but when you plug a heating pad into that same socket, you only feel warmth. Without energy, we couldn't turn on lights, we couldn't brush our teeth, we couldn't make our lunch, and we couldn't travel to school. In fact, without energy, we couldn't even wake up because our bodies require energy to function. We use energy for every single thing that we do, whether we are awake or asleep. Ability to Do Work or Produce Heat When we speak of using energy, we are really referring to transferring energy from one place to another. When you use energy to throw a ball, you transfer energy from your body to the ball, and this causes the ball to fly through the air. When you use energy to warm your house, you transfer energy from the furnace to the air in your home, and this causes the temperature in your house to rise. Although energy is used in many kinds of different situations, all of these uses rely on energy being transferred in one of two ways. Energy can be transferred as heat or as work . When scientists speak of heat , they are referring to energy that is transferred from an object with a higher temperature to an object with a lower temperature, as a result of the temperature difference. Heat will "flow" from the hot object to the cold object until both end up at the same temperature. When you cook with a metal pot, you witness energy being transferred in the form of heat. Initially, only the stove element is hot—the pot and the food inside the pot are cold. As a result, heat moves from the hot stove element to the cold pot. After a while, enough heat is transferred from the stove to the pot, raising the temperature of the pot and all of its contents (Figure \(\PageIndex{1}\)). Heat is only one way in which energy can be transferred. Energy can also be transferred as work . The scientific definition of work is force (any push or pull) applied over a distance . When you push an object and cause it to move, you do work, and you transfer some of your energy to the object. At this point, it's important to warn you of a common misconception. Sometimes we think that the amount of work done can be measured by the amount of effort put in. This may be true in everyday life, but it is not true in science. By definition, scientific work requires that force be applied over a distance . It does not matter how hard you push or how hard you pull. If you have not moved the object, you haven't done any work. So far, we've talked about the two ways in which energy can be transferred from one place, or object, to another. Energy can be transferred as heat, and energy can be transferred as work. But the question still remains— what IS energy ? Kinetic Energy Machines use energy, our bodies use energy, energy comes from the sun, energy comes from volcanoes, energy causes forest fires, and energy helps us to grow food. With all of these seemingly different types of energy, it's hard to believe that there are really only two different forms of energy: kinetic energy and potential energy. Kinetic energy is energy associated with motion. When an object is moving, it has kinetic energy. When the object stops moving, it has no kinetic energy. While all moving objects have kinetic energy, not all moving objects have the same amount of kinetic energy. The amount of kinetic energy possessed by an object is determined by its mass and its speed. The heavier an object is and the faster it is moving, the more kinetic energy it has. Kinetic energy is very common, and it's easy to spot examples of it in the world around you. Sometimes we even try to capture kinetic energy and use it to power things like our home appliances. If you are from California, you might have driven through the Tehachapi Pass near Mojave or the Montezuma Hills in Solano County and seen the windmills lining the slopes of the mountains (Figure \(\PageIndex{2}\)). These are two of the larger wind farms in North America. As wind rushes along the hills, the kinetic energy of the moving air particles turns the windmills, trapping the wind's kinetic energy so that people can use it in their houses and offices. Potential Energy Potential energy is stored energy. It is energy that remains available until we choose to use it. Think of a battery in a flashlight. If left on, the flashlight battery will run out of energy within a couple of hours, and the flashlight will die. If, however, you only use the flashlight when you need it, and turn it off when you don’t, the battery will last for days or even months. The battery contains a certain amount of energy, and it will power the flashlight for a certain amount of time, but because the battery stores potential energy, you can choose to use the energy all at once, or you can save it and only use a small amount at a time. Any stored energy is potential energy. There are a lot of different ways in which energy can be stored, and this can make potential energy very difficult to recognize. In general, an object has potential energy because of its position relative to another object . For example, when a rock is held above the earth, it has potential energy because of its position relative to the ground. This is potential energy because the energy is stored for as long as the rock is held in the air. Once the rock is dropped, though, the stored energy is released as kinetic energy as the rock falls. Chemical Energy There are other common examples of potential energy. A ball at the top of a hill stores potential energy until it is allowed to roll to the bottom. When two magnets are held next to one another, they store potential energy too. For some examples of potential energy, though, it's harder to see how "position" is involved. In chemistry, we are often interested in what is called chemical potential energy . Chemical potential energy is energy stored in the atoms, molecules, and chemical bonds that make up matter. How does this depend on position? As you learned earlier, the world, and all of the chemicals in it are made up of atoms and molecules. These store potential energy that is dependent on their positions relative to one another. Of course, you can't see atoms and molecules. Nevertheless, scientists do know a lot about the ways in which atoms and molecules interact, and this allows them to figure out how much potential energy is stored in a specific quantity (like a cup or a gallon) of a particular chemical. Different chemicals have different amounts of potential energy because they are made up of different atoms, and those atoms have different positions relative to one another. Since different chemicals have different amounts of potential energy, scientists will sometimes say that potential energy depends not only on position , but also on composition . Composition affects potential energy because it determines which molecules and atoms end up next to one another. For example, the total potential energy in a cup of pure water is different than the total potential energy in a cup of apple juice, because the cup of water and the cup of apple juice are composed of different amounts of different chemicals. At this point, you may wonder just how useful chemical potential energy is. If you want to release the potential energy stored in an object held above the ground, you just drop it. But how do you get potential energy out of chemicals? It's actually not difficult. Use the fact that different chemicals have different amounts of potential energy . If you start with chemicals that have a lot of potential energy and allow them to react and form chemicals with less potential energy, all the extra energy that was in the chemicals at the beginning, but not at the end, is released. Units of Energy Energy is measured in one of two common units: the calorie and the joule. The joule \(\left( \text{J} \right)\) is the SI unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A calorie \(\left( \text{cal} \right)\) is the quantity of heat required to raise the temperature of 1 gram of water by \(1^\text{o} \text{C}\). For example, raising the temperature of \(100 \: \text{g}\) of water from \(20^\text{o} \text{C}\) to \(22^\text{o} \text{C}\) would require \(100 \times 2 = 200 \: \text{cal}\). Calories contained within food are actually kilocalories \(\left( \text{kcal} \right)\). In other words, if a certain snack contains 85 food calories, it actually contains \(85 \: \text{kcal}\) or \(85,000 \: \text{cal}\). In order to make the distinction, the dietary calorie is written with a capital C. \[1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories} \nonumber \] To say that the snack "contains" 85 Calories means that \(85 \: \text{kcal}\) of energy are released when that snack is processed by your body. Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below. \[1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J} \nonumber \] We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested. \[400 \: \text{Cal} = 400 \: \text{kcal} \times \dfrac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ} \nonumber \] Summary Any time we use energy, we transfer energy from one object to another. Energy can be transferred in one of two ways: as heat, or as work. Heat is the term given to energy that is transferred from a hot object to a cooler object due to the difference in their temperatures. Work is the term given to energy that is transferred as a result of a force applied over a distance. Energy comes in two fundamentally different forms: kinetic energy and potential energy. Kinetic energy is the energy of motion. Potential energy is stored energy that depends on the position of an object relative to another object. Chemical potential energy is a special type of potential energy that depends on the positions of different atoms and molecules relative to one another. Chemical potential energy can also be thought of according to its dependence on chemical composition. Energy can be converted from one form to another. The total amount of mass and energy in the universe is conserved. Contributions & Attributions Wikibooks |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Mechanics__in_Chemistry_(Simons_and_Nichols)/04%3A_Atomic_Orbitals | Valence atomic orbitals on neighboring atoms combine to form bonding, non-bonding and antibonding molecular orbitals. In Section 1 the Schrödinger equation for the motion of a single electron moving about a nucleus of charge Z was explicitly solved. The energies of these orbitals relative to an electron infinitely far from the nucleus with zero kinetic energy were found to depend strongly on Z and on the principal quantum number n, as were the radial "sizes" of these hydrogenic orbitals. Closed analytical expressions for the \(r\),\(θ\), and \(ϕ\) dependence of these orbitals are given in Appendix B. The reader is advised to also review this material before undertaking study of this section. 4.1: Shapes of Atomic Orbitals Shapes of atomic orbitals play central roles in governing the types of directional bonds an atom can form. 4.2: Directions of Atomic Orbitals Atomic orbital directions also determine what directional bonds an atom will form. 4.3: Sizes and Energies Orbital energies and sizes go hand-in-hand; small 'tight' orbitals have large electron binding energies (i.e., low energies relative to a detached electron). For orbitals on neighboring atoms to have large (and hence favorable to bond formation) overlap, the two orbitals should be of comparable size and hence of similar electron binding energy. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Molecular_and_Atomic_Spectroscopy/04_Learning_Objectives/02_Ultraviolet_Visible_Spectroscopy | After completing this unit the student will be able to: Compare and contrast atomic and molecular spectra. Explain why atomic spectra consist of lines whereas molecular spectra at room temperature are broad and continuous. Justify the difference in molecular spectra at room temperature and 10K. Describe the cause of Doppler broadening. Determine the effect of conjugation on a UV/Vis absorption spectrum. Determine the effect of non-bonding electrons on a UV/Vis absorption spectrum. Determine the effect of solvent on the energy of n- \(\pi\) * and \(\pi\) - \(\pi\) * transitions. Evaluate the utility of UV/Vis spectroscopy as a qualitative and quantitative method. Describe a procedure by which UV/Vis spectroscopy can be used to determine the pKa of a weak acid. |
Courses/De_Anza_College/CHEM_10%3A_Introduction_to_Chemistry_(Parajon_Puenzo)/08%3A_Acids_and_Bases/8.02%3A_Acids_and_Bases | Learning Objectives Identify an Arrhenius acid and an Arrhenius base. Identify a Brønsted-Lowry acid and a Brønsted-Lowry base. Write chemical reactions between an Arrhenius acid and an Arrhenius base and between Brønsted-Lowry acid and a Brønsted-Lowry base. There are three major classifications of substances known as acids or bases. The Arrhenius definition states that an acid produces H + in solution and a base produces OH - . This theory was developed by Svante Arrhenius in 1883. Later, two more sophisticated and general theories were proposed. These are the Brønsted-Lowry and the Lewis definitions of acids and bases. The Lewis theory is discussed elsewhere. The Arrhenius Theory of Acids and Bases In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds, termed acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. An Arrhenius acid is a compound that increases the concentration of \(\ce{H^{+}}\) ions that are present when added to water. These H + ions form the hydronium ion (H 3 O + ) when they combine with water molecules. This process is represented in a chemical equation by adding H 2 O to the reactants side. \[ \ce{HCl(aq) \rightarrow H^{+}(aq) + Cl^{-}(aq) } \nonumber \] In this reaction, hydrochloric acid (\(HCl\)) dissociates completely into hydrogen (H + ) and chlorine (Cl - ) ions when dissolved in water, thereby releasing H + ions into solution. Formation of the hydronium ion equation: \[\ce{ HCl(aq) + H_2O(l) \rightarrow H_3O^{+}(aq) + Cl^{-}(aq)} \nonumber \] An Arrhenius base is a compound that increases the concentration of \(\ce{OH^{-}}\) ions that are present when added to water. The dissociation is represented by the following equation: \[\ce{ NaOH \; (aq) \rightarrow Na^{+} \; (aq) + OH^{-} \; (aq) } \nonumber \] In this reaction, sodium hydroxide (NaOH) disassociates into sodium (\(Na^+\)) and hydroxide (\(OH^-\)) ions when dissolved in water, thereby releasing OH - ions into solution. Arrhenius acids are substances which produce hydrogen ions in solution and Arrhenius bases are substances which produce hydroxide ions in solution. Limitations to the Arrhenius Theory The Arrhenius theory has many more limitations than the other two theories. The theory does not explain the weak base ammonia (NH 3 ), which in the presence of water, releases hydroxide ions into solution, but does not contain OH- itself. Also, the Arrhenius definition of acid and base is limited to aqueous (i.e., water) solutions. The Brønsted-Lowry Theory of Acids and Bases In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed new definitions for acids and bases, ones that focus on proton transfer. A Brønsted-Lowry acid is any species that can donate a proton (H + ) to another molecule. A Brønsted-Lowry base is any species that can accept a proton from another molecule. In short, a Brønsted-Lowry acid is a proton donor (PD) , while a Brønsted-Lowry base is a proton acceptor (PA) . A Brønsted-Lowry acid is a proton donor, while a Brønsted-Lowry base is a proton acceptor. Let us use the reaction of ammonia in water to demonstrate the Brønsted-Lowry definitions of an acid and a base. Ammonia and water molecules are reactants, while the ammonium ion and the hydroxide ion are products: \[\ce{NH3(aq) + H2O (ℓ) <=> NH^{+}4(aq) + OH^{−}(aq) }\label{Eq1} \] What has happened in this reaction is that the original water molecule has donated a hydrogen ion to the original ammonia molecule, which in turn has accepted the hydrogen ion. We can illustrate this as follows (Figure \(\PageIndex{2}\)): Because the water molecule donates a hydrogen ion to the ammonia, it is the Brønsted-Lowry acid, while the ammonia molecule—which accepts the hydrogen ion—is the Brønsted-Lowry base. Thus, ammonia acts as a base in both the Arrhenius sense and the Brønsted-Lowry sense. Is an Arrhenius acid like hydrochloric acid still an acid in the Brønsted-Lowry sense? Yes, but it requires us to understand what really happens when HCl is dissolved in water. Recall that the hydrogen atom is a single proton surrounded by a single electron. To make the hydrogen ion , we remove the electron, leaving a bare proton. Do we really have bare protons floating around in aqueous solution? No, we do not. What really happens is that the H + ion attaches itself to H 2 O to make H 3 O + , which is called the hydronium ion . For most purposes, H + and H 3 O + represent the same species, but writing H 3 O + instead of H + shows that we understand that there are no bare protons floating around in solution. Rather, these protons are actually attached to solvent molecules. The Hydronium Ion A proton in aqueous solution may be surrounded by more than one water molecule, leading to formulas like \(\ce{H5O2^{+}}\) or \(\ce{H9O4^{+}}\) rather than \(\ce{H3O^{+}}\). It is simpler, however, to use \(\ce{H3O^{+}}\) to represent the hydronium ion . With this in mind, how do we define HCl as an acid in the Brønsted-Lowry sense? Consider what happens when HCl is dissolved in H 2 O: \[\ce{HCl(g) + H_2O (ℓ) \rightarrow H_3O^{+}(aq) + Cl^{−}(aq) }\label{Eq2} \] We can depict this process using Lewis electron dot diagrams (Figure \(\PageIndex{4}\)): Now we see that a hydrogen ion is transferred from the HCl molecule to the H 2 O molecule to make chloride ions and hydronium ions. As the hydrogen ion donor, HCl acts as a Brønsted-Lowry acid; as a hydrogen ion acceptor, H 2 O is a Brønsted-Lowry base. So HCl is an acid not just in the Arrhenius sense but also in the Brønsted-Lowry sense. Moreover, by the Brønsted-Lowry definitions, H 2 O is a base in the formation of aqueous HCl. So the Brønsted-Lowry definitions of an acid and a base classify the dissolving of HCl in water as a reaction between an acid and a base—although the Arrhenius definition would not have labeled H 2 O a base in this circumstance. A summary of the acid-base definitions based on the Arrhenius and Bronstes-Lowry theories is given in Table \(\PageIndex{1}\). Type Acid Base Arrhenius \(\ce{H^+}\) ions in solution \(\ce{OH^-}\) ions in solution Brønsted-Lowry \(\ce{H^+}\) donor \(\ce{H^+}\) acceptor All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases. Example \(\PageIndex{1}\) Aniline (C 6 H 5 NH 2 ) is slightly soluble in water. It has a nitrogen atom that can accept a hydrogen ion from a water molecule just like the nitrogen atom in ammonia does. Write the chemical equation for this reaction and identify the Brønsted-Lowry acid and base. Solution C 6 H 5 NH 2 and H 2 O are the reactants. When C 6 H 5 NH 2 accepts a proton from H 2 O, it gains an extra H and a positive charge and leaves an OH − ion behind. The reaction is as follows: \[\ce{C6H5NH2(aq) + H2O(ℓ) <=> C6H5NH3^{+}(aq) + OH^{−}(aq)} \nonumber \] Because C 6 H 5 NH 2 accepts a proton, it is the Brønsted-Lowry base. The H 2 O molecule, because it donates a proton, is the Brønsted-Lowry acid. Exercise \(\PageIndex{1}\) Identify the Brønsted-Lowry acid and the Brønsted-Lowry base in this chemical equation. \[\ce{H2PO4^{-} + H_2O <=> HPO4^{2-} + H3O^{+}} \nonumber \] Answer Brønsted-Lowry acid: H 2 PO 4 - ; Brønsted-Lowry base: H 2 O Exercise \(\PageIndex{2}\) Which of the following compounds is a Bronsted-Lowry base? HCl HPO 4 2 - H 3 PO 4 NH 4 + CH 3 NH 3 + Answer: A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H + . This eliminates \(\ce{HCl}\), \(\ce{H3PO4}\), \(\ce{NH4^{+}}\) and \(\ce{CH_3NH_3^{+}}\) because they are Bronsted-Lowry acids. They all give away protons. In the case of \(\ce{HPO4^{2-}}\), consider the following equation: \[\ce{HPO4^{2-} (aq) + H2O (l) \rightarrow PO4^{3-} (aq) + H3O^{+}(aq) } \nonumber \] Here, it is clear that HPO 4 2 - is the acid since it donates a proton to water to make H 3 O + and PO 4 3 - . Now consider the following equation: \[ \ce{ HPO4^{2-}(aq) + H2O(l) \rightarrow H2PO4^{-} + OH^{-}(aq)} \nonumber \] In this case, HPO 4 2 - is the base since it accepts a proton from water to form H 2 PO 4 - and OH - . Thus, HPO 4 2 - is an acid and base together, making it amphoteric. Since HPO 4 2 - is the only compound from the options that can act as a base, the answer is (b) HPO 4 2 - . Summary An Arrhenius acid is a compound that increases the H + ion concentration and an Arrhenius base is a compound that increases the OH − ion concentration in aqueous solution. A Brønsted-Lowry acid is a proton donor; a Brønsted-Lowry base is a proton acceptor. All Arrhenius acids and bases are Brønsted-Lowry acids and bases as well. However, not all Brønsted-Lowry acids and bases are Arrhenius acids and bases. |
Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_5%3A_Rates_of_Chemical_and_Physical_Processes/19%3A_Nuclear_Chemistry/19.1%3A_Mass-Energy_Relationships_in_Nuclei | Learning Objectives To understand the factors that affect nuclear stability. Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or decay, at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus. The Atomic Nucleus Each element can be represented by the notation \(^A_Z \textrm X\), where A , the mass number, is the sum of the number of protons and the number of neutrons, and Z , the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons , and an atom with a particular number of protons and neutrons is called a nuclide . Nuclides with the same number of protons but different numbers of neutrons are called isotopes . Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways: \(^A_Z \textrm X\) \(\ce{^{16}_8 O}\) \(\ce{^{17}_8 O}\) \(\ce{^{18}_8 O}\) \(^A \textrm X\) \(\ce{^{16} O}\) \(\ce{^{17} O}\) \(\ce{^{18} O}\) \(\textrm{element-A:}\) \(\textrm{oxygen-16}\) \(\textrm{oxygen-17}\) \(\textrm{oxygen-18}\) Because the number of neutrons is equal to A − Z , we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s natural abundance . For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive , emitting subatomic particles and electromagnetic radiation. The emissions are collectively called radioactivity and can be measured. Isotopes that emit radiation are called radioisotopes . Nuclear Stability The nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force , an extremely powerful but very short-range attractive force between nucleons ( Figure \(\PageIndex{1}\) ). All stable nuclei except the hydrogen-1 nucleus ( 1 H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability. The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in Figure \(\PageIndex{2}\) . The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, 1 H and 3 He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., \(^4_2 \textrm{He}\), \(^{10}_5 \textrm{B}\), and \(^{40}_{20} \textrm{Ca}\)). All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with Z > 83 are unstable and radioactive. As shown in Figure \(\PageIndex{3}\) , more than half of the stable nuclei (166 out of 279) have even numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called magic numbers 2, 8, 20, 50, 82, and 126. For example, tin ( Z = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium ( Z = 49) and antimony ( Z = 51), have only 2 stable isotopes each. Nuclei with magic numbers of both protons and neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are \(^4_2 \textrm{He}\), with 2 protons and 2 neutrons, and \(^{208}_{82} \textrm{Pb}\), with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element. Most stable nuclei contain even numbers of both neutrons and protons The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in Figure \(\PageIndex{2}\) , the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei. Origin of the Magic Numbers Multiple models have been formulated to explain the origin of the magic numbers and two popular ones are the Nuclear Shell Model and the Liquid Drop Model . Unfortuneatly, both require advanced quantum mechanics to fully understand and are beyond the scope of this text. Example \(\PageIndex{1}\) Classify each nuclide as stable or radioactive. \(\ce{_{15}^{30} P}\) \(\ce{_{43}^{98} Tc}\) tin-118 \(\ce{_{94}^{239} Pu}\) Given : mass number and atomic number Asked for : predicted nuclear stability Strategy : Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide. Solution : a. This isotope of phosphorus has 15 neutrons and 15 protons, giving a neutron-to-proton ratio of 1.0. Although the atomic number, 15, is much less than the value of 83 above which all nuclides are unstable, the neutron-to-proton ratio is less than that expected for stability for an element with this mass. As shown in Figure \(\PageIndex{2}\) , its neutron-to-proton ratio should be greater than 1. Moreover, this isotope has an odd number of both neutrons and protons, which also tends to make a nuclide unstable. Consequently, \(_{15}^{30} \textrm P\) is predicted to be radioactive, and it is. b. This isotope of technetium has 55 neutrons and 43 protons, giving a neutron-to-proton ratio of 1.28, which places \(_{43}^{98} \textrm{Tc}\) near the edge of the band of stability. The atomic number, 55, is much less than the value of 83 above which all isotopes are unstable. These facts suggest that \(_{43}^{98} \textrm{Tc}\) might be stable. However, \(_{43}^{98} \textrm{Tc}\) has an odd number of both neutrons and protons, a combination that seldom gives a stable nucleus. Consequently, \(_{43}^{98} \textrm{Tc}\) is predicted to be radioactive, and it is. c. Tin-118 has 68 neutrons and 50 protons, for a neutron-to-proton ratio of 1.36. As in part b, this value and the atomic number both suggest stability. In addition, the isotope has an even number of both neutrons and protons, which tends to increase nuclear stability. Most important, the nucleus has 50 protons, and 50 is one of the magic numbers associated with especially stable nuclei. Thus \(_{50}^{118} \textrm{Sn}\)should be particularly stable. d. This nuclide has an atomic number of 94. Because all nuclei with Z > 83 are unstable, \(_{94}^{239} \textrm{Pu}\) must be radioactive. Exercise \(\PageIndex{1}\) Classify each nuclide as stable or radioactive. \(\ce{_{90}^{232} Th}\) \(\ce{_{20}^{40} Ca}\) \(\ce{_8^{15} O}\) \(\ce{_{57}^{139} La}\) Answer a radioactive Answer b stable Answer c radioactive Answer d stable Superheavy Elements In addition to the “peninsula of stability” there is a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements , with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with Z = 114 and Z = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with A = 292 that was found in 232 Th. With an estimated half-life greater than 10 8 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with Z = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements. Summary Subatomic particles of the nucleus (protons and neutrons) are called nucleons . A nuclide is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is radioactive , and its emissions are collectively called radioactivity . Isotopes that emit radiation are called radioisotopes . Each nucleon is attracted to other nucleons by the strong nuclear force . Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain magic numbers of protons and neutrons are often especially stable. Superheavy elements , with atomic numbers near 126, may even be stable enough to exist in nature. |
Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/06%3A_Acid-Base_and_Donor-Acceptor_Chemistry/6.03%3A_Brnsted-Lowry_Concept/6.3.01%3A_Brnsted-Lowry_Concept | The Brønsted-Lowry acid-base concept The Brønsted-Lowry acid-base concept overcomes the Arrhenius system's inability to describe reactions that take place outside of aqueous solution by moving the focus away from the solution and onto the acid and base themselves. It does this by redefining acid-base reactivity as involving the transfer of a hydrogen ion, \(H^+\), between an acid and a base. Specifically, a Brønsted acid is a substance that loses an \(H^+\) ion by donating it to a base. This means that a Brønsted base is defined as a substance which accepts \(H^+\) from an acid when it reacts. Because the Brønsted-Lowry concept is concerned with \(H^+\) ion transfer rather than the creation of a particular chemical species, it is able to handle a diverse array of acid-base concepts. In fact, from the viewpoint of the Brønsted-Lowry concept, Arrhenius acids and bases are just a special case involving hydrogen ion donation and acceptance involving water. Arrhenius acids donate \(H^+\) ion to water, which acts as a Brønsted base in accepting it to give \(H_3O^+\): \[\label{ 6.3.1} \] Similarly, Arrhenius bases act as Brønsted bases in accepting a hydrogen ion from the Brønsted acid water: \[ \nonumber \] In this way it can be seen that Arrhenius acids and bases are defined in terms of their causing hydrogen ions to be donated to and abstracted from water, respectively, while Brønsted acids and bases are defined in terms of their ability to donate and accept hydrogen ions to and from anything. Becasue the Brønsted-Lowry concept can handle any sort of hydrogen ion transfer it readily accommodates many reactions that Arrhenius theory cannot, including those that take place outside of water, such as the reaction between gaseous hydrochloric acid and ammonia: \[ \nonumber \] The classification of acids as strong or weak usually refers to their ability to donate or abstract hydrogen ions to or from water to give \(H_3O^+\) and \(OH^-\), respectively, i.e., their Arrhenius acidity and basicity. However, acids and bases may be classified as strong and weak under the Brønsted-Lowry definition based on whether they completely transfer or accept hydrogen ions; it is just that in this case it is important to specify the conditions under which a given acid or base acts strong or weak. For example, acetic acid acts as a weak acid in water but is a strong acid in triethylamine, since in the latter case it completely transfers a hydrogen ion to triethylamine to give triethylammonium acetate. Alternatively, the acidity or basicity of a compound may be specified using a thermodynamic scale like the Hammett acidity . Conjugate Acids and Bases By redefining acids and bases in terms of hydrogen ion donation and acceptance, the Brønsted-Lowry system makes it easy to recognize that when an acid loses its hydrogen ion it becomes a substance that is capable of receiving it back again, namely, a base. Consider, for example, the base dissociation of ammonia in water. When ammonia acts as a Brønsted base and receives a hydrogen ion from water, ammonium ion and hydroxide are formed: \[ \nonumber \] The ammonium ion is itself a weak acid that can undergo dissociation: \[ \nonumber \] In this case ammonia and ammonium ion are acid-base conjugates. In general acids and bases that differ by a single ionizable hydrogen ion are said to be conjugates of one another. The strengths of conjugates vary reciprocally with one another, so the stronger the acid the weaker the base and vice versa. For example, in water, acetic acid acts as a weak Brønsted acid: \[ \nonumber \] and acetic acid's conjugate base, acetate, acts as a weak Brønsted base. \[ \nonumber \] However, in liquid ammonia acetic acid acts as a strong Brønsted acid: \[ \nonumber \] while its conjugate base, acetate, is neutral. \[ \nonumber \] The reciprocal relationship between the strengths of acids and their conjugate bases has several consequences: Under conditions when an acid or base acts as a weak acid or base its conjugate acts as weak as well. Conversely, when an acid or base acts as a strong acid or base its conjugate acts as a neutral species. When a Brønsted acid and base react with one another, the equilibrium favors formation of the weakest acid-base pair. That is why the acid-base reaction between acetic acid and ammonia in liquid ammonia proceeded to give the weak acid ammonium ion and neutral acetate. This consequence is particularly important for understanding the behavior of acids and bases in nonaqueous solvents, as illustrated by the following example. Example \(\PageIndex{1}\) Can a solution of lithium diisopropylamide in heptane be used to form lithium cyclopentadienide? The \(pK_a\) of cyclopentadiene and diisopropylamine are ~15 and 40, respectively, and the proposed reaction is as follows: Solution: Since cyclopentadiene is a stronger acid than diisopropylamine (the stronger the acid the lower the \(pK_a\)) the equilibrium will favor protonation of the diisopropylamine by cyclopentadiene. Consequently addition of a heptane solution of lithium diisopropylamide to monomeric cyclopentadiene should give lithium cyclopentadienide. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.05%3A_Chemical_Potentials/1.5.15%3A_Chemical_Potentials-_Solute-_Concentration_and_Molality_Scales | For a given solution we can express the chemical potential of solute \(j\), \(\mu_{\mathrm{j}}(\mathrm{aq})\) in an aqueous solution at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\left(\approx \mathrm{p}^{0}\right)\) using two equations. Therefore, at fixed \(\mathrm{T}\) and \(\mathrm{p}\), \[\begin{aligned} &\mu_{\mathrm{j}}^{0}(\mathrm{aq})+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{m}_{\mathrm{j}} \, \gamma_{\mathrm{j}} / \mathrm{m}^{0}\right)= \\ &\mu_{\mathrm{j}}^{0}(\mathrm{aq} ; \mathrm{c}-\text { scale })+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{c}_{\mathrm{j}} \, \mathrm{y}_{\mathrm{j}} / \mathrm{c}_{\mathrm{r}}\right) \end{aligned} \nonumber \] Therefore, \[\ln \left(\mathrm{y}_{\mathrm{j}}\right)=\ln \left(\gamma_{\mathrm{j}}\right)+\ln \left(\mathrm{m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right) +(1 / \mathrm{R} \, \mathrm{T}) \,\left[\mu_{\mathrm{j}}^{0}(\mathrm{aq})-\mu_{\mathrm{j}}^{0}(\mathrm{aq} ; \mathrm{c}-\text { scale })\right] \nonumber \] In the latter two equations the composition variables \(\mathrm{m}_{j}\) and \(\mathrm{c}_{j}\) are expressed in the units ‘\(\mathrm{mol kg}^{-1}\)’ and ‘\(\mathrm{mol dm}^{-3}\)’ respectively [1]. The ratio ‘\(\mathrm{c}_{\mathrm{j}} / \mathrm{m}_{\mathrm{j}}\)’ equals the density expressed in the unit ‘\(\mathrm{kg dm}^{-3}\)’. For dilute solutions, \(\mathrm{c}_{\mathrm{j}} / \mathrm{m}_{\mathrm{j}}=\rho_{1}^{*}(\ell)\), the density of the pure solvent. \[\text { Also, } \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0}=\left[\mathrm{mol} \mathrm{d \textrm {dm } ^ { - 3 }}\right] /\left[\mathrm{mol} \mathrm{kg}^{-1}\right]=\left[\mathrm{kg} \mathrm{dm}^{-3}\right] \nonumber \] For dilute aqueous solutions at ambient pressure and \(298.2 \mathrm{~K}\) [2,3], \[\ln \left(\mathrm{m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right)=-\ln (0.997) \nonumber \] With reference to equation (b), with increasing dilution, \(\mathrm{y}_{\mathrm{j}} \rightarrow 1, \gamma_{\mathrm{j}} \rightarrow 1,\left(\mathrm{~m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right) \rightarrow \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \rho_{1}^{*}(\ell)\) Hence, \[\mu_{\mathrm{j}}^{0}(\mathrm{aq} ; \mathrm{c}-\mathrm{scale})-\mu_{\mathrm{j}}^{0}(\mathrm{aq})=\mathrm{R} \, \mathrm{T} \, \ln \left[\mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \rho_{1}^{*}(\ell)\right] \nonumber \] We combine equations (b) and (e). \[\ln \left(\mathrm{y}_{\mathrm{j}}\right)=\ln \left(\gamma_{\mathrm{j}}\right)+\ln \left(\mathrm{m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right)-\ln \left[\mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \rho_{1}^{*}(\ell)\right] \nonumber \] \[\ln \left(\mathrm{y}_{\mathrm{j}}\right)=\ln \left(\gamma_{\mathrm{j}}\right)+\ln \left(\mathrm{m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right)-\ln \left[\mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \rho_{1}^{*}(\ell)\right] \nonumber \] Footnotes [1] A given solution is prepared by adding \(\mathrm{n}_{j}\) moles of solute \(j\) to \(\mathrm{w}_{1} \mathrm{~kg}\) of solvent. Molality of solute \(\mathrm{j} / \mathrm{mol} \mathrm{kg}{ }^{-1}=\left(\mathrm{n}_{\mathrm{j}} / \mathrm{w}_{1}\right)\) Total mass of solution/kg \(=w_{1}+n_{j} \, M_{j}\) where molar mass of solute/kg \(\mathrm{mol}^{-1}=\mathrm{M}_{\mathrm{j}}\) Volume of solution/\(\mathrm{m}^{3} = \mathrm{V}\) Density of solution \(\rho / \mathrm{kg} \mathrm{m}^{-3}=\left[\frac{\mathrm{w}_{1}+\mathrm{n}_{\mathrm{j}} \, \mathrm{M}_{\mathrm{j}}}{\mathrm{V}}\right]\) By convention chemists express the composition of solutions in terms of (i) concentration using the unit ‘\(\mathrm{mol dm}^{-3}\)’ and (ii) molality using the unit, ‘\(\mathrm{mol kg}^{-1}\)’. These composition scales stem from the fact that at \(298.15 \mathrm{~K}\), \(1 \mathrm{~dm}^{3}\) of water has a mass of approx. \(1 \mathrm{~kg}\). So as we swap composition scales a conversion factor is often required . For dilute solutions \(w_{1}>n_{j} \, M_{j}\) and density of solution \(\rho\) equals the density of the pure solvent (at same temperature and pressure), i.e. density \(\rho=\rho 1(\ell) \mathrm{kg} \mathrm{m} \mathrm{m}^{-3}\) [2] A typical conversion takes the following form for water at \(298.2 \mathrm{~K}\) and ambient pressure. \(\begin{aligned} \text { Density }=0.997 \mathrm{~g} \mathrm{~cm}^{-3} &=0.997\left(10^{-3} \mathrm{~kg}\right)\left(10^{-2} \mathrm{~m}^{-3}\right.\\ &=0.997 \mathrm{X} \mathrm{} 10^{3} \mathrm{~kg} \mathrm{~m}^{-3} \\ =& 997 \mathrm{~kg} \mathrm{~m}^{-3}=0.997 \mathrm{~kg} \mathrm{\textrm {dm } ^ { - 3 }} \end{aligned}\) \(\text { Then } \frac{\mathrm{c}_{\mathrm{j}} / \mathrm{mol} \mathrm{dm}^{-3}}{\mathrm{~m}_{\mathrm{j}} / \mathrm{mol} \mathrm{kg}^{-1}}=\frac{\mathrm{n}_{\mathrm{j}} / \mathrm{mol}}{\mathrm{V} / \mathrm{dm}^{3}} \, \frac{\mathrm{w}_{1} / \mathrm{kg}}{\mathrm{n}_{\mathrm{j}} / \mathrm{mol}}=\frac{\mathrm{w}_{1} / \mathrm{kg}}{\mathrm{V} / \mathrm{dm}^{3}}=\rho / \mathrm{kg} \mathrm{dm}^{-3}\) [3] \(\begin{aligned} \ln \left(\mathrm{m}_{\mathrm{j}} \, \mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0} \, \mathrm{c}_{\mathrm{j}}\right) &=\ln \left[\left(\mathrm{c}_{\mathrm{r}} / \mathrm{m}^{0}\right) /\left(\mathrm{c}_{\mathrm{j}} / \mathrm{m}_{\mathrm{j}}\right)\right] \\ =& \ln \left[\left(\mathrm{kg} \mathrm{d \textrm {m } ^ { - 3 } ) / \rho ]}=-\ln \left(\rho / \mathrm{kg} \mathrm{d \textrm {dm } ^ { - 3 } )}\right.\right.\right. \end{aligned}\) |
Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/03%3A_Using_Chemical_Equations_in_Calculations/3.07%3A_Energy | Energy is usually defined as the capability for doing work. For example, a billiard ball can collide with a second ball, changing the direction or speed of motion of the latter. In such a process the motion of the first ball would also be altered. We would say that one billiard ball did work on (transferred energy to) the other. Kinetic Energy I mage source: Smart Learning for All Energy due to motion is called kinetic energy and is represented by E k . For an object moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed: \[ E_{k} = \frac{1}{2} mu^{2} \label{1} \] where m = mass of the object u = speed of object If the two billiard balls mentioned above were studied in outer space, where friction due to their collisions with air molecules or the surface of a pool table would be negligible, careful measurements would reveal that their total kinetic energy would be the same before and after they collided. This is an example of the law of conservation of energy , which states that energy cannot be created or destroyed under the usual conditions of everyday life. Whenever there appears to be a decrease in energy somewhere, there is a corresponding increase somewhere else. Example \(\PageIndex{1}\) : Kinetic Energy Calculate the kinetic energy of a Volkswagen Beetle of mass 844 kg (1860 lb) which is moving at 13.4 m s –1 (30 miles per hour). Solution: \(\large E_{k} = \frac{1}{2} m u^{2} = \frac{1}{2} \times 844 \text{ kg} \times ( 13.4 \text{ m} \text{ s}^{-1} )^{2} = 7.58 \times 10^{4} \text{ kg}\text{ m}^{2} \text{ s}^{-2}\) In other words the units for energy are derived from the SI base units kilogram for mass, meter for length, and second for time. A quantity of heat or any other form of energy may be expressed in kilogram meter squared per second squared. In honor of Joule’s pioneering work this derived unit 1 kg m 2 s –2 called the joule , abbreviated J. The Volkswagen in question could do nearly 76 000 J of work on anything it happened to run into. Potential Energy I mage source: Smart Learning for All Potential Energy is energy that is stored by rising in height, or by other means. It frequently comes from separating things that attract, like rising birds are being separated from the Earth that attracts them, or by pulling magnets apart, or pulling an electrostatically charged balloon from an oppositely charged object to which it has clung. Potential Energy is abbreviated E P and gravitational potential energy is calculated as follows: \[\large E_{P} = mgh \tag{2} \] where m = mass of the object in kg g = gravitational constant, 9.8 m s 2 h = height in m Notice that E P has the same units, kg m 2 s –2 or Joule as kinetic energy. Example \(\PageIndex{2}\): Kinetic Energy Application How high would the VW weighing 844 kg and moving at 30 mph need to rise (vertically) on a hill to come to a complete stop, if none of the stopping power came from friction? Solution : The car's kinetic energy is 7.58 × 10 4 kg m 2 s –2 (from EXAMPLE \(\PageIndex{1}\) ), so all of this would have to be converted to E P . Then we could calculate the vertical height: \(\large E_{P} = mgh = 7.58 \times 10^{4} \text{ kg} \text{ m}^{2} \text{ s}^{-2} = 844 \text{ kg} \times 9.8 \text{m} \text {s}^{-2} \times h \) \( \large h = 9.2 \text{ m} \) Even when there is a great deal of friction, the law of conservation of energy still applies. If you put a milkshake on a mixer and leave it there for 10 min, you will have a warm, rather unappetizing drink. The whirling mixer blades do work on (transfer energy to) the milkshake, raising its temperature. The same effect could be produced by heating the milkshake, a fact which suggests that heating also involves a transfer of energy. The first careful experiments to determine how much work was equivalent to a given quantity of heat were done by the English physicist James Joule (1818 to 1889) in the 1840s. In an experiment very similar to our milkshake example, Joule connected falling weights through a pulley system to a paddle wheel immersed in an insulated container of water. This allowed him to compare the temperature rise which resulted from the work done by the weights with that which resulted from heating. Units with which to measure energy may be derived from the SI base units of Table 1 from The International System of Units (SI)(opens in new window) by using Eq. \(\ref{1}\). Another unit of energy still widely used by chemists is the calorie . The calorie used to be defined as the energy needed to raise the temperature of one gram of water from 14.5°C to 15.5°C but now it is defined as exactly 4.184 J. |
Courses/Lumen_Learning/Book%3A_US_History_I_(OS_Collection)_(Lumen)/16%3A_Troubled_Times%3A_The_Tumultuous_1850s/16.7%3A_Assignment%3A_Pre-Civil_War_Perspectives | History contains many people with different circumstances and perspectives. To understand actions in the past, a historian mustunderstand their perspective and concerns and how they interacted. Imagine you are one of the following people from the decades before the US Civil War: An abolitionist (white or black, male or female) A women’s rights activist (white or black, male or female) A free black person An inslaved black person A white person in the US south (male or female, poor, middle-class, or rich.) A white person in the US north (male or female, poor, middle-class, or rich.) Tell your (imaginary) life story? How do you make your living? How do you feel about the major issues of the day? What are your primary concerns (economic, social, or political)? Why? Write about 200 words. CC licensed content, Original Authored by : Thomas deMayo. Provided by : Reynolds Community College. License : CC BY: Attribution |
Courses/Brevard_College/CHE_201%3A_Organic_Chemistry_I/03%3A_Unsaturated_Hydrocarbons/3.09%3A_Alkynes | Learning Objectives Describe the general physical and chemical properties of alkynes. Name alkynes given formulas and write formulas for alkynes given names. The simplest alkyne—a hydrocarbon with carbon-to-carbon triple bond—has the molecular formula C 2 H 2 and is known by its common name—acetylene (Figure \(\PageIndex{1}\)). Its structure is H–C≡C–H. Acetylene is used in oxyacetylene torches for cutting and welding metals. The flame from such a torch can be very hot. Most acetylene, however, is converted to chemical intermediates that are used to make vinyl and acrylic plastics, fibers, resins, and a variety of other products. Alkynes are similar to alkenes in both physical and chemical properties. For example, alkynes undergo many of the typical addition reactions of alkenes. Nomenclature The International Union of Pure and Applied Chemistry (IUPAC) names for alkynes parallel those of alkenes, except that the family ending is - yne rather than - ene . The IUPAC name for acetylene is ethyne. The names of other alkynes are illustrated in the following exercises. Rule 1. Find the longest carbon chain that includes both carbons of the triple bond. Rule 2. Number the longest chain starting at the end closest to the triple bond. A 1-alkyne is referred to as a terminal alkyne and alkynes at any other position are called internal alkynes. Rule 3. After numbering the longest chain with the lowest number assigned to the alkyne, label each of the substituents at its corresponding carbon. While writing out the name of the molecule, arrange the substituents in alphabetical order. If there are more than one of the same substituent use the prefixes di, tri, and tetra for two, three, and four substituents respectively. These prefixes are not taken into account in the alphabetical order. Examples: Concept Review Exercises Briefly identify the important differences between an alkene and an alkyne. How are they similar? The alkene (CH 3 ) 2 CHCH 2 CH=CH 2 is named 4-methyl-1-pentene. What is the name of (CH 3 ) 2 CHCH 2 C≡CH? Do alkynes show cis-trans isomerism? Explain. Answers Alkenes have double bonds; alkynes have triple bonds. Both undergo addition reactions. 4-methyl-1-pentyne No; a triply bonded carbon atom can form only one other bond. It would have to have two groups attached to show cis-trans isomerism. Key Takeaway Alkynes are hydrocarbons with carbon-to-carbon triple bonds and properties much like those of alkenes. Exercises Draw the structure for each compound. acetylene 3-methyl-1-hexyne Draw the structure for each compound. 4-methyl-2-hexyne 3-octyne Name each alkyne. CH 3 CH 2 CH 2 C≡CH CH 3 CH 2 CH 2 C≡CCH 3 Answers H–C≡C–H 1-pentyne 2-hexyne Attributions and citations Naming Alkynes. (2021, February 15). Retrieved May 20, 2021, from https://chem.libretexts.org/@go/page/31481 Alkynes. (2020, August 17). Retrieved May 20, 2021, from https://chem.libretexts.org/@go/page/16058 |
Courses/Honolulu_Community_College/CHEM_100%3A_Chemistry_and_Society/14%3A_Earth/14.01%3A_Spaceship_Earth-_Structure_and_Composition | Learning Objective Describe the chemical composition and structure of Earth. The layers found inside Earth are divided by composition into core, mantle, and crust or by mechanical properties into lithosphere and asthenosphere. Scientists use information from earthquakes and computer modeling to learn about Earth’s interior. THE EARTH’S LAYERS The layers scientists recognize are pictured in Figure \(\PageIndex{1}\). Core, mantle, and crust are divisions based on chemical composition: Crust: The Earth's surface is the crust. Generally speaking, the crust is predominately silicon oxide and aluminum oxide. Continental crust is thicker and less dense than oceanic crust. Earth’s crust varies in thickness from less than 5 km (under mid-ocean spreading ridges) to more than 70 km (beneath the highest mountain range). Mantle: The next layer down chemically is the mantle. The mantle has an ultramafic composition – it contains more iron, magnesium, less aluminum and somewhat less silicon than the crust. The mantle is roughly 2,900 km thick. In terms of volume, the mantle is the largest of earth’s three chemical layers. Core:The final layer is the core, which is mostly iron and nickel. The core is about 3,500 km thick. Table \(\PageIndex{1}\) summarizes the chemical layers of the earth. Crust Mantle Core composition: high Si, Al, & O composition: moderate Si, high Mg & Fe composition: Fe & Ni thickness: 5 to 70 km thickness: 2,900 km thickness: 3,500 km Table \(\PageIndex{2}\) provides the elemental composition of the Earth's crust. Lithosphere and asthenosphere are divisions based on mechanical properties: The lithosphere is composed of both the crust and the portion of the upper mantle that behaves as a brittle, rigid solid. The asthenosphere is partially molten upper mantle material that behaves plastically and can flow. Compositional and Mechanical Layers of the Earth https://www.khanacademy.org/science/cosmology-and-astronomy/earth-history-topic/plate-techtonics/v/compositional-and-mechanical-layers-of-the-earth Video \(\PageIndex{1}\) A comparison of the compositional and mechanical layers of the earth. Crust and Lithosphere Earth’s outer surface is its crust; a cold, thin, brittle outer shell made of rock. The crust is very thin, relative to the radius of the planet. There are two very different types of crust, each with its own distinctive physical and chemical properties, which are summarized in Table \(\PageIndex{3}\). Crust Thickness Density Composition Rock types Oceanic 5-12 km (3-8 mi) 3.0 g/cm3 Mafic Basalt and gabbro Continental Avg. 35 km (22 mi) 2.7 g/cm3 Felsic All types Oceanic crust is composed of mafic magma that erupts on the seafloor to create basalt lava flows or cools deeper down to create the intrusive igneous rock gabbro (Figure \(\PageIndex{2}\)). Sediments, primarily muds and the shells of tiny sea creatures, coat the sea floor. Sediment is thickest near the shore where it comes off the continents in rivers and on wind currents. Continental crust is made up of many different types of igneous, metamorphic, and sedimentary rocks. The average composition is granite, which is much less dense than the mafic rocks of the oceanic crust (Figure \(\PageIndex{3}\)). Because it is thick and has relatively low density, continental crust rises higher on the mantle than oceanic crust, which sinks into the mantle to form basins. When filled with water, these basins form the planet’s oceans. The lithosphere is the outermost mechanical layer, which behaves as a brittle, rigid solid. The lithosphere is about 100 kilometers thick. Look at Figure \(\PageIndex{1}\). Can you find where the crust and the lithosphere are located? How are they different from each other? The definition of the lithosphere is based on how earth materials behave, so it includes the crust and the uppermost mantle, which are both brittle. Since it is rigid and brittle, when stresses act on the lithosphere, it breaks. This is what we experience as an earthquake. Mantle The two most important things about the mantle are: (1) it is made of solid rock, and (2) it is hot. Scientists know that the mantle is made of rock based on evidence from seismic waves, heat flow, and meteorites. The properties fit the ultramafic rock peridotite, which is made of the iron- and magnesium-rich silicate minerals (Figure \(\PageIndex{4}\)). Peridotite is rarely found at Earth’s surface. Scientists know that the mantle is extremely hot because of the heat flowing outward from it and because of its physical properties. Core At the planet’s center lies a dense metallic core. Scientists know that the core is metal because: The density of Earth’s surface layers is much less than the overall density of the planet, as calculated from the planet’s rotation. If the surface layers are less dense than average, then the interior must be denser than average. Calculations indicate that the core is about 85% iron metal with nickel metal making up much of the remaining 15%. Metallic meteorites are thought to be representative of the core. The 85% iron/15% nickel calculation above is also seen in metallic meteorites (Figure \(\PageIndex{5}\)). Figure \(\PageIndex{5}\) An iron meteorite is the closest thing to the Earth’s core that we can hold in our hands. If Earth’s core were not metal, the planet would not have a magnetic field. Metals such as iron are magnetic, but rock, which makes up the mantle and crust, is not. Summary The core, mantle, and crust are divisions of Earth based on chemical composition. The lithosphere and asthenosphere are divisions of Earth based on mechanical properties. The three most abundant elements of Earth's crust are oxygen (46.6% by weight), silicon (27.7%), and aluminum (8.1%). Contributors Libretext: Fundamentals of Geology (Schulte) Marisa Alviar-Agnew ( Sacramento City College ) Wikipedia |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/3%3AStuff_to_Review_from_General_Chemistry/05%3A_Electronic_Structure_and_Periodic_Properties/5.03%3A_Periodic_Variations_in_Element_Properties | Learning Objectives Describe and explain the observed trends in atomic size, ionization energy, and electron affinity of the elements The elements in groups (vertical columns) of the periodic table exhibit similar chemical behavior. This similarity occurs because the members of a group have the same number and distribution of electrons in their valence shells. However, there are also other patterns in chemical properties on the periodic table. For example, as we move down a group, the metallic character of the atoms increases. Oxygen, at the top of Group 16 (6A), is a colorless gas; in the middle of the group, selenium is a semiconducting solid; and, toward the bottom, polonium is a silver-grey solid that conducts electricity. As we go across a period from left to right, we add a proton to the nucleus and an electron to the valence shell with each successive element. As we go down the elements in a group, the number of electrons in the valence shell remains constant, but the principal quantum number increases by one each time. An understanding of the electronic structure of the elements allows us to examine some of the properties that govern their chemical behavior. These properties vary periodically as the electronic structure of the elements changes. They are (1) size (radius) of atoms and ions, (2) ionization energies, and (3) electron affinities. Variation in Covalent Radius The quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, to determine their relative sizes that give roughly similar values. We will use the covalent radius (Figure \(\PageIndex{1}\)), which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity). We know that as we scan down a group, the principal quantum number, n , increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its covalent radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the covalent radii of the halogens in Table \(\PageIndex{1}\) and Figure \(\PageIndex{1}\). The trends for the entire periodic table can be seen in Figure \(\PageIndex{2}\). Atom Covalent radius (pm) Nuclear charge F 64 9 Cl 99 17 Br 114 35 I 133 53 At 148 85 As shown in Figure \(\PageIndex{2}\), as we move across a period from left to right, we generally find that each element has a smaller covalent radius than the element preceding it. This might seem counterintuitive because it implies that atoms with more electrons have a smaller atomic radius. This can be explained with the concept of effective nuclear charg e, \(Z_{eff}\). This is the pull exerted on a specific electron by the nucleus, taking into account any electron–electron repulsions. For hydrogen, there is only one electron and so the nuclear charge ( Z ) and the effective nuclear charge ( Z eff ) are equal. For all other atoms, the inner electrons partially shield the outer electrons from the pull of the nucleus, and thus: \[Z_\ce{eff}=Z−shielding \nonumber \] Shielding is determined by the probability of another electron being between the electron of interest and the nucleus, as well as by the electron–electron repulsions the electron of interest encounters. Core electrons are adept at shielding, while electrons in the same valence shell do not block the nuclear attraction experienced by each other as efficiently. Thus, each time we move from one element to the next across a period, Z increases by one, but the shielding increases only slightly. Thus, Z eff increases as we move from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by electrons on the right side of the periodic table draws them closer to the nucleus, making the covalent radii smaller. Thus, as we would expect, the outermost or valence electrons are easiest to remove because they have the highest energies, are shielded more, and are farthest from the nucleus. As a general rule, when the representative elements form cations, they do so by the loss of the ns or np electrons that were added last in the Aufbau process. The transition elements, on the other hand, lose the ns electrons before they begin to lose the ( n – 1) d electrons, even though the ns electrons are added first, according to the Aufbau principle. Example \(\PageIndex{1}\): Sorting Atomic Radii Predict the order of increasing covalent radius for Ge, Fl, Br, Kr. Solution Radius increases as we move down a group, so Ge < Fl (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so Kr < Br < Ge. Putting the trends together, we obtain Kr < Br < Ge < Fl. Exercise \(\PageIndex{1}\) Give an example of an atom whose size is smaller than fluorine. Answer Ne or He Variation in Ionic Radii Ionic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; it is smaller than the atom from which it is derived (Figure \(\PageIndex{3}\)). For example, the covalent radius of an aluminum atom (1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 ) is 118 pm, whereas the ionic radius of an Al 3 + (1 s 2 2 s 2 2 p 6 ) is 68 pm. As electrons are removed from the outer valence shell, the remaining core electrons occupying smaller shells experience a greater effective nuclear charge Z eff (as discussed) and are drawn even closer to the nucleus. Cations with larger charges are smaller than cations with smaller charges (e.g., V 2+ has an ionic radius of 79 pm, while that of V 3+ is 64 pm). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, n . An anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in \(Z_{eff}\) per electron. Both effects (the increased number of electrons and the decreased Z eff ) cause the radius of an anion to be larger than that of the parent atom ( Figure \(\PageIndex{3}\)). For example, a sulfur atom ([Ne]3 s 2 3 p 4 ) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3 s 2 3 p 6 ) is 170 pm. For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii. Atoms and ions that have the same electron configuration are said to be isoelectronic . Examples of isoelectronic species are N 3– , O 2– , F – , Ne, Na + , Mg 2 + , and Al 3 + (1 s 2 2 s 2 2 p 6 ). Another isoelectronic series is P 3– , S 2– , Cl – , Ar, K + , Ca 2 + , and Sc 3 + ([Ne]3 s 2 3 p 6 ). For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms. Variation in Ionization Energies The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy (IE 1 ). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge: \[\ce{X}(g)⟶\ce{X+}(g)+\ce{e-}\hspace{20px}\ce{IE_1} \nonumber \] The energy required to remove the second most loosely bound electron is called the second ionization energy (IE 2 ). \[\ce{X+}(g)⟶\ce{X^2+}(g)+\ce{e-}\hspace{20px}\ce{IE_2} \nonumber \] The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period. Figure \(\PageIndex{4}\) graphs the relationship between the first ionization energy and the atomic number of several elements. Within a period, the values of first ionization energy for the elements (IE 1 ) generally increases with increasing Z . Down a group, the IE 1 value generally decreases with increasing Z . There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2 s 2 ) is an s electron, whereas the electron removed during the ionization of boron ([He]2 s 2 2 p 1 ) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins. Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE 1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2 p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure \(\PageIndex{4}\). Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table \(\PageIndex{2}\), there is a large increase in the ionization energies (color change) for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization. Element IE1 IE2 IE3 IE4 IE5 IE6 IE7 K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343 Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9 Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0 Ga 578.8 1979.4 2964.6 6180.0 8298.7 10873.9 13594.8 Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available Example \(\PageIndex{2}\): Ranking Ionization Energies Predict the order of increasing energy for the following processes: IE 1 for Al, IE 1 for Tl, IE 2 for Na, IE 3 for Al. Solution Removing the 6 p 1 electron from Tl is easier than removing the 3 p 1 electron from Al because the higher n orbital is farther from the nucleus, so IE 1 (Tl) < IE 1 (Al). Ionizing the third electron from \[\ce{Al}\hspace{20px}\ce{(Al^2+⟶Al^3+ + e- )} \nonumber \] requires more energy because the cation Al 2 + exerts a stronger pull on the electron than the neutral Al atom, so IE 1 (Al) < IE 3 (Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain: IE 1 (Tl) < IE 1 (Al) < IE 3 (Al) < IE 2 (Na). Exercise \(\PageIndex{2}\) Which has the lowest value for IE 1 : O, Po, Pb, or Ba? Answer Ba Variation in Electron Affinities The electron affinity [EA] is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion). \[\ce{X}(g)+\ce{e-}⟶\ce{X-}(g)\hspace{20px}\ce{EA_1} \nonumber \] This process can be either endothermic or exothermic, depending on the element. The EA of some of the elements is given in Figure \(\PageIndex{6}\). You can see that many of these elements have negative values of EA, which means that energy is released when the gaseous atom accepts an electron. However, for some elements, energy is required for the atom to become negatively charged and the value of their EA is positive. Just as with ionization energy, subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a –2 ion, and so on. As we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 (5A), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher n level, which is more difficult to do. Group 2 (2A) has a filled ns subshell, and so the next electron added goes into the higher energy np , so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled np subshell and the next electron must be paired with an existing np electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA. We also might expect the atom at the top of each group to have the largest EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the greatest EA. The reduction of the EA of the first member can be attributed to the small size of the n = 2 shell and the resulting large electron–electron repulsions. For example, chlorine, with an EA value of –348 kJ/mol, has the highest value of any element in the periodic table. The EA of fluorine is –322 kJ/mol. When we add an electron to a fluorine atom to form a fluoride anion (F – ), we add an electron to the n = 2 shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the n = 3 shell, it occupies a considerably larger region of space and the electron–electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily. The properties discussed in this section (size of atoms and ions, effective nuclear charge, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus. Summary Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller, because the number of valence electrons has changed while the nuclear charge has remained constant. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Electron affinity (the energy associated with forming an anion) is more favorable (exothermic) when electrons are placed into lower energy orbitals, closer to the nucleus. Therefore, electron affinity becomes increasingly negative as we move left to right across the periodic table and decreases as we move down a group. For both IE and electron affinity data, there are exceptions to the trends when dealing with completely filled or half-filled subshells. Glossary covalent radius one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond effective nuclear charge charge that leads to the Coulomb force exerted by the nucleus on an electron, calculated as the nuclear charge minus shielding electron affinity energy required to add an electron to a gaseous atom to form an anion ionization energy energy required to remove an electron from a gaseous atom or ion. The associated number (e.g., second ionization energy) corresponds to the charge of the ion produced (X 2+ ) isoelectronic group of ions or atoms that have identical electron configurations |
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/05%3A__Alkanes_and_Conformations/5.02%3A_Generic_(Abbreviated)_Structures_(aka_R_Groups) | learning objective use R groups to draw generic functional groups - refer to section 3.1 Drawing Generic (abbreviated) Organic Structures In chapter 2, we learned to recognize and distinguish between organic functional groups. Often when drawing organic structures, chemists find it convenient to use the letter 'R' to designate part of a molecule outside of the region of interest. "R" represents the "Rest of the Molecule". If we just want to refer in general to a functional group without drawing a specific molecule, for example, we can use 'R groups' to focus attention on the group of interest: The 'R' group is a convenient way to abbreviate the structures of large biological molecules, especially when we are interested in something that is occurring specifically at one location on the molecule. For example, in chapter 15 when we look at biochemical oxidation-reduction reactions involving the flavin molecule, we will abbreviate a large part of the flavin structure which does not change at all in the reactions of interest: As an alternative, we can use a 'break' symbol to indicate that we are looking at a small piece or section of a larger molecule. This is used commonly in the context of drawing groups on large polymers such as proteins or DNA. Finally, 'R' groups can be used to concisely illustrate a series of related compounds, such as the family of penicillin-based antibiotics. Using abbreviations appropriately is a very important skill to develop when studying organic chemistry in a biological context, because although many biomolecules are very large and complex (and take forever to draw!), usually we are focusing on just one small part of the molecule where a change is taking place. As a rule, you should never abbreviate any atom involved in a bond-breaking or bond-forming event that is being illustrated: only abbreviate that part of the molecule which is not involved in the reaction of interest. For example, carbon #2 in the reactant/product below most definitely is involved in bonding changes, and therefore should not be included in the 'R' group. If you are unsure whether to draw out part of a structure or abbreviate it, the safest thing to do is to draw it out. Exercise 1. a) If you intend to draw out the chemical details of a reaction in which the methyl ester functional group of cocaine (see earlier figure) was converted to a carboxylate plus methanol, what would be an appropriate abbreviation to use for the cocaine structure (assuming that you only wanted to discuss the chemistry specifically occurring at the ester group)? b) Below is the (somewhat complicated) reaction catalyzed by an enzyme known as 'Rubisco', by which plants 'fix' carbon dioxide. Carbon dioxide and the oxygen of water are colored red and blue respectively to help you see where those atoms are incorporated into the products. Propose an appropriate abbreviation for the starting compound (ribulose 1,5-bisphosphate), using two different 'R' groups, R 1 and R 2 . Solutions to exercises |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(LibreTexts)/15%3A_Chemical_Equilibrium/15.08%3A_The_Effect_of_a_Concentration_Change_on_Equilibrium | Consider the following system under equilibrium: \[ \underbrace{\ce{Fe^{3+}(aq)}}_{\text{colorless}} + \underbrace{ \ce{SCN^{-}(aq)}}_{\text{colorless}} \rightleftharpoons \underbrace{\ce{FeSCN^{2+}(aq)}}_{\text{red}} \nonumber \] If more \(Fe^{3+}\) is added to the reaction, what will happen? According to Le Chatelier's Principle, the system will react to minimize the stress. Since Fe 3 + is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right , producing more FeSCN 2 + . For this particular reaction, we will be able to see that this has happened, as the solution will become a darker red color. There are a few different ways to state what happens here when more Fe 3 + is added, all of which have the same meaning: equilibrium shifts to the right equilibrium shifts to the product side the forward reaction is favored What changes does this cause in the concentrations of the reaction participants? \(\ce{Fe^{3+}}\) Since this is what was added to cause the stress, the concentration of \(\ce{Fe^{3+}}\) will increase. (A shorthand way to indicate this: \(\ce{[Fe]^{3+}\: \uparrow}\) (Reminder: the square brackets represent "concentration") \(\ce{SCN^{-}(aq)}\) Equilibrium will shift to the right, which will use up the reactants. The concentration of \(\ce{SCN^{-}(aq)}\) will decrease \(\ce{[SCN]^{-}\: \downarrow}\) as the rate of the forward reaction increases. \(\ce{FeSCN^{2+}}\) When the forward reaction rate increases, more products are produced, and the concentration of \(\ce{FeSCN^{2+}}\) will increase. \(\ce{[FeSCN]^{2+}} \uparrow \) How about the value of K eq ? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of K eq will be unchanged. The value of K eq does not change when changes in concentration cause a shift in equilibrium. What if more FeSCN 2 + is added? Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favor the reverse reaction, since the reverse reaction will use up the additional FeSCN 2 + . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}(aq)}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts (in a side reaction) with something already in the reaction. Let's remove SCN - from the system (perhaps by adding some Pb 2 + ions—the lead(II) ions will form a precipitate with SCN - , removing them from the solution). What will happen now? Equilibrium will shift to replace SCN - —the reverse reaction will be favored because that is the direction that produces more SCN - . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored (but also ↓ because it was removed) \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/13%3A_Structure_Determination_-_Nuclear_Magnetic_Resonance_Spectroscopy/13.11%3A_DEPT_C_NMR_Spectroscopy | Objectives After completing this section, you should be able to Understand why the peaks do not have splitting like in 1 H NMR. Using DEPT, distinguish whether a methyl (CH 3 ), methylene (CH 2 ), methine (CH) or quarternary C is present in the molecule and how many. Propose a structure based on 13 C spectral data. DEPT (Distortionless Enhancement by Polarization Transfer) ¹³C NMR Spectroscopy is a powerful technique used in organic chemistry to elucidate the structure of organic molecules. Unlike traditional ¹³C NMR spectroscopy, which provides only basic information about carbon environments, DEPT enhances signals from specific types of carbon atoms, allowing for more detailed analysis. DEPT works by selectively enhancing signals from three types of carbon atoms: methyl (CH₃), methylene (CH₂), and quaternary carbons (C with no attached hydrogens). This enhancement is achieved through a series of pulse sequences that manipulate the nuclear spins of carbon atoms, resulting in distinct peaks in the NMR spectrum corresponding to each carbon type. DEPT experiments are used for distinguishing between a CH 3 group (methyl), a CH 2 group (methylene), and a CH group (methine). The proton pulse is set at 45°, 90°, or 135° in the three separate experiments. The different pulses depend on the number of protons attached to a carbon atom. Figure \(\PageIndex{1}\) is an example about DEPT spectrum of n-isobutlybutrate. While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH 2 ) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH 3 ) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13 C spectral data in examples and problems. Below are two more examples of 13 C NMR spectra of simple organic molecules, along with the type of substitution for that carbon which was obtained from a DEPT experiment. Exercise \(\PageIndex{1}\) 13 C-NMR (and DEPT) data for some common biomolecules are shown below (data are from the Aldrich Library of 1 H and 13 C NMR). Match the NMR data to the correct structure, and make complete peak assignments. spectrum a: 168.10 ppm (C), 159.91 ppm (C), 144.05 ppm (CH), 95.79 ppm (CH) spectrum b: 207.85 ppm (C), 172.69 ppm (C), 29.29 ppm (CH 3 ) spectrum c: 178.54 ppm (C), 53.25 ppm (CH), 18.95 ppm (CH 3 ) spectrum d: 183.81 ppm (C), 182. 63 ppm (C), 73.06 ppm (CH), 45.35 ppm (CH 2 ) Answer |
Courses/Fullerton_College/Introductory_Biochemistry/01%3A_Organic_Chemistry_-_Alkanes_and_Halogenated_Hydrocarbons/1.07%3A_Halogenated_Hydrocarbons | Learning Objectives To name halogenated hydrocarbons given formulas and write formulas for these compounds given names. Many organic compounds are closely related to the alkanes. As we noted previously, alkanes react with halogens to produce halogenated hydrocarbons, the simplest of which have a single halogen atom substituted for a hydrogen atom of the alkane. Even more closely related are the cycloalkanes, compounds in which the carbon atoms are joined in a ring, or cyclic fashion. The reactions of alkanes with halogens produce halogenated hydrocarbons, compounds in which one or more hydrogen atoms of a hydrocarbon have been replaced by halogen atoms: The replacement of only one hydrogen atom gives an alkyl halide (or haloalkane). The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending -ide . The IUPAC system uses the name of the parent alkane with a prefix indicating the halogen substituents, preceded by number indicating the substituent’s location. The prefixes are fluoro -, chloro -, bromo -, and iodo -. Thus CH 3 CH 2 Cl has the common name ethyl chloride and the IUPAC name chloroethane. Alkyl halides with simple alkyl groups (one to four carbon atoms) are often called by common names. Those with a larger number of carbon atoms are usually given IUPAC names. Example \(\PageIndex{1}\) Give the IUPAC names for each compound. CH 3 CH 2 CH 2 Br Solution The molecule is a propane with a halogen bromine (Br). For the IUPAC name, the prefix for bromine (bromo) is combined with the name for a three-carbon chain (propane), preceded by a number identifying the carbon atom to which the Br atom is attached, so the IUPAC name is 1-bromopropane. The molecule is a propane with a halogen chlorine (Cl) atom attached to the middle carbon atom. For the IUPAC name, the Cl atom (prefix chloro -) attached to the middle (second) carbon atom of a propane chain results in 2-chloropropane. Exercise \(\PageIndex{1}\) Give the IUPAC names for each compound. CH 3 CH 2 I CH 3 CH 2 CH 2 CH 2 F Example \(\PageIndex{2}\) Give the IUPAC name for each compound. Solution The parent alkane has five carbon atoms in the longest continuous chain; it is pentane. A bromo (Br) group is attached to the second carbon atom of the chain. The IUPAC name is 2-bromopentane. The parent alkane is hexane. Methyl (CH 3 ) and bromo (Br) groups are attached to the second and fourth carbon atoms, respectively. Listing the substituents in alphabetical order gives the name 4-bromo-2-methylhexane. Exercise \(\PageIndex{2}\) Give the IUPAC name for each compound. A wide variety of interesting and often useful compounds have one or more halogen atoms per molecule. For example, methane (CH 4 ) can react with chlorine (Cl 2 ), replacing one, two, three, or all four hydrogen atoms with Cl atoms. Several halogenated products derived from methane and ethane (CH 3 CH 3 ) are listed in Table \(\PageIndex{1}\), along with some of their uses. Formula Common Name IUPAC Name Some Important Uses Derived from CH4 Derived from CH4 Derived from CH4 Derived from CH4 CH3Cl methyl chloride chloromethane refrigerant; the manufacture of silicones, methyl cellulose, and synthetic rubber CH2Cl2 methylene chloride dichloromethane laboratory and industrial solvent CHCl3 chloroform trichloromethane industrial solvent CCl4 carbon tetrachloride tetrachloromethane dry-cleaning solvent and fire extinguishers (but no longer recommended for use) CBrF3 halon-1301 bromotrifluoromethane fire extinguisher systems CCl3F chlorofluorocarbon-11 (CFC-11) trichlorofluoromethane foaming plastics CCl2F2 chlorofluorocarbon-12 (CFC-12) dichlorodifluoromethane refrigerant Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 Derived from CH3CH3 CH3CH2Cl ethyl chloride chloroethane local anesthetic ClCH2CH2Cl ethylene dichloride 1,2-dichloroethane solvent for rubber CCl3CH3 methylchloroform 1,1,1-trichloroethane solvent for cleaning computer chips and molds for shaping plastics To Your Health: Halogenated Hydrocarbons Once widely used in consumer products, many chlorinated hydrocarbons are suspected carcinogens (cancer-causing substances) and also are known to cause severe liver damage. An example is carbon tetrachloride (CCl 4 ), once used as a dry-cleaning solvent and in fire extinguishers but no longer recommended for either use. Even in small amounts, its vapor can cause serious illness if exposure is prolonged. Moreover, it reacts with water at high temperatures to form deadly phosgene (COCl 2 ) gas, which makes the use of CCl 4 in fire extinguishers particularly dangerous. Ethyl chloride, in contrast, is used as an external local anesthetic. When sprayed on the skin, it evaporates quickly, cooling the area enough to make it insensitive to pain. It can also be used as an emergency general anesthetic. Bromine-containing compounds are widely used in fire extinguishers and as fire retardants on clothing and other materials. Because they too are toxic and have adverse effects on the environment, scientists are engaged in designing safer substitutes for them, as for many other halogenated compounds. To Your Health: Chlorofluorocarbons and the Ozone Layer Alkanes substituted with both fluorine (F) and chlorine (Cl) atoms have been used as the dispersing gases in aerosol cans, as foaming agents for plastics, and as refrigerants. Two of the best known of these chlorofluorocarbons (CFCs) are listed in Table \(\PageIndex{1}\). Chlorofluorocarbons contribute to the greenhouse effect in the lower atmosphere. They also diffuse into the stratosphere, where they are broken down by ultraviolet (UV) radiation to release Cl atoms. These in turn break down the ozone (O 3 ) molecules that protect Earth from harmful UV radiation. Worldwide action has reduced the use of CFCs and related compounds. The CFCs and other Cl- or bromine (Br)-containing ozone-destroying compounds are being replaced with more benign substances. Hydrofluorocarbons (HFCs), such as CH 2 FCF 3 , which have no Cl or Br to form radicals, are one alternative. Another is hydrochlorofluorocarbons (HCFCs), such as CHCl 2 CF 3 . HCFC molecules break down more readily in the troposphere, and fewer ozone-destroying molecules reach the stratosphere. Concept Review Exercises What is the IUPAC name for the HFC that has the formula ?(Hint: you must use a number to indicate the location of each substituent F atom.) What is the IUPAC name for the HCFC that has the formula ? Answers 1,1,1,2-tetrafluoroethane 2,2-dichloro-1,1,1- trifluoroethane Key Takeaway The replacement of an hydrogen atom on an alkane by a halogen atom—F, Cl, Br, or I—forms a halogenated compound. |
Courses/Howard_University/General_Chemistry%3A_An_Atoms_First_Approach/Unit_4%3A__Thermochemistry/09%3A_Thermochemistry/Chapter_9.04%3A__Heats_of_Formation | 0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Learning Objectives To understand Enthalpies of Formation and be able to use them to calculated Enthalpies of Reaction Enthalpies of Formation Chapter 7 and Chapter 8 presented a wide variety of chemical reactions, and you learned how to write balanced chemical equations that include all the reactants and the products except heat. One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data starting from the elemental forms of each atom at 25 o C and 1 atm pressure. Enthalpy of formation (Δ H f ) The enthalpy change for the formation of 1 mol of a compound from its component elements. : The enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The corresponding relationship is \( elements \rightarrow compound \;\;\;\;\ \Delta H_{rxn} = \Delta H_{f} \tag{9.4.1} \) For example, \( C\left (s \right ) + O_{2}\left (g \right ) \rightarrow CO_{2}\left (g \right ) \; \; \; \; \Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \) The sign convention for Δ H f is the same as for any enthalpy change: Δ H f < 0 if heat is released when elements combine to form a compound and Δ H f > 0 if heat is absorbed. Note the Pattern The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. Standard Enthalpies of Formation The magnitude of Δ H for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane. The standard conditions The conditions under which most thermochemical data are tabulated: 1 atm for all gases and a concentration of 1.0 M for all species in solution. for which most thermochemical data are tabulated are a pressure of 1 atmosphere (atm) for all gases and a concentration of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state The most stable form of a pure substance at a pressure of 1 atm at a specified temperature. . This is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called standard enthalpies of formation ( ΔH o f ) The enthalpy change for the formation of 1 mol of a compound from its component elements when the component elements are each in their standard states. The standard enthalpy of formation of any element in its most stable form is zero by definition. (which is pronounced “delta H eff naught”). The standard enthalpy of formation of any element in its standard state is zero by definition . For example, although oxygen can exist as ozone (O 3 ), atomic oxygen (O), and molecular oxygen (O 2 ), O 2 is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H 2 (g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite ( Figure 9.4.1 ). Therefore, O 2 (g), H 2 (g), and graphite have ΔH o f values of zero. Figure 9.4.1 Elemental Carbon Although graphite and diamond are both forms of elemental carbon, graphite is more stable at 1 atm pressure and 25°C than diamond is. Given enough time, diamond will revert to graphite under these conditions. Hence graphite is the standard state of carbon. The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction: \( 6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \tag{9.4.2} \) It is not possible to measure the value of ΔH o f for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, O 2 , and H 2 and measuring the heat evolved as glucose is formed; the reaction shown in Equation 9.4.2 does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of are obtained using Hess’s law and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of Δ H o f for an extensive list of compounds are given in the Reference Tables . Note that Δ H o f values are always reported in kilojoules per mole of the substance of interest. Also notice in the Reference Tables that the standard enthalpy of formation of O 2 (g) is zero because it is the most stable form of oxygen in its standard state. Example 4 For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. HCl(g) MgCO 3 (s) CH 3 (CH 2 ) 14 CO 2 H(s) (palmitic acid) Given: compound Asked for: balanced chemical equation for its formation from elements in standard states Strategy: Use Appendix A to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made. Solution: To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in the Reference Tables by a ΔH o f value of 0 kJ/mol. Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are H 2 (g) and Cl 2 (g), respectively, the unbalanced chemical equation is H 2 (g) + Cl 2 (g) → HCl(g) Fractional coefficients are required in this case because ΔH o f values are reported for 1 mol of the product, HCl. Multiplying both H 2 (g) and Cl 2 (g) by 1/2 balances the equation: \( \dfrac{1}{2}H_{2}\left ( g \right )+ \dfrac{1}{2}Cl_{2}\left ( g \right ) \)\rightarrow HCl\left ( g \right ) \) The standard states of the elements in this compound are Mg(s), C(s, graphite), and O 2 (g). The unbalanced chemical equation is thus Mg(s) + C(s, graphite) + O 2 (g) → MgCO 3 (s) This equation can be balanced by inspection to give \( Mg\left ( g \right )+ C\left ( s, graphite \right ) \dfrac{3}{2}O_{2}\left ( g \right )\rightarrow MgCO_{3}\left ( s \right ) \) Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows: C(s, graphite) + H 2 (g) + O 2 (g) → CH 3 (CH 2 ) 14 CO 2 H(s) There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is 16C(s, graphite) + 16H 2 (g) + O 2 (g) → CH 3 (CH 2 ) 14 CO 2 H(s) Exercise For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. NaCl(s) H 2 SO 4 (l) CH 3 CO 2 H(l) (acetic acid) Answer: \( Na\left ( g \right )+ \dfrac{1}{2}Cl_{2}\left ( g \right )\rightarrow NaCl\left ( s \right ) \) \( H_{2}\left ( g \right ) + \dfrac{1}{8}S_{8}\left ( s \right ) + 2O_{2}\left ( g \right ) \rightarrow H_{2}SO_{4}\left ( l \right ) \) 2C(s) + O 2 (g) + 2H 2 (g) → CH 3 CO 2 H(l) Standard Enthalpies of Reaction Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for any reaction involving substances whose ΔH f o values are known. The standard enthalpy of reaction ( ΔH rxn ) The enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard state. is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction \( aA + bB \rightarrow cC + dD \tag{9.4.3}\) where A, B, C, and D are chemical substances and a , b , c , and d are their stoichiometric coefficients. The magnitude of ΔH ο is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients: \( \begin{matrix} \Delta H_{rxn}^{o} = \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] - & \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ] \\ reactants & products \end{matrix} \tag{9.4.4} \) More generally, we can write \( \Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \tag{9.4.5} \) where the symbol Σ means “sum of” and m and n are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation 9.4.5 arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters. Note the Pattern Products minus reactants” summations are typical of state functions. To demonstrate the use of tabulated ΔH ο values, we will use them to calculate ΔH rxn for the combustion of glucose, the reaction that provides energy for your brain: \( C_{6}H_{12}O_{6} \left ( s \right ) + O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right ) + 6H_{2}O\left ( l \right ) \tag{9.4.6} \) Using Equation 9.4.5 , we write \( \Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ 6\Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \tag{9.4.7} \) From the Reference Tables , the relevant ΔH ο f values are ΔH ο f [CO 2 (g)] = - 393.5 kJ/mol , ΔH ο f [H 2 O(l)] = - 285.8 kJ/mol , and ΔH ο f [C 6 H 12 O 6 (s) ] = - 1273.3 kJ/mol . Because O 2 (g) is a pure element in its standard state, ΔH ο f [O 2 (g) ] = 0 kJ/mol. Inserting these values into Equation 9.4.7 and changing the subscript to indicate that this is a combustion reaction, we obtain \( \begin{matrix} \Delta H_{comb}^{o} = \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] \\ - \left [-1273.3 + 6\left ( 0 \; kJ\mol \right ) \right ] = -2802.5 \; kJ/mol \end{matrix} \tag{9.4.8} \) As illustrated in Figure 9.4.2 , we can use Equation 9.4.8 to calculate ΔH ο f for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled ΔH ο comb The alternative hypothetical pathway consists of four separate reactions that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the ΔH ο f values of the reactants. Consequently, the enthalpy changes are \( \begin{matrix} \Delta H_{1}^{o} = \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] = -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) = +1273.3 \; kJ \\ \Delta H_{2}^{o} = 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) = 0 \; kJ \end{matrix} \tag{9.4.9} \) (Recall that when we reverse a reaction, we must also reverse the sign of the accompanying enthalpy change.) The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O 2 ) to the elements is therefore +1273.3 kJ. Figure 9.4.2 A Thermochemical Cycle for the Combustion of Glucose Two hypothetical pathways are shown from the reactants to the products. The green arrow labeled ΔH ο comb indicates the combustion reaction. Alternatively, we could first convert the reactants to the elements via the reverse of the equations that define their standard enthalpies of formation (the upward arrow, labeled ΔH ο 1 and ΔH ο 2 ). Then we could convert the elements to the products via the equations used to define their standard enthalpies of formation (the downward arrows, labeled ΔH ο 3 and ΔH ο 4 ). Because enthalpy is a state function, ΔH ο comb is equal to the sum of the enthalpy changes ΔH ο 1 + ΔH ο 2 + ΔH ο 3 + ΔH ο 4 . The reactions that convert the elements to final products (downward purple arrows in Figure 9.4.2 ) are identical to those used to define the ΔH ο f values of the products. Consequently, the enthalpy changes (from the Reference Tables ) are \( \begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \\ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix} \) The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using any pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ): \( \Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \tag{9.4.10} \) This is the same result we obtained using the “products minus reactants” rule and ΔH ο f values. The two results must be the same because Equation 9.4.10 is just a more compact way of describing the thermochemical cycle shown in Figure 9.4.2 . Example 5 Long-chain fatty acids such as palmitic acid [CH 3 (CH 2 ) 14 CO 2 H] are one of the two major sources of energy in our diet ( ΔH ο f =−891.5 kJ/mol ). Use the data in the Reference Table to calculate ΔH ο comb for the combustion of palmitic acid. Based on the energy released in combustion per gram , which is the better fuel — glucose or palmitic acid? Given: compound and ΔH ο f values Asked for: ΔH ο comb per mole and per gram Strategy: A After writing the balanced chemical equation for the reaction, use Equation 9.4.5 and the values from the Reference Table to calculate ΔH ο comb the energy released by the combustion of 1 mol of palmitic acid. B Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of palmitic acid. Compare this value with the value calculated in Equation 9.4.8 for the combustion of glucose to determine which is the better fuel. Solution: A To determine the energy released by the combustion of palmitic acid, we need to calculate its ΔH ο f As always, the first requirement is a balanced chemical equation: C 16 H 32 O 2 (s) + 23O 2 (g) → 16CO 2 (g) + 16H 2 O(l) Using Equation 9.4.5 (“products minus reactants”) with ΔH ο f values from the Reference Table (and omitting the physical states of the reactants and products to save space) gives \( \Delta H_{comb}^{o} = \sum m \Delta {H^o}_f\left( {products} \right) - \sum n \Delta {H^o}_f\left( {reactants} \right) \) \( = \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \) \( - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \) \( = -9977.3 \; kJ/mol \) This is the energy released by the combustion of 1 mol of palmitic acid. B The energy released by the combustion of 1 g of palmitic acid is \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \) As calculated in Equation 9.4.8 , ΔH ο f of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \) The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight. Exercise Use the data in Appendix A to calculate ΔH ο rxn for the water–gas shift reaction , which is used industrially on an enormous scale to obtain H 2 (g): \( \begin{pmatrix} CO\left ( g \right )+H_{2}O\left ( g \right )\rightarrow CO_{2} \left ( g \right )+H_{2}\left ( g \right ) \\ water-gas \; shift \; reaction \end{pmatrix} \) Answer: −41.2 kJ/mol We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s ΔH ο f which we cannot obtain otherwise. This procedure is illustrated in Example 3. Example 6 Beginning in 1923, tetraethyllead [(C 2 H 5 ) 4 Pb] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are CO 2 (g), H 2 O(l), and red PbO(s). What is the standard enthalpy of formation of tetraethyllead, given that ΔH ο f is −19.29 kJ/g for the combustion of tetraethyllead and ΔH ο f of red PbO(s) is −219.0 kJ/mol? Given: reactant, products, and ΔH ο comb values Asked for: ΔH ο f of the reactants Strategy: A Write the balanced chemical equation for the combustion of tetraethyl lead. Then insert the appropriate quantities into Equation 9.4.4 to get the equation for ΔH ο f of tetraethyl lead. B Convert ΔH ο comb per gram given in the problem to ΔH ο comb per mole by multiplying ΔH ο comb per gram by the molar mass of tetraethyllead. C Use the Reference Table to obtain values of ΔH ο f for the other reactants and products. Insert these values into the equation for ΔH ο f of tetraethyl lead and solve the equation. Solution: A The balanced chemical equation for the combustion reaction is as follows: 2(C 2 H 5 ) 4 Pb(l) + 27O 2 (g) → 2PbO(s) + 16CO 2 (g) + 20H 2 O(l) Using Equation 9.4.5 gives \( \Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \) Solving for ΔH ο f [(C 2 H 5 )4Pb] gives \( \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \) The values of all terms other than ΔH ο f [(C 2 H 5 )4Pb] are given in the Reference Table B The magnitude of ΔH ο comb is given in the problem in kilojoules per gram of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get ΔH ο comb for 1 mol of tetraethyl lead: \( \Delta H_{comb}^{o} = \left ( \dfrac{-1929 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) = -6329 \; kJ/mol \) Because the balanced chemical equation contains 2 mol of tetraethyllead, ΔH ο rxn is \( \Delta H_{rxn}^{o} = 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6929 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) = -12,480 \; kJ \) C Inserting the appropriate values into the equation for ΔH ο f [(C 2 H 5 )4Pb] gives \( \begin{matrix} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = & \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] \\ & & +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \\ & & \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\\ & = & -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{matrix} \) Exercise Ammonium sulfate [(NH 4 ) 2 SO 4 ] is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid: 2NH 3 (g) + H 2 SO 4 (aq) → (NH 4 ) 2 SO 4 (s) The value of ΔH o rxn is −2805 kJ/g H 2 SO 4 . Use the data in Appendix A to calculate the standard enthalpy of formation of ammonium sulfate (in kilojoules per mole). Answer: −1181 kJ/mol Key Equations relationship between Δ H o rxn and ΔH ο f Equation 9.4.5 : ΔH o rxn = ΣΔH o f (products) − Δ H o f (reactants) Summary The enthalpy of formation (Δ H f ) is the enthalpy change that accompanies the formation of a compound from its elements. Standard enthalpies of formation ( Δ H o f ) are determined under standard conditions : a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their standard states (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The standard enthalpy of reaction ( Δ H o rxn ) can be calculated from the sum of the standard enthalpies of formation of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The enthalpy of solution (Δ H soln ) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. Key Takeaways The standard state for measuring and reporting enthalpies of formation or reaction is 25 o C and 1 atm. The elemental form of each atom is that with the lowest enthalpy in the standard state. The standard state heat of formation for the elemental form of each atom is zero. Conceptual Problems Please be sure you are familiar with the topics discussed in Essential Skills 4 ( Section 9.9 ) before proceeding to the Conceptual Problems. Describe how Hess’s law can be used to calculate the enthalpy change of a reaction that cannot be observed directly. When you apply Hess’s law, what enthalpy values do you need to account for each change in physical state? What is the difference between ΔH o f and ΔH f ? How can ΔH o f of a compound be determined if the compound cannot be prepared by the reactions used to define its standard enthalpy of formation? For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. a. HBr b. CH 3 OH c. NaHCO 3 Describe the distinction between Δ H soln and Δ H f . The following table lists ΔH o soln values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest. Compound ΔHosoln(kJ/mol) KOH −57.61 LiNO3 −2.51 KMnO4 43.56 NaC2H3O2 −17.32 Numerical Problems Please be sure you are familiar with the topics discussed in Essential Skills 4 ( Section 9.9 ) before proceeding to the Numerical Problems. Using "Appendix A , calculate ΔH o rxn for each chemical reaction. a. 2Mg(s) + O 2 (g) → 2MgO(s) b. CaCO 3 (s, calcite) → CaO(s) + CO 2 (g) c. AgNO 3 (s) + NaCl(s) → AgCl(s) + NaNO 3 (s) Using "Appendix A , determine ΔH o rxn for each chemical reaction. a. 2Na(s) + Pb(NO 3 ) 2 (s) → 2NaNO 3 (s) + Pb(s) b. Na 2 CO 3 (s) + H 2 SO 4 (l) → Na 2 SO 4 (s) + CO 2 (g) + H 2 O(l) c. 2KClO 3 (s) → 2KCl(s) + 3O 2 (g) Calculate ΔH o rxn for each chemical equation. If necessary, balance the chemical equations. a. Fe(s) + CuCl 2 (s) → FeCl 2 (s) + Cu(s) b. (NH 4 ) 2 SO 4 (s) + Ca(OH) 2 (s) → CaSO 4 (s) + NH 3 (g) + H 2 O(l) c. Pb(s) + PbO 2 (s) + H 2 SO 4 (l) → PbSO 4 (s) + H 2 O(l) Calculate ΔH o rxn for each reaction. If necessary, balance the chemical equations. a. 4HBr(g) + O 2 (g) → 2H 2 O(l) + 2Br 2 (l) b. 2KBr(s) + H 2 SO 4 (l) → K 2 SO 4 (s) + 2HBr(g) c. 4Zn(s) + 9HNO 3 (l) → 4Zn(NO 3 ) 2 (s) + NH 3 (g) + 3H 2 O(l) Use the data in "Appendix A to calculate ΔH o f for the reaction Sn(s, white) + 4HNO 3 (l) → SnO 2 (s) + 4NO 2 (g) + 2H 2 O(l). Use the data in "Appendix A to calculate ΔH o f for the reaction P 4 O 10 (s) + 6H 2 O(l) → 4H 3 PO 4 (l). How much heat is released or required in the reaction of 0.50 mol of HBr(g) with 1.0 mol of chlorine gas to produce bromine gas? How much energy is released or consumed if 10.0 g of N 2 O 5 is completely decomposed to produce gaseous nitrogen dioxide and oxygen? In the mid-1700s, a method was devised for preparing chlorine gas from the following reaction: NaCl(s) + H 2 SO 4 (l) + MnO 2 (s) → Na 2 SO 4 (s) + MnCl 2 (s) + H 2 O(l) + Cl 2 (g) Calculate ΔH o rxn for this reaction. Is the reaction exothermic or endothermic? Would you expect heat to be evolved during each reaction? solid sodium oxide with gaseous sulfur dioxide to give solid sodium sulfite solid aluminum chloride reacting with water to give solid aluminum oxide and hydrogen chloride gas How much heat is released in preparing an aqueous solution containing 6.3 g of calcium chloride, an aqueous solution containing 2.9 g of potassium carbonate, and then when the two solutions are mixed together to produce potassium chloride and calcium carbonate? Answers a. −1203 kJ/mol O 2 b. 179.2 kJ c. −59.3 kJ −174.1 kJ/mol −20.3 kJ −34.3 kJ/mol Cl 2 ; exothermic Δ H = −2.86 kJ CaCl 2 : −4.6 kJ; K 2 CO 3 , −0.65 kJ; mixing, −0.28 kJ Contributors Anonymous Modified by Joshua Halpern |
Bookshelves/Introductory_Chemistry/Map%3A__Introductory_Chemistry_(Corwin)/09%3A_The_Mole_Concept/9.02%3A_Mole_Calculations_I-_atom_conversions | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Courses/American_River_College/CHEM_309%3A_Applied_Chemistry_for_the_Health_Sciences/11%3A_Lipids_-_An_Introduction/11.10%3A_Fatty_Acid_Catabolism | Learning Objectives To describe the reactions needed to completely oxidize a fatty acid to carbon dioxide and water. Like glucose, the fatty acids released in the digestion of triglycerides and other lipids are broken down in a series of sequential reactions accompanied by the gradual release of usable energy. Some of these reactions are oxidative and require nicotinamide adenine dinucleotide (NAD + ) and flavin adenine dinucleotide (FAD). The enzymes that participate in fatty acid catabolism are located in the mitochondria, along with the enzymes of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. This localization of enzymes in the mitochondria is of the utmost importance because it facilitates efficient utilization of energy stored in fatty acids and other molecules. Fatty acid catabolism is initiated in the cytoplasm. There the fatty acid is activated by conversion to an energy-rich fatty acid derivative of coenzyme A called fatty acyl-coenzyme A (CoA). The activation is catalyzed by acyl-CoA synthetase . For each molecule of fatty acid activated, one molecule of coenzyme A and one molecule of adenosine triphosphate (ATP) are used, equaling a net utilization of the two high-energy bonds in one ATP molecule (which is therefore converted to adenosine monophosphate [AMP] rather than adenosine diphosphate [ADP]). For this reason the activation of one fatty acid is considered to require the energy equivalent of 2 ATP's. Figure \(\PageIndex{1}\): Activation of the fatty acid by CoA for catabolism. The fatty acyl-CoA diffuses to the inner mitochondrial membrane for the catabolism. Catabolism of Fatty Acids Catabolism of the fatty acyl-CoA occurs in the mitochondrial matrix via a sequence of four reactions known collectively as β-oxidation because the β-carbon undergoes successive oxidations in the progressive removal of two carbon atoms (Figure \(\PageIndex{2}\)). The fatty acyl-CoA formed in the final step becomes the substrate for the first step in the next round of the spiral. STEP 1: First oxidation of an alkane (dehydrogenation) A fatty acyl-CoA is oxidized to yield a trans double bond between the α- and β- carbons (the second and third carbons). FAD is oxidizing agent and is converted to FADH 2 , which moves into the electron transport chain. STEP 2: Hydration (addition of water across the double bond) The trans alkene is then hydrated to form a secondary alcohol. The hydroxyl group is placed on the β-carbon. STEP 3: Second oxidation (oxidation of an alcohol) The secondary alcohol is then oxidized to a ketone by NAD + acting as the oxidizing agent. STEP 4: Cleavage Acetyl-CoA cleaves off to yield a fatty acid that is two carbons shorter than before. The chain shortened fatty acid will start back at step 1 and go through another spiral. Note In each spiral, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH 2 are produced. The final spiral yields two acetyl-CoA molecules. Because each shortened fatty acyl-CoA cycles back to the beginning of the pathway, β-oxidation is referred to as the fatty acid spiral. The fate of the acetyl-CoA obtained from fatty acid oxidation depends on the needs of an organism. It may enter the citric acid cycle and be oxidized to produce energy, it may be used for the formation of water-soluble derivatives known as ketone bodies, or it may serve as the starting material for the synthesis of fatty acids. Looking Closer: Ketone Bodies In the liver, most of the acetyl-CoA obtained from fatty acid oxidation is oxidized by the citric acid cycle. However, some of the acetyl-CoA is used to synthesize a group of compounds known as ketone bodies : acetoacetate, β-hydroxybutyrate, and acetone shown in figure 11.10.3. Figure \(\PageIndex{3}\): Ketone bodies are produced from acetyl-CoA not consumed by the citric acid cycle. The acetoacetate and β-hydroxybutyrate synthesized by the liver are released into the blood for use as a metabolic fuel (to be converted back to acetyl-CoA) by other tissues, particularly the kidney and the heart. Thus, during prolonged starvation, ketone bodies provide about 70% of the energy requirements of the brain. Under normal conditions, the kidneys excrete about 20 mg of ketone bodies each day, and the blood levels are maintained at about 1 mg of ketone bodies per 100 mL of blood. In starvation, diabetes mellitus, and certain other physiological conditions in which cells do not receive sufficient amounts of carbohydrate, the rate of fatty acid oxidation increases to provide energy. This leads to an increase in the concentration of acetyl-CoA. The increased acetyl-CoA cannot be oxidized by the citric acid cycle because of a decrease in the concentration of oxaloacetate, which is diverted to glucose synthesis. In response, the rate of ketone body formation in the liver increases further, to a level much higher than can be used by other tissues. The excess ketone bodies accumulate in the blood and the urine, a condition referred to as ketosis . When the acetone in the blood reaches the lungs, its volatility causes it to be expelled in the breath. The sweet smell of acetone, a characteristic of ketosis, is frequently noticed on the breath of severely diabetic patients. Because two of the three kinds of ketone bodies are weak acids, their presence in the blood in excessive amounts overwhelms the blood buffers and causes a marked decrease in blood pH (to 6.9 from a normal value of 7.4). This decrease in pH leads to a serious condition known as acidosis . One of the effects of acidosis is a decrease in the ability of hemoglobin to transport oxygen in the blood. In moderate to severe acidosis, breathing becomes labored and very painful. The body also loses fluids and becomes dehydrated as the kidneys attempt to get rid of the acids by eliminating large quantities of water. The lowered oxygen supply and dehydration lead to depression; even mild acidosis leads to lethargy, loss of appetite, and a generally run-down feeling. Untreated patients may go into a coma. At that point, prompt treatment is necessary if the person’s life is to be saved. Note The number of times a spiral is repeated for a fatty acid containing n carbon atoms is (n/2 – 1) because the final spiral yields two acetyl-CoA molecules. ATP Yield from Fatty Acid Oxidation The amount of ATP obtained from fatty acid oxidation depends on the size of the fatty acid being oxidized. As an example, we’ll study octanoic acid, a saturated long chain carboxylic acid acid with 8 carbon atoms. Calculating its ATP yield provides a model for determining the ATP yield of all other fatty acids. Figure \(\PageIndex{4}\): Catabolism of octanoic acid. Activation and Spirals The catabolism of 1 molecule of octanoic acid requires 2 molecules of ATP (for activation) and forms 4 molecules of acetyl-CoA, 3 molecules of NADH, and 3 molecules of FADH 2 . This complete conversion requires 3 spirals to be repeated. Citric Acid Cycle The 4 molecules of acetyl-CoA are then processed by the citric acid cycle. Recall from earlier that each molecule of acetyl-CoA metabolized by the citric acid cycle yields 3 molecules of NADH, 1 molecule of FADH 2 , and 1 molecule of GTP. The 4 molecules of acetyl-CoA require 4 citric acid cycles and produce 12 molecules of NADH, 4 molecules of FADH 2 , and 4 molecules of GTP. Electron Transport Chain The reduced coenzymes NADH and FADH 2 from the spirals and the citric acid cycle then enter the electron transport chain and are converted to ATP. Each NADH molecule is responsible for the formation of 2.5 ATP and each FADH 2 forms 1.5 ATP. There are 12 + 3 = 15 NADH and 4 + 3 = 7 FADH 2 molecules from the 3 spirals and 4 Citric Acid Cycles that enter the ETC to produce 48 ATP (15 x 2.5 + 7 x 1.5). Total ATP count The 4 molecules of GTP from the citric acid cycle are equivalent in energy to ATP and so are added to 48 ATP from the ETC. Two ATP molecules are required for the activation step and hence subtracted. The total ATP yield from octanoic acid is 48+4-2= 50 ATP. The oxidation of fatty acids produces large quantities of water. This water, sustains migratory birds and animals (such as the camel) for long periods of time. Summary Fatty acids, obtained from the breakdown of triglycerides and other lipids, are oxidized through a series of four reactions known in a spiral. In each round of the spiral, 1 molecule of acetyl-CoA, 1 molecule of NADH, and 1 molecule of FADH 2 are produced. The acetyl-CoA, NADH, and FADH 2 are used in the citric acid cycle, the electron transport chain, and oxidative phosphorylation to produce ATP. |
Courses/Lumen_Learning/Book%3A_US_History_I_(OS_Collection)_(Lumen)/04%3A_Early_Globalization%3A_The_Atlantic_World%2C_1492-1650/04.8%3A_Primary_Source_Reading%3A_Vasco_De_Gama | Introduction to the Source Vasco da Gama was born about 1460 at Sines, Portugal. Both Prince John and Prince Manuel continued the efforts of Prince Henry to find a sea route to India, and in 1497 Manuel placed Vasco da Gama, who already had some reputation as a warrior and navigator, in charge of four vessels built especially for the expedition. They set sail July 8, 1497, rounded the Cape of Good Hope four months later, and reached Calicut May 20, 1498. The Moors in Calicut instigated the Zamorin of Calicut against him, and he was compelled to return with the bare discovery and the few spices he had bought there at inflated prices [but still he made a 3000% profit!]. A force left by a second expedition under Cabral (who discovered Brazil by sailing too far west), left behind some men in a “factory” or trading station, but these were killed by the Moors in revenge for Cabral’s attacks on Arab shipping in the Indian Ocean. Vasco da Gama was sent on a mission of vengeance in 1502, he bombarded Calicut (virtually destroying the port), and returned with great spoil. His expedition turned the commerce of Europe from the Mediterranean cities to the Atlantic Coast, and opened up the east to European enterprise. Source Vasco da Gama was born about 1460 at Sines, Portugal. Both Prince John and Prince Manuel continued the efforts of Prince Henry to find a sea route to India, and in 1497 Manuel placed Vasco da Gama, who already had some reputation as a warrior and navigator, in charge of four vessels built especially for the expedition. They set sail July 8, 1497, rounded the Cape of Good Hope four months later, and reached Calicut May 20, 1498. The Moors in Calicut instigated the Zamorin of Calicut against him, and he was compelled to return with the bare discovery and the few spices he had bought there at inflated prices [but still he made a 3000% profit!]. A force left by a second expedition under Cabral (who discovered Brazil by sailing too far west), left behind some men in a “factory” or trading station, but these were killed by the Moors in revenge for Cabral’s attacks on Arab shipping in the Indian Ocean. Vasco da Gama was sent on a mission of vengeance in 1502, he bombarded Calicut (virtually destroying the port), and returned with great spoil. His expedition turned the commerce of Europe from the Mediterranean cities to the The city of Calicut is inhabited by Christians. [The first voyagers to India mistook the Hindus for Christians.] They are of tawny complexion. Some of them have big beards and long hair, whilst others clip their hair short or shave the head, merely allowing a tuft to remain on the crown as a sign that they are Christians. They also wear moustaches. They pierce the ears and wear much gold in them. They go naked down to the waist, covering their lower extremities with very fine cotton stuffs. But it is only the most respectable who do this, for the others manage as best they are able. The women of this country, as a rule, are ugly and of small stature. They wear many jewels of gold round the neck, numerous bracelets on their arms, and rings set with precious stones on their toes. All these people are well-disposed and apparently of mild temper. At first sight they seem covetous and ignorant. When we arrived at Calicut the king was fifteen leagues away. The captain-major sent two men to him with a message, informing him that an ambassador had arrived from the King of Portugal with letters, and that if he desired it he would take them to where the king then was. The king presented the bearers of this message with much fine cloth. He sent word to the captain-major bidding him welcome, saying that he was about to proceed to Calicut. As a matter of fact, he started at once with a large retinue. A pilot accompanied our two men, with orders to take us to a place called Pandarani, below the place (Capna) where we anchored at first. At this time we were actually in front of the city of Calicut. We were told that the anchorage at the place to which we were to go was good, whilst at the place we were then it was bad, with a stony bottom, which was quite true; and, moreover, that it was customary for the ships which came to this country to anchor there for the sake of safety. We ourselves did not feel comfortable, and the captain-major had no sooner received this royal message than he ordered the sails to be set, and we departed. We did not, however, anchor as near the shore as the king’s pilot desired. When we were at anchor, a message arrived informing the captain-major that the king was already in the city. At the same time the king sent a bale, with other men of distinction, to Pandarani, to conduct the captain-major to where the king awaited him. This bale is like an alcaide, and is always attended by two hundred men armed with swords and bucklers. As it was late when this message arrived, the captain-major deferred going. On the following morning, which was Monday, May 28th, the captain-major set out to speak to the king, and took with him thirteen men. On landing, the captain-major was received by the alcaide, with whom were many men, armed and unarmed. The reception was friendly, as if the people were pleased to see us, though at first appearances looked threatening, for they carried naked swords in their hands. A palanquin was provided for the captain-major, such as is used by men of distinction in that country, as also by some of the merchants, who pay something to the king for this privilege. The captain-major entered the palanquin, which was carried by six men by turns. Attended by all these people we took the road of Calicut, and came first to another town, called Capna. The captain-major was there deposited at the house of a man of rank, whilst we others were provided with food, consisting of rice, with much butter, and excellent boiled fish. The captain-major did not wish to eat, and as we had done so, we embarked on a river close by, which flows between the sea end the mainland, close to the coast. The two boats in which we embarked were lashed together, so that we were not separated. There were numerous other boats, all crowded with people. As to those who were on the banks I say nothing; their number was infinite, and they had all come to see us. We went up that river for about a league, and saw many large ships drawn up high and dry on its banks, for there is no port here. When we disembarked, the captain-major once more entered his palanquin. The road was crowded with a countless multitude anxious to see us. Even the women came out of their houses with children in their arms and followed us. When we arrived (at Calicut) they took us to a large church, and this is what we saw: The body of the church is as large as a monastery, all built of hewn stone and covered with tiles. At the main entrance rises a pillar of bronze as high as a mast, on the top of which was perched a bird, apparently a cock. In addition to this, there was another pillar as high as a man, and very stout. In the center of the body of the church rose a chapel, all built of hewn stone, with a bronze door sufficiently wide for a man to pass, and stone steps leading up to it. Within this sanctuary stood a small image which they said represented Our Lady. Along the walls, by the main entrance, hung seven small bells. In this church the captain-major said his prayers, and we with him. We did not go within the chapel, for it is the custom that only certain servants of the church, called quafees, should enter. These quafees wore some threads passing over the left shoulder and under the right arm, in the same manner as our deacons wear the stole. They threw holy water over us, and gave us some white earth, which the Christians of this country are in the habit of putting on their foreheads, breasts, around the neck, and on the forearms. They threw holy water upon the captain-major and gave him some of the earth, which he gave in charge of someone, giving them to understand that he would put it on later. Many other saints were painted on the walls of the church, wearing crowns. They were painted variously, with teeth protruding an inch from the mouth, and four or five arms. Below this church there was a large masonry tank, similar to many others which we had seen along the road. After we had left that place, and had arrived at the entrance to the city (of Calicut) we were shown another church, where we saw things like those described above. Here the crowd grew so dense that progress along the street became next to impossible, and for this reason they put the captain-major into a house, and us with him. The king sent a brother of the bale, who was a lord of this country, to accompany the captain-major, and he was attended by men beating drums, blowing arafils and bagpipes, and firing off matchlocks. In conducting the captain-major they showed us much respect, more than is shown in Spain to a king. The number of people was countless, for in addition to those who surrounded us, and among whom there were two thousand armed men, they crowded the roofs and houses. The further we advanced in the direction of the king’s palace, the more did they increase in number. And when we arrived there, men of much distinction and great lords came out to meet the captain-major, and joined those who were already in attendance upon him. It was then an hour before sunset. When we reached the palace we passed through a gate into a courtyard of great size, and before we arrived at where the king was, we passed four doors, through which we had to force our way, giving many blows to the people. When, at last, we reached the door where the king was, there came forth from it a little old man, who holds a position resembling that of a bishop, and whose advice the king acts upon in all affairs of the church. This man embraced the captain-major when he entered the door. Several men were wounded at this door, and we only got in by the use of much force. Public domain content Authored by : Oliver J. Thatcher. Provided by : The Library of Original Sources. Located at : http://legacy.fordham.edu/halsall/mod/1497degama.asp . Project : The Modern Internet History Sourcebook. License : Public Domain: No Known Copyright |
Courses/National_Yang_Ming_Chiao_Tung_University/Chemical_Principles_for_Medical_Students/01%3A_Electronic_Structure_of_Atoms/1.01%3A_The_Wave_Nature_of_Light | Learning Objectives To learn about the characteristics of electromagnetic waves. Light, X-Rays, infrared and microwaves are among the types of electromagnetic waves. Scientists discovered much of what we know about the structure of the atom by observing the interaction of atoms with various forms of radiant, or transmitted, energy, such as the energy associated with the visible light we detect with our eyes, the infrared radiation we feel as heat, the ultraviolet light that causes sunburn, and the x-rays that produce images of our teeth or bones. All these forms of radiant energy should be familiar to you. We begin our discussion of the development of our current atomic model by describing the properties of waves and the various forms of electromagnetic radiation. Properties of Waves A wave is a periodic oscillation that transmits energy through space . Anyone who has visited a beach or dropped a stone into a puddle has observed waves traveling through water (Figure \(\PageIndex{1}\)). These waves are produced when wind, a stone, or some other disturbance, such as a passing boat, transfers energy to the water, causing the surface to oscillate up and down as the energy travels outward from its point of origin. As a wave passes a particular point on the surface of the water, anything floating there moves up and down. Waves have characteristic properties (Figure \(\PageIndex{2}\)). As you may have noticed in Figure \(\PageIndex{1}\), waves are periodic, that is, they repeat regularly in both space and time. The distance between two corresponding points in a wave—between the midpoints of two peaks, for example, or two troughs—is the wavelength (\(λ\), lowercase Greek lambda). Wavelengths are described by a unit of distance, typically meters. The frequency (\( u\), lowercase Greek nu) of a wave is the number of oscillations that pass a particular point in a given period of time. The usual units are oscillations per second (1/s = s −1 ), which in the SI system is called the hertz (Hz). It is named after German physicist Heinrich Hertz (1857–1894), a pioneer in the field of electromagnetic radiation. The amplitude , or vertical height, of a wave is defined as half the peak-to-trough height; as the amplitude of a wave with a given frequency increases, so does its energy. As you can see in Figure \(\PageIndex{2}\), two waves can have the same amplitude but different wavelengths and vice versa. The distance traveled by a wave per unit time is its speed (\(v\)), which is typically measured in meters per second (m/s). The speed of a wave is equal to the product of its wavelength and frequency: \[\begin{align} (\text{wavelength})(\text{frequency}) &= \text{speed} \nonumber \\[4pt] \lambda u &=v \label{6.1.1a} \\[4pt] \left ( \dfrac{meters}{\cancel{wave}} \right )\left ( \dfrac{\cancel{\text{wave}}}{\text{second}} \right ) &=\dfrac{\text{meters}}{\text{second}} \label{6.1.1b} \end{align} \] Be careful not to confuse the symbols for the speed, \(v\), with the frequency, \( u\). Different types of waves may have vastly different possible speeds and frequencies. Water waves are slow compared to sound waves, which can travel through solids, liquids, and gases. Whereas water waves may travel a few meters per second, the speed of sound in dry air at 20°C is 343.5 m/s. Ultrasonic waves, which travel at an even higher speed (>1500 m/s) and have a greater frequency, are used in such diverse applications as locating underwater objects and the medical imaging of internal organs. Electromagnetic Radiation Water waves transmit energy through space by the periodic oscillation of matter (the water). In contrast, energy that is transmitted, or radiated, through space in the form of periodic oscillations of electric and magnetic fields is known as electromagnetic radiation . (Figure \(\PageIndex{3}\)). Some forms of electromagnetic radiation are shown in Figure \(\PageIndex{4}\). In a vacuum, all forms of electromagnetic radiation—whether microwaves, visible light, or gamma rays—travel at the speed of light ( c ), which turns out to be a fundamental physical constant with a value of 2.99792458 × 10 8 m/s (about 3.00 ×10 8 m/s or 1.86 × 10 5 mi/s). This is about a million times faster than the speed of sound. Because the various kinds of electromagnetic radiation all have the same speed ( c ), they differ in only wavelength and frequency. As shown in Figure \(\PageIndex{4}\) and Table \(\PageIndex{1}\), the wavelengths of familiar electromagnetic radiation range from 10 1 m for radio waves to 10 −12 m for gamma rays, which are emitted by nuclear reactions. By replacing \(v\) with \( c\) in Equation \(\ref{6.1.1a}\), we can show that the frequency of electromagnetic radiation is inversely proportional to its wavelength: \[ \begin{align} c&=\lambda u \\[4pt] u &=\dfrac{c}{\lambda } \label{6.1.2} \end{align} \] For example, the frequency of radio waves is about 10 8 Hz, whereas the frequency of gamma rays is about 10 20 Hz. Visible light, which is electromagnetic radiation that can be detected by the human eye, has wavelengths between about 7 × 10 −7 m (700 nm, or 4.3 × 10 14 Hz) and 4 × 10 −7 m (400 nm, or 7.5 × 10 14 Hz). Note that when frequency increases, wavelength decreases; c being a constant stays the same. Similarly, when frequency decreases, the wavelength increases. Within the visible range our eyes perceive radiation of different wavelengths (or frequencies) as light of different colors, ranging from red to violet in order of decreasing wavelength. The components of white light—a mixture of all the frequencies of visible light—can be separated by a prism, as shown in part (b) in Figure \(\PageIndex{4}\). A similar phenomenon creates a rainbow, where water droplets suspended in the air act as tiny prisms. Unit Symbol Wavelength (m) Type of Radiation picometer pm 10−12 gamma ray angstrom Å 10−10 x-ray nanometer nm 10−9 UV, visible micrometer μm 10−6 infrared millimeter mm 10−3 infrared centimeter cm 10−2 microwave meter m 100 radio As you will soon see, the energy of electromagnetic radiation is directly proportional to its frequency and inversely proportional to its wavelength: \[ \begin{align} E\; &\propto\; u \label{6.1.3} \\[4pt] & \propto\; \dfrac{1}{\lambda } \label{6.1.4} \end{align} \] Whereas visible light is essentially harmless to our skin, ultraviolet light, with wavelengths of ≤ 400 nm, has enough energy to cause severe damage to our skin in the form of sunburn. Because the ozone layer of the atmosphere absorbs sunlight with wavelengths less than 350 nm, it protects us from the damaging effects of highly energetic ultraviolet radiation. The energy of electromagnetic radiation increases with increasing frequency and decreasing wavelength. Electromagnetic Radiation: https://youtu.be/TZy7a69pP-w Example \(\PageIndex{1}\): Wavelength of Radiowaves Your favorite FM radio station, WXYZ, broadcasts at a frequency of 101.1 MHz. What is the wavelength of this radiation? Given: frequency Asked for: wavelength Strategy: Substitute the value for the speed of light in meters per second into Equation \(\ref{6.1.2}\) to calculate the wavelength in meters. Solution: From Equation \(\ref{6.1.2}\), we know that the product of the wavelength and the frequency is the speed of the wave, which for electromagnetic radiation is 2.998 × 10 8 m/s: \[ \begin{align*} λν &= c \\[4pt] &= 2.998 \times 10^8 m/s \end{align*} \nonumber \] Thus the wavelength \(λ\) is given by \[ \begin{align*} \lambda &=\dfrac{c}{ u } \\[4pt] &=\left ( \dfrac{2.988\times 10^{8}\; m/\cancel{s}}{101.1\; \cancel{MHz}} \right )\left ( \dfrac{1\; \cancel{MHz}}{10^{6}\; \cancel{s^{-1}}} \right ) \\[4pt] &=2.965\; m \end{align*} \nonumber \] Exercise \(\PageIndex{1}\) As the police officer was writing up your speeding ticket, she mentioned that she was using a state-of-the-art radar gun operating at 35.5 GHz. What is the wavelength of the radiation emitted by the radar gun? Answer 8.45 mm Summary Understanding the electronic structure of atoms requires an understanding of the properties of waves and electromagnetic radiation. A wave is a periodic oscillation by which energy is transmitted through space. All waves are periodic , repeating regularly in both space and time. Waves are characterized by several interrelated properties: wavelength (\(λ\)), the distance between successive waves; frequency (\( u\)), the number of waves that pass a fixed point per unit time; speed (\(v\)), the rate at which the wave propagates through space; and amplitude , the magnitude of the oscillation about the mean position. The speed of a wave is equal to the product of its wavelength and frequency. Electromagnetic radiation consists of two perpendicular waves, one electric and one magnetic, propagating at the speed of light (\(c\)). Electromagnetic radiation is radiant energy that includes radio waves, microwaves, visible light, x-rays, and gamma rays, which differ in their frequencies and wavelengths. |
Courses/University_of_Connecticut/Organic_Chemistry_-_Textbook_for_Chem_2443/06%3A_Stereochemistry/6.01%3A_All_Stereochemistry_Topics/6.1.07%3A_Stereochemistry_in_Chemical__Reactions | Objectives After completing this section, you should be able to identify a compound as being prochiral. identify the Re and Si faces of prochiral sp 2 centre. identify atoms (or groups of atoms) as pro - R or pro - S on a prochiral sp 3 centre. Key Terms Make certain that you can define, and use in context, the key terms below. prochiral pro - R pro - S Re Si Prochiral Carbons When a tetrahedral carbon can be converted to a chiral center by changing only one of the attached groups, it is referred to as a ‘ prochiral' carbon. The two hydrogens on the prochiral carbon can be described as 'prochiral hydrogens'. Note that if, in a 'thought experiment', we were to change either one of the prochiral hydrogens on a prochiral carbon center to a deuterium (the 2 H isotope of hydrogen), the carbon would now have four different substituents and thus would be a chiral center. Prochirality is an important concept in biological chemistry, because enzymes can distinguish between the two ‘identical’ groups bound to a prochiral carbon center due to the fact that they occupy different regions in three-dimensional space . Consider the isomerization reaction below, which is part of the biosynthesis of isoprenoid compounds. We do not need to understand the reaction itself (it will be covered in chapter 14); all we need to recognize at this point is that the isomerase enzyme is able to distinguish between the prochiral 'red' and the 'blue' hydrogens on the isopentenyl diphosphate (IPP) substrate. In the course of the left to right reaction, IPP specifically loses the 'red' hydrogen and keeps the 'blue' one. Prochiral hydrogens can be unambiguously designated using a variation on the R/S system for labeling chiral centers. For the sake of clarity, we'll look at a very simple molecule, ethanol, to explain this system. To name the 'red' and 'blue' prochiral hydrogens on ethanol, we need to engage in a thought experiment. If we, in our imagination, were to arbitrarily change red H to a deuterium, the molecule would now be chiral and the chiral carbon would have the R configuration (D has a higher priority than H). For this reason, we can refer to the red H as the pro-R hydrogen of ethanol, and label it H R . Conversely, if we change the blue H to D and leave red H as a hydrogen, the configuration of the molecule would be S , so we can refer to blue H as the pro-S hydrogen of ethanol, and label it H S . Looking back at our isoprenoid biosynthesis example, we see that it is specifically the pro-R hydrogen that the isopentenyl diphosphate substrate loses in the reaction. Prochiral hydrogens can be designated either enantiotopic or diastereotopic. If either H R or H S on ethanol were replaced by a deuterium, the two resulting isomers would be enantiomers (because there are no other stereocenters anywhere on the molecule). Thus, these two hydrogens are referred to as enantiotopic . In ( R )-glyceraldehyde-3-phosphate (( R )-GAP), however, we see something different: R )-GAP already has one chiral center. If either of the prochiral hydrogens H R or H S is replaced by a deuterium, a second chiral center is created, and the two resulting molecules will be diastereomers (one is S,R, one is R,R ). Thus, in this molecule, H R and H S are referred to as diastereotopic hydrogens. Finally, hydrogens that can be designated neither enantiotopic nor diastereotopic are called homotopic . If a homotopic hydrogen is replaced by deuterium, a chiral center is not created. The three hydrogen atoms on the methyl (CH 3 ) group of ethanol (and on any methyl group) are homotopic. An enzyme cannot distinguish among homotopic hydrogens. Example \(\PageIndex{1}\) Identify in the molecules below all pairs/groups of hydrogens that are homotopic, enantiotopic, or diastereotopic. When appropriate, label prochiral hydrogens as H R or H S . Answer Groups other than hydrogens can be considered prochiral. The alcohol below has two prochiral methyl groups - the red one is pro-R , the blue is pro-S . How do we make these designations? Simple - just arbitrarily assign the red methyl a higher priority than the blue, and the compound now has the R configuration - therefore red methyl is pro-R . Citrate is another example. The central carbon is a prochiral center with two 'arms' that are identical except that one can be designated pro-R and the other pro-S . In an isomerization reaction of the citric acid (Krebs) cycle, a hydroxide is shifted specifically to the pro-R arm of citrate to form isocitrate: again, the enzyme catalyzing the reaction distinguishes between the two prochiral arms of the substrate (we will study this reaction in chapter 13). Exercise \(\PageIndex{1}\) Assign pro-R and pro-S designations to all prochiral groups in the amino acid leucine. ( Hint : there are two pairs of prochiral groups!). Are these prochiral groups diastereotopic or enantiotopic? Answer Prochiral Carbonyl and Imine Groups Trigonal planar, sp 2 -hybridized carbons are not, as we well know, chiral centers– but they can be prochiral centers if they are bonded to three different substitutuents. We (and the enzymes that catalyze reactions for which they are substrates) can distinguish between the two planar ‘faces’ of a prochiral sp 2 - hybridized group. These faces are designated by the terms re and si . To determine which is the re and which is the si face of a planar organic group, we simply use the same priority rankings that we are familiar with from the R/S system, and trace a circle: re is clockwise and si is counterclockwise. When the two groups adjacent to a carbonyl (C=O) are not the same, we can distinguish between the re and si 'faces' of the planar structure. The concept of a trigonal planar group having two distinct faces comes into play when we consider the stereochemical outcome of a nucleophilic addition reaction. Nucleophilic additions to carbonyls will be covered in greater detail in Chapter 19 . Notice that in the course of a carbonyl addition reaction, the hybridization of the carbonyl carbon changes from sp 2 to sp 3 , meaning that the bond geometry changes from trigonal planar to tetrahedral. If the two R groups are not equivalent, then a chiral center is created upon addition of the nucleophile. The configuration of the new chiral center depends upon which side of the carbonyl plane the nucleophile attacks from. Reactions of this type often result in a 50:50 racemic mixture of stereoisomers, but it is also possible that one stereoisomer may be more abundant, depending on the structure of the reactants and the conditions under which the reaction takes place. Below, for example, we are looking down on the re face of the ketone group in pyruvate. If we flipped the molecule over, we would be looking at the si face of the ketone group. Note that the carboxylate group does not have re and si faces, because two of the three substituents on that carbon are identical (when the two resonance forms of carboxylate are taken into account). As we will see in chapter 10, enzymes which catalyze reactions at carbonyl carbons act specifically from one side or the other. We need not worry about understanding the details of the reaction pictured above at this point, other than to notice the stereochemistry involved. The pro-R hydrogen (along with the two electrons in the C-H bond) is transferred to the si face of the ketone (in green), forming, in this particular example, an alcohol with the R configuration. If the transfer had taken place at the re face of the ketone, the result would have been an alcohol with the S configuration. Exercise \(\PageIndex{2}\) For each of the carbonyl groups in uracil, state whether we are looking at the re or the si face in the structural drawing below. Answer Exercise \(\PageIndex{3}\) a) State which of the following hydrogen atoms are pro-R or pro-S . b) Identify which side is Re or Si. Answer a) Left compound: H a = pro-S and H b = pro-R ; Right compound: H a = pro-R and H b = pro-S b) A – Re; B – Si; C – Re; D – Si \) Exercise \(\PageIndex{4}\) State whether the H's indicated below are pro- R or pro- S for the following structures. Answer a) H a is pro- R ; H b is pro- S H a is pro- R ; H b is pro- S Exercise \(\PageIndex{5}\) In the structures below, determine if the H's are homotopic, enantiotopic, or diastereotopic. Answer In a), the CH 2 is diastereotopic since there is another chiral center on the molecule. Both CH 3 's are homotopic since replacing one of them doesn't create a chiral center. IN b), the CH 2 's are enantiotopic since it would create the only chiral center on the molecule. Both CH 3 's are homotopic since replacing one of them doesn't create a chiral center. Exercise \(\PageIndex{6}\) State whether you are looking down at the molecule from the re face or si face. Answer You are looking at the si face . The re face would be if you were facing the molecule from the back. You are looking at the re face . The si face would be if you were facing the molecule from the back. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/03%3A_Chemical_Bonding/3.03%3A_The_Covalent_Bond_Clarified_Through_the_Use_of_the_Virial_Theorem | Abstract Slater's method of analyzing the covalent bond using the virial theorem is presented for the hydrogen molecule. The overall conclusion reached is the same as that reached with Ruedenberg's well-known ab initio quantum mechanical study - electron kinetic energy plays an important role in chemical bond formation. Introduction In the second edition of his classic monograph, Molecular Quantum Mechanics , Peter Atkins begins the chapter on molecular structure with the following sentences ( 1 ), Now we come to the heart of chemistry. If we can understand what holds atoms together as molecules we may also start to understand why, under certain conditions, old arrangements change in favor of new ones. We shall understand structure, and through structure, the mechanism of change. Few will argue with Atkins' eloquent assertion that the chemical bond is at the heart of chemistry, however finding a satisfactory discussion of chemistry's central concept is difficult. For example, a survey of widely adopted general chemistry texts showed the following errors in the description of the covalent bond to be prevalent. Surprisingly the origin of some of these errors can be traced to two authoritative monographs on chemical bonding of enormous influence ( 2 ). The covalent bond is presented as a purely electrostatic phenomenon. Electron kinetic energy is never mentioned, even though the total energy of a molecule is a sum of kinetic and potential energy contributions, and atomic and molecular stability cannot be understood solely in terms of potential energy. Closely related to this is that what is actually an energy curve is called a potential energy curve. What is shown in introductory texts is the total molecular energy as a function of internuclear separation under the Born-Oppenheimer approximation. In other words, nuclear kinetic energy is frozen, but electron kinetic energy contributes to the total molecular energy. It is claimed that the electron density in the inter-nuclear (bond) region has a lower potential energy because it is attracted to two nuclei. Actually using simple electrostatic arguments it is easy to show that the electron-nuclear potential energy is higher in the internuclear region than it is closer to the nuclei. On the basis of potential energy alone the electrons would prefer to be in the nucleus. An energy minimum, or molecular ground state, is achieved because of increases in nuclear-nuclear and electron-electron repulsions as the internuclear separation decreases. As will be shown later, the immediate cause of the molecular ground state is a sharp increase in electron kinetic energy. The amount of electron density transferred to the bonding region is greatly overstated, sometimes implying that a pair of electrons is shared between two nuclei rather than by two nuclei. Unfortunately, to find accurate analyses of the physical origin of the covalent bond one must go to the research or pedagogical literature; chemistry textbooks will not, in general, provide enlightenment. In the 1960s and 70s Ruedenberg and his collaborators carried out a detailed quantum mechanical study of the covalent bond in H 2 + ( 4 - 6 ). The most important conclusion of this thorough and insightful study was that electron kinetic energy plays a crucial role in the formation of a chemical bond. Ruedenberg's contributions to the understanding of the chemical bond have been summarized in the pedagogical literature ( 7 - 12 ) and in review articles ( 13 - 15 ). There are also at least two encyclopedia entries which give accurate and clear interpretations of covalent bond formation ( 16, 17 ). It is surprising that none of these efforts to make Ruedenberg's work accessible to the non-specialist have had any noticeable impact on the way chemical bonding is presented by the authors of chemistry textbooks currently used in the undergraduate curriculum. While physical chemistry texts avoid the errors cited above, they generally do not attempt to provide an "explanation" of the chemical bond. For example, after outlining the mathematical techniques required to solve Schr�dinger's equation for H 2 + and H 2 , physical chemistry texts do not interpret the calculations other than to say something to the effect that the stability of the chemical bond is a quantum mechanical effect that has no classical analog or explanation. Atkins' physical chemistry text is a notable exception; in a footnote he alerts the reader to the subtleties of the chemical bond and briefly outlines the interplay between kinetic and potential energy during the formation of the covalent bond in H 2 + ( 3 ). It is not widely appreciated that John C. Slater used the virial theorem to come to a similar conclusion about the importance of electron kinetic energy in chemical bond formation in the early days of quantum mechanics. In a paper published in the inaugural volume of the Journal of Chemical Physics Slater pioneered the use of the virial theorem in interpreting the chemical bond. With regard to the virial theorem he said ( 18 ), "... this theorem gives a means of finding kinetic and potential energy separately for all configurations of the nuclei, as soon as the total energy is known, from experiment or theory ." (emphasis added) The purpose of this paper is to outline Slater's method of analysis for the hydrogen molecule, the simplest example of the traditional two-electron chemical bond. Background Theory The Hamiltonian energy operator (in atomic units) for the hydrogen molecule consists of the following ten contributions, \[ \hat{H} = - \frac{1}{2M} \nabla_a^2 - \frac{1}{2M} \nabla_b^2 - \frac{1}{2} \nabla_1^2 - \frac{1}{2} \nabla_2^2 - \frac{1}{r_{a1}} - \frac{1}{r_{a2}} - \frac{1}{r_{b1}} - \frac{1}{r_{b2}} + \frac{1}{r_{12}} + \frac{1}{R} \nonumber \] where a and b label the nuclei, 1 and 2 label the electrons, and R = R ab , the internuclear separation. Under the Born-Oppenheimer approximation the nuclear kinetic energy operators drop out, and the molecular electronic energy as a function of internuclear separation, E(R) , can be obtained by solving Schr�dinger's equation variationally for the resulting eight-term electronic Hamiltonian, H el , which in this analysis includes nuclear-nuclear repulsion. \[ E(R) = \frac{ \langle \Psi (R;~1,2) | \hat{H}_{el} | \Psi (R;~1,2) \rangle}{ \langle \Psi (R;~1,2) | \Psi (R;~1,2) \rangle} \nonumber \] Now one can return to equation (1) with the calculated E(R) and treat it as the molecular "potential" field in which the nuclei move. The nuclear kinetic energy operators are re-written in terms of the nuclear center-of-mass coordinate and the internal coordinate R . After discarding the term involving the motion of the center of mass, the Hamiltonian energy operator becomes \[ \hat{H} = - \frac{1}{2 \mu} \nabla^2 + E(R) \nonumber \] Following Slater, a more empirical approach is taken here and the ab initio E(R) is replaced by an analytical surrogate such as the Morse function ( 19 ) which has a similar R-dependence ( 20 ). \[ E(R) = D_e \left[ 1 - \text{exp} (- \beta (R-R_e )) \right]^2 - D_e \nonumber \] In other words the Morse function represents the total energy of a diatomic molecule assuming the Born-Oppenheimer approximation. Substitution of equation (4) into (3) yields an energy Hamiltonian for which Schrödinger's equation has an exact solution with the following quantized nuclear vibrational states ( 2 ), \[ G(v) = \left( v + \frac{1}{2} \right) \overline{ \nu}_e - \left( v + \frac{1}{2} \right)^2 \overline{ \nu}_e x_e \nonumber \] where v = 0, 1, 2, ..., and the Morse parameters, D e and b, are defined as, \[ \begin{matrix} D_e = \frac{hc \overline{ \nu}_e (1- x_e^2)}{4x_e} & \beta = \sqrt{ \frac{ \mu}{D_e}} \pi \overline{ \nu}_e c\end{matrix} \nonumber \] Fitting equation (5) to spectroscopic data for H 2 yields the following values for the Morse parameters: D e = 7.92x10 -19 joule; b = 0.0191 pm -1 ; and R e = 74.1 pm ( 21 ). The Virial Theorem At this point the virial theorem is used to obtain a total energy profile for covalent bond formation in H 2 . The first step in this approach is to acknowledge that the energy of a molecule is a sum of kinetic and potential energy contributions. \[ E(R) = T(R) + V(R) \nonumber \] The virial theorem for a diatomic molecule as a function of internuclear separation, R , is ( 14 ) \[ 2T(R) + V(R) + R \frac{dE(R)}{dR} = 0 \nonumber \] Equations (7) and (8) can now be used to obtain expressions for the kinetic and potential energy in terms of the total energy and its first derivative. \[ T(R) = -E(R) - R \frac{dE(R)}{dR} \nonumber \] \[ V(R) = 2E(R) + R \frac{dE(R)}{dR} \nonumber \] Quoting Slater again ( 18 ) These important equations determine the mean kinetic and potential energies as a function of r, one might almost say, experimentally, directly from the curves of E as a function of r which can be found from band spectra. The theory is so simple and direct that one can accept the results without question, remembering only the limitation of accuracy mentioned above. (Here Slater is referring to the neglect of zero point vibration in his analysis.) Thus, when equation (4), parameterized as indicated above, is used in equations (9) and (10) the energy profile for covalent bond formation in H 2 shown in Figure 1 is obtained. Analysis This energy profile shows that as the inter-nuclear separation decreases, the potential energy rises, falls, and then rises again. The kinetic energy first decreases and then increases at about the same value of R that the potential energy begins to decrease. The quantum mechanical interpretation of the energy profile is as follows ( 4, 18 ). As the molecular orbital is formed at large R constructive interference between the two overlapping atomic orbitals on the hydrogen atoms draws electron density away from the nuclear centers into the inter-nuclear region. The potential energy rises as electron density is removed from the region around the nuclei, but the total energy decreases because of a larger decrease in kinetic energy due to charge delocalization - the electrons now belong to the molecule and not the individual atoms. Thus, a decrease in kinetic energy funds the initial build-up of charge between the nuclei that is traditionally associated with chemical bond formation. Following this initial phase, at an inter-nuclear separation of about 150 pm, the potential energy begins to decrease and the kinetic energy increases, both sharply eventually, while the total energy continues to decrease gradually. This is an atomic affect, not a molecular one, as Ruedenberg clearly showed. The initial transfer of charge away from the nuclei and into the bond region allows the atomic orbitals to contract causing a large decrease in potential energy because the electron density is moved, on average, closer to the nuclei. The kinetic energy increases because the atomic orbitals are smaller and kinetic energy is inversely proportional to the square of the average orbital radius. This is an atomic affect because the orbital contraction actually causes some electron density to be withdrawn from the bonding region and returned to the nuclei. An energy minimum is reached while the potential energy is still in a significant decline, indicating that kinetic energy, which is increasing rapidly, is the immediate cause of a stable bond and the molecular ground state in H 2 . The final increase in potential energy which is mainly due to nuclear-nuclear repulsion doesn't begin until the inter-nuclear separation is less than 50 pm, while the equilibrium bond length is 74 pm. Thus the common explanation that an energy minimum is reached because of nuclear-nuclear and electron-electron repulsion does not have merit. As Ruedenberg ( 4 ) has noted "there are no ground states in classical mechanics or electrostatics." Conclusion In conclusion a review of some basic principles and some additional observations are offered. Under the Born-Oppenheimer approximation the quantum mechanical energy operator for H 2 consists of the final eight terms in equation (1), which fall into four types of energy contributions: electron kinetic energy, electron-electron, nuclear-nuclear, and electron-nuclear potential energy. Electron-nuclear potential energy is the only negative contribution to the total energy and is essential in understanding the stability of the chemical bond. H 2 is a stable molecule because it has a lower total energy than its constituent atoms, not simply because there is a build-up of charge in the inter-nuclear region. As shown above, by itself this charge build-up actually increases the energy rather than decreasing it as is popularly, but incorrectly, believed. We all recognize that electron kinetic energy is important at the computational level through its presence in the energy operator given in equation (1). In other words ground states are calculated by minimizing the total energy, the sum of the kinetic and potential energy contributions. However, many ignore kienticenergy when it comes to interpretation because they believe it is irrelevant. The origin of this error is a common misapplication of the virial theorem. For example, it is common to argue from the virial theorem in the form, \[ \Delta E = \frac{ \Delta V}{2} = - \Delta T \nonumber \] that bond stability is due solely to potential energy, because it decreases with bond formation and kinetic energy actually increases by twice as much. However, this form of the virial theorem is valid only for R = R �� and R = R e , the initial and final states. It tells us nothing about what is occuring when the bond is actually being formed - for that you need equations (7 - 10). As has been shown above, what is actually occuring during bond formation is a rather subtle role reversal between kinetic and potential energy. It is ironic that Slater correctly used the virial theorem to show that electron kinetic energy is essential to understanding the chemical bond in the 1930s, while today many use it incorrectly to show that electron kinetic energy is irrelevant. It has been known since the early years of the 20 th century that no dynamic or static array of charged particles is stable on the basis of classical electrostatic principles alone. Therefore, the quantum mechanical picture of a wave-particle duality for the electron and the peculiar quantum mechanical nature of kinetic energy are essential in understanding atomic and molecular stability and structure. The importance of kinetic energy runs counter to conventional opinion regarding the covalent bond in two seemingly paradoxical ways. First, a decrease in kinetic energy due to incipient molecular orbital formation funds the transfer of charge density into the internuclear region, lowering the total energy. Second, a large increase in kinetic energy associated with the subsequent atomic orbital contraction prevents the collapse of the molecule and causes an energy minimum and a stable molecular ground state. In a previous publication ( 10b ) the formation of the covalent bond in H 2 was analyzed in terms of a two-step mechanism involving: a) orbital contraction ("hybridization") followed by b) orbital overlap (charge delocalization and redistribution). It was shown that step a) is an atomic and endothermic process, while b) is a molecular and exothermic effect. The present study is consistent with this prior analysis, but reverses the order of the contributing effects; atomic orbital overlap precedes atomic orbital contraction. Both mechanisms clearly reveal the important role that electron kinetic energy plays in chemical bond formation. In summary I concur with Kutzelnigg's observation ( 22 ), "The chemical bond is a highly complex phenomenon which eludes all attempts at simple description." This means we must be careful how we teach the chemical bond to our students. It is not acceptable to present incorrect models of bonding to undergraduates because they are easier to understand and therefore easier to teach. Literature Cited: Atkins, P. W. Molecular Quantum Mechanics , Oxford University Press, Oxford, UK, 1983, p. 250. Pauling, L. The Nature of the Chemical Bond , 3 rd ed., Cornell University Press, Ithaca, 1960, pp. 19-21. Coulson, C. A. Valence , 2 nd ed., Oxford University Press, London, 1961, pp. 85-86. Atkins, P. W. Physical Chemistry , 6 th ed., W. H. Freeman and Co., New York, 1998, p. 397. Ruedenberg, K. Rev. Mod. Phys. 1962 , 34 , 326-352. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971 , 54 , 1495-1511. Feinberg, M. J.; Ruedenberg, K. J. Chem. Phys. 1971 , 55 , 5804-5818. Ruedenberg, K. In Localization and Delocalization in Quantum Chemistry; Chalvet, O. et al., Eds.; Reidel: Dordrecht, The Netherlands, 1975 ; Vol. I, pp 223-245. Baird, N. C. J. Chem. Educ. 1986 , 63 , 660-664. Harcourt, R. D. Am. J. Phys. 1988 , 56 , 660-661. Nordholm, S. J. Chem. Educ. 1988 , 65 , 581-584. Bacskay, G. G.; Reimers, J. R.; Nordholm, S. J. Chem. Educ. 1997 , 74 , 1494-1502. Rioux, F. a) Chem. Educator 1997 , 2(6) , 1-14; b) Chem. Educator 2001 , 6(5) , 288-290. Harcourt, R. D.; Solomon, H.; Beckworth, J.; Chislett, L. Am. J. Phys. 1982 , 50 , 557-559. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1973 , 12 , 546-562. Melrose, M. P.; Chauhan, M.; Kahn, F. Theor. Chim . Acta 1994 , 88 , 311-324. Gordon, M. S.; Jensen, J. H. Theor . Chem. Acc. 2000 , 103 , 248-251. Gordon, M. S.; Jensen, J. H. Encyclopedia of Computational Chemistry , Schleyer, P. v. R., Ed.; John Wiley & Sons: New York, 1998, pp 3198-3214. Harris, F. E. In Encyclopedia of Physics , 2 nd Edition; Lerner, R. G.; Trig, G. L., Eds.; VCH Publishers, Inc.: New York, 1991, pp 762-764. Slater, J. C. J. Chem. Phys. 1933 , 1 , 687-691. Morse, P. M. Phys. Rev. 1929 , 34 , 57. For a virial theorem analysis using the best ab initio results for H 2 see, Winn, J. S. J. Chem. Phys. 1981 , 74 , 608-611. Huber, K. P.; Herzberg Molecular Spectra and Molecular Structure , vol IV, Constants of Diatomic Molecules ; Van Nostrand Reinhold: New York, 1979. Kutzelnigg, W. Angew. Chem. Int. Ed. Eng. 1984 , 23 , 292. |
Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.05%3A_Molecular_Orbital_Theory | Learning Objectives Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals Describe traits of bonding and antibonding molecular orbitals Calculate bond orders based on molecular electron configurations Write molecular electron configurations for first- and second-row diatomic molecules Relate these electron configurations to the molecules’ stabilities and magnetic properties For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O 2 , presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O 2 : This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O 2 is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called paramagnetism , and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O 2 indicates that all electrons are paired. How do we account for this discrepancy? Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure \(\PageIndex{1}\)), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight. Experiments show that each O 2 molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion. Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table \(\PageIndex{1}\) summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure. Valence Bond Theory Molecular Orbital Theory considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*) forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules needs multiple structures to describe resonance NaN Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ , analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital ( Ψ 2 ) . Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin. We will consider the molecular orbitals in molecules composed of two identical atoms (H 2 or Cl 2 , for example). Such molecules are called homonuclear diatomic molecules . In these diatomic molecules, several types of molecular orbitals occur. The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO) . The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure \(\PageIndex{2}\)). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density. There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure \(\PageIndex{3}\). The in-phase combination produces a lower energy σ s molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy \(σ^∗_s\) m olecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σ s orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals . Electrons in the \(σ^∗_s\) orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals . Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals. In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure \(\PageIndex{4}\)). If two atoms are located along the x -axis in a Cartesian coordinate system, the two p x orbitals overlap end to end and form σ px (bonding) and \(σ^∗_{px}\) (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s -orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital. The side-by-side overlap of two p orbitals gives rise to a pi (\(π\)) bonding molecular orbital and a \( π^*\) antibonding molecular orbital , as shown in Figure \(\PageIndex{5}\). In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei. In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side ( p y and p z ), so these four atomic orbitals combine pairwise to create two π orbitals and two \(π^*\) orbitals. The \(π_{py}\) and \(π^∗_{py}\) orbitals are oriented at right angles to the \(π_{pz}\) and \(π^∗_{pz}\) orbitals. Except for their orientation, the π py and π pz orbitals are identical and have the same energy; they are degenerate orbitals . The \(π^∗_{py}\) and \(π^∗_{pz}\) antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: \(σ_{px}\) and \(σ^∗_{px}\), \(π_{py}\) and \(π^∗_{py}\), \(π_{pz}\) and \(π^∗_{pz}\). Example \(\PageIndex{1}\): M olecular Orbitals Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy. Solution This is an in-phase combination, resulting in a σ 3 p orbital This will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine. This is an out-of-phase combination, resulting in a \(π^∗_{3p}\) orbital. Exercise \(\PageIndex{1}\) Label the molecular orbital shown as \(σ\) or \(π\), bonding or antibonding and indicate where the node occurs. Answer The orbital is located along the internuclear axis, so it is a \(σ\) orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital. Application: Computational Chemistry in Drug Design While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (Figure \(\PageIndex{6}\)). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway. Molecular Orbital Energy Diagrams The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure \(\PageIndex{7}\)). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2 p atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*). We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure \(\PageIndex{7}\)). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as \(\ce{Be2+}\)) would have the molecular electron configuration \((σ_{1s})^2(σ^∗_{1s})^2(σ_{2s})^2(σ^∗_{2s})^1\). It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons. Bond Order The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons. When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon. In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation: \[\textrm{bond order}=\dfrac{(\textrm{number of bonding electrons})−(\textrm{number of antibonding electrons})}{2} \nonumber \] The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders. Bonding in Diatomic Molecules A dihydrogen molecule (H 2 ) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ 1 s bonding orbital. A dihydrogen molecule, H 2 , readily forms because the energy of a H 2 molecule is lower than that of two H atoms. The σ 1 s orbital that contains both electrons is lower in energy than either of the two 1 s atomic orbitals. A molecular orbital can hold two electrons, so both electrons in the H 2 molecule are in the σ 1 s bonding orbital; the electron configuration is \((σ_{1s})^2\). We represent this configuration by a molecular orbital energy diagram (Figure \(\PageIndex{8}\)) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin. A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have \[\ce{bond\: order\: in\: H2}=\dfrac{(2−0)}{2}=1 \nonumber \] Because the bond order for the H–H bond is equal to 1, the bond is a single bond. A helium atom has two electrons, both of which are in its 1 s orbital. Two helium atoms do not combine to form a dihelium molecule, He 2 , with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He 2 as \((σ_{1s})^2(σ^∗_{1s})^2\) as in Figure \(\PageIndex{9}\). The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero. \[\ce{bond\: order\: in\: He2}=\dfrac{(2−2)}{2}=0 \nonumber \] A bond order of zero indicates that no bond is formed between two atoms. The Diatomic Molecules of the Second Period Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li 2 , Be 2 , B 2 , C 2 , N 2 , O 2 , F 2 , and Ne 2 . However, we can predict that the Be 2 molecule and the Ne 2 molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table \(\PageIndex{1}\)). Molecule Electron Configuration Bond Order Li2 \((σ_{2s})^2\) 1 Be2 (unstable) \((σ_{2s})^2(σ^∗_{2s})^2\) 0 B2 \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^2\) 1 C2 \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4\) 2 N2 \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4(σ_{2px})^2\) 3 O2 \((σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},π^∗_{2pz})^2\) 2 F2 \((σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},\:π^∗_{2pz})^4\) 1 Ne2 (unstable) \((σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},π^∗_{2pz})^4(σ^∗_{2px})^2\) 0 We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place. As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure \(\PageIndex{10}\). Looking at Ne 2 molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the p orbitals (Li through N) we observe a different pattern, in which the σ p orbital is higher in energy than the π p set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule. This switch in orbital ordering occurs because of a phenomenon called s-p mixing . s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σ s wavefunction mathematically combines with the σ p wavefunction, with the result that the σ s orbital becomes more stable, and the σ p orbital becomes less stable (Figure \(\PageIndex{11}\)). Similarly, the antibonding orbitals also undergo s-p mixing, with the σ s * becoming more stable and the σ p * becoming less stable. s-p mixing occurs when the s and p orbitals have similar energies. The energy difference between 2 s and 2 p orbitals in O, F, and Ne is greater than that in Li, Be, B, C, and N. Because of this, O 2 , F 2 , and Ne exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure \(\PageIndex{7}\). All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σ p orbital is raised above the π p set. Using the MO diagrams shown in Figure \(\PageIndex{11}\), we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table \(\PageIndex{1}\), Be 2 and Ne 2 molecules would have a bond order of 0, and these molecules do not exist. The combination of two lithium atoms to form a lithium molecule, Li 2 , is analogous to the formation of H 2 , but the atomic orbitals involved are the valence 2 s orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ 2 s bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li 2 molecule to be stable. The molecule is, in fact, present in an appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table \(\PageIndex{1}\) with a bond order greater than zero are also known. The O 2 molecule has enough electrons to half fill the \((π^∗_{2py},\:π^∗_{2pz})\) level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O 2 is in accord with the fact that the oxygen molecule has two unpaired electrons ( Figure \(\PageIndex{10}\)). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory. Application: Band Theory in Extended Systems When two identical atomic orbitals on different atoms combine, two molecular orbitals result (e.g., \(H_2\) in Figure \(\PageIndex{8}\)). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are in-phase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase. In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically > 10 23 atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When N valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, N /2 (filled) bonding orbitals and N /2 (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure \(\PageIndex{12}\)) shows the bands for three important classes of materials: insulators, semiconductors, and conductors. In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is “easy” to overcome, so they are good conductors of electricity. In an insulator, the band gap is so “large” that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when “moderate” amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics. Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to 46% of the energy in sunlight could be converted into electricity using solar cells. Example \(\PageIndex{2}\): M olecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons Draw the molecular orbital diagram for the oxygen molecule, O 2 . From this diagram, calculate the bond order for O 2 . How does this diagram account for the paramagnetism of O 2 ? Solution We draw a molecular orbital energy diagram similar to that shown in Figure \(\PageIndex{7}\). Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure \(\PageIndex{7}\). We calculate the bond order as \[\ce{O2}=\dfrac{(8−4)}{2}=2 \nonumber \] Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the (π 2 py , π 2 pz )* molecular orbitals. Exercise \(\PageIndex{2}\) The main component of air is N 2 . From the molecular orbital diagram of N 2 , predict its bond order and whether it is diamagnetic or paramagnetic. Answer N 2 has a bond order of 3 and is diamagnetic. Example \(\PageIndex{3}\): Ion Predictions with MO Diagrams Give the molecular orbital configuration for the valence electrons in \(\ce{C2^2-}\). Will this ion be stable? Solution Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σ p orbital. The valence electron configuration for C 2 is \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4\). Adding two more electrons to generate the \(\ce{C2^2-}\) anion will give a valence electron configuration of \((σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4(σ_{2px})^2\) Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable. Exercise \(\PageIndex{3}\) How many unpaired electrons would be present on a \(\ce{Be2^2-}\) ion? Would it be paramagnetic or diamagnetic? Answer two, paramagnetic Key Concepts and Summary Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from s orbitals or from p orbitals oriented in an end-to-end fashion. Molecular orbitals formed from p orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called π orbitals. We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory. |
Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_II%3A_Practical_Aspects_of_Structure_-_Purification_and_Spectroscopy/07%3A_Purification_of_Molecular_Compounds/7.15%3A_Instrumentation_Using_Chromatography | Liquid chromatography is often done with more sophisticated equipment. One kind of method is called "high performance liquid chromatography" or HPLC. Rather than packing stationary phase into a glass column, a steel column containing the stationary phase can be purchased. The column can be plumbed into a system that contains a solvent pump to push eluent through the column. After passing through the column, the liquid may go into a UV spectrometer so you can detect when compounds are eluting from the column. The whole apparatus is controlled by a computer. By clicking a button, you can change how quickly the solvent flows. You can easily change the ratio of solvents in the eluent by clicking a button, too. In addition to a UV spectrometer, other instruments can be used with an HPLC system to get information about compounds being eluted . One of the most important is mass spectrometry (MS). Liquid chromatography-mass spectrometry (LC-MS) can be used to determine the molecular weights of the compounds as they elute. That information can be used to help identify the compound. Gas chromatography is another important variation that you should know about. Instead of passing a liquid over the stationary phase, an inert gas moves over the stationary phase. The inert gas may be helium or nitrogen. The equilibrium here is between compounds absorbed onto the stationary phase and compounds moving in the gas phase. Intermolecular attractions with the stationary phase play a role in GC, but so does the boiling point of the compounds. Because most compounds are not very volatile, they would spend all their time sitting on the solid phase under normal conditions. For that reason, the column in a gas chromatograph is placed inside an oven. The temperature in this oven is carefully controlled so that compounds will spend a greater fraction of time in the gas phase. A heating element in the oven can increase the temperature, and a fan is present to help cool it down when. The eluent can't be varied in GC. It is just an inert gas. To control separation of compounds in GC, we can change the pressure of the inert gas, which controls how quickly the gas flows. We can also control the temperature, which influences how much time compounds spend moving along in the gas phase . We can also choose different kinds of columns with different stationary phases. |
Courses/SUNY_Potsdam/Book%3A_Organic_Chemistry_I_(Walker)/06%3A_Acids_Bases_and_Electron_Flow/6.03%3A_Bronsted-Lowry_acids_and_bases | A: The Brønsted-Lowry definition of acidity and basicity We’ll begin our discussion of acid-base chemistry with a couple of essential definitions. The first of these was proposed in 1923 by the Danish chemist Johannes Brønsted and the English chemist Thomas Lowry, and has come to be known as the Brønsted-Lowry definition of acidity and basicity . An acid, by the Brønsted-Lowry definition, is a species which acts as a proton donor (i.e., it gives away an H + ), while a base is a proton (H + ) acceptor. One of the most familiar examples of a Brønsted-Lowry acid-base reaction is between hydrochloric acid and hydroxide ion: In this reaction, a proton is transferred from HCl (the acid, or proton donor ) to hydroxide ion (the base, or proton acceptor ). As we learned in the previous chapter, curved arrows depict the movement of electrons in this bond-breaking and bond-forming process. After a Brønsted-Lowry acid donates a proton, what remains is called the conjugate base . Chloride ion is thus the conjugate base of hydrochloric acid. Conversely, when a Brønsted-Lowry base accepts a proton it is converted into its conjugate acid form: water is thus the conjugate acid of hydroxide ion. Here is an organic acid-base reaction between acetic acid and methylamine: In the reverse of this reaction, acetate ion is the base and methylammonium ion (protonated methylamine) is the acid. What makes a compound acidic (likely to donate a proton) or basic (likely to accept a proton)? Answering that question is one of our main jobs in this chapter, and will require us to put to use much of what we learned about organic structure in the first two chapters, as well as the ideas about thermodynamics that we reviewed in chapter 5. For now, let’s just consider one common property of bases: in order to act as a base, a molecule must have a reactive pair of electrons . In all of the acid-base reactions we’ll see in this chapter, the basic species has an atom with a lone pair of electrons. When methylamine acts as a base, for example, the lone pair of electrons on the nitrogen atom is used to form a new bond to a proton. A negative charge often (but not always!) indicates that a structure (in this case, an anion) is likely to act as a base. Clearly, methylammonium ion cannot act as a base – it does not have a reactive pair of electrons with which to accept a proton. Later, in chapter 10, we will study reactions in which a pair of electrons in a pi bond of an alkene acts in a basic fashion – but for now, will concentrate on the basicity of non-bonding (lone pair) electrons. Exercise Complete the reactions below – in other words, draw structures for the missing conjugate acids and conjugate bases that result from the curved arrows provided. [reveal-answer q=”191829″]Show Solution[/reveal-answer] [hidden-answer a=”191829″] [/hidden-answer] In summary, A Brønsted-Lowry acid is a proton (hydrogen ion) donor. A Brønsted-Lowry base is a proton (hydrogen ion) acceptor. When a Brønsted acid HA dissociates in water, it increases the concentration of hydrogen ions in the solution, $$[H^+]$$; conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to generate $$[OH^-]$$. Acid dissociation \[HA_{(aq)} \rightleftharpoons A^-_{(aq)} + H^+_{(aq)}\] Acid Ionization Constant: \[K_a=\dfrac{[A^-][H^+]}{[HA]}\] Base dissociation: \[B_{(aq)} + H_2O_{(l)} \rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}\] Base Ionization Constant \[K_b = \dfrac{[HB^+][OH^-]}{[B]}\] The determination of a substance as a Brønsted-Lowry acid or base can only be done by observing the reaction. In the case of the HOH it is a base in the first case and an acid in the second case. Water does not need to be involved in a Bronsted-Lowry reaction. In general, for an acid HA and a base Z, we have \[ HA + Z \rightleftharpoons A^- + HZ^+ \] A Donates H to form HZ + . Z Accepts H from A which forms HZ + A – becomes conjugate base of HA and in the reverse reaction it accepts a H from HZ to recreate HA in order to remain in equilibrium HZ + becomes a conjugate acid of Z and in the reverse reaction it donates a H to A – recreating Z in order to remain in equilibrium Questions Why is $$HA$$ an Acid? Why is $$Z^-$$ a Base? How can A – be a base when HA was and Acid? How can HZ + be an acid when Z used to be a Base? Now that we understand the concept, let’s look at an an example with actual compounds! \[ HCl + H_2O \rightleftharpoons H_3O^+ + Cl^¯ \] HCl is the acid because it is donating a proton to H 2 O H 2 O is the base because H 2 O is accepting a proton from HCL H 3 O + is the conjugate acid because it is donating an acid to CL turn into it’s conjugate acid H 2 O Cl¯ is the conjugate base because it accepts an H from H 3 O to return to it’s conjugate acid HCl How can H 2 O be a base? I thought it was neutral? Answers [reveal-answer q=”454720″]Show Solution[/reveal-answer] [hidden-answer a=”454720″] It has a proton that can be transferred It receives a proton from HA A – is a conjugate base because it is in need of a H in order to remain in equilibrium and return to HA HZ + is a conjugate acid because it needs to donate or give away its proton in order to return to it’s previous state of Z In the Brønsted -Lowry Theory what makes a compound an element or a base is whether or not it donates or accepts protons. If the H 2 O was in a different problem and was instead donating an H rather than accepting an H it would be an acid! [/hidden-answer] Overview of acid-base reactions A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ochem1walker/?p=97 An acid-base (proton transfer) reaction For our first example of chemical reactivity, let’s look at a very simple reaction that occurs between hydroxide ion and hydrochloric acid: $$HCl+OH^- \Rightarrow H_20+Cl^-$$ This is an acid-base reaction: a proton is transferred from HCl, the acid, to hydroxide, the base. The product is water (the conjugate acid of hydroxide) and chloride ion (the conjugate base of HCl). You have undoubtedly seen this reaction before in general chemistry. Despite its simplicity (and despite the fact that the reactants and products are inorganic rather than organic), this reaction allows us to consider for the first time many of the fundamental ideas of organic chemistry that we will be exploring in various contexts throughout this text. Key to understanding just about any reaction mechanism is the concept of electron density , and how it is connected to the electron movement (bond-breaking and bond-forming) that occurs in a reaction. The hydroxide ion – specifically, the electronegative oxygen atom in the hydroxide ion – has high electron density due to negative charge and the polarity of the hydrogen-oxygen bond. The hydroxide oxygen is electron-rich . The hydrogen atom in HCl, on the other hand, has low electron density: it is electron-poor . As you might expect, something that is electron-rich is attracted to something that is electron-poor. As hydroxide and HCl move closer to each other, a lone pair of electrons on the electron-rich hydroxide oxygen is attracted by the electron-poor proton of HCl, and electron movement occurs towards the proton. The two electrons in the hydrogen-chlorine sigma bond are repelled by this approaching hydroxide electron density, and therefore move even farther away from the proton and towards the chlorine nucleus. The consequence of all of this electron movement is that the hydrogen-chlorine bond is broken, as the two electrons from that bond completely break free from the 1s orbital of the hydrogen and become a lone pair in the 3p orbital of a chloride anion. At the same time that the hydrogen-chlorine bond is breaking, a new sigma bond forms between hydrogen and oxygen, containing the two electrons that previously were a lone pair on hydroxide. The result of this bond formation is, of course, a water molecule. Previously ( section 6.1. ), we saw how curved arrows were used to depict electron movement that occurs in chemical reactions, where bonds are broken and new bonds are formed. The HCl + OH – reaction, for example, is depicted by drawing two curved arrows. The first arrow originates at one of the lone pairs on the hydroxide oxygen and points to the ‘H’ symbol in the hydrogen bromide molecule, illustrating the ‘attack’ of the oxygen lone pair and subsequent formation of the new hydrogen-oxygen bond. The second curved arrow originates at the hydrogen-bromine bond and points to the ‘Br’ symbol, indicating that this bond is breaking – the two electrons are ‘leaving’ and becoming a lone pair on bromide ion. It is very important to emphasize at this point that these curved, two-barbed arrows always represent the movement of two electrons. Most of this book will be devoted to the description of reaction mechanisms involving two-electron movement, so these full-headed arrows will become very familiar. In the second semester, however, we will look at radical reaction mechanisms, where single-electron movement occurs. For these processes, a curved, single-barbed (‘ fish-hook ‘) arrow will be used. Exercise Draw electron movement arrows to illustrate the acid-base reaction between acetic acid, CH 3 COOH, and ammonia, NH 3 . Draw out the full Lewis structures of reactants and products. [reveal-answer q=”893749″]Show Solution[/reveal-answer] [hidden-answer a=”893749″] [/hidden-answer] CC licensed content, Shared previously 1.8.2. Bronsted-Lowry Acids and Bases. Authored by : Sarah Rundle (UCD), Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook. Located at : https://chem.libretexts.org/LibreTexts/Purdue/Purdue_Chem_26100%3A_Organic_Chemistry_I_(Wenthold)/Chapter_01%3A_Introduction_and_Review/1.8_Acids_and_Bases/1.8.2._Bronsted-Lowry_Acids_and_Bases . Project : Chemistry LibreTexts. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike 7.1: Overview of acid-base reactions. Authored by : u00a0Tim Soderbergu00a0(University of Minnesota, Morris). Located at : https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_07%3A_Organic_compounds_as_acids_and_bases/7.1%3A_Overview_of_acid-base_reactions . Project : Chemistry LibreTexts. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Organic Chemistry With a Biological Emphasis. Authored by : Tim Soderberg (University of Minnesota, Morris). Located at : https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_06%3A_Introduction_to_organic_reactivity_and_catalysis/6.1%3A_A_first_look_at_reaction_mechanisms#6.1A:_An_acid-base_(proton_transfer)_reaction . Project : Chemistry LibreTexts. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike |
Courses/Modesto_Junior_College/Chemistry_142%3A_Pre-General_Chemistry_(Brzezinski)/zz%3A_Back_Matter/20%3A_Glossary | Words (or words that have the same definition) The definition is case sensitive (Optional) Image to display with the definition [Not displayed in Glossary, only in pop-up on pages] (Optional) Caption for Image (Optional) External or Internal Link (Optional) Source for Definition (Eg. "Genetic, Hereditary, DNA ...") (Eg. "Relating to genes or heredity") NaN The infamous double helix https://bio.libretexts.org/ CC-BY-SA; Delmar Larsen Word(s) Definition Image Caption Link Source Sample Word 1 Sample Definition 1 NaN NaN NaN NaN |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/2%3ALab_Textbook_(Nichols)/07%3A_Technique_Summaries/7.02%3A_Using_Calibrated_Glass_Pipettes | 0 1 2 3 NaN NaN NaN NaN Place pipette tip in reagent bottle, squeeze pipette bulb, and connect to the pipette. Partially release your hand to create suction. Do not let to completely or liquid will withdraw forcibly and possibly into the bulb. Apply suction until liquid is withdrawn to just past the desired mark. Remove the pipette bulb and place your finger atop the pipette. Allow tiny amounts of air to be let into the top of the pipette by wiggling your finger or a slight release of pressure. Drain the liquid to the desired mark. Tightly hold the pipette with your finger, bring it to the transfer flask and deliver the reagent to the desired mark. Touch the pipette to the side of the container to dislodge the drip at the end of the pipette. NaN If a pipette is drained to the tip, To-deliver (T.D.) pipettes and volumetric pipettes should not be blown out. To-contain (T.C.) pipettes should be "blown out". NaN Note: If pipette is wet with a different solution before use, obtain a fresh one or "condition" the pipette with two rinses of the reagent. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/States_of_Matter/Properties_of_Liquids/Cohesive_and_Adhesive_Forces | Cohesive and adhesive forces are associated with bulk (or macroscopic) properties, and hence the terms are not applicable to the discussion of atomic and molecular properties . When a liquid comes into contact with a surface (such as the walls of a graduated cylinder or a tabletop), both cohesive and adhesive forces will act on it. These forces govern the shape which the liquid takes on. Due to the effects of adhesive forces, liquid on a surface can spread out to form a thin, relatively uniform film over the surface, a process known as wetting . Alternatively, in the presence of strong cohesive forces, the liquid can divide into a number of small, roughly spherical beads that stand on the surface, maintaining minimal contact with the surface. Adhesive and Cohesive Forces The term "cohesive forces" is a generic term for the collective intermolecular forces (e.g., hydrogen bonding and van der Waals forces ) responsible for the bulk property of liquids resisting separation. Specifically, these attractive forces exist between molecules of the same substance. For instance, rain falls in droplets, rather than a fine mist, because water has strong cohesion, which pulls its molecules tightly together, forming droplets. This force tends to unite molecules of a liquid, gathering them into relatively large clusters due to the molecules' dislike for its surroundings. Similarly, the term "adhesive forces" refers to the attractive forces between unlike substance, such as mechanical forces (sticking together) and electrostatic forces (attraction due to opposing charges). In the case of a liquid wetting agent, adhesion causes the liquid to cling to the surface on which it rests. When water is poured on clean glass, it tends to spread, forming a thin, uniform film over the glasses surface. This is because the adhesive forces between water and glass are strong enough to pull the water molecules out of their spherical formation and hold them against the surface of the glass, thus avoiding the repulsion between like molecules. Macroscopic Effects of Cohesive and Adhesive Forces When a liquid is placed on a smooth surface, the relative strengths of the cohesive and adhesive forces acting on that liquid determine the shape it will take (and whether or not it will wet the surface). If the adhesive forces between a liquid and a surface are stronger, they will pull the liquid down, causing it to wet the surface. However, if the cohesive forces among the liquid itself are stronger, they will resist such adhesion and cause the liquid to retain a spherical shape and bead the surface. Case I: The Meniscus The meniscus is the curvature of a liquid's surface within a container, such as a graduated cylinder. However, before we explain why some liquid have a concave up meniscus while others share a concave down meniscus, we have to understand the adhesive forces at work of surface tension . Water, for example, is a polar molecule that consists of a partial positive charge on the hydrogens and a partial negative charge on the oxygen. Thus, within liquid water, each molecule's partial positive charge is attracted to its neighbor's partial negative charge. This is the origin of the cohesive forces within the water. Water molecules buried inside the liquid is then being pulled and pushed evenly in every direction, producing no net pull. Meanwhile, the molecules on the surface of the liquid, lacking pulling forces in the upward direction, thus encompass a net downward pull. How does this cohesive force create both a concave up and concave down surface then? The answer is in its relationship to the adhesive forces between the water molecules and the container's surface. When the cohesive force of the liquid is stronger than the adhesive force of the liquid to the wall, the liquid concaves down in order to reduce contact with the surface of the wall. When the adhesive force of the liquid to the wall is stronger than the cohesive force of the liquid, the liquid is more attracted to the wall than its neighbors, causing the upward concavity. Case II: Tears of Wine In agitated glasses of wine, droplets of wine seemingly "float" above the meniscus of the liquid and form "tears." This age-old phenomenon is the result of surface tension and cohesive and adhesive forces. Alcohol is more volatile than water. As a result, "evaporation of alcohol produces a surface tension gradient driving a thin film up along the side of a wine glass" (Adamson). This process is called the "solutal Marangoni effect." 2 Due to adhesive forces, some waters clings to the walls of the glass. The "tears" form from the cohesive forces within the water holding it together. It is important to note that the surface tension gradient is "the driving force for the motion of the liquid" (Gugliotti), but the actual formation of the tears is a result of cohesive and adhesive forces. Problems Name two examples where the cohesive force dominates over the adhesive force and vice versa. When in a glass graduated cylinder, water presents an upwardly concave meniscus. However, when water is filled to the tip of the cylinder, the water level could maintain higher than the wall of the cylinder without pouring out resembling a concave down meniscus. Use the principles of cohesive and adhesive forces to explain this situation. Explain why a water strider can glide on the water with the knowledge of cohesion in water. Propose different types of forces that adhesive forces can build on. Answers When cohesive force is stronger than the adhesive force: concave up meniscus, water forms droplets on the surface. When adhesive force is stronger than the cohesive force: concave down meniscus, the surfaces are covered by the wetting agent, the last drops of liquid in the bottle always refuse to come out. Since water forms a concave up meniscus, the adhesion of the molecules to the glass is stronger than the cohesion among the molecules. However, in the absence of the adhesive force (when water reaches the tip of the glass), the cohesive force remains present. Thus cohesive force alone proves that it can still hold itself in place without pouring out of the cylinder. This example emphasizes the importance of that cohesive force and adhesive forces do not simply cancel each other out, yet it is the difference between the two that determines the characteristic of the liquid. This problem addresses once again the concept of surface tension. Because the cohesion of the water is built on the weak intermolecular forces of the water, when a water strider stepson to the surface, an extra energy will be necessary to overcome to break those bonds to increase surface area. Moreover, since the gravitational pull down on the water strider cannot overcome the activation energy to break these intermolecular forces, the water strider is able to glide freely on the water. Complementary shape, chemical bonds form, weak intermolecular forces such as H-bonding or Van der Waals forces. References Petrucci, et al. General Chemistry: Principles & Modern Applications: AIE (Hardcover). Upper Saddle River: Pearson/Prentice Hall, 2007. Gugliotti, Marcos. "Tears of Wine." Journal of Chemical Education 81.1 (2004): 67-68. Web. 9 Mar. 2010. Adamson, A. W.; Gast, A. P. Physical Chemistry of Surfaces , 6th ed.; John Wiley and Sons: New York, 1997; p 371. |
Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/10%3A_Organohalides/10.07%3A_Organometallic_Coupling_Reactions | Objectives After completing this section, you should be able to write an equation for the formation of an alkyllithium from an alkyl halide. write an equation for the formation of a lithium dialkylcopper (Gilman) reagent from an alkyllithium and copper(I) iodide. write an equation for the coupling of a lithium dialkylcopper reagent with an alkyl halide (i.e., a Corey-House synthesis). draw the structure of the product formed from a given Corey-House synthesis. identify the reagents needed to convert two given organohalides to a specified hydrocarbon through a Corey-House synthesis. Key Terms Make certain that you can define, and use in context, the key terms below. Corey-House synthesis pheromone Study Notes A pheromone is a chemical released by members of one species to cause specific behavioural or physiological changes in other members of the same species. Examples include sex pheromones, alarm pheromones and trail pheromones. The Corey-House synthesis provides us with a method of coupling together two alkyl groups through the formation of a new carbon-carbon bond. The product of such a reaction is an alkane, and this synthetic method gives us a route for the preparation of unsymmetrical alkanes. The method was developed during the late 1960s by E. J. Corey and Herbert House working independently at Harvard University and Massachusetts Institute of Technology, respectively. The overall synthetic route is shown on the next page. Note that R and R′ represent alkyl groups, which can be the same or different, and X represents a halogen (preferably bromine or iodine). In order to obtain a good yield of alkane, both R′X and RX should be primary alkyl halides. However, the experimental procedure can be modified so that this synthesis can be carried out using a wide range of alkyl, aryl, vinyl, benzyl and allyl halides. A detailed discussion of these modifications is beyond the scope of this course, but you should be aware of possible limitations in the use of the Corey-House synthesis. Note: In some textbooks, lithium diorganocoppers are referred to as lithium dialkylcoppers, or cuprates. Gilman Reagents Another important reaction exhibited by organometallic reagents is metal exchange. Organolithium reagents react with cuprous iodide to give a lithium dimethylcopper reagent, which is referred to as a Gilman reagent. Gilman reagents are a source of carbanion like nucleophiles similar to Grignard and Organo lithium reagents. However, the reactivity of organocuprate reagents is slightly different and this difference will be exploited in different situations. In the case of α, β unsaturated carbonyls organocuprate reagents allow for an 1,4 addition of an alkyl group. As we will see later Grignard and Organolithium reagents add alkyl groups 1,2 to α, β unsaturated carbonyls Organocuprate reagents are made from the reaction of organolithium reagents and \(CuI\) \[ 2RLi + CuI \rightarrow R_2CuLi + LiI\] This acts as a source of R: - \[2 Ch_3Li + CuI \rightarrow (CH_3)_2CuLi + LiI\] Example Coupling Reactions with Gilman Reagents The coupling of Giman reactions with organochlorides, organobromides, and organoiodides is useful in organic synthesis because it forms a carbon bond. During the reaction one of the alkyl groups fromt he Gilman reagent replaces the halogen for the organohalide. Examples Exercise Exercise \(\PageIndex{1}\) Starting with alkyl halides containing no more than four carbon atoms, how would you synthesize each of the following alkanes? 2,5-dimethylhexane 2-methylhexane Answer Notice that in ( a ), both the alkyl halides are primary. This fact should ensure a good yield of product. In ( b ) we have the choice of using 2-bromopropane and 1-bromobutane, or 1-bromo-2-methylpropane and 1-bromopropane. We chose the latter as it enables us to use two primary alkyl halides, and hence a simpler procedure |
Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria | Complex ions typically exist in complex equilibrium involving its central metal ion and the ligands. Chelation A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Complex-Ion Equilibria In general, chemical equilibrium is reached when the forward reaction rate is equal to the reverse reaction rate and can be described using an equilibrium constant, K. Complex ion equilibria are no exception to this and have their own unique equilibrium constant. This formation constant, Kf , describes the formation of a complex ion from its central ion and attached ligands. This constant may be caled a stability constant or association constant. Complex-Ion Equilibrian II Hard and Soft Acids and Bases The thermodynamic stability of a metal complex depends greatly on the properties of the ligand and the metal ion and on the type of bonding. Metal–ligand interaction is an example of a Lewis acid–base interaction. Lewis bases can be divided into two categories: hard bases contain small, relatively nonpolarizable donor atoms (such as N, O, and F), and soft bases contain larger, relatively polarizable donor atoms (such as P, S, and Cl). Stability of Metal Complexes and Chelation Ligands like chloride, water, and ammonia are said to be monodentate (one-toothed, from the Greek mono, meaning “one,” and the Latin dent-, meaning “tooth”): they are attached to the metal via only a single atom. Ligands can, however, be bidentate (two-toothed, from the Greek di, meaning “two”), tridentate (three-toothed, from the Greek tri, meaning “three”), or, in general, polydentate (many-toothed, from the Greek poly, meaning “many”), indicating that they are attached to the metal at two, th |
Courses/Purdue/Chem_26505%3A_Organic_Chemistry_I_(Lipton)/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/Grinnell_College/CHM_364%3A_Physical_Chemistry_2_(Grinnell_College)/01%3A_The_Dawn_of_the_Quantum_Theory/1.08%3A_The_Bohr_Theory_of_the_Hydrogen_Atom | Introduce the fundamentals behind the Bohr Atom and demonstrate it can predict the Rydberg's equation for the atomic spectrum of hydrogen Rutherford's Failed Planetary Atom Ernest Rutherford had proposed a model of atoms based on the \(\alpha\)-particle scattering experiments of Hans Geiger and Ernest Marsden. In these experiments helium nuclei (\(\alpha\)-particles) were shot at thin gold metal foils. Most of the particles were not scattered; they passed unchanged through the thin metal foil. Some of the few that were scattered were scattered in the backward direction; i.e. they recoiled. This backward scattering requires that the foil contain heavy particles. When an \(\alpha\)-particle hits one of these heavy particles it simply recoils backward, just like a ball thrown at a brick wall. Since most of the \(α\)-particles don’t get scattered, the heavy particles (the nuclei of the atoms) must occupy only a very small region of the total space of the atom. Most of the space must be empty or occupied by very low-mass particles. These low-mass particles are the electrons that surround the nucleus. There are some basic problems with the Rutherford model. The Coulomb force that exists between oppositely charge particles means that a positive nucleus and negative electrons should attract each other, and the atom should collapse. To prevent the collapse, the electron was postulated to be orbiting the positive nucleus. The Coulomb force (discussed below) is used to change the direction of the velocity, just as a string pulls a ball in a circular orbit around your head or the gravitational force holds the moon in orbit around the Earth. The origin for this hypothesis that suggests this perspective is plausible is the similarity of gravity and Coulombic interactions. The expression for the force of gravity between two masses ( Newton's Law of gravity ) is \[F_{gravity} \propto \dfrac{m_1m_2}{r^2}\label{1.8.1} \] with \(m_1\) and \(m_2\) representing the mass of object 1 and 2, respectively and \(r\) representing the distance between the objects centers The expression for the Coulomb force between two charged species is \[F_{Coulomb} \propto \dfrac{Q_1Q_2}{r^2}\label{1.8.2} \] with \(Q_1\) and \(Q_2\) representing the charge of object 1 and 2, respectively and \(r\) representing the distance between the objects centers. However, this analogy has a problem too. An electron going around in a circle is constantly being accelerated because its velocity vector is changing. A charged particle that is being accelerated emits radiation. This property is essentially how a radio transmitter works. A power supply drives electrons up and down a wire and thus transmits energy (electromagnetic radiation) that your radio receiver picks up. The radio then plays the music for you that is encoded in the waveform of the radiated energy. If the orbiting electron is generating radiation, it is losing energy. If an orbiting particle loses energy, the radius of the orbit decreases. To conserve angular momentum, the frequency of the orbiting electron increases. The frequency increases continuously as the electron collapses toward the nucleus. Since the frequency of the rotating electron and the frequency of the radiation that is emitted are the same, both change continuously to produce a continuous spectrum and not the observed discrete lines. Furthermore, if one calculates how long it takes for this collapse to occur, one finds that it takes about \(10^{‑11}\) seconds. This means that nothing in the world based on the structure of atoms could exist for longer than about \(10^{-11}\) seconds. Clearly something is terribly wrong with this classical picture, which means that something was missing at that time from the known laws of physics. A conservative force is dependent only on the position of the object. If a force is conservative, it is possible to assign a numerical value for the potential at any point. When an object moves from one location to another, the force changes the potential energy of the object by an amount that does not depend on the path taken. The potential can be constructed as simple derivatives for 1-D forces: \[F = -\dfrac{dV}{dx} \nonumber \] or as gradients in 3-D forces \[F = -\nabla V \nonumber \] where \(\nabla\) is the vector of partial derivatives \[\nabla = \left ( \dfrac{\partial}{\partial x}, \dfrac{\partial}{\partial y}, \dfrac{\partial}{\partial z} \right) \nonumber \] The most familiar conservative forces are gravity and Coulombic forces. The Coulomb force law (Equation \(\ref{1.8.2}\)) comes from the corresponding Coulomb potential (sometimes call electrostatic potential) \[V(r)=\dfrac{kQ_1 Q_2}{r} \label{1.8.5} \] and it can be easily verified that the Coulombic force from this interaction (\(F(r)\)) is \[F(r)=-\dfrac{dV}{dr} \label{1.8.6} \] As \(r\) is varied, the energy will change, so that we have an example of a potential energy curve \(V(r)\) (Figure \(\PageIndex{2; left}\)) . If \(Q_1\) and \(Q_2\) are the same sign, then the curve which is a purely repulsive potential , i.e., the energy increases monotonically as the charges are brought together and decreases monotonically as they are separated. From this, it is easy to see that like charges (charges of the same sign) repel each other. If the charges are of opposite sign, then the curve appears roughly Figure \(\PageIndex{2; right}\) and this is a purely attractive potential . Thus, the energy decreases as the charges are brought together, implying that opposite charges attract The Bohr Model It is observed that line spectra discussed in the previous sections show that hydrogen atoms absorb and emit light at only discrete wavelengths. This observation is connected to the discrete nature of the allowed energies of a quantum mechanical system. Quantum mechanics postulates that, in contrast to classical mechanics, the energy of a system can only take on certain discrete values. This leaves us with the question: How do we determine what these allowed discrete energy values are? After all, it seems that Planck's formula for the allowed energies came out of nowhere. The model we will describe here, due to Niels Bohr in 1913, is an early attempt to predict the allowed energies for single-electron atoms such as \(\ce{H}\), \(\ce{He^{+}}\), \(\ce{Li^{2+}}\), \(\ce{Be^{3+}}\), etc. Although Bohr's reasoning relies on classical concepts and hence, is not a correct explanation, the reasoning is interesting, and so we examine this model for its historical significance. Consider a nucleus with charge \(+Ze\) and one electron orbiting the nucleus. In this analysis, we will use another representation of the constant \(k\) in Coulomb's law (Equation \(\ref{1.8.5}\)), which is more commonly represented in the form: \[k=\dfrac{1}{4\pi \epsilon_0} \label{1.8.7} \] where \(\epsilon_0\) is known as the permittivity of free space with the numerical value \(\epsilon_0 = 8.8541878\times 10^{-12} \ C^2 J^{-1} m^{-1}\) . The total energy of the electron (the nucleus is assumed to be fixed in space at the origin) is the sum of kinetic and potential energies: \[E_{total}=\underset{\text{kinetic energy}}{\dfrac{p^2}{2m_e}} - \underset{\text{potential energy}}{\dfrac{Ze^2}{4\pi \epsilon_0 r}} \nonumber \] The force on the electron is \[\vec{F}=-\dfrac{Ze^2}{4\pi \epsilon_0 r^3}r \nonumber \] and its magnitude is \[F=|\vec{F}|=\dfrac{Ze^2}{4\pi \epsilon_0 r^3}|r|=\dfrac{Ze^2}{4\pi \epsilon_0 r^2} \nonumber \] since \(\vec{F}=m_e \vec{a}\) , the magnitude, it follows that \(|\vec{F}|=m_e |\vec{a}|\) . If we assume that the orbit is circular, then the acceleration is purely centripetal , so \[|a|=\dfrac{v^2}{r} \nonumber \] where \(v\) is the velocity of the electron. Equating force \(|F|\) to \(m_e |a|\) , we obtain \[\dfrac{Ze^2}{4\pi \epsilon_0 r^2}=m_e\dfrac{v^2}{r} \nonumber \] or \[\dfrac{Ze^2}{4\pi \epsilon_0}=m_e v^2 r \nonumber \] or \[\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=(m_e vr)^2 \label{1.8.14} \] The reason for writing the equation this way is that the quantity \(m_e vr\) is the classical orbital angular momentum of the electron. Bohr was familiar with Maxwell's theory of classical electromagnetism and knew that in a classical theory, the orbiting electron should radiate energy away and eventually collapse into the nucleus (Figure 1.8.1 ). He circumvented this problem by following Planck's idea underlying blackbody radiation and positing that the orbital angular momentum \(m_e vr\) of the electron could only take on specific values \[m_e vr=n\hbar\label{1.8.15} \] with \(n=1,2,3,...\). Note that the electron must be in motion, so \(n=0\) is not allowed. Substituting Equation \(\ref{1.8.15}\) into the Equation \(\ref{1.8.14}\), we find \[\dfrac{Ze^2 m_e r}{4\pi \epsilon_0}=n^2 (\hbar)^2\label{1.8.16} \] Equation \ref{1.8.16} implies that orbits could only have certain allowed radii \[\begin{align}r_n &= \dfrac{4\pi \epsilon_0 \hbar^2}{Ze^2 m_e}n^2 \\[4pt] &=\dfrac{a_0}{Z}n^2 \label{1.8.16B} \end{align} \] with \(n=1,2,3,...\). T he collection of constants has been defined to be \(a_0\) \[a_0=\dfrac{4\pi \epsilon_0 \hbar^2}{e^2 m_e} \nonumber \] a quantity that is known as th e Bohr radius . We can also calculate the allowed momenta since \(m_e vr=n\hbar\), and \(p=m_e v\). Thus, \[\begin{align*}p_n r_n &=n\hbar \\[4pt] p_n &=\dfrac{n\hbar}{r_n}\\[4pt] &=\dfrac{\hbar Z}{a_0 n} \\[4pt] &= \dfrac{Ze^2 m_e}{4\pi \epsilon_0 \hbar n}\end{align*} \] From \(p_n\) and \(r_n\) , we can calculate the allowed energies from \[E_n=\dfrac{p^2_n}{2m_e}-\dfrac{Ze^2}{4\pi \epsilon_0 r_n} \nonumber \] Substituting in the expressions for \(p_n\) and \(r_n\) and simplifying gives \[E_n=-\dfrac{Z^2 e^4 m_e}{32\pi^2 \epsilon_{0}^{2}\hbar^2}\dfrac{1}{n^2}=-\dfrac{e^4 m_e}{8 \epsilon_{0}^{2}h^2}\dfrac{Z^2}{n^2} \label{1.8.20} \] W e can redefine a new energy scale by defining the Rydberg as \[1 \ Ry = \dfrac{e^4 m_e}{8\epsilon_{0}^{2} h^2} =2.18\times 10^{-18} \ J. \nonumber \] and this simplifies the allowed energies predicted by the Bohr model (Equation \ref{1.8.20}) as \[E_n=-(2.18\times 10^{-18})\dfrac{Z^2}{n^2} \ J=-\dfrac{Z^2}{n^2} \ R_y \label{1.8.21} \] Hence, the energy of the electron in an atom also is quantized. Equation \(\ref{1.8.21}\) gives the energies of the electronic states of the hydrogen atom. It is very useful in analyzing spectra to represent these energies graphically in an energy-level diagram. An energy-level diagram has energy plotted on the vertical axis with a horizontal line drawn to locate each energy level (Figure 1.8.4 ) . These turn out to be the correct energy levels, apart from small corrections that cannot be accounted for in this pseudo-classical treatment. Despite the fact that the energies are essentially correct, the Bohr model masks the true quantum nature of the electron, which only emerges from a fully quantum mechanical analysis. Calculate a value for the Bohr radius using Equation \(\ref{1.8.16}\) to check that this equation is consistent with the value 52.9 pm. What would the radius be for \(n = 1\) in the \(\ce{Li^{2+}}\) ion? Answer Starting from Equation \ref{1.8.16} and solving for \(r\): \[ \begin{align*} \dfrac{Ze^2m_er}{4πϵ_0} &=n^2ℏ^2 \\ r &=\dfrac{4 n^2 \hbar^2 πϵ_0}{Z e^2 m_e} \end{align*} \nonumber \] with \(e\) is the fundamental charge: \(e=1.60217662 \times 10^{-19}C^2\) \(m_e\) is the mass of an electron: \(m_e= 9.10938356 \times 10^{-31}kg\) \(\epsilon_o\) is the permittivity of free space: \(\epsilon_o = 8.854 \times 10^{-12}C^2N^{-1}m^{-2}\) \(\hbar\) is the reduced planks constant: \(\hbar=1.0546 \times 10^{-34}m^2kg/s\) For the ground-state of the hydrogen atom: \(Z=1\) and \(n=1\). \[ \begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{e^2m_e} \\ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854 \times 10^{-12}C^2N^{-1}m^{-2}}{(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \\ &=5.29 \times 10^{-11}m = 52.9\, pm\end{align*} \nonumber \] For the ground-state of the lithium +2 ion: \(Z=3\) and \(n=1\) \[ \begin{align*} r &=\dfrac{4 \hbar^2 πϵ_0}{3 e^2m_e} \\ &= \dfrac{4 (1.0546 \times 10^{-34}m^2kg/s)^2 \times π \times 8.854\times10^{-12}C^2N^{-1}m^{-2}}{3(1.60217662 \times 10^{-19}C)^2(9.10938356 \times 10^{-31}kg)} \\ &=1.76 \times 10^{-11}m = 17.6 \,pm\end{align*} \nonumber \] As expected, the \(\ce{Li^{2+}}\) has a smaller radius than the \(\ce{H}\) atoms because of the increased nuclear charge. How do the radii of the hydrogen orbits vary with \(n\)? Prepare a graph showing \(r\) as a function of \(n\). States of hydrogen atoms with \(n = 200\) have been prepared (called Rydberg states). What is the diameter of the atoms in these states? Answer This is a straightforward application of Equation of \ref{1.8.16B}. The hydrogen atom has only certain allowable radii and these radii can be predicted from the equation that relates them with each \(n\). Note that the electron must be in motion so \(n = 0\) is not allowed. \(4 \pi \epsilon_{0}=1.113 \times 10^{-10} \mathrm{C}^{2} \mathrm{J}^{-1} \mathrm{m}^{-1}\) and \(\hbar=1.054 \times 10^{-34} \mathrm{J} \mathrm{s},\) also knowing \[\begin{aligned} e &=1.602 \times 10^{-19} \mathrm{C} \text { with } \\ m_{e} &=9.109 \times 10^{-31} \mathrm{kg} \end{aligned} \nonumber \] and \(Z\) is the nuclear charge, we use this equation directly. A simplification can be made by taking advantage of the fact that \[a_{0}=\frac{4 \pi \epsilon_{0} \hbar^{2}}{e^{2} m_{e}} \nonumber \] resulting in \[r_{n}=\frac{a_{0}}{Z} n^{2} \nonumber \] where \(a_{0}=5.292 \times 10^{-11} \mathrm{m}\) which is the Bohr Radius. Suppose we want to find the radius where \(n=200 . n^{2}=40000\) so plugging in directly we have \[ \begin{align*} r_{n} &=\frac{\left(5.292 \times 10^{-11}\right)}{(1)}(40000) \\[4pt] &=2.117 \times 10^{-6} m \end{align*} \nonumber \] for the radius of a hydrogen atom with an electron excited to the \(\mathrm{n}=200\) state. The diameter is then \(4.234 \times 10^{-6} \mathrm{m}\). The Wave Argument for Quantization The above discussion is based off of a classical picture of an orbiting electron with the quantization from the angular momentum (Equation \(\ref{1.8.15}\)) requirement lifted from Planck's quantization arguments. Hence, only allows certain trajectories are stable (with differing radii). However, as discussed previously, the electron will have a wavelike property also with a de Broglie wavelength \(\lambda\) \[\lambda = \dfrac{h}{p} \nonumber \] Hence, a larger momentum \(p\) implies a shorter wavelength. That means as \(n\) increases (Equation \(\ref{1.8.21}\)), the wavelength must also increase; this is a common feature in quantum mechanics and will be often observed. In the Bohr atom, the circular symmetry and the wave property of the electron requires that the electron waves have an integer number of wavelengths (Figure \(\PageIndex{1A}\)). If not, then the waves will overlap imperfectly and cancel out (i.e., the electron will cease to exist) as demonstrated in Figure \(\PageIndex{1B}\). A more detailed discussion of the effect of electron waves in atoms will be discuss in the following chapters. Derivation of the Rydberg Equation from Bohr Model Given a prediction of the allowed energies of a system, how could we go about verifying them? The general experimental technique known as spectroscopy permits us to probe the various differences between the allowed energies. Thus, if the prediction of the actual energies, themselves, is correct, we should also be able to predict these differences. Let us assume that we are able to place the electron in Bohr's hydrogen atom into an energy state \(E_n\) for \(n>1\), i.e. one of its so-called excited states . The electron will rapidly return to its lowest energy state, known as the ground state and, in doing so, emit light. The energy carried away by the light is determined by the condition that the total energy is conserved (Figure 1.8.6 ). Thus, if \(n_i\) is the integer that characterizes the initial (excited) state of the electron, and \(n_f\) is the final state (here we imagine that \(n_f =1\), but is applicable in cases that \(n_f <n_i\), i.e., emission) \[E_{nf}=E_{ni}-h\nu \nonumber \] or \[\nu=\dfrac{E_{ni}-E_{nf}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2} h^3}\left ( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right ) \label{1.8.23} \] We can now identify the Rydberg constant \(R_H\) with the ratio of constants on the right hand side of Equation \(\ref{1.8.23}\) \[ R_H = \dfrac {m_ee^4}{8 \epsilon ^2_0 h^3 } \nonumber \] Evaluating \(R_H\) from the fundamental constants in this formula gives a value within 0.5% of that obtained experimentally from the hydrogen atom spectrum. Thus, by observing the emitted light, we can determine the energy difference between the initial and final energy levels, which results in the emission spectra discussed in Sections 1.4 and 1.5. Different values of \(n_f\) determine which emission spectrum is observed, and the examples shown in the figure are named after the individuals who first observed them. The figure below shows some of the transitions possible for different \(n_f\) and \(n_i\) values discussed previously. If the atom absorbs light it ends up in an excited state as a result of the absorption. The absorption is only possible for light of certain frequencies, and again, conservation of energy determines what these frequencies are. If light is absorbed, then the final energy \(E_{nf}\) will be related to the initial energy \(E_{ni}\) with \(n_f >n_i\) by \[E_{nf}=E_{ni}+h\nu \nonumber \] or \[\nu=\dfrac{E_{nf}-E_{ni}}{h}=\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2}h^3}\left ( \dfrac{1}{n_{i}^{2}}-\dfrac{1}{n_{f}^{2}}\right ) \nonumber \] Calculate the energy of a photon that is produced when an electron in a hydrogen atom goes from an orbit with \(n = 4\) to an orbit with \(n = 1\). What happens to the energy of the photon as the initial value of \(n\) approaches infinity? Answer a: \[\begin{align*} E_{\text{nf}} &= E_{ni} - h\nu \\ E_{photon} = h\nu &= E_{nf} - E_{ni}\\ &= \frac{Z^2e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\\ &=\frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{1^2} - \frac{1}{4^2}\right)\\ &=2.18 \times 10^{-18}\left(1 - \frac{1}{16} \right)\\ &=2.04 \times 10^{-18} J \end{align*} \nonumber \] b: As \(n_i \rightarrow \infty\) \[\begin{align*} E_{photon} &= \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\\ \frac{1}{n_i^2} &\rightarrow 0\\ E_{photon} &\rightarrow \frac{e^4m_e}{8\epsilon_o^2h^2}\left(\frac{1}{n_f^2}\right)\\ \end{align*} \nonumber \] Bohr’s proposal explained the hydrogen atom spectrum, the origin of the Rydberg formula, and the value of the Rydberg constant. Specifically it demonstrated that the integers in the Rydberg formula are a manifestation of quantization. The energy, the angular momentum, and the radius of the orbiting electron all are quantized. This quantization also parallels the concept of stable orbits in the Bohr model. Only certain values of \(E\), \(M\), and \(r\) are possible, and therefore the electron cannot collapse onto the nucleus by continuously radiating energy because it can only have certain energies, and it cannot be in certain regions of space. The electron can only jump from one orbit (quantum state) to another. The quantization means that the orbits are stable, and the electron cannot spiral into the nucleus in spite of the attractive Coulomb force. Although Bohr’s ideas successfully explained the hydrogen spectrum, they failed when applied to the spectra of other atoms. In addition a profound question remained. Why is angular momentum quantized in units of \(\hbar\)? As we shall see, de Broglie had an answer to this question, and this answer led Schrödinger to a general postulate that produces the quantization of angular momentum as a consequence. This quantization is not quite as simple as proposed by Bohr, and we will see that it is not possible to determine the distance of the electron from the nucleus as precisely as Bohr thought. In fact, since the position of the electron in the hydrogen atom is not at all as well defined as a classical orbit (such as the moon orbiting the earth) it is called an orbital. An electron orbital represents or describes the position of the electron around the nucleus in terms of a mathematical function called a wavefunction that yields the probability of positions of the electron. |
Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/15%3A_Chemical_Equilibrium/15.08%3A_Le_Chateliers_Principle-_How_a_System_at_Equilibrium_Responds_to_Disturbances | Learning Objectives Describe the ways in which an equilibrium system can be stressed Predict the response of a stressed equilibrium using Le Chatelier’s principle As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient (\(Q\)) is equal to the equilibrium constant (\(K\)). We next address what happens when a system at equilibrium is disturbed so that \(Q\) is no longer equal to \(K\). If a system at equilibrium is subjected to a perturbance or stress (such as a change in concentration) the position of equilibrium changes. Since this stress affects the concentrations of the reactants and the products, the value of \(Q\) will no longer equal the value of \(K\). To re-establish equilibrium, the system will either shift toward the products (if \((Q \leq K)\) or the reactants (if \((Q \geq K)\) until \(Q\) returns to the same value as \(K\). This process is described by Le Chatelier's principle. Le Chatelier's principle When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in \(Q\); the reaction will shift to re-establish \(Q = K\). Predicting the Direction of a Reversible Reaction Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of \(Q\) and \(K\) for the system to predict the changes. A chemical system at equilibrium can be temporarily shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium. The stress on the system in Figure \(\PageIndex{1}\) is the reduction of the equilibrium concentration of SCN − (lowering the concentration of one of the reactants would cause \(Q\) to be larger than K ). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN) 2+ should decrease, increasing the concentration of SCN − part way back to its original concentration, and increasing the concentration of Fe 3 + above its initial equilibrium concentration. The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction: \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} \label{15.7.1a} \] \[K_c=\mathrm{50.0 \; at\; 400°C} \label{15.7.1b} \] The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with \(\mathrm{[H_2] = [I_2]} = 0.221\; M\) and \(\ce{[HI]} = 1.563 \;M\) is at equilibrium; for this mixture, \(Q_c = K_c = 50.0\). If \(\ce{H_2}\) is introduced into the system so quickly that its concentration doubles before it begins to react (new \(\ce{[H_2]} = 0.442\; M\)), the reaction will shift so that a new equilibrium is reached, at which \(\ce{[H_2]} = 0.374\; M\), \(\ce{[I_2]} = 0.153\; M\), and \(\ce{[HI]} = 1.692\; M\). This gives: \[\begin{align*} Q_c &=\mathrm{\dfrac{[HI]^2}{[H_2][I_2]}} \\[4pt] &=\dfrac{(1.692)^2}{(0.374)(0.153)} \\[4pt] &= 50.0 =K_c \label{15.7.2} \end{align*} \] We have stressed this system by introducing additional \(\ce{H_2}\). The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess \(\ce{H_2}\), reducing the amount of uncombined \(\ce{I_2}\), and forming additional \(\ce{HI}\). Le Chatelier’s Principle (Changing Concentrations): Le Chatelier’s Principle (Changing Concentrations)(opens in new window) [youtu.be] A Video Discussing Le Chatelier’s Principle (Changing Concentrations): Le Chatelier’s Principle (Changing Concentrations)(opens in new window) [youtu.be] (opens in new window) Effect of Change in Pressure on Equilibrium Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for K c ) or partial pressure (for K P ). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium. As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure. Consider what happens when we increase the pressure on a system in which \(\ce{NO}\), \(\ce{O_2}\), and \(\ce{NO_2}\) are at equilibrium: \[\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)} \label{15.7.3} \] The formation of additional amounts of \(\ce{NO2}\) decreases the total number of molecules in the system because each time two molecules of \(\ce{NO_2}\) form, a total of three molecules of \(\ce{NO}\) and \(\ce{O_2}\) are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of \(\ce{NO_2}\) into \(\ce{NO}\) and \(\ce{O_2}\), which tends to restore the pressure. Now consider this reaction: \[\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g)} \label{15.7.4} \] Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide. Le Chatelier’s Principle (Changes in Pressure or Volume): Le Chatelier’s Principle (Changes in Pressure or Volume)(opens in new window) [youtu.be] Effect of Change in Temperature on Equilibrium Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle. When hydrogen reacts with gaseous iodine, heat is evolved. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{15.7.5} \] Because this reaction is exothermic, we can write it with heat as a product. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{15.7.6} \] Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H 2 and I 2 and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide. When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H 2 and I 2 decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C. Temperature affects the equilibrium between \(\ce{NO_2}\) and \(\ce{N_2O_4}\) in this reaction \[\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{15.7.7} \] The positive ΔH value tells us that the reaction is endothermic and could be written \[\ce{heat}+\ce{N_2O4(g) \rightleftharpoons 2NO2(g)} \label{15.7.8} \] At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown \(\ce{NO_2}\) molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless \(\ce{N_2O_4}\) increases, and the concentration of brown \(\ce{NO_2}\) decreases, causing the brown color to fade. The overview of how different disturbances affect the reaction equilibrium properties is tabulated in Table \(\PageIndex{1}\). Disturbance Observed Change as Equilibrium is Restored Direction of Shift Effect on K reactant added added reactant is partially consumed toward products none product added added product is partially consumed toward reactants none decrease in volume/increase in gas pressure pressure decreases toward side with fewer moles of gas none increase in volume/decrease in gas pressure pressure increases toward side with more moles of gas none temperature increase heat is absorbed toward products for endothermic, toward reactants for exothermic changes temperature decrease heat is given off toward reactants for endothermic, toward products for exothermic changes Example \(\PageIndex{1}\) Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium. \(2HgO_{(s)} \rightleftharpoons 2Hg_{(l)} + \mathbf{O}_{2(g)}\): the amount of HgO is doubled. \(NH_4HS_{(s)} \rightleftharpoons \mathbf{NH}_{3(g)} + H_2S_{(g)}\): the concentration of \(H_2S\) is tripled. \( \textbf{n-butane}_{(g)} \rightleftharpoons isobutane_{(g)}\): the concentration of isobutane is halved. Given : equilibrium systems and changes Asked for : equilibrium constant expressions and effects of changes Strategy : Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made. Solution : Because \(HgO_{(s)}\) and \(Hg_{(l)}\) are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, \(K = [O_2]\). The equilibrium concentration of \(O_2\) is a constant and does not depend on the amount of \(HgO\) present. Hence adding more \(HgO\) will not affect the equilibrium concentration of \(O_2\), so no compensatory change is necessary. \(NH_4HS\) does not appear in the equilibrium constant expression because it is a solid. Thus \(K = [NH_3][H_2S]\), which means that the concentrations of the products are inversely proportional. If adding \(H_2S\) triples the \(H_2S\) concentration, for example, then the \(NH_3\) concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals \(K\). For this reaction, \(K = \frac{[isobutane]}{[\textit{n-butane}]}\), so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium. Exercise \(\PageIndex{1}\) Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium? \(\ce{HBr (g) + NaH (s) \rightleftharpoons NaBr (s)} + \mathbf{H_2(g)}\): the concentration of \(\ce{HBr}\) is decreased by a factor of 3. \(\ce{6Li (s)} + \mathbf{N_2(g)} \ce{ \rightleftharpoons 2Li3N(s)}\): the amount of \(\ce{Li}\) is tripled. \(\mathbf{SO_2(g)} + \ce{ Cl2(g) \rightleftharpoons SO2Cl2(l)}\): the concentration of \(\ce{Cl2}\) is doubled. Answer a \(K = \dfrac{[H_2]}{[HBr]}\); \([H_2]\) must decrease by about a factor of 3. Answer b \(K = \dfrac{1}{[N_2]}\); solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary. Answer c \(K = \dfrac{1}{[SO_2][Cl_2]}\); \([SO_2]\) must decrease by about half. Le Chatelier’s Principle (Changes in Temperature): Le Chatelier’s Principle (Changes in Temperature)(opens in new window) [youtu.be] Catalysts Do Not Affect Equilibrium As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations. The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation \[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{15.7.9} \] A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry. Fritz Haber Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements (Equation \ref{15.7.9}). The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (\(\ce{N_2}\)) is nutritionally unavailable to a majority of plants due the tremendous stability of the nitrogen-nitrogen triple bond. Therefore, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation). Legumes achieve this conversion at ambient temperature by exploiting bacteria equipped with suitable enzymes. In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.” 1 Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science. Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995. Summary Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side. Footnotes 1 Herrlich, P. “The Responsibility of the Scientist: What Can History Teach Us About How Scientists Should Handle Research That Has the Potential to Create Harm?” EMBO Reports 14 (2013): 759–764. Glossary Le Chatelier's principle when a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance position of equilibrium concentrations or partial pressures of components of a reaction at equilibrium (commonly used to describe conditions before a disturbance) stress change to a reaction's conditions that may cause a shift in the equilibrium |
Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Theoretical_and_Actual_Yields | Key Terms (Excess reagent, limiting reagent) Theoretical and actual yields Percentage or actual yield Learning Objectives Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction and explain why. Calculate theoretical yields of products formed in reactions that involve limiting reagents. Evaluate percentage or actual yields from known amounts of reactants Theoretical and Actual Yields Reactants not completely used up are called excess reagents, and the reactant that completely reacts is called the limiting reagent. This concept has been illustrated for the reaction: \[\mathrm{2 Na + Cl_2 \rightarrow 2 NaCl} \nonumber \] Amounts of products calculated from the complete reaction of the limiting reagent are called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield. \(\mathrm{percent\: yield = \dfrac{actual\: yield}{theoretical\: yield}\times100}\) Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction. Example \(\PageIndex{1}\) Methyl alcohol can be produced in a high-pressure reaction \(\mathrm{CO_{\large{(g)}} + 2 H_{2\large{(g)}} \rightarrow CH_3OH_{\large{(l)}}}\) If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess amount of \(\ce{CO}\), estimate the theoretical and the percentage yield? Solution To calculate the theoretical yield, consider the reaction \(\begin{alignat}{2} \ce{&CO_{\large{(g)}} +\, &&2 H_{2\large{(g)}} \rightarrow \, &&CH_3OH_{\large{(l)}}}\\ &\:28.0 &&\:4.0 &&\:\:\:32.0 \hspace{45px}\ce{(stoichiometric\: masses\: in\: g,\: kg,\: or\: tons)} \end{alignat}\) \(\mathrm{1.2\: tons\: H_2 \times\dfrac{32.0\: CH_3OH}{4.0\: H_2}= 9.6\: tons\: CH_3OH}\) Thus, the theoretical yield from 1.2 metric tons (1.2x10 6 g) of hydrogen gas is 9.6 tons. The actual yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is \(\mathrm{\%\: yield =\dfrac{6.1\: tons}{9.6\: tons}\times 100 = 64 \%}\) Due to chemical equilibrium or the mass action law , the limiting reagent may not be completely consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the product will cause an even lower actual yield. Example \(\PageIndex{2}\) A solution containing silver ion, \(\ce{Ag+}\), has been treated with excess of chloride ions \(\ce{Cl-}\). When dried, 0.1234 g of \(\ce{AgCl}\) was recovered. Assuming the percentage yield to be 98.7%, how many grams of silver ions were present in the solution? HINT The reaction and relative masses of reagents and product are: \(\begin{alignat}{2}\ce{ Ag^+_{\large{(aq)}}} &+ \mathrm{Cl^-_{\large{(aq)}}} &&\rightarrow \ce{AgCl_{\large{(s)}}} \\ 107.868 &+ 35.453 &&= 143.321 \end{alignat}\) The calculation, \(\mathrm{0.1234\: g\: AgCl \times \dfrac{107.868\: g\: Ag^+}{143.321\: g\: AgCl}= 0.09287\: g\: Ag^+}\) shows that 0.1234 g dry \(\ce{AgCl}\) comes from 0.09287 g \(\ce{Ag+}\) ions. Since the actual yield is only 98.7%, the actual amount of \(\ce{Ag+}\) ions present is therefore \(\mathrm{\dfrac{0.09287\: g\: Ag^+}{0.987}= 0.09409\: g\: Ag^+}\) DISCUSSION One can also calculate the theoretical yield of \(\ce{AgCl}\) from the percentage yield of 98.7% to be \(\mathrm{\dfrac{0.1234\: g\: AgCl}{0.987}= 0.1250\: g\: AgCl}\) From 0.1250 g \(\ce{AgCl}\), the amount of \(\ce{Ag+}\) present is also 0.09409 g. Skill Developing Problems In an analytical experiment, you are asked to determine the amount of iodide ion \(\ce{I+}\) in 10.00 mL of a solution that does not contain any other ions that will form a precipitate with silver ions. You have learned that \(\ce{Ag+}\) ions precipitate all the iodide ions in a solution. In performing the experiment, shall you treat \(\ce{AgNO3}\) as the excess reagent or limiting reagent? Molar mass or atomic weight: \(\ce{Ag}\), 107.868; \(\ce{I}\), 126.904 (You should know where to find them). Hint: \(\ce{AgNO3}\) is the excess reagent Skill - Apply the concept of excess and limiting reagents for work. You can add \(\ce{AgNO3}\) slowly until the clear portion of the solution gives no precipitate when a drop of \(\ce{AgNO3}\) solution is added. This indicates that all the \(\ce{I-}\) ions are consumed. From the 10.00 mL iodide solution, you have added \(\ce{AgNO3}\) solution or solid. How do you know that you have added an excess amount of \(\ce{AgNO3}\) to precipitate the iodide ions? Hint: Test for excess or limiting reagent. Skill - Excess reagent can be tested for its presence, and limiting reagent can be tested for its absence. In an analytical experiment, 0.1234 g of \(\ce{AgI}\) was obtained from a 10.00-mL solution with excess silver nitrate. How much (in g) iodide ions are present? Hint: 0.05670 g \(\ce{Ag}\) and 0.06670 g \(\ce{I}\) Skill - Calculate the amount of limiting reagent from the amount of products. In an analytical experiment, 0.1234 g of \(\ce{AgI}\) was obtained from a 10.00-mL solution with excess silver nitrate. What is the iodide concentration (mol/L or M) in the solution? Hint: 0.05256 mol/L Skill - Calculate the concentration when the amount of solute is known. The concept of concentration will be covered in the unit dealing with solution, but you should be able to convert; see the following relationship. 0.1234 g \(\ce{AgI}\) = 0.0005256 mol = 0.5256 mili-mol \(\ce{AgI}\) or \(\ce{Ag}\) or \(\ce{I}\). |
Courses/Rutgers_University/General_Chemistry/Chapter_7._Chemical_Reactions_and_Chemical_Quantities/7.2%3A_Classifying_Chemical_Reactions | Learning Objectives Define three common types of chemical reactions (precipitation, acid-base, and oxidation-reduction) Classify chemical reactions as one of these three types given appropriate descriptions or chemical equations Identify common acids and bases Predict the solubility of common inorganic compounds by using solubility rules Compute the oxidation states for elements in compounds Humans interact with one another in various and complex ways, and we classify these interactions according to common patterns of behavior. When two humans exchange information, we say they are communicating. When they exchange blows with their fists or feet, we say they are fighting. Faced with a wide range of varied interactions between chemical substances, scientists have likewise found it convenient (or even necessary) to classify chemical interactions by identifying common patterns of reactivity. This module will provide an introduction to three of the most prevalent types of chemical reactions: precipitation, acid-base, and oxidation-reduction. Precipitation Reactions and Solubility Rules A precipitation reaction is one in which dissolved substances react to form one (or more) solid products. Many reactions of this type involve the exchange of ions between ionic compounds in aqueous solution and are sometimes referred to as double displacement , double replacement , or metathesis reactions. These reactions are common in nature and are responsible for the formation of coral reefs in ocean waters and kidney stones in animals. They are used widely in industry for production of a number of commodity and specialty chemicals. Precipitation reactions also play a central role in many chemical analysis techniques, including spot tests used to identify metal ions and gravimetric methods for determining the composition of matter (see the last module of this chapter). The extent to which a substance may be dissolved in water, or any solvent, is quantitatively expressed as its solubility , defined as the maximum concentration of a substance that can be achieved under specified conditions. Substances with relatively large solubilities are said to be soluble . A substance will precipitate when solution conditions are such that its concentration exceeds its solubility. Substances with relatively low solubilities are said to be insoluble , and these are the substances that readily precipitate from solution. More information on these important concepts is provided in the text chapter on solutions. For purposes of predicting the identities of solids formed by precipitation reactions, one may simply refer to patterns of solubility that have been observed for many ionic compounds (Table \(\PageIndex{1}\)). 0 1 Soluble compounds contain Exceptions to these solubility rules include group 1 metal cations (Li+, Na+, K+, Rb+, and Cs+) and ammonium ion \(\left(\ce{NH4+}\right)\) the halide ions (Cl−, Br−, and I−) the acetate \(\ce{(C2H3O2- )}\), bicarbonate \(\ce{(HCO3- )}\), nitrate \(\ce{(NO3- )}\), and chlorate \(\ce{(ClO3- )}\) ions the sulfate \(\ce{(SO4- )}\) ion halides of Ag+, \(\ce{Hg2^2+}\), and Pb2+ sulfates of Ag+, Ba2+, Ca2+, \(\ce{Hg2^2+}\), Pb2+, and Sr2+ Insoluble compounds contain Exceptions to these insolubility rules include carbonate \(\ce{(CO3^2- )}\), chromate \(\ce{(CrO4^2- )}\), phosphate \(\ce{(PO4^3- )}\), and sulfide (S2−) ions hydroxide ion (OH−) compounds of these anions with group 1 metal cations and ammonium ion hydroxides of group 1 metal cations and Ba2+ A vivid example of precipitation is observed when solutions of potassium iodide and lead nitrate are mixed, resulting in the formation of solid lead iodide: \[\ce{2KI}(aq)+\ce{Pb(NO3)2}(aq)\rightarrow \ce{PbI2}(s)+\ce{2KNO3}(aq) \nonumber \] This observation is consistent with the solubility guidelines: The only insoluble compound among all those involved is lead iodide, one of the exceptions to the general solubility of iodide salts. The net ionic equation representing this reaction is: \[\ce{Pb^2+}(aq)+\ce{2I-}(aq)\rightarrow \ce{PbI2}(s) \nonumber \] Lead iodide is a bright yellow solid that was formerly used as an artist’s pigment known as iodine yellow (Figure \(\PageIndex{1}\)). The properties of pure PbI 2 crystals make them useful for fabrication of X-ray and gamma ray detectors. The solubility guidelines in Table \(\PageIndex{1}\) may be used to predict whether a precipitation reaction will occur when solutions of soluble ionic compounds are mixed together. One merely needs to identify all the ions present in the solution and then consider if possible cation/anion pairing could result in an insoluble compound. For example, mixing solutions of silver nitrate and sodium fluoride will yield a solution containing Ag + , \(\ce{NO3-}\), Na + , and F − ions. Aside from the two ionic compounds originally present in the solutions, AgNO 3 and NaF, two additional ionic compounds may be derived from this collection of ions: NaNO 3 and AgF. The solubility guidelines indicate all nitrate salts are soluble but that AgF is one of the exceptions to the general solubility of fluoride salts. A precipitation reaction, therefore, is predicted to occur, as described by the following equations: \[\ce{NaF}(aq)+\ce{AgNO3}(aq)\rightarrow \ce{AgF}(s)+\ce{NaNO3}(aq)\hspace{20px}\ce{(molecular)} \nonumber \] \[\ce{Ag+}(aq)+\ce{F-}(aq)\rightarrow \ce{AgF}(s)\hspace{20px}\ce{(net\: ionic)} \nonumber \] Example \(\PageIndex{1}\): P redicting Precipitation Reactions Predict the result of mixing reasonably concentrated solutions of the following ionic compounds. If precipitation is expected, write a balanced net ionic equation for the reaction. potassium sulfate and barium nitrate lithium chloride and silver acetate lead nitrate and ammonium carbonate Solution (a) The two possible products for this combination are KNO 3 and BaSO 4 . The solubility guidelines indicate BaSO 4 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Ba^2+}(aq)+\ce{SO4^2-}(aq)\rightarrow \ce{BaSO4}(s) \nonumber \] (b) The two possible products for this combination are LiC 2 H 3 O 2 and AgCl. The solubility guidelines indicate AgCl is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Ag+}(aq)+\ce{Cl-}(aq)\rightarrow \ce{AgCl}(s) \nonumber \] (c) The two possible products for this combination are PbCO 3 and NH 4 NO 3 . The solubility guidelines indicate PbCO 3 is insoluble, and so a precipitation reaction is expected. The net ionic equation for this reaction, derived in the manner detailed in the previous module, is \[\ce{Pb^2+}(aq)+\ce{CO3^2-}(aq)\rightarrow \ce{PbCO3}(s) \nonumber \] Exercise \(\PageIndex{1}\) Which solution could be used to precipitate the barium ion, Ba 2 + , in a water sample: sodium chloride, sodium hydroxide, or sodium sulfate? What is the formula for the expected precipitate? Answer sodium sulfate, BaSO 4 Acid-Base Reactions An acid-base reaction is one in which a hydrogen ion, H + , is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion, and a full chapter is devoted to this topic later in the text. For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an acid is a substance that will dissolve in water to yield hydronium ions, H 3 O + . As an example, consider the equation shown here: \[\ce{HCl}(aq)+\ce{H2O}(aq)\rightarrow \ce{Cl-}(aq)+\ce{H3O+}(aq) \nonumber \] The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H 3 O + ions are produced by a chemical reaction in which H + ions are transferred from HCl molecules to H 2 O molecules (Figure \(\PageIndex{2}\)). The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called strong acids , and HCl is one among just a handful of common acid compounds that are classified as strong (Table \(\PageIndex{1}\)). A far greater number of compounds behave as weak acids and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions. Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{CH3CO2-}(aq)+\ce{H3O+}(aq) \nonumber \] When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, \(\ce{CH3CO2-}\) (Figure \(\PageIndex{3}\)). (The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.) Compound Formula Name in Aqueous Solution HBr hydrobromic acid HCl hydrochloric acid HI hydroiodic acid HNO3 nitric acid HClO4 perchloric acid H2SO4 sulfuric acid A base is a substance that will dissolve in water to yield hydroxide ions, OH − . The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion—for example, NaOH and Ca(OH) 2 . When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH) 2 dissolve in water and dissociate completely to produce cations (K + and Ba 2 + , respectively) and hydroxide ions, OH − . These bases, along with other hydroxides that completely dissociate in water, are considered strong bases . Consider as an example the dissolution of lye (sodium hydroxide) in water: \[\ce{NaOH}(s)\rightarrow \ce{Na+}(aq)+\ce{OH-}(aq) \nonumber \] This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na + and OH − ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases. Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as weak bases . These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (Figure \(\PageIndex{4}\)). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here: \[\ce{NH3}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{NH4+}(aq)+\ce{OH-}(aq) \nonumber \] This is, by definition, an acid-base reaction, in this case involving the transfer of H + ions from water molecules to ammonia molecules. Under typical conditions, only about 1% of the dissolved ammonia is present as \(\ce{NH4+}\) ions. The chemical reactions described in which acids and bases dissolved in water produce hydronium and hydroxide ions, respectively, are, by definition, acid-base reactions. In these reactions, water serves as both a solvent and a reactant. A neutralization reaction is a specific type of acid-base reaction in which the reactants are an acid and a base, the products are often a salt and water, and neither reactant is the water itself: \[\mathrm{acid+base\rightarrow salt+water} \nonumber \] To illustrate a neutralization reaction, consider what happens when a typical antacid such as milk of magnesia (an aqueous suspension of solid Mg(OH) 2 ) is ingested to ease symptoms associated with excess stomach acid (HCl): \[\ce{Mg(OH)2}(s)+\ce{2HCl}(aq)\rightarrow \ce{MgCl2}(aq)+\ce{2H2O}(l). \nonumber \] Note that in addition to water, this reaction produces a salt, magnesium chloride. Example \(\PageIndex{2}\): Writing Equations for Acid-Base Reactions Write balanced chemical equations for the acid-base reactions described here: the weak acid hydrogen hypochlorite reacts with water a solution of barium hydroxide is neutralized with a solution of nitric acid Solution (a) The two reactants are provided, HOCl and H 2 O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H + from HOCl to H 2 O to generate hydronium ions, H 3 O + and hypochlorite ions, OCl − . \[\ce{HOCl}(aq)+\ce{H2O}(l)\rightleftharpoons \ce{OCl-}(aq)+\ce{H3O+}(aq) \nonumber \] A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely. (b) The two reactants are provided, Ba(OH) 2 and HNO 3 . Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba 2 + ) and the anion generated when the acid transfers its hydrogen ion \(\ce{(NO3- )}\). \[\ce{Ba(OH)2}(aq)+\ce{2HNO3}(aq)\rightarrow \ce{Ba(NO3)2}(aq)+\ce{2H2O}(l) \nonumber \] Exercise \(\PageIndex{21}\) Write the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. (Hint: Consider the ions produced when a strong acid is dissolved in water.) Answer \[\ce{H3O+}(aq)+\ce{OH-}(aq)\rightarrow \ce{2H2O}(l) \nonumber \] Explore the microscopic view of strong and weak acids and bases. Oxidation-Reduction Reactions Earth’s atmosphere contains about 20% molecular oxygen, O 2 , a chemically reactive gas that plays an essential role in the metabolism of aerobic organisms and in many environmental processes that shape the world. The term oxidation was originally used to describe chemical reactions involving O 2 , but its meaning has evolved to refer to a broad and important reaction class known as oxidation-reduction (redox) reactions . A few examples of such reactions will be used to develop a clear picture of this classification. Some redox reactions involve the transfer of electrons between reactant species to yield ionic products, such as the reaction between sodium and chlorine to yield sodium chloride: \[\ce{2Na}(s)+\ce{Cl2}(g)\rightarrow \ce{2NaCl}(s) \nonumber \] It is helpful to view the process with regard to each individual reactant, that is, to represent the fate of each reactant in the form of an equation called a half-reaction : \[ \begin{align*} \ce{2Na}(s) &\rightarrow \ce{2Na+}(s)+\ce{2e-} \\[4pt] \ce{Cl2}(g)+\ce{2e-} &\rightarrow \ce{2Cl-}(s) \end{align*} \nonumber \] These equations show that Na atoms lose electrons while Cl atoms (in the Cl 2 molecule) gain electrons , the “ s ” subscripts for the resulting ions signifying they are present in the form of a solid ionic compound. For redox reactions of this sort, the loss and gain of electrons define the complementary processes that occur: \(\begin{align} \textbf{oxidation}&=\textrm{loss of electrons}\\ \textbf{reduction}&=\textrm{gain of electrons} \end{align}\) In this reaction, then, sodium is oxidized and chlorine undergoes reduction . Viewed from a more active perspective, sodium functions as a reducing agent (reductant) , since it provides electrons to (or reduces) chlorine. Likewise, chlorine functions as an oxidizing agent (oxidant) , as it effectively removes electrons from (oxidizes) sodium. \(\begin{align} \textbf{reducing agent}&=\textrm{species that is oxidized}\\ \textbf{oxidizing agent}&=\textrm{species that is reduced} \end{align}\) Some redox processes, however, do not involve the transfer of electrons. Consider, for example, a reaction similar to the one yielding \(\ce{NaCl}\): \[\ce{H2}(g)+\ce{Cl2}(g)\rightarrow \ce{2HCl}(g) \nonumber \] The product of this reaction is a covalent compound, so transfer of electrons in the explicit sense is not involved. To clarify the similarity of this reaction to the previous one and permit an unambiguous definition of redox reactions, a property called oxidation number has been defined. The oxidation number (or oxidation state ) of an element in a compound is the charge its atoms would possess if the compound was ionic . The following guidelines are used to assign oxidation numbers to each element in a molecule or ion. The oxidation number of an atom in an elemental substance is zero. The oxidation number of a monatomic ion is equal to the ion’s charge. Oxidation numbers for common nonmetals are usually assigned as follows: Hydrogen: +1 when combined with nonmetals, −1 when combined with metals Oxygen: −2 in most compounds, sometimes −1 (so-called peroxides, \(\ce{O2^2-}\)), very rarely \(-\dfrac{1}{2}\) (so-called superoxides, \(\ce{O2-}\)), positive values when combined with F (values vary) Halogens: −1 for F always, −1 for other halogens except when combined with oxygen or other halogens (positive oxidation numbers in these cases, varying values) The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the molecule or ion. Note: The proper convention for reporting charge is to write the number first, followed by the sign (e.g., 2+), while oxidation number is written with the reversed sequence, sign followed by number (e.g., +2). This convention aims to emphasize the distinction between these two related properties. Example \(\PageIndex{3}\): Assigning Oxidation Numbers Follow the guidelines in this section of the text to assign oxidation numbers to all the elements in the following species: H 2 S \(\ce{SO3^2-}\) Na 2 SO 4 Solution (a) According to guideline 1, the oxidation number for H is +1. Using this oxidation number and the compound’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur: \(\ce{charge\: on\: H2S}=0=(2\times +1)+(1\times x)\) \(x=0-(2\times +1)=-2\) (b) Guideline 3 suggests the oxidation number for oxygen is −2. Using this oxidation number and the ion’s formula, guideline 4 may then be used to calculate the oxidation number for sulfur: \(\ce{charge\: on\: SO3^2-}=-2=(3\times -2)+(1\times x)\) \(x=-2-(3\times -2)=+4\) (c) For ionic compounds, it’s convenient to assign oxidation numbers for the cation and anion separately. According to guideline 2, the oxidation number for sodium is +1. Assuming the usual oxidation number for oxygen (−2 per guideline 3), the oxidation number for sulfur is calculated as directed by guideline 4: \(\ce{charge\: on\: SO4^2-}=-2=(4\times -2)+(1\times x)\) \(x=-2-(4\times -2)=+6\) Exercise \(\PageIndex{3}\) Assign oxidation states to the elements whose atoms are underlined in each of the following compounds or ions: K N O 3 Al H 3 \(\mathrm{\underline{N}H_4^+}\) \(\mathrm{\sideset{ }{_{\large{4}}^{-}}{H_2\underline{P}O}}\) Answer a N, +5 Answer b Al, +3 Answer c N, −3 Answer d P, +5 Using the oxidation number concept, an all-inclusive definition of redox reaction has been established. Oxidation-reduction (redox) reactions are those in which one or more elements involved undergo a change in oxidation number. While the vast majority of redox reactions involve changes in oxidation number for two or more elements, a few interesting exceptions to this rule do exist as shown below\). Definitions for the complementary processes of this reaction class are correspondingly revised as shown here: \[\begin{align} \textbf{oxidation}&=\textrm{increase in oxidation number}\\ \textbf{reduction}&=\textrm{decrease in oxidation number} \end{align} \nonumber \] Returning to the reactions used to introduce this topic, they may now both be identified as redox processes. In the reaction between sodium and chlorine to yield sodium chloride, sodium is oxidized (its oxidation number increases from 0 in Na to +1 in NaCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in NaCl). In the reaction between molecular hydrogen and chlorine, hydrogen is oxidized (its oxidation number increases from 0 in H 2 to +1 in HCl) and chlorine is reduced (its oxidation number decreases from 0 in Cl 2 to −1 in HCl). Several subclasses of redox reactions are recognized, including combustion reactions in which the reductant (also called a fuel ) and oxidant (often, but not necessarily, molecular oxygen) react vigorously and produce significant amounts of heat, and often light, in the form of a flame. Solid rocket-fuel reactions such as the one depicted below are combustion processes. A typical propellant reaction in which solid aluminum is oxidized by ammonium perchlorate is represented by this equation: \[\ce{10Al}(s)+\ce{6NH4ClO4}(s)\rightarrow \ce{4Al2O3}(s)+\ce{2AlCl3}(s)+\ce{12H2O}(g)+\ce{3N2}(g) \nonumber \] Watch a brief video showing the test firing of a small-scale, prototype, hybrid rocket engine planned for use in the new Space Launch System being developed by NASA . The first engines firing at 3 s (green flame) use a liquid fuel/oxidant mixture, and the second, more powerful engines firing at 4 s (yellow flame) use a solid mixture. Single-displacement (replacement) reactions are redox reactions in which an ion in solution is displaced (or replaced) via the oxidation of a metallic element. One common example of this type of reaction is the acid oxidation of certain metals: \[\ce{Zn}(s)+\ce{2HCl}(aq)\rightarrow \ce{ZnCl2}(aq)+\ce{H2}(g) \nonumber \] Metallic elements may also be oxidized by solutions of other metal salts; for example: \[\ce{Cu}(s)+\ce{2AgNO3}(aq)\rightarrow \ce{Cu(NO3)2}(aq)+\ce{2Ag}(s) \nonumber \] This reaction may be observed by placing copper wire in a solution containing a dissolved silver salt. Silver ions in solution are reduced to elemental silver at the surface of the copper wire, and the resulting Cu 2 + ions dissolve in the solution to yield a characteristic blue color (Figure \(\PageIndex{4}\)). Example \(\PageIndex{4}\): Describing Redox Reactions Identify which equations represent redox reactions, providing a name for the reaction if appropriate. For those reactions identified as redox, name the oxidant and reductant. \(\ce{ZnCO3}(s)\rightarrow \ce{ZnO}(s)+\ce{CO2}(g)\) \(\ce{2Ga}(l)+\ce{3Br2}(l)\rightarrow \ce{2GaBr3}(s)\) \(\ce{2H2O2}(aq)\rightarrow \ce{2H2O}(l)+\ce{O2}(g)\) \(\ce{BaCl2}(aq)+\ce{K2SO4}(aq)\rightarrow \ce{BaSO4}(s)+\ce{2KCl}(aq)\) \(\ce{C2H4}(g)+\ce{3O2}(g)\rightarrow \ce{2CO2}(g)+\ce{2H2O}(l)\) S olution Redox reactions are identified per definition if one or more elements undergo a change in oxidation number. This is not a redox reaction, since oxidation numbers remain unchanged for all elements. This is a redox reaction. Gallium is oxidized, its oxidation number increasing from 0 in Ga( l ) to +3 in GaBr 3 ( s ). The reducing agent is Ga( l ). Bromine is reduced, its oxidation number decreasing from 0 in Br 2 ( l ) to −1 in GaBr 3 ( s ). The oxidizing agent is Br 2 ( l ). This is a redox reaction. It is a particularly interesting process, as it involves the same element, oxygen, undergoing both oxidation and reduction (a so-called disproportionation reaction) . Oxygen is oxidized, its oxidation number increasing from −1 in H 2 O 2 ( aq ) to 0 in O 2 ( g ). Oxygen is also reduced, its oxidation number decreasing from −1 in H 2 O 2 ( aq ) to −2 in H 2 O( l ). For disproportionation reactions, the same substance functions as an oxidant and a reductant. This is not a redox reaction, since oxidation numbers remain unchanged for all elements. This is a redox reaction (combustion). Carbon is oxidized, its oxidation number increasing from −2 in C 2 H 4 ( g ) to +4 in CO 2 ( g ). The reducing agent (fuel) is C 2 H 4 ( g ). Oxygen is reduced, its oxidation number decreasing from 0 in O 2 ( g ) to −2 in H 2 O( l ). The oxidizing agent is O 2 ( g ). Exercise \(\PageIndex{4}\) This equation describes the production of tin(II) chloride: \[\ce{Sn}(s)+\ce{2HCl}(g)\rightarrow \ce{SnCl2}(s)+\ce{H2}(g) \nonumber \] Is this a redox reaction? If so, provide a more specific name for the reaction if appropriate, and identify the oxidant and reductant. Answer Yes, a single-replacement reaction. Sn( s ) is the reductant, HCl( g ) is the oxidant. Balancing Redox Reactions via the Half-Reaction Method Redox reactions that take place in aqueous media often involve water, hydronium ions, and hydroxide ions as reactants or products. Although these species are not oxidized or reduced, they do participate in chemical change in other ways (e.g., by providing the elements required to form oxyanions). Equations representing these reactions are sometimes very difficult to balance by inspection, so systematic approaches have been developed to assist in the process. One very useful approach is to use the method of half-reactions, which involves the following steps: Write the two half-reactions representing the redox process. Balance all elements except oxygen and hydrogen. Balance oxygen atoms by adding H 2 O molecules. Balance hydrogen atoms by adding H + ions. Balance charge 1 by adding electrons. If necessary, multiply each half-reaction’s coefficients by the smallest possible integers to yield equal numbers of electrons in each. Add the balanced half-reactions together and simplify by removing species that appear on both sides of the equation. For reactions occurring in basic media (excess hydroxide ions), carry out these additional steps: Add OH − ions to both sides of the equation in numbers equal to the number of H + ions. On the side of the equation containing both H + and OH − ions, combine these ions to yield water molecules. Simplify the equation by removing any redundant water molecules. Finally, check to see that both the number of atoms and the total charges 2 are balanced. Example \(\PageIndex{5}\): Balancing Redox Reactions in Acidic Solution Write a balanced equation for the reaction between dichromate ion and iron(II) to yield iron(III) and chromium(III) in acidic solution. \[\ce{Cr2O7^2- + Fe^2+ \rightarrow Cr^3+ + Fe^3+} \nonumber \] S olution Write the two half-reactions . Each half-reaction will contain one reactant and one product with one element in common. \(\ce{Fe^2+ \rightarrow Fe^3+}\) \(\ce{Cr2O7^2- \rightarrow Cr^3+}\) Balance all elements except oxygen and hydrogen . The iron half-reaction is already balanced, but the chromium half-reaction shows two Cr atoms on the left and one Cr atom on the right. Changing the coefficient on the right side of the equation to 2 achieves balance with regard to Cr atoms. \(\ce{Fe^2+ \rightarrow Fe^3+}\) \(\ce{Cr2O7^2- \rightarrow 2Cr^3+}\) Balance oxygen atoms by adding H 2 O molecules . The iron half-reaction does not contain O atoms. The chromium half-reaction shows seven O atoms on the left and none on the right, so seven water molecules are added to the right side. \(\ce{Fe^2+ \rightarrow Fe^3+}\) \(\ce{Cr2O7^2- \rightarrow 2Cr^3+ + 7H2O}\) Balance hydrogen atoms by adding H + ions . The iron half-reaction does not contain H atoms. The chromium half-reaction shows 14 H atoms on the right and none on the left, so 14 hydrogen ions are added to the left side. \(\ce{Fe^2+ \rightarrow Fe^3+}\) \(\ce{Cr2O7^2- + 14H+ \rightarrow 2Cr^3+ + 7H2O}\) Balance charge by adding electrons . The iron half-reaction shows a total charge of 2+ on the left side (1 Fe 2 + ion) and 3+ on the right side (1 Fe 3 + ion). Adding one electron to the right side bring that side’s total charge to (3+) + (1−) = 2+, and charge balance is achieved. The chromium half-reaction shows a total charge of (1 × 2−) + (14 × 1+) = 12+ on the left side (\(\ce{1 Cr2O7^2-}\) ion and 14 H + ions). The total charge on the right side is (2 × 3+) = 6 + (2 Cr 3 + ions). Adding six electrons to the left side will bring that side’s total charge to (12+ + 6−) = 6+, and charge balance is achieved. \(\ce{Fe^2+ \rightarrow Fe^3+ + e-}\) \(\ce{Cr2O7^2- + 14H+ + 6e- \rightarrow 2Cr^3+ + 7H2O}\) Multiply the two half-reactions so the number of electrons in one reaction equals the number of electrons in the other reaction . To be consistent with mass conservation, and the idea that redox reactions involve the transfer (not creation or destruction) of electrons, the iron half-reaction’s coefficient must be multiplied by 6. \(\ce{6Fe^2+ \rightarrow 6Fe^3+ + 6e-}\) \(\ce{Cr2O7^2- + 6e- + 14H+ \rightarrow 2Cr^3+ + 7H2O}\) Add the balanced half-reactions and cancel species that appear on both sides of the equation . \[\ce{6Fe^2+ + Cr2O7^2- + 6e- + 14H+ \rightarrow 6Fe^3+ + 6e- + 2Cr^3+ + 7H2O} \nonumber \] Only the six electrons are redundant species. Removing them from each side of the equation yields the simplified, balanced equation here: \[\ce{6Fe^2+ + Cr2O7^2- + 14H+ \rightarrow 6Fe^3+ + 2Cr^3+ + 7H2O} \nonumber \] A final check of atom and charge balance confirms the equation is balanced. Unnamed: 0 Reactants Products Fe 6 6 Cr 2 2 O 7 7 H 14 14 charge 24+ 24+ Exercise \(\PageIndex{5}\) In acidic solution, hydrogen peroxide reacts with Fe 2 + to produce Fe 3 + and H 2 O. Write a balanced equation for this reaction. Answer \[\ce{H2O2}(aq)+\ce{2H+}(aq)+\ce{2Fe^2+} \rightarrow \ce{2H2O}(l)+\ce{2Fe^3+} \nonumber \] Summary Chemical reactions are classified according to similar patterns of behavior. A large number of important reactions are included in three categories: precipitation, acid-base, and oxidation-reduction (redox). Precipitation reactions involve the formation of one or more insoluble products. Acid-base reactions involve the transfer of hydrogen ions between reactants. Redox reactions involve a change in oxidation number for one or more reactant elements. Writing balanced equations for some redox reactions that occur in aqueous solutions is simplified by using a systematic approach called the half-reaction method. Footnotes 1 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. 2 The requirement of “charge balance” is just a specific type of “mass balance” in which the species in question are electrons. An equation must represent equal numbers of electrons on the reactant and product sides, and so both atoms and charges must be balanced. Glossary acid substance that produces H 3 O + when dissolved in water acid-base reaction reaction involving the transfer of a hydrogen ion between reactant species base substance that produces OH − when dissolved in water combustion reaction vigorous redox reaction producing significant amounts of energy in the form of heat and, sometimes, light half-reaction an equation that shows whether each reactant loses or gains electrons in a reaction. insoluble of relatively low solubility; dissolving only to a slight extent neutralization reaction reaction between an acid and a base to produce salt and water oxidation process in which an element’s oxidation number is increased by loss of electrons oxidation-reduction reaction (also, redox reaction) reaction involving a change in oxidation number for one or more reactant elements oxidation number (also, oxidation state) the charge each atom of an element would have in a compound if the compound were ionic oxidizing agent (also, oxidant) substance that brings about the oxidation of another substance, and in the process becomes reduced precipitate insoluble product that forms from reaction of soluble reactants precipitation reaction reaction that produces one or more insoluble products; when reactants are ionic compounds, sometimes called double-displacement or metathesis reduction process in which an element’s oxidation number is decreased by gain of electrons reducing agent (also, reductant) substance that brings about the reduction of another substance, and in the process becomes oxidized salt ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide single-displacement reaction (also, replacement) redox reaction involving the oxidation of an elemental substance by an ionic species soluble of relatively high solubility; dissolving to a relatively large extent solubility the extent to which a substance may be dissolved in water, or any solvent strong acid acid that reacts completely when dissolved in water to yield hydronium ions strong base base that reacts completely when dissolved in water to yield hydroxide ions weak acid acid that reacts only to a slight extent when dissolved in water to yield hydronium ions weak base base that reacts only to a slight extent when dissolved in water to yield hydroxide ions |
Bookshelves/Environmental_Chemistry/Green_Chemistry_and_the_Ten_Commandments_of_Sustainability_(Manahan)/03%3A_The_Elements_-_Basic_Building_Blocks_of_Green_Chemicals/3.01%3A_Elements_Atoms_and_Atomic_Theory | Chemistry is the science of matter. The fundamental building blocks of matter are the atoms of the various elements, which are composed of subatomic particles , the positively charged proton (+), the negatively charged electron (-), and the electrically neutral neutron (n). It is the properties of these atoms that determine matter’s chemical behavior. More specifically, it is the arrangement and energy levels of electrons in atoms that direct how they interact with each other, thus dictating all chemical behavior. One of the most fundamental aspects of chemistry is that elemental behavior varies periodically with increasing atomic number. This has enabled placement of elements in an orderly arrangement with increasing atomic number known as the periodic table . The periodic behavior of elements’ chemical properties is due to the fact that, as atomic number increases, electrons are added incrementally to atoms and occupy so-called shells, each filled with a specific number of electrons. After each shell is filled, a new shell is started, thus beginning a new period (row) of the periodic table. This sounds complicated, and indeed may be so, occupying the full-time computational activities of banks of computers to explain the behavior of electrons in matter. However, this behavior can be viewed in simplified models and is most easily understood for the first 20 elements using dots to represent electrons, enabling construction of an abbreviated 20-element periodic table. Although simple, this table helps understand and explain most of the chemical phenomena discussed in this book. The chapter also emphasizes some of the green aspects of the first 20 elements and how they relate to sustainability. Included among these elements are the nitrogen, oxygen, carbon (contained in carbon dioxide), and hydrogen and oxygen (in water vapor) that make up most of the air in the “green” atmosphere; the hydrogen and oxygen in water, arguably the greenest compound of all; the sodium and chlorine in common table salt; the silicon, calcium, and oxygen that compose most mineral matter, including the soil that grows plants supplying food to most organisms; and the hydrogen, oxygen, carbon, nitrogen, phosphorus, and sulfur that are the predominant elements in all living material. Long Before Subatomic Particles Were Known, There Was Dalton’s Atomic Theory Atomic theory describes the atoms in relation to chemical behavior. With the sophisticated tools now available to chemists, the nature of atoms, largely based upon the subatomic particles of which they are composed, especially the negatively charged electrons, is well known. But long before these sophisticated tools were even dreamed about, more than two centuries ago in 1808, an English schoolteacher named John Dalton came up with the atomic theory that bears his name. To a large extent, this theory is the conceptual basis of modern chemistry. Key aspects of Dalton’s atomic theory are the following: The matter in each element is composed of extremely small particles called atoms. (Dalton regarded atoms as indivisible, unchanging bodies. We now know that they exchange and share electrons, which is the basis of chemical bonding.) Atoms of different elements have different chemical properties. (These differences may range from very slight, such as those between the noble gases neon and argon, to vastly different, such as those between highly metallic sodium and strongly nonmetallic chlorine.) Atoms cannot be created, destroyed, or changed to atoms of other elements. (In modern times, the provision is added that these things do not happen in ordinary chemical processes, since atoms can be changed to atoms of other elements by nuclear reactions, such as those that occur in nuclear reactors.) Chemical compounds are formed by the combination of atoms of different elements in definite, constant ratios that usually can be expressed as integers or simple fractions. Chemical reactions involve the separation and combination of atoms. (This phenomenon was surmised before anything was known about the nature of chemical bonds that are broken and formed as part of the process of chemical reactions.) Three Important Laws Dalton’s atomic theory explains the three important laws listed below. Evidence for these laws had been found prior to the publishing of Dalton’s atomic theory, and the atomic theory is largely based upon them. Law of Conservation of Mass : There is no detectable change in mass in an ordinary chemical reaction . (This law, which was first stated in 1798 by “the father of chemistry,” the Frenchman Antoine Lavoisier, follows from the fact that in ordinary chemical reactions no atoms are lost, gained, or changed; in chemical reactions, mass is conserved.) Law of Constant Composition: A specific chemical compound always contains the same elements in the same proportions by mass . Law of Multiple Proportions : When two elements combine to form two or more compounds, the masses of one combining with a fixed mass of the other are in ratios of small whole numbers . A common illustration of this law is provided by the simple hydrocarbon compounds of carbon and hydrogen, which include \(\ce{CH4}\), \(\ce{C2H2}\), \(\ce{C2H4}\), and \(\ce{C2H6}\). In these compounds the relative masses of \(\ce{C}\) and \(\ce{H}\) are in ratios of small whole numbers. The Nature of Atoms At this point it is useful to note several characteristics of atoms, which were introduced in Section 2.11. Atoms are extremely small and extremely light. Their individual masses are expressed by the minuscule atomic mass unit, u. The sizes of atoms are commonly expressed in picometers, where a picometer is 0.000 000 001 millimeters (a millimeter is the smallest division on the metric side of a ruler). Atoms may be regarded as spheres with diameters between 100 and 300 picometers. As noted at the beginning of this chapter, atoms are composed of three basic subatomic particles , the positively charged proton , the electrically neutral neutron , and the much lighter negatively charged electron . Each proton and neutron has a mass of essentially 1 atomic mass unit, whereas the mass of the electron is only about 1/2000 as much The protons and neutrons are located in the nucleus at the center of the atom and the electrons compose a “fuzzy cloud” of negative charge around the nucleus. Essentially all the mass of an atom is in the nucleus and virtually all the volume is in the cloud of electrons. Each atom of a specific element has the same number of protons in its nucleus. This is the atomic number of the element. Each element has a name and is represented by a chemical symbol consisting of one or two letters. Atoms of the same element that have different numbers of neutrons and, therefore, different masses, are called isotopes . Isotopes may be represented by symbols such as \(\ce{^{12}_6C}\) where the subscript is the atomic number and the superscript is the mass number , which is the sum of the numbers of protons and neutrons in an atom. The average mass of all the atoms of an element is the atomic mass . Atomic masses are expressed relative to the carbon \(\ce{^{12}_6C}\) the isotope, which contains 6 protons and 6 neutrons in its nucleus. The mass of this isotope is taken as exactly 12 u. Atomic masses normally are not integers, in part because atoms of most elements consist of two or more isotopes with different masses. Electrons in Atoms The behavior of electrons in the cloud of negative charge making up most of the volume of atoms, particularly their energy levels and orientations in space, are what determine chemical behavior. Arrangements of electrons are described by electron configuration . A detailed description of electron configuration is highly mathematical and sophisticated, but is represented in a very simplified fashion in this chapter. Because of their opposite charges, electrons are strongly attracted to positively charged nuclei, but they do not come to rest on it. The placement of electrons in atoms determines the configuration of the periodic table, a complete version of which is printed at the end of this chapter. Elements are listed across this table in periods such that elements located in the same vertical groups have generally similar chemical behavior. The derivation of the complete periodic table showing more than 100 elements is too complicated for this book. So, in the remainder of this chapter, the first 20 elements will be discussed in order and the placement of electrons in the atoms of these elements will illustrate how these elements can be placed in the periodic table. From this information a brief 20-element periodic table will be constructed that should be very useful in explaining chemical behavior. |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence | The aim of the book is to give a comprehensive account of the chemiluminescence subject suitable for students, research workers, professional analytical chemists and life scientists. The chemistry of each reagent that has been used in chemiluminescence is explained and the techniques for increasing the magnitude of the emitted signal are discussed. Further sections describe the different instrumental methods that have been used and examine the sort of work that has been carried out with them. The final sections discuss the applications of chemiluminescence in some of the major areas of analytical chemistry. 1: Introduction to Chemiluminescence 2: Chemiluminescence Reagents 3: Enhancement of Chemiluminescence 4: Instrumentation Thumbnail: Chemiluminescence after a reaction of hydrogen peroxide and luminol. This is an image from video youtu.be/8_82cNtZSQE. (CC BY-SA 4.0; Tavo Romann ). |
Courses/Rio_Hondo/Chemistry_110%3A_An_Introduction_to_General_Organic_and_Biological_Chemistry_(Garg)/06%3A_Chemical_Quantities_and_Reactions/6.11%3A_Combustion_Reactions | How do you cook the perfect marshmallow? Roasting marshmallows over an open fire is a favorite past-time for campers, outdoor cook-outs, and just gathering around a fire in the back yard. The trick is to get the marshmallow a nice golden brown without catching it on fire. Too often we are not successful and we see the marshmallow burning on the stick – a combustion reaction taking place right in front of us. Combustion Reactions A combustion reaction is a reaction in which a substance reacts with oxygen gas, releasing energy in the form of light and heat. Combustion reactions must involve \(\ce{O_2}\) as one reactant. The combustion of hydrogen gas produces water vapor: \[2 \ce{H_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{H_2O} \left( g \right)\nonumber \] Notice that this reaction also qualifies as a combination reaction. The Hindenberg was a hydrogen-filled airship that suffered an accident upon its attempted landing in New Jersey in 1937. The hydrogen immediately combusted in a huge fireball, destroying the airship and killing 36 people. The chemical reaction was a simple one: hydrogen combining with oxygen to produce water. Many combustion reactions occur with a hydrocarbon, a compound made up solely of carbon and hydrogen. The products of the combustion of hydrocarbons are carbon dioxide and water. Many hydrocarbons are used as fuel because their combustion releases very large amounts of heat energy. Propane \(\left( \ce{C_3H_8} \right)\) is a gaseous hydrocarbon that is commonly used as the fuel source in gas grills. \[\ce{C_3H_8} \left( g \right) + 5 \ce{O_2} \left( g \right) \rightarrow 3 \ce{CO_2} \left( g \right) + 4 \ce{H_2O} \left( g \right)\nonumber \] Example \(\PageIndex{1}\): Combustion Reactions Ethanol can be used as a fuel source in an alcohol lamp. The formula for ethanol is \(\ce{C_2H_5OH}\). Write the balanced equation for the combustion of ethanol. Solution Step 1: Plan the problem . Ethanol and oxygen are the reactants. As with a hydrocarbon, the products of the combustion of an alcohol are carbon dioxide and water. Step 2: Solve . Write the skeleton equation: \[\ce{C_2H_5OH} \left( l \right) + \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + \ce{H_2O} \left( g \right)\nonumber \] Balance the equation. \[\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( g \right)\nonumber \] Step 3: Think about your result. Combustion reactions must have oxygen as a reactant. Note that the water produced is in the gas state, rather than the liquid state, because of the high temperatures that accompany a combustion reaction. Summary Combustion reaction is defined and examples are given. Review What is needed for a combustion reaction to take place? What is formed in any combustion reaction? Mercury reacts with oxygen to form mercuric oxide. Is this a combustion reaction? What are the products of any combustion reaction involving a hydrocarbon? |
Courses/Colorado_State_University_Pueblo/Elementary_Concepts_in_Physics_and_Chemistry/06%3A_Chapter_6_-_Survey_of_Chemical_Reactions/6.01%3A_Prelude_to_Chemical_Reactions | The space shuttle—and any other rocket-based system—uses chemical reactions to propel itself into space and maneuver itself when it gets into orbit. The rockets that lift the orbiter are of two different types. The three main engines are powered by reacting liquid hydrogen with liquid oxygen to generate water. Then there are the two solid rocket boosters, which use a solid fuel mixture that contains mainly ammonium perchlorate and powdered aluminum. The chemical reaction between these substances produces aluminum oxide, water, nitrogen gas, and hydrogen chloride. Although the solid rocket boosters each have a significantly lower mass than the liquid oxygen and liquid hydrogen tanks, they provide over 80% of the lift needed to put the shuttle into orbit—all because of chemical reactions. |
Courses/Southeast_Missouri_State_University/CH185%3A_General_Chemistry_(Ragain)/13%3A_Solutions/13.04%3A_Solution_Equilibrium_and_Factors_Affecting_Solubility | ↵ Learning Objectives To understand the relationship among temperature, pressure, and solubility. The understand that the solubility of a solid may increase or decrease with increasing temperature, To understand that the solubility of a gas decreases with an increase in temperature and a decrease in pressure. Experimentally it is found that the solubility of most compounds depends strongly on temperature and, if a gas, on pressure as well. As we shall see, the ability to manipulate the solubility by changing the temperature and pressure has several important consequences. Effect of Temperature on the Solubility of Solids Figure \(\PageIndex{1}\) shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and \(\ce{CH_3CO_2Na}\)) exhibit a dramatic increase in solubility with increasing temperature. Others (such as \(\ce{NaCl}\) and \(\ce{K_2SO_4}\)) exhibit little variation, and still others (such as \(\ce{Li_2SO_4}\)) become less soluble with increasing temperature. Notice in particular the curves for \(\ce{NH4NO3}\) and \(\ce{CaCl2}\). The dissolution of ammonium nitrate in water is endothermic (\(ΔH_{soln} = +25.7\; kJ/mol\)), whereas the dissolution of calcium chloride is exothermic (\(ΔH_{soln} = −68.2 \;kJ/mol\)), yet Figure \(\PageIndex{1}\) shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature. The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization, the separation of compounds on the basis of their solubilities in a given solvent. For example, if we have a mixture of 150 g of sodium acetate (\(\ce{CH_3CO_2Na}\)) and 50 g of \(\ce{KBr}\), we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure \(\PageIndex{1}\), both compounds dissolve in water at 80°C, and all 50 g of \(\ce{KBr}\) remains in solution at 0°C. Only about 36 g of \(\ce{CH3CO2Na}\) are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of \(\ce{CH_3CO_2Na}\) crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original \(\ce{CH_3CO_2Na}\) in essentially pure form in only one step. Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure \(\PageIndex{1}\) and from antibiotics to enzymes. For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was \(\ce{KBr}\) in this example) and preferably present in relatively small amounts. Effect of Temperature on the Solubility of Gases The solubility of gases in liquids decreases with increasing temperature, as shown in Figure \(\PageIndex{2}\). Attractive intermolecular interactions in the gas phase are essentially zero for most substances. When a gas dissolves, it does so because its molecules interact with solvent molecules. Because heat is released when these new attractive interactions form, dissolving most gases in liquids is an exothermic process (\(ΔH_{soln} < 0\)). Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas. The phenomenon is similar to that involved in the increase in the vapor pressure of a pure liquid with increasing temperature. In the case of vapor pressure, however, it is attractive forces between solvent molecules that are being overcome by the added thermal energy when the temperature is increased. The decrease in the solubilities of gases at higher temperatures has both practical and environmental implications. Anyone who routinely boils water in a teapot or electric kettle knows that a white or gray deposit builds up on the inside and must eventually be removed. The same phenomenon occurs on a much larger scale in the giant boilers used to supply hot water or steam for industrial applications, where it is called “boiler scale,” a deposit that can seriously decrease the capacity of hot water pipes (Figure \(\PageIndex{3}\)). The problem is not a uniquely modern one: aqueducts that were built by the Romans 2000 years ago to carry cold water from alpine regions to warmer, drier regions in southern France were clogged by similar deposits. The chemistry behind the formation of these deposits is moderately complex and will be described elsewhere, but the driving force is the loss of dissolved \(\ce{CO2}\) from solution. Hard water contains dissolved \(\ce{Ca^{2+}}\) and \(\ce{HCO3^{-}}\) (bicarbonate) ions. Calcium bicarbonate (\(\ce{Ca(HCO3)2}\) is rather soluble in water, but calcium carbonate (\(\ce{CaCO3}\)) is quite insoluble. A solution of bicarbonate ions can react to form carbon dioxide, carbonate ion, and water: \[\ce{2HCO3^{-}(aq) -> CO3^{2-}(aq) + H2O(l) + CO2(aq)} \label{13.9} \] Heating the solution decreases the solubility of \(\ce{CO2}\), which escapes into the gas phase above the solution. In the presence of calcium ions, the carbonate ions precipitate as insoluble calcium carbonate, the major component of boiler scale. Thermal Pollution In thermal pollution , lake or river water that is used to cool an industrial reactor or a power plant is returned to the environment at a higher temperature than normal. Because of the reduced solubility of \(\ce{O2}\) at higher temperatures (Figure \(\PageIndex{2}\)), the warmer water contains less dissolved oxygen than the water did when it entered the plant. Fish and other aquatic organisms that need dissolved oxygen to live can literally suffocate if the oxygen concentration of their habitat is too low. Because the warm, oxygen-depleted water is less dense, it tends to float on top of the cooler, denser, more oxygen-rich water in the lake or river, forming a barrier that prevents atmospheric oxygen from dissolving. Eventually even deep lakes can be suffocated if the problem is not corrected. Additionally, most fish and other nonmammalian aquatic organisms are cold-blooded, which means that their body temperature is the same as the temperature of their environment. Temperatures substantially greater than the normal range can lead to severe stress or even death. Cooling systems for power plants and other facilities must be designed to minimize any adverse effects on the temperatures of surrounding bodies of water. A similar effect is seen in the rising temperatures of bodies of water such as the k0oi89Chesapeake Bay, the largest estuary in North America, where \lobal warming has been implicated as the cause. For each 1.5°C that the bay’s water warms, the capacity of water to dissolve oxygen decreases by about 1.1%. Many marine species that are at the southern limit of their distributions have shifted their populations farther north. In 2005, the eelgrass, which forms an important nursery habitat for fish and shellfish, disappeared from much of the bay following record high water temperatures. Presumably, decreased oxygen levels decreased populations of clams and other filter feeders, which then decreased light transmission to allow the eelsgrass to grow. The complex relationships in ecosystems such as the Chesapeake Bay are especially sensitive to temperature fluctuations that cause a deterioration of habitat quality. Effect of Pressure on the Solubility of Gases: Henry’s Law External pressure has very little effect on the solubility of liquids and solids. In contrast, the solubility of gases increases as the partial pressure of the gas above a solution increases. This point is illustrated in Figure \(\PageIndex{4}\), which shows the effect of increased pressure on the dynamic equilibrium that is established between the dissolved gas molecules in solution and the molecules in the gas phase above the solution. Because the concentration of molecules in the gas phase increases with increasing pressure, the concentration of dissolved gas molecules in the solution at equilibrium is also higher at higher pressures. The relationship between pressure and the solubility of a gas is described quantitatively by Henry’s law, which is named for its discoverer, the English physician and chemist, William Henry (1775–1836): \[C = kP \label{13.3.1} \] where \(C\) is the concentration of dissolved gas at equilibrium, \(P\) is the partial pressure of the gas, and \(k\) is the Henry’s law constant, which must be determined experimentally for each combination of gas, solvent, and temperature. Although the gas concentration may be expressed in any convenient units, we will use molarity exclusively. The units of the Henry’s law constant are therefore mol/(L·atm) = M/atm. Values of the Henry’s law constants for solutions of several gases in water at 20°C are listed in Table \(\PageIndex{1}\). As the data in Table \(\PageIndex{1}\) demonstrate, the concentration of a dissolved gas in water at a given pressure depends strongly on its physical properties. For a series of related substances, London dispersion forces increase as molecular mass increases. Thus among the Group 18 elements, the Henry’s law constants increase smoothly from \(\ce{He}\) to \(\ce{Ne}\) to \(\ce{Ar}\). Gas Henry’s Law Constant [mol/(L·atm)] × 10−4 \(\ce{He}\) 3.9 \(\ce{Ne}\) 4.7 \(\ce{Ar}\) 15.0 \(\ce{H_2}\) 8.1 \(\ce{N_2}\) 7.1 \(\ce{O_2}\) 14.0 \(\ce{CO_2}\) 392.0 Oxygen is Especially Soluble Nitrogen and oxygen are the two most prominent gases in the Earth’s atmosphere and they share many similar physical properties. However, as Table \(\PageIndex{1}\) shows, \(\ce{O2}\) is twice as soluble in water as \(\ce{N2}\). Many factors contribute to solubility including the nature of the intermolecular forces at play. For a details discussion, see "The O 2 /N 2 Ratio Gas Solubility Mystery" by Rubin Battino and Paul G. Seybold ( J. Chem. Eng. Data 2011, 56, 5036–5044 ), Gases that react chemically with water, such as \(\ce{HCl}\) and the other hydrogen halides, \(\ce{H2S}\), and \(\ce{NH3}\), do not obey Henry’s law; all of these gases are much more soluble than predicted by Henry’s law. For example, \(\ce{HCl}\) reacts with water to give \(\ce{H^{+}(aq)}\) and \(\ce{Cl^{-}(aq)}\), not dissolved \(\ce{HCl}\) molecules, and its dissociation into ions results in a much higher solubility than expected for a neutral molecule. Gases that react with water do not obey Henry’s law. Henry’s law has important applications. For example, bubbles of \(\ce{CO2}\) form as soon as a carbonated beverage is opened because the drink was bottled under \(\ce{CO2}\) at a pressure greater than 1 atm. When the bottle is opened, the pressure of \(\ce{CO2}\) above the solution drops rapidly, and some of the dissolved gas escapes from the solution as bubbles. Henry’s law also explains why scuba divers have to be careful to ascend to the surface slowly after a dive if they are breathing compressed air. At the higher pressures under water, more N2 from the air dissolves in the diver’s internal fluids. If the diver ascends too quickly, the rapid pressure change causes small bubbles of \(\ce{N2}\) to form throughout the body, a condition known as “the bends.” These bubbles can block the flow of blood through the small blood vessels, causing great pain and even proving fatal in some cases. Due to the low Henry’s law constant for \(\ce{O2}\) in water, the levels of dissolved oxygen in water are too low to support the energy needs of multicellular organisms, including humans. To increase the \(\ce{O2}\) concentration in internal fluids, organisms synthesize highly soluble carrier molecules that bind \(\ce{O2}\) reversibly. For example, human red blood cells contain a protein called hemoglobin that specifically binds \(\ce{O2}\) and facilitates its transport from the lungs to the tissues, where it is used to oxidize food molecules to provide energy. The concentration of hemoglobin in normal blood is about 2.2 mM, and each hemoglobin molecule can bind four \(\ce{O2}\) molecules. Although the concentration of dissolved \(\ce{O2}\) in blood serum at 37°C (normal body temperature) is only 0.010 mM, the total dissolved \(\ce{O2}\) concentration is 8.8 mM, almost a thousand times greater than would be possible without hemoglobin. Synthetic oxygen carriers based on fluorinated alkanes have been developed for use as an emergency replacement for whole blood. Unlike donated blood, these “blood substitutes” do not require refrigeration and have a long shelf life. Their very high Henry’s law constants for \(\ce{O2}\) result in dissolved oxygen concentrations comparable to those in normal blood. A Video Discussing Henry's Law. Video Link: Henry's Law (The Solubility of Gases in Solvents), YouTube(opens in new window) [youtu.be] Example \(\PageIndex{1}\) The Henry’s law constant for \(\ce{O2}\) in water at 25°C is \(1.27 \times 10^{-3} M/atm\), and the mole fraction of \(\ce{O2}\) in the atmosphere is 0.21. Calculate the solubility of \(\ce{O2}\) in water at 25°C at an atmospheric pressure of 1.00 atm. Given : Henry’s law constant, mole fraction of \(\ce{O2}\), and pressure Asked for: solubility Strategy : Use Dalton’s law of partial pressures to calculate the partial pressure of oxygen. (For more information about Dalton’s law of partial pressures) Use Henry’s law to calculate the solubility, expressed as the concentration of dissolved gas. Solution : A According to Dalton’s law , the partial pressure of \(\ce{O2}\) is proportional to the mole fraction of \(\ce{O2}\): \[\begin{align*} P_A &= \chi_A P_t \\[4pt] &= (0.21)(1.00\; atm) \\[4pt] &= 0.21\; atm \end{align*} \nonumber \] B From Henry’s law, the concentration of dissolved oxygen under these conditions is \[\begin{align*} [\ce{CO2}] &= k P_{\ce{O2}} \\[4pt] &=(1.27 \times 10^{-3}\; M/\cancel{atm}) (0.21\; \cancel{atm}) \\[4pt] &=2.7 \times 10^{-4}\; M \end{align*} \nonumber \] Exercise \(\PageIndex{1}\) To understand why soft drinks “fizz” and then go “flat” after being opened, calculate the concentration of dissolved \(\ce{CO2}\) in a soft drink: bottled under a pressure of 5.0 atm of \(\ce{CO2}\) in equilibrium with the normal partial pressure of \(\ce{CO_2}\) in the atmosphere (approximately \(3 \times 10^{-4} \;atm\)). The Henry’s law constant for \(\ce{CO2}\) in water at 25°C is \(3.4 \times 10^{-2}\; M/atm\). Answer a \(0.17 M\) Answer b \(1 \times 10^{-5} M\) Summary The solubility of most substances depends strongly on the temperature and, in the case of gases, on the pressure. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. Henry’s law describes the relationship between the pressure and the solubility of a gas. |
Courses/Modesto_Junior_College/Chemistry_142%3A_Pre-General_Chemistry_(Brzezinski)/CHEM_142%3A_Text_(Brzezinski)/02%3A_Numbers_and_Measurement/2.02%3A_Significant_Figures_in_Calculations | Note from Dr. B. It is not obvious in the last example and exercise that the number 2 and the number 5 are meant to be exact numbers. In problems that I write, I would specify that a number is counted or exact. If it is not specified, you should assume that the number 2 has only one significant figure. Learning Objectives Use significant figures correctly in arithmetical operations. Rounding Before dealing with the specifics of the rules for determining the significant figures in a calculated result, we need to be able to round numbers correctly. To round a number, first decide how many significant figures the number should have. Once you know that, round to that many digits, starting from the left. If the number immediately to the right of the last significant digit is less than 5, it is dropped and the value of the last significant digit remains the same. If the number immediately to the right of the last significant digit is greater than or equal to 5, the last significant digit is increased by 1. Consider the measurement \(207.518 \: \text{m}\). Right now, the measurement contains six significant figures. How would we successively round it to fewer and fewer significant figures? Follow the process as outlined in Table \(\PageIndex{1}\). Number of Significant Figures Rounded Value Reasoning 6 207.518 All digits are significant 5 207.520 8 rounds the 1 up to 2 4 207.500 2 is dropped 3 208.000 5 rounds the 7 up to 8 2 210.000 8 is replaced by a 0 and rounds the 0 up to 1 1 200.000 1 is replaced by a 0 Notice that the more rounding that is done, the less reliable the figure is. An approximate value may be sufficient for some purposes, but scientific work requires a much higher level of detail. It is important to be aware of significant figures when you are mathematically manipulating numbers. For example, dividing 125 by 307 on a calculator gives 0.4071661238… to an infinite number of digits. But do the digits in this answer have any practical meaning, especially when you are starting with numbers that have only three significant figures each? When performing mathematical operations, there are two rules for limiting the number of significant figures in an answer—one rule is for addition and subtraction, and one rule is for multiplication and division. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the least precise operation. An answer is no more precise than the least precise number used to get the answer. Multiplication and Division For multiplication or division, the rule is to count the number of significant figures in each number being multiplied or divided and then limit the significant figures in the answer to the lowest count. An example is as follows: The final answer, limited to four significant figures, is 4,094. The first digit dropped is 1, so we do not round up. Scientific notation provides a way of communicating significant figures without ambiguity. You simply include all the significant figures in the leading number. For example, the number 450 has two significant figures and would be written in scientific notation as 4.5 × 10 2 , whereas 450.0 has four significant figures and would be written as 4.500 × 10 2 . In scientific notation, all significant figures are listed explicitly. Example \(\PageIndex{1}\) Write the answer for each expression using scientific notation with the appropriate number of significant figures. 23.096 × 90.300 125 × 9.000 Solution a Explanation Answer The calculator answer is 2,085.5688, but we need to round it to five significant figures. Because the first digit to be dropped (in the tenths place) is greater than 5, we round up to 2,085.6. \(2.0856 \times 10^3\) b Explanation Answer The calculator gives 1,125 as the answer, but we limit it to three significant figures. \(1.13 \times 10^3\) Addition and Subtraction How are significant figures handled in calculations? It depends on what type of calculation is being performed. If the calculation is an addition or a subtraction, the rule is as follows: limit the reported answer to the rightmost column that all numbers have significant figures in common. For example, if you were to add 1.2 and 4.71, we note that the first number stops its significant figures in the tenths column, while the second number stops its significant figures in the hundredths column. We therefore limit our answer to the tenths column. We drop the last digit—the 1—because it is not significant to the final answer. The dropping of positions in sums and differences brings up the topic of rounding. Although there are several conventions, in this text we will adopt the following rule: the final answer should be rounded up if the first dropped digit is 5 or greater, and rounded down if the first dropped digit is less than 5. Example \(\PageIndex{2}\) 13.77 + 908.226 1,027 + 611 + 363.06 Solution a Explanation Answer The calculator answer is 921.996, but because 13.77 has its farthest-right significant figure in the hundredths place, we need to round the final answer to the hundredths position. Because the first digit to be dropped (in the thousandths place) is greater than 5, we round up to 922.00 \(922.00 = 9.2200 \times 10^2\) b Explanation Answer The calculator gives 2,001.06 as the answer, but because 611 and 1027 has its farthest-right significant figure in the ones place, the final answer must be limited to the ones position. \(2,001.06 = 2.001 \times 10^3\) Exercise \(\PageIndex{2}\) Write the answer for each expression using scientific notation with the appropriate number of significant figures. 217 ÷ 903 13.77 + 908.226 + 515 255.0 − 99 0.00666 × 321 Answer a: \(0.240 = 2.40 \times 10^{-1}\) Answer b: \(1,437 = 1.437 \times 10^3\) Answer c: \(156 = 1.56 \times 10^2\) Answer d: \(2.14 = 2.14 \times 10^0\) Remember that calculators do not understand significant figures. You are the one who must apply the rules of significant figures to a result from your calculator. Calculations Involving Multiplication/Division and Addition/Subtraction In practice, chemists generally work with a calculator and carry all digits forward through subsequent calculations. When working on paper, however, we often want to minimize the number of digits we have to write out. Because successive rounding can compound inaccuracies, intermediate rounding needs to be handled correctly. When working on paper, always round an intermediate result so as to retain at least one more digit than can be justified and carry this number into the next step in the calculation. The final answer is then rounded to the correct number of significant figures at the very end. In the worked examples in this text, we will often show the results of intermediate steps in a calculation. In doing so, we will show the results to only the correct number of significant figures allowed for that step, in effect treating each step as a separate calculation. This procedure is intended to reinforce the rules for determining the number of significant figures, but in some cases it may give a final answer that differs in the last digit from that obtained using a calculator, where all digits are carried through to the last step. Example \(\PageIndex{3}\) 2(1.008 g) + 15.99 g 137.3 s + 2(35.45 s) \( {118.7 g \over 2} - 35.5 g \) Solution a. Empty DataFrame Columns: [(Explanation, 2(1.008 g) + 15.99 g = Perform multiplication first. 2 (1.008 g 4 sig figs) = 2.016 g 4 sig figs The number with the least number of significant figures is 1.008 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition. 2.016 g thousandths place + 15.99 g hundredths place (least precise) = 18.006 g Round the final answer. Round the final answer to the hundredths place since 15.99 has its farthest right significant figure in the hundredths place (least precise).), (Answer, 18.01 g (rounding up))] Index: [] b. Explanation Answer 137.3 s + 2(35.45 s) = Perform multiplication first. 2(35.45 s 4 sig figs) = 70.90 s 4 sig figs The number with the least number of significant figures is 35.45; the number 2 is an exact number and therefore has an infinite number of significant figures. Then, perform the addition. 137.3 s tenths place (least precise) + 70.90 s hundredths place = 208.20 s Round the final answer. Round the final answer to the tenths place based on 137.3 s. 208.2 s c. Empty DataFrame Columns: [(Explanation, \( {118.7 g \over 2} - 35.5 g \) = Perform division first. \( {118.7 g \over 2} \) 4 sig figs = 59.35 g 4 sig figs The number with the least number of significant figures is 118.7 g; the number 2 is an exact number and therefore has an infinite number of significant figures. Perform subtraction next. 59.35 g hundredths place − 35.5 g tenths place (least precise) = 23.85 g Round the final answer. Round the final answer to the tenths place based on 35.5 g.), (Answer, 23.9 g (rounding up))] Index: [] Exercise \(\PageIndex{3}\) Complete the calculations and report your answers using the correct number of significant figures. 5(1.008s) - 10.66 s 99.0 cm+ 2(5.56 cm) Answer a -5.62 s Answer b 110.2 cm Summary Rounding If the number to be dropped is greater than or equal to 5, increase the number to its left by 1 (e.g. 2.9699 rounded to three significant figures is 2.97). If the number to be dropped is less than 5, there is no change (e.g. 4.00443 rounded to four significant figures is 4.004). The rule in multiplication and division is that the final answer should have the same number of significant figures as there are in the number with the fewest significant figures. The rule in addition and subtraction is that the answer is given the same number of decimal places as the term with the fewest decimal places. |
Courses/can/CHEM_232_-_Organic_Chemistry_II_(Puenzo)/14%3A_Biomolecules_-_Nucleic_Acids/14.01%3A_Chapter_Objectives | Learning Objectives Understand the components and chemical structure of nucleotides, and their role in forming DNA and RNA molecules. Analyze the principles of base pairing and hydrogen bonding in nucleic acids, and how they contribute to the stability and specificity of DNA and RNA structures. Examine the chemical mechanisms involved in nucleic acid synthesis (polymerization) and degradation (hydrolysis), and the enzymatic processes of DNA replication, RNA transcription, and repair. Explore the different conformations of nucleic acids, including their impact on function and interactions with other molecules. Learn about analytical techniques used to study nucleic acids, and how they provide insights into structure, function, and interactions. Nucleic acids are biopolymers found in all living organisms, serving as the essential molecules for the storage, transmission, and expression of genetic information. These complex macromolecules are composed of repeating units called nucleotides, which consist of three main components: a nitrogenous base, a five-carbon sugar (ribose or deoxyribose), and a phosphate group. There are two primary types of nucleic acids: deoxyribonucleic acid (DNA) and ribonucleic acid (RNA). DNA is the genetic blueprint of an organism, encoding the instructions for the synthesis of proteins and the regulation of cellular activities. It exists in the form of a double helix, with two complementary strands held together by hydrogen bonds between the nitrogenous bases adenine (A) and thymine (T), and guanine (G) and cytosine (C). RNA, on the other hand, plays diverse roles in gene expression, including messenger RNA (mRNA) which carries genetic information from DNA to the ribosomes for protein synthesis, transfer RNA (tRNA) which brings amino acids to the ribosomes during protein synthesis, and ribosomal RNA (rRNA) which forms the structural and catalytic core of ribosomes. The structure and function of nucleic acids are intricately linked. The sequence of nitrogenous bases along the nucleic acid chain determines the genetic code, with each triplet of bases (codon) encoding a specific amino acid or serving as a signal for regulatory functions. The ability of nucleic acids to undergo replication, transcription, and translation enables the faithful transmission and expression of genetic information from one generation to the next. Beyond their role in genetics, nucleic acids also participate in various cellular processes, including DNA repair, RNA processing, and gene regulation. Moreover, advances in biotechnology have led to the manipulation and engineering of nucleic acids for applications such as genetic engineering, molecular diagnostics, and gene therapy. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.05%3A_Chemical_Potentials/1.5.03%3A_Chemical_Potentials-_Solutions-_General_Properties | A key quantity in chemical thermodynamics is the chemical potential of chemical substance \(j\), \(\mu_{j}\). The latter is the differential dependence of Gibbs energy on amount of substance \(j\) at fixed \(\mathrm{T}\), \(\mathrm{p}\) and amounts of all other substances in the system [1]. \[\mu_{\mathrm{j}}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{i} \neq \mathrm{j})} \nonumber \] An important point to note is that the conditions ‘fixed \(T\) and fixed \(p\)’ on the partial differential refer to intensive variables. These conditions are called Gibbsian in recognition of the development by Gibbs of the concept of thermodynamic potential for changes in the properties of a closed system at fixed \(\mathrm{T}\) and fixed \(\mathrm{p}\). In general terms, the chemical potential of substance \(j\) is defined using analogous partial derivatives of the thermodynamic internal energy \(\mathrm{U}\), enthalpy \(\mathrm{H}\) and Helmholtz energy \(\mathrm{F}\). \[\mu_{\mathrm{j}}=\left(\frac{\partial \mathrm{G}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{p}, \mathrm{n}(\mathrm{i} \neq \mathrm{j})}=\left(\frac{\partial \mathrm{U}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{s}, \mathrm{V}, \mathrm{n}(\mathrm{i} \neq \mathrm{j})}=\left(\frac{\partial \mathrm{H}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{s}, \mathrm{p}, \mathrm{n}(\mathrm{i} \neq \mathrm{j})}=\left(\frac{\partial \mathrm{F}}{\partial \mathrm{n}_{\mathrm{j}}}\right)_{\mathrm{T}, \mathrm{V}, \mathrm{n}(\mathrm{i} \neq \mathrm{j})} \nonumber \] With reference to a given closed system, thermodynamics defines macroscopic properties including volume \(\mathrm{V}\), Gibbs energy \(\mathrm{G}\), enthalpy \(\mathrm{H}\) and entropy \(\mathrm{S}\). Nevertheless we need to “tell” these thermodynamic variables that a given system probably comprises different chemical substances. The analysis is reasonably straightforward if we define the system under consideration by the ‘Gibbsian’ set of independent variables; i.e. \(\mathrm{T}\), \(\mathrm{p}\) and amounts of each chemical substance [2]. The analysis leads to the definition of a chemical potential for each substance \(j\), \(\mu_{j}\), in a closed system [3,4]. Footnotes [1] It might be argued that we have switched our attention from closed to open systems because we are considering a change in Gibbs energy when we add \(\partial n_{j}\) moles of substance to the system. This comment is true in part. But what we actually envisage is something a little different. We take a closed system containing \(n_{1}\) and \(n_{j}\) moles of substances \(1\) and \(j\) respectively. We open the system, rapidly pop in \(\delta \mathrm{n}_{\mathrm{j}}\) moles of substance \(j\) and put the lid back on the system to return it to the closed state. Then the closed system contains \(\left(\mathrm{n}_{\mathrm{j}}+\delta \mathrm{n}_{\mathrm{j}}\right)\) moles of substance j so changes in chemical composition and molecular organisation follow producing a change in Gibbs energy at, say, fixed \(\mathrm{T}\) and fixed \(\mathrm{p}\). [2] G. N. Lewis, (with possibly one of the key papers in chemistry) Z.Phys.Chem.,1907, 61 ,129. Proc. Acad. Arts Sci.,1907, 43 ,259. [Comment: Paper (a) is the German translation of paper (b).] [3] The analysis presented here (a) is confined to bulk systems in the absence of magnetic and electric fields and (b) ignores surface effects. [4] To quote E. Grunwald [J. Am. Chem. Soc., 1984, 106 , 5414] “any first derivative with respect to any variable of state at equilibrium is isodelphic”; see also E. Grunwald, Thermodynamics of Molecular Species, Wiley, New York, 1997. |
Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/03%3A_Ionic_Bonding_and_Simple_Ionic_Compounds | There are only 118 known chemical elements but tens of millions of known chemical compounds. Compounds can be very complex combinations of atoms, but many important compounds are fairly simple. Table salt, as we have seen, consists of only two elements: sodium and chlorine. Nevertheless, the compound has properties completely different from either elemental sodium (a chemically reactive metal) or elemental chlorine (a poisonous, green gas). We will see additional examples of such differences in this chapter as we consider how atoms combine to form compounds. 3.1: Prelude to Ionic Bonding and Simple Ionic Compounds We will see that the word salt has a specific meaning in chemistry, but to most people, this word refers to table salt. This kind of salt is used as a condiment throughout the world, but it was not always so abundant. Two thousand years ago, Roman soldiers received part of their pay as salt, which explains why the words salt and salary come from the same Latin root (salarium). Today, table salt is either mined or obtained from the evaporation of saltwater. 3.2: Two Types of Bonding Atoms have a tendency to have eight electrons in their valence shell. The attraction of oppositely charged ions is what makes ionic bonds. 3.3: Ions Ions can be positively charged or negatively charged. A Lewis diagram is used to show how electrons are transferred to make ions and ionic compounds. 3.4: Formulas for Ionic Compounds Proper chemical formulas for ionic compounds balance the total positive charge with the total negative charge. Groups of atoms with an overall charge, called polyatomic ions, also exist. 3.5: Ionic Nomenclature Each ionic compound has its own unique name that comes from the names of the ions. After learning a few more details about the names of individual ions, you will be a step away from knowing how to name ionic compounds. This section begins the formal study of nomenclature, the systematic naming of chemical compounds. 3.E: Ionic Bonding and Simple Ionic Compounds (Exercises) These are homework exercises to accompany Chapter 3 of the Ball et al. "The Basics of GOB Chemistry" Textmap. 3.S: Ionic Bonding and Simple Ionic Compounds (Summary) To ensure that you understand the material in this chapter, you should review the meanings of the following bold terms and ask yourself how they relate to the topics in the chapter. Template:HideTOC |
Courses/Lumen_Learning/Book%3A_English_Composition_I-3_(Lumen)/22%3A_Research_Process%3A_Writing_Ethically/21.3%3A_CRAAP_Analysis | Let’s take a closer look at how analyzing the C.R.A.A.P. in a source can serve as a valuable source evaluation tool. Currency: The timeliness of the information Key Question : When was the item of information published or produced? Determining when an item of information was published or produced is an aspect of evaluating information. The date information was published or produced tells you how current it is or how contemporaneous it is with the topic you are researching. There are two facets to the issue of currency. Is the information the most recent version? Is the information the original research, description, or account? The question of most recent version of information versus an original or primary version can be a critical one. For example: If you were doing a project on the survival of passengers in car crashes, you would need the most recent information on automobile crash tests, structural strength of materials, car wreck mortality statistics, etc. If, on the other hand, you were doing a project on the feelings of college students about the VietNam War during the 1960s, you would need information written in the 1960s by college students (primary sources) as well as materials written since then about college students in the 1960s (secondary sources). Key indicators of the currency of the information are: date of copyright date of publication date of revision or edition dates of sources cited date of patent or trademark Relevance: The importance of the information for your needs Key Question: How does this source contribute to my research paper? The discussion of suitability above is essentially the same thing as relevance. When you read through your source, consider how the source will effectively support your argument and how you can utilize the source in your paper. You should also consider whether the source provides sufficient coverage of the topic. Information sources with broad, shallow coverage mean that you need to find other sources of information to obtain adequate details about your topic. Information sources with a very narrow focus or a distinct bias mean that you need to find additional sources to obtain the information on other aspects of your topic. Some questions to consider are: Does the information relate to my topic or answer my question? Who is the intended audience? Is the information at an appropriate level (i.e. not too simple or advanced) for my needs? Did I look at a variety of sources before deciding to use this one? Would I be comfortable using this source for my college research paper? Authority: The source of the information Key Question : Is the person, organization, or institution responsible for the intellectual content of the information knowledgeable in that subject? Determining the knowledge and expertise of the author of information is an important aspect of evaluating the reliability of information. Anyone can make an assertion or a statement about some thing, event, or idea, but only someone who knows or understands what that thing, event, or idea is can make a reasonably reliable statement or assertion about it. Some external indications of knowledge of or expertise are: a formal academic degree in a subject area professional or work-related experience–businessmen, government agency personnel, sports figures, etc. have expertise on their area of work active involvement in a subject or organization by serious amateurs who spend substantial amounts of personal time researching and studying that subject area. organizations, agencies, institutions, corporations with active involvement or work in a particular subject area. HINT: Be careful of opinions stated by professionals outside of their area of work expertise. Accuracy: The reliability, truthfulness, and correctness of the information Key Question : How free from error is this piece of information? Establishing the accuracy, or relative accuracy, of information is an important part of evaluating the reliability of information. It is easier to establish the accuracy of facts than it is opinions, interpretations, or ideas. The more an idea, opinion, or other piece of information varies from the accepted point of view on a particular topic the harder it is to establish its accuracy. It may be completely accurate but corroborating it is both more necessary and more difficult. An important aspect of accuracy is the intellectual integrity of the item. Are the sources appropriately cited in the text and listed in the references? Are quotations cited correctly and in context? Out of context quotations can be misleading and sometimes completely erroneous. Are there exaggerations, omissions, or errors? These are difficulty to identify if you use only one source of information. Always use several different sources of information on your topic. Analyzing what different sources say about a topic is one way to understand that topic. In addition to errors of fact and integrity, you need to watch for errors of logic. Errors of logic occur primarily in the presentation of conclusions, opinions, interpretations, editorials, ideas, etc. Some indications that information is accurate are: the same information can be found in other reliable sources the experiment can be replicated and returns the same results the documentation provided in support of the information is substantive the sources used for documentation are known to be generally reliable the author of the information is known to have expertise on that subject the presentation is free from logical fallacies or errors quotations are “in context”-the meaning of the original work is kept in the work which quotes the original quotations are correctly cited acronyms are clearly defined at the beginning Some indications that information may not be accurate are: facts cannot be verified or are contradicted in other sources sources used are known to be unreliable or highly biased bibliography of sources used is inadequate or non-existent quotations are taken out of context and given a different meaning acronyms are not defined and the intended audience is a general one presence of one or more logical fallacies authority cited is another part of the same organization Purpose: The reason the information exists Key Question : Who is this information written for or this product developed for? Identifying the intended audience of the information or product is another aspect of evaluating information. The intended audience of an item generally determines the style of presentation, the level of technical detail, and the depth of coverage. You should also consider the author’s objectivity. Are they trying to persuade? Do they present any bias? While it is unlikely that anything humans do is ever absolutely objective, it is important to establish that the information you intend to use is reasonably objective, or if it is not, to establish exactly what the point of view or bias is. There are times when information expressing a particular point of view or bias is useful, but you must use it consciously. You must know what the point of view is and why that point of view is important to your project. For example , books on food sanitation written for children, for restaurant workers, or for research microbiologists will be very different even though they all cover the same topic. Determining the intended audience of a particular piece of information will help you decide whether or not the information will be too basic, too technical, too general, or just right for your needs. The intended audience can also indicate the potential reliability of the item because some audiences require more documentation than others. For example , items produced for scholarly or professional audiences are generally produced by experts and go through a peer evaluation process. Items produced for the mass market frequently are not produced by experts and generally do not go through an evaluation process. Some indications of the intended audience are: highly technical language, complex analysis, very sophisticated/technical tools can indicate a technical, professional, or scholarly audience how-to information or current practices in “X” are frequently written by experts for practitioners in that field substantive and serious presentations of a topic with not too much technical language are generally written for the educated lay audience popular language, fairly simple presentations of a topic, little or no analysis, inexpensive tools can indicate a general or popular audience bibliographies, especially long bibliographies, are generally compiled by and for those doing research on that topic The CRAAP Test Review the steps of the C.R.A.A.P method and practice evaluating sources in this tutorial from Eastern Michigan University. Be sure to complete the practice exercises at the end of the tutorial. CC licensed content, Original Revision and Adaptation. Provided by : Lumen Learning. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike CC licensed content, Shared previously Evaluating Information. Provided by : The University of Rhode Island. Located at : http://uri.libguides.com/start/craap . License : CC BY-SA: Attribution-ShareAlike Provided by : Virginia Tech. Located at : http://info-skills.lib.vt.edu/evaluating_info/2.html . Project : University Libraries Information Skills Modules. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike CRAAP Tutorial. Authored by : Bill Marino. Provided by : Eastern Michigan University. Located at : http://www.emich.edu/library/help/reliability/ . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike |
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Old_Textbook_for_Chemistry_143_-_2019_to_2021/12%3A_Acid-Base_Equilibria/12.05%3A_Buffers | Learning Objectives Describe the composition and function of acid–base buffers Calculate the pH of a buffer before and after the addition of added acid or base A mixture of a weak acid and its conjugate base (or a mixture of a weak base and its conjugate acid) is called a buffer solution, or a buffer . Buffer solutions resist a change in pH when small amounts of a strong acid or a strong base are added (Figure \(\PageIndex{1}\)). A solution of acetic acid (\(\ce{CH3COOH}\) and sodium acetate \(\ce{CH3COONa}\)) is an example of a buffer that consists of a weak acid and its salt. An example of a buffer that consists of a weak base and its salt is a solution of ammonia (\(\ce{NH3(aq)}\)) and ammonium chloride (\(\ce{NH4Cl(aq)}\)). How Buffers Work A mixture of acetic acid and sodium acetate is acidic because the K a of acetic acid is greater than the K b of its conjugate base acetate. It is a buffer because it contains both the weak acid and its salt. Hence, it acts to keep the hydronium ion concentration (and the pH) almost constant by the addition of either a small amount of a strong acid or a strong base. If we add a base such as sodium hydroxide, the hydroxide ions react with the few hydronium ions present. Then more of the acetic acid reacts with water, restoring the hydronium ion concentration almost to its original value: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \] The pH changes very little. If we add an acid such as hydrochloric acid, most of the hydronium ions from the hydrochloric acid combine with acetate ions, forming acetic acid molecules: \[\ce{H3O+}(aq)+\ce{CH3CO2-}(aq)⟶\ce{CH3CO2H}(aq)+\ce{H2O}(l) \nonumber \] Thus, there is very little increase in the concentration of the hydronium ion, and the pH remains practically unchanged (Figure \(\PageIndex{2}\)). A mixture of ammonia and ammonium chloride is basic because the K b for ammonia is greater than the K a for the ammonium ion. It is a buffer because it also contains the salt of the weak base. If we add a base (hydroxide ions), ammonium ions in the buffer react with the hydroxide ions to form ammonia and water and reduce the hydroxide ion concentration almost to its original value: \[\ce{NH4+}(aq)+\ce{OH-}(aq)⟶\ce{NH3}(aq)+\ce{H2O}(l) \nonumber \] If we add an acid (hydronium ions), ammonia molecules in the buffer mixture react with the hydronium ions to form ammonium ions and reduce the hydronium ion concentration almost to its original value: \[\ce{H3O+}(aq)+\ce{NH3}(aq)⟶\ce{NH4+}(aq)+\ce{H2O}(l) \nonumber \] The three parts of the following example illustrate the change in pH that accompanies the addition of base to a buffered solution of a weak acid and to an unbuffered solution of a strong acid. Example \(\PageIndex{1}\): pH Changes in Buffered and Unbuffered Solutions Acetate buffers are used in biochemical studies of enzymes and other chemical components of cells to prevent pH changes that might change the biochemical activity of these compounds. Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate. Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL. Solution Calculate the pH of an acetate buffer that is a mixture with 0.10 M acetic acid and 0.10 M sodium acetate. To determine the pH of the buffer solution we use a typical equilibrium calculation (as illustrated in earlier Examples): Determine the direction of change. The equilibrium in a mixture of H 3 O + , \(\ce{CH3CO2-}\), and CH 3 CO 2 H is: \[\ce{CH3CO2H}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \nonumber \] The equilibrium constant for CH 3 CO 2 H is not given, so we look it up in Table E1: K a = 1.8 × 10 −5 . With [CH 3 CO 2 H] = \(\ce{[CH3CO2- ]}\) = 0.10 M and [H 3 O + ] = ~0 M , the reaction shifts to the right to form H 3 O + . Determine x and equilibrium concentrations . A table of changes and concentrations follows: Solve for x and the equilibrium concentrations. We find: \[x=1.8×10^{−5}\:M \nonumber \] and \[\ce{[H3O+]}=0+x=1.8×10^{−5}\:M \nonumber \] Thus: \[\mathrm{pH=−log[H_3O^+]=−log(1.8×10^{−5})} \nonumber \] \[=4.74 \nonumber \] 4. Check the work . If we calculate all calculated equilibrium concentrations, we find that the equilibrium value of the reaction coefficient, Q = K a . (b) Calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of this buffer, giving a solution with a volume of 101 mL. First, we calculate the concentrations of an intermediate mixture resulting from the complete reaction between the acid in the buffer and the added base. Then we determine the concentrations of the mixture at the new equilibrium: Determine the moles of NaOH. One milliliter (0.0010 L) of 0.10 M NaOH contains: \[\mathrm{0.0010\cancel{L}×\left(\dfrac{0.10\:mol\: NaOH}{1\cancel{L}}\right)=1.0×10^{−4}\:mol\: NaOH} \nonumber \] Determine the moles of CH 2 CO 2 H. Before reaction, 0.100 L of the buffer solution contains: \[\mathrm{0.100\cancel{L}×\left(\dfrac{0.100\:mol\:CH_3CO_2H}{1\cancel{L}}\right)=1.00×10^{−2}\:mol\:CH_3CO_2H} \nonumber \] Solve for the amount of NaCH 3 CO 2 produced. The 1.0 × 10 −4 mol of NaOH neutralizes 1.0 × 10 −4 mol of CH 3 CO 2 H, leaving: \[\mathrm{(1.0×10^{−2})−(0.01×10^{−2})=0.99×10^{−2}\:mol\:CH_3CO_2H} \nonumber \] and producing 1.0 × 10 −4 mol of NaCH 3 CO 2 . This makes a total of: [\mathrm{(1.0×10^{−2})+(0.01×10^{−2})=1.01×10^{−2}\:mol\:NaCH_3CO_2} \nonumber \] 4. Find the molarity of the products. After reaction, CH 3 CO 2 H and NaCH 3 CO 2 are contained in 101 mL of the intermediate solution, so: \[\ce{[CH3CO2H]}=\mathrm{\dfrac{9.9×10^{−3}\:mol}{0.101\:L}}=0.098\:M \nonumber \] \[\ce{[NaCH3CO2]}=\mathrm{\dfrac{1.01×10^{−2}\:mol}{0.101\:L}}=0.100\:M \nonumber \] Now we calculate the pH after the intermediate solution, which is 0.098 M in CH 3 CO 2 H and 0.100 M in NaCH 3 CO 2 , comes to equilibrium. The calculation is very similar to that in part (a) of this example: This series of calculations gives a pH = 4.75. Thus the addition of the base barely changes the pH of the solution. (c) For comparison, calculate the pH after 1.0 mL of 0.10 M NaOH is added to 100 mL of a solution of an unbuffered solution with a pH of 4.74 (a 1.8 × 10 −5 - M solution of HCl). The volume of the final solution is 101 mL. Solution This 1.8 × 10 −5 - M solution of HCl has the same hydronium ion concentration as the 0.10- M solution of acetic acid-sodium acetate buffer described in part (a) of this example. The solution contains: \(\mathrm{0.100\:L×\left(\dfrac{1.8×10^{−5}\:mol\: HCl}{1\:L}\right)=1.8×10^{−6}\:mol\: HCl} \) As shown in part (b), 1 mL of 0.10 M NaOH contains 1.0 × 10 −4 mol of NaOH. When the NaOH and HCl solutions are mixed, the HCl is the limiting reagent in the reaction. All of the HCl reacts, and the amount of NaOH that remains is: \( (1.0×10^{−4})−(1.8×10^{−6})=9.8×10^{−5}\:M \) The concentration of NaOH is: \(\dfrac{9.8×10^{−5}\:M\:\ce{NaOH}}{0.101\:\ce{L}}=9.7×10^{−4}\:M \) The pOH of this solution is: \(\mathrm{pOH=−log[OH^- ]=−log(9.7×10^{−4})=3.01} \) The pH is: \(\mathrm{pH=14.00−pOH=10.99} \) The pH changes from 4.74 to 10.99 in this unbuffered solution. This compares to the change of 4.74 to 4.75 that occurred when the same amount of NaOH was added to the buffered solution described in part (b). Exercise \(\PageIndex{1}\) Show that adding 1.0 mL of 0.10 M HCl changes the pH of 100 mL of a 1.8 × 10 −5 M HCl solution from 4.74 to 3.00. Answer Initial pH of 1.8 × 10 −5 M HCl; pH = −log[H 3 O + ] = −log[1.8 × 10 −5 ] = 4.74 Moles of H 3 O + added by addition of 1.0 mL of 0.10 M HCl: 0.10 moles/L × 0.0010 L = 1.0 × 10 −4 moles; final pH after addition of 1.0 mL of 0.10 M HCl: \[\mathrm{pH=−log[H_3O^+]=−log\left(\dfrac{total\: moles\:H_3O^+}{total\: volume}\right)=−log\left(\dfrac{1.0×10^{−4}\:mol+1.8×10^{−6}\:mol}{101\:mL\left(\dfrac{1\:L}{1000\:mL}\right)}\right)=3.00} \nonumber \] Buffer Capacity Buffer solutions do not have an unlimited capacity to keep the pH relatively constant (Figure \(\PageIndex{3}\)). If we add so much base to a buffer that the weak acid is exhausted, no more buffering action toward the base is possible. On the other hand, if we add an excess of acid, the weak base would be exhausted, and no more buffering action toward any additional acid would be possible. In fact, we do not even need to exhaust all of the acid or base in a buffer to overwhelm it; its buffering action will diminish rapidly as a given component nears depletion. The buffer capacity is the amount of acid or base that can be added to a given volume of a buffer solution before the pH changes significantly, usually by one unit. Buffer capacity depends on the amounts of the weak acid and its conjugate base that are in a buffer mixture. For example, 1 L of a solution that is 1.0 M in acetic acid and 1.0 M in sodium acetate has a greater buffer capacity than 1 L of a solution that is 0.10 M in acetic acid and 0.10 M in sodium acetate even though both solutions have the same pH. The first solution has more buffer capacity because it contains more acetic acid and acetate ion. Selection of Suitable Buffer Mixtures There are two useful rules of thumb for selecting buffer mixtures: A good buffer mixture should have about equal concentrations of both of its components. A buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other. Figure \(\PageIndex{4}\) shows an acetic acid-acetate ion buffer as base is added. The initial pH is 4.74. A change of 1 pH unit occurs when the acetic acid concentration is reduced to 11% of the acetate ion concentration. Weak acids and their salts are better as buffers for pHs less than 7; weak bases and their salts are better as buffers for pHs greater than 7. Blood is an important example of a buffered solution, with the principal acid and ion responsible for the buffering action being carbonic acid, H 2 CO 3 , and the bicarbonate ion, \(\ce{HCO3-}\). When an excess of hydrogen ion enters the blood stream, it is removed primarily by the reaction: \[\ce{H3O+}(aq)+\ce{HCO3-}(aq)⟶\ce{H2CO3}(aq)+\ce{H2O}(l) \nonumber \] When an excess of the hydroxide ion is present, it is removed by the reaction: \[\ce{OH-}(aq)+\ce{H2CO3}(aq)⟶\ce{HCO3-}(aq)+\ce{H2O}(l) \nonumber \] The pH of human blood thus remains very near 7.35, that is, slightly basic. Variations are usually less than 0.1 of a pH unit. A change of 0.4 of a pH unit is likely to be fatal. The Henderson-Hasselbalch Approximation The ionization-constant expression for a solution of a weak acid can be written as: \[K_\ce{a}=\ce{\dfrac{[H3O+][A- ]}{[HA]}} \nonumber \] Rearranging to solve for [H 3 O + ], we get: \[\ce{[H3O+]}=K_\ce{a}×\ce{\dfrac{[HA]}{[A- ]}} \nonumber \] Taking the negative logarithm of both sides of this equation, we arrive at: \[\mathrm{−log[H_3O^+]=−log\mathit{K}_a − log\dfrac{[HA]}{[A^- ]}} \nonumber \] which can be written as \[\mathrm{pH=p\mathit{K}_a+log\dfrac{[A^- ]}{[HA]}} \nonumber \] where p K a is the negative of the common logarithm of the ionization constant of the weak acid (p K a = −log K a ). This equation relates the pH, the ionization constant of a weak acid, and the concentrations of the weak acid and its salt in a buffered solution. Scientists often use this expression, called the Henderson-Hasselbalch approximation , to calculate the pH of buffer solutions. It is important to note that the “ x is small” assumption must be valid to use this equation. Lawrence Joseph Henderson and Karl Albert Hasselbalch Lawrence Joseph Henderson (1878–1942) was an American physician, biochemist and physiologist, to name only a few of his many pursuits. He obtained a medical degree from Harvard and then spent 2 years studying in Strasbourg, then a part of Germany, before returning to take a lecturer position at Harvard. He eventually became a professor at Harvard and worked there his entire life. He discovered that the acid-base balance in human blood is regulated by a buffer system formed by the dissolved carbon dioxide in blood. He wrote an equation in 1908 to describe the carbonic acid-carbonate buffer system in blood. Henderson was broadly knowledgeable; in addition to his important research on the physiology of blood, he also wrote on the adaptations of organisms and their fit with their environments, on sociology and on university education. He also founded the Fatigue Laboratory, at the Harvard Business School, which examined human physiology with specific focus on work in industry, exercise, and nutrition. In 1916, Karl Albert Hasselbalch (1874–1962), a Danish physician and chemist, shared authorship in a paper with Christian Bohr in 1904 that described the Bohr effect, which showed that the ability of hemoglobin in the blood to bind with oxygen was inversely related to the acidity of the blood and the concentration of carbon dioxide. The pH scale was introduced in 1909 by another Dane, Sørensen, and in 1912, Hasselbalch published measurements of the pH of blood. In 1916, Hasselbalch expressed Henderson’s equation in logarithmic terms, consistent with the logarithmic scale of pH, and thus the Henderson-Hasselbalch equation was born. Medicine: The Buffer System in Blood The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction: \[\ce{CO2}(g)+\ce{2H2O}(l)⇌\ce{H2CO3}(aq)⇌\ce{HCO3-}(aq)+\ce{H3O+}(aq) \nonumber \] The concentration of carbonic acid, H 2 CO 3 is approximately 0.0012 M , and the concentration of the hydrogen carbonate ion, \(\ce{HCO3-}\), is around 0.024 M . Using the Henderson-Hasselbalch equation and the p K a of carbonic acid at body temperature, we can calculate the pH of blood: \[\mathrm{pH=p\mathit{K}_a+\log\dfrac{[base]}{[acid]}=6.1+\log\dfrac{0.024}{0.0012}=7.4} \nonumber \] The fact that the H 2 CO 3 concentration is significantly lower than that of the \(\ce{HCO3-}\) ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded. Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the \(\ce{HCO3-}\) ion, producing H 2 CO 3 . An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO 2 from the blood through the lungs driving the equilibrium reaction such that [H 3 O + ] is lowered. If the blood is too alkaline, a lower breath rate increases CO 2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H + ] and restoring an appropriate pH. Summary A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base). Key Equations p K a = −log K a p K b = −log K b \(\mathrm{pH=p\mathit{K}_a+\log\dfrac{[A^- ]}{[HA]}}\) Glossary buffer capacity amount of an acid or base that can be added to a volume of a buffer solution before its pH changes significantly (usually by one pH unit) buffer mixture of a weak acid or a weak base and the salt of its conjugate; the pH of a buffer resists change when small amounts of acid or base are added Henderson-Hasselbalch equation equation used to calculate the pH of buffer solutions |
Courses/Diablo_Valley_College/DVC_Chem_106%3A_Rusay/Vocabulary_II | The assignment is to define and build a comprehensive list of the Key Vocabulary Terms, which are pertinent to and used in Chem 106. Phase I and Phase II have been completed, and open to the public. This is an ongoing, expanding list and will be added to throughout the course. Acid: A chemical species that donates protons or hydrogen ions and/or accepts electrons . Activation Energy: The minimum amount of energy required to initiate a reaction and is denoted by E a . Amino acids: An amino acid is a type of organic acid that contains an acid functional group and an amine functional group Anabolic : Chemical reactions that use simpler substances to form more complex molecules or to store energy, these usually require energy. Anion: An ion that has a negative charge because it has a surplus of electrons. Base: An anionic substance that accepts a proton during an acid-base reaction. Catabolic: Is the breakdown of complex molecules into simpler substances which causes a release of energy that is used to drive chemical reactions like anabolic reactions. Catalyst: A substance that increases the rate of a chemical reaction without itself undergoing any permanent chemical change Catenane: A class of chemical compounds that contains two of more interlocking rings that aren't bonded chemically. Cation: postive charged ions. Chemical formula: states the number and type of atoms is a molecule of a given substance. Chromatography:The separation of a mixture by passing it in solution through a medium in which the components move at different rates. ( color separation) Condensation: The conversion of vapor gas to liquid. Covalent bond:a kind of chemical bond where atoms share the outermost valence electrons. Decomposition: It is the analysis of the separation of a chemical compound into elements or simpler compounds. Usually these aren't desired chemical reactions. Deposition: the direct solidification of a vapor by cooling; the reserve of sublimation ( chen wang ) Electrolyte: A substance which when dissolved produces a solution that conducts electricity. Endothermic (Endergonic): a process which absorbs thermal (heat) energy. Energy: The potential or capacity to do work. It is expressed in joules. Enzyme: A substance produced by a living organism that acts as a catalyst to bring about a specific biochemical reaction Exothermic (Exergonic): a reaction or process that releases energy in the form of heat. Ionic bond: A bond in which electrons are transferred between atoms Ionization:The molecule has the charge by ions getting or losing the electrons in the chemical change. Isomers: E ach of two or more compounds with the same formula but a different arrangment of atoms in the molecute and different properties. (AK) Joule: an SI unit of energy or work represented by "J". Kinetics: The rate of a chemical or biochemical reaction. Lewis structure: a 2D/flat representation (using lines, dots and symbols) of the bonds between atoms of a molecule and the pairs of electrons that are present. Melting point: a temperature at which a solid will begin to melt. molality: a measure of the concentration of a solute in a solution in terms of amount of substance in a specified amount of mass of the solvent Molar Mass: The mass of an element or compound in grams (g); it also equals to molecular weight (amu) Molarity: Is the number of moles of solute divided by the number of liters of solution. Molecular formula: This is the written formula of a compound, including the number of atoms of each element in one molecule. For example, sugar's molecular formula is C 12 H 22 O 11 Oxidation: A reaction in which the atoms of an element lose electrons and the oxidation number of the element is correspondingly increased. Percent yield: The amount of product obtained in a chemical reaction relative to the theoretical yield. Photosynthesis: The process by which green plants and some other organisms use sunlight to synthesize carbohydrates from carbon dioxide and water. Photosynthesis in plants generally involves the green pigment chlorophyll and generates oxygen as a byproduct Protein: any group of complex organic macromolecules that contain carbon, hydrogen, oxygen, nitrogen, and usually sulfur are composed of one or more chains of amino acids Rate of reaction: The speed in which a chemical process or reaction takes place. Reduction: a half reaction Rotaxane: A dumbbell shaped molecule. Solute: The substance that dissolves. For example, sugar is the solute dissolved in a liquid, like water. Solvent: The substance doing the dissolving. For example, water is the solvent of sugar, because it dissolves sugar to become a simple solution of sugar water. Sublimation: a change in the state of matter where a substance changes directly to a gas from a solid state without becoming liquid. For ex, dry ice. Tetrahedral: a tetrahedral molecular shape is when an atom is located at the center with four sub-atoms that are located at the corners. Creating space between each other. Vaporization: Evaporation of a liquid that occurs from the surface of a liquid into a gaseous phase that is not saturated with the evaporating substance. VSEPR: “Valence Shell Electron Pair Repulsion” theory aka “ Gillespie-Nyholm theory” that is based on the repulsive behavior of electron pairs. In conjunction with a Lewis Dot Structure representation, it is used to predict the shape of a molecule. Carvones: ??-carvone ??-carvone __??__ __HCN__ _H2O__ __??__ __C2H4__ __SCN-__ __C2__ __CH20__ __??__ _CO2__ __??__ __??__ __??__ __??__ __B2__ __??__ __??__ __??__ __??__ __??__ __??__ |
Courses/DePaul_University/Physical_Chemistry_for_Biological_Sciences/01%3A_The_Dawn_of_the_Quantum_Theory/1.02%3A_Quantum_Hypothesis_Used_for_Blackbody_Radiation_Law | To understand how energy is quantized in blackbody radiation By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as Maxwell’s equations , developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could not be explained by the theories available at the time. One experimental phenomenon that could not be adequately explained by classical physics was blackbody radiation (Figure 1.2.1 ). Attempts to explain or calculate this spectral distribution from classical theory were complete failures. A theory developed by Rayleigh and Jeans predicted that the intensity should go to infinity at short wavelengths. Since the intensity actually drops to zero at short wavelengths, the Rayleigh-Jeans result was called the ultraviolet catastrophe (Figure 1.2.1 dashed line). There was no agreement between theory and experiment in the ultraviolet region of the blackbody spectrum. Quantizing Electrons in the Radiator In 1900, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is quantized rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in Figure 1.2.1 , so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy). Energy can be gained or lost only in integral multiples of a quantum. Although quantization may seem to be an unfamiliar concept, we encounter it frequently in quantum mechanics (hence the name). For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. It is also similar to going up and down a hill using discrete stair steps rather than being able to move up and down a continuous slope. Your potential energy takes on discrete values as you move from step to step. Even electrical charge is quantized: an ion may have a charge of −1 or −2, but not −1.33 electron charges. Planck's quantization of energy is described by the his famous equation: \[ E=h \nu \label{Eq1.2.1} \] where the proportionality constant \(h\) is called Planck’s constant , one of the most accurately known fundamental constants in science \[h=6.626070040(81) \times 10^{−34}\, J\cdot s \nonumber \] However, for our purposes, its value to four significant figures is sufficient: \[h = 6.626 \times 10^{−34} \,J\cdot s \nonumber \] As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of \(hν\), Planck devised an equation that fit the experimental data shown in Figure 1.2.1 . We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting a large number of lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in Figure 1.2.1 , and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature. At the time he proposed his radical hypothesis, Planck could not explain why energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law. In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew. Max Planck explain the spectral distribution of blackbody radiation as result from oscillations of electrons. Similarly, oscillations of electrons in an antenna produce radio waves. Max Planck concentrated on modeling the oscillating charges that must exist in the oven walls, radiating heat inwards and—in thermodynamic equilibrium—themselves being driven by the radiation field. He found he could account for the observed curve if he required these oscillators not to radiate energy continuously, as the classical theory would demand, but they could only lose or gain energy in chunks, called quanta , of size \(h\nu\), for an oscillator of frequency \(\nu\) (Equation \(\ref{Eq1.2.1} \)). With that assumption, Planck calculated the following formula for the radiation energy density inside the oven: \[ \begin{align} d\rho(\nu,T) &= \rho_\nu (T) d\nu \\[4pt] &= \dfrac {2 h \nu^3}{c^2} \cdot \dfrac {1 }{\exp \left( \dfrac {h\nu}{k_B T}\right)-1} d\nu \label{Eq2a} \end{align} \] with \(\pi = 3.14159\) \(h\) = \(6.626 \times 10^{-34} J\cdot s\) \(c\) = \(3.00 \times 10^{8}\, m/s\) \(\nu\) = \(1/s\) \(k_B\) = \(1.38 \times 10^{-23} J/K\) \(T\) is absolute temperature (in Kelvin) Planck's radiation energy density (Equation \(\ref{Eq2a}\)) can also be expressed in terms of wavelength \(\lambda\). \[\rho (\lambda, T) = \dfrac {2 hc^2}{\lambda ^5} \left(\dfrac {1}{ e^{\dfrac {hc}{\lambda k_B T}} - 1}\right) \label{Eq2b} \] With a wavelength of maximum energy density at: \[ \lambda_{max}=\frac{hc}{4.965kT} \nonumber \] Planck's equation (Equation \(\ref{Eq2b}\)) gave an excellent agreement with the experimental observations for all temperatures (Figure 1.2.2 ). Planck made many substantial contributions to theoretical physics, but his fame as a physicist rests primarily on his role as the originator of quantum theory. In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government. One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler and bombing during the last weeks of World War II destroyed Planck’s home. After WWII, the major German scientific research organization was renamed the Max Planck Society. Use Equation \(\ref{Eq2b}\) to show that the units of \(ρ(λ,T)\,dλ\) are \(J/m^3\) as expected for an energy density. The near perfect agreement of this formula with precise experiments (e.g., Figure 1.2.3 ), and the consequent necessity of energy quantization, was the most important advance in physics in the century. His blackbody curve was completely accepted as the correct one: more and more accurate experiments confirmed it time and again, yet the radical nature of the quantum assumption did not sink in. Planck was not too upset—he didn’t believe it either, he saw it as a technical fix that (he hoped) would eventually prove unnecessary. Part of the problem was that Planck’s route to the formula was long, difficult and implausible—he even made contradictory assumptions at different stages, as Einstein pointed out later. However, the result was correct anyway! The mathematics implied that the energy given off by a blackbody was not continuous, but given off at certain specific wavelengths, in regular increments. If Planck assumed that the energy of blackbody radiation was in the form \[E = nh \nu \nonumber \] where \(n\) is an integer, then he could explain what the mathematics represented. This was indeed difficult for Planck to accept, because at the time, there was no reason to presume that the energy should only be radiated at specific frequencies. Nothing in Maxwell’s laws suggested such a thing. It was as if the vibrations of a mass on the end of a spring could only occur at specific energies. Imagine the mass slowly coming to rest due to friction, but not in a continuous manner. Instead, the mass jumps from one fixed quantity of energy to another without passing through the intermediate energies. To use a different analogy, it is as if what we had always imagined as smooth inclined planes were, in fact, a series of closely spaced steps that only presented the illusion of continuity. Summary The agreement between Planck’s theory and the experimental observation provided strong evidence that the energy of electron motion in matter is quantized. In the next two sections, we will see that the energy carried by light also is quantized in units of \(h \bar {\nu}\). These packets of energy are called “photons.” |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/18%3A_Ethers_and_Epoxides_Thiols_and_Sulfides/18.01%3A_Names_and_Properties_of_Ethers | After completing this section, you should be able to write two acceptable names for a simple dialkyl ether, given its Kekulé, shorthand or condensed structure. name a complicated ether by the IUPAC system, given its Kekulé, shorthand or condensed structure. draw the Kekulé, condensed or shorthand structure of an ether, given an acceptable name. explain why the boiling point of an ether is generally higher than that of an alkane of similar molecular mass. Structure of Ethers Ethers are a class of organic compounds that contain an sp 3 hybridized oxygen between two alkyl groups and have the formula R-O-R'. These compounds are used in dyes, perfumes, oils, waxes and other industrial uses. Aliphatic ethers have no aryl groups directly attached to the ether oxygen. Examples of Aliphatic Ethers Aromatic ethers have at least one aryl ring directly attached to the ether oxygen. In aryl ethers, the lone pair electrons on oxygen are conjugated with the aromatic ring which significantly changes the properties of the ether. Example of Aromatic Ethers The sp 3 hybridization of oxygen gives ethers roughly the same geometry as alcohols and water. The R-O-R' bond angle is close to what is expected in a tetrahedral geometry. The bond angle of dimethyl ether is 112 o which is larger than the H-O-H bond angle in water (104.5 o ) due to the steric repulsion of the methyl groups. The presence of an electronegative oxygen atom gives ethers a small dipole moment with the electron density primarily on oxygen (red and orange in the electrostatic potential map). Comparisons of Physical Properties of Alcohols and Ethers Ethers, unlike alcohols, have no hydrogen atom on the oxygen atom (that is, no OH group). Therefore, there is no intermolecular hydrogen bonding between ether molecules, which makes their boiling points much lower than an alcohol with similar mass. Despite the presence of a small dipole moment, ethers have boiling points that are about the same as alkanes of comparable molar mass. (Table 18.1.1 ). Condensed Structural Formula Name Molar Mass Boiling Point (°C) Intermolecular Hydrogen Bonding in Pure Liquid? CH3CH2CH3 propane 44 –42 no CH3OCH3 dimethyl ether 46 –25 no CH3CH2OH ethyl alcohol 46 78 yes CH3CH2CH2CH2CH3 pentane 72 36 no CH3CH2OCH2CH3 diethyl ether 74 35 no CH3CH2CH2CH2OH butyl alcohol 74 117 yes Ether molecules do have an oxygen atom, however, and engage in hydrogen bonding with water molecules. Consequently, an ether has about the same solubility in water as the alcohol that is isomeric with it. For example, dimethyl ether and ethanol (both having the molecular formula C 2 H 6 O) are completely soluble in water, whereas diethyl ether and 1-butanol (both C 4 H 10 O) are barely soluble in water (8 g/100 mL of water). Peroxide Formation Many ethers can react with oxygen to form explosive peroxide compounds n a free radical process called autoxidation. For this reason ethers should not be stored for long periods of time and should not be stored in glass bottles. The danger is particularly acute when ether solutions are distilled to near dryness. The hydroperoxides can become more concentrated during a distillation because they tend to have a slightly higher boiling point than the corresponding ether. Before performing an ether distillation great care should be taken to test for the presence of peroxides. Naming Ethers When no other functional group is present, simple ethers are often given common functional class names. Both alkyl groups attached to the oxygen atom are named as substituents (in alphabetical order) and then the word ether is added. The common names for alkyl substituents discussed in Section 3.3 are often used. IUPAC nomenclature for ethers should be used for complicated ethers, compounds with more than one ether linkage, and compounds where other functional groups are present with an ether. In these cases, an RO group of the ether is named as an alkoxy substituent. Common alkoxy substituents are given names derived from their alkyl component. The suffix -yl is replaced with -oxy . (Table 18.1.2 ): Alkyl Group Name Unnamed: 2 Alkoxy Group Name.1 CH3– Methyl NaN CH3O– Methoxy CH3CH2– Ethyl NaN CH3CH2O– Ethoxy (CH3)2CH– Isopropyl NaN (CH3)2CHO– Isopropoxy (CH3)3C– tert-Butyl NaN (CH3)3CO– tert-Butoxy C6H5– Phenyl NaN C6H5O– Phenoxy Cyclic Ethers Cyclic ethers are a type of heterocycle with one or more oxygens located in the ring. Many cyclic ethers have common names and are often used as solvents due to their inert nature. These ring structures are also found in many biological molecules such as sugars and DNA. The rings are numbered so that an oxygen gets position 1. Name the following ethers: b) c) d) e) f) g) Answers a) 3-Isopropoxypentane b) 1,3-Dimethoxybenzene c) 2-Methyltetrahydropyran d) Cyclopentyl ethyl ether e) 4-Bromo-1-ethoxybenzene f) Dicyclohexyl ether g) 4-Butoxycyclohexene |
Bookshelves/Inorganic_Chemistry/Organometallic_Chemistry_(Evans)/03%3A_Structural_Fundamentals/3.02%3A_Open_Coordination_Site | The concept of coordinative unsaturation can be confusing for the student of organometallic chemistry, but recognizing open coordination sites in OM complexes is a critical skill. Introduction Let’s begin with a famous example of coordinative unsaturation from organic chemistry. An analogy from organic chemistry. The reactivity of the carbene flows from its open coordination site. Carbenes are both nucleophilic and electrophilic, but the essence of their electrophilicity comes from the fact that they don’t have their fair share of electrons (8). They have not been saturated with electrons—carbenes want more ! To achieve saturation, carbenes may inherit a pair of electrons from a σ bond (σ-bond insertion), π bond (cyclopropanation), or lone pair (ylide formation). Notice that, simply by spotting coordinative unsaturation, we’ve been able to fully describe the carbene’s reactivity! We can do the same with organometallic complexes—open coordination sites suggest specific reactivity patterns. That’s why understanding coordinative unsaturation and recognizing its telltale sign (the open coordination site) are essential skills for the organometallic chemist. Coordinative unsaturation is not just the possession of a coordination number less than 6, or an apparent space in which a ligand might be able to approach. Sixteen or fewer total electrons on the metal center are a second necessity—just as, in the organic case, 6 or fewer electrons on the unsaturated atom are essential. Sixteen or fewer total electrons and coordination number less than 6 add up to a more fundamental synonym for an open coordination site: an empty metal-centered orbital! (<= 16 total electrons on M) + (< 6 coordination number on M) = empty d orbital on M The lesson here is that we can’t just look to the geometry of a complex to determine whether it bears an open coordination site—electron counting is essential too. Like carbenes and carbocations, metal complexes containing open coordination sites don’t just hang around. They react rapidly with all kinds of electron sources. Furthermore, you won’t see them in stable starting materials or products. Thus, recognizing when a complex has the potential for an open site is important. What are some structural signs that point to the possibility of an open coordination site? 1. Weakly coordinating ligands Ligand dissociation from an 18-electron complex produces an open coordination site. Solvent ligands and side-on σ ligands—both of which bind relatively weakly to metals—often engage in this process. Try counting electrons in the two pairs of cationic iridium complexes below. Dissociation of weakly bound ligands reveals open coordination sites. Dissociation of a π-system ligand may also reveal an open coordination site. 2. Reaction conditions encouraging dissociation In this category, we might file away photochemical and amine-oxide-mediated dissociations of CO. Conditions like these encourage the loss of a ligand and subsequent replacement with something else, àla S N 1 substitution. 3. Potential for reductive elimination Reductive elimination is the open coordination site’s dream come true: two (sites) for the price of one (step)! Factors affecting the favorability of reductive elimination are beyond the scope of this post, but we can mention a couple here: steric crowding and an electron-poor metal. Finally, it’s important to note that open coordination sites often show up in fragments of OM complexes examined for one reason or another. We can relate these fragments to organic or main-group intermediates using isolobal analogies, powerful conceptual tools that we’ll explore in detail in another post. For example, the fragment (CO) 5 Cr is isolobal with the organic carbocation—the two sets of frontier MOs are analogous, and both structures have an open coordination site. All three of these analogous fragments bear an open coordination site. |
Courses/De_Anza_College/CHEM_10%3A_Introduction_to_Chemistry_(Parajon_Puenzo)/02%3A_Atoms/2.04%3A_The_Mole_and_Molar_Mass | Learning Objectives Define mole and molar mass. Perform calculations to convert between moles and mass of a substance. Perform calculations to convert between mass of a substance and number of particles. The Mole and Avogadro's Number It certainly is easy to count bananas or to count elephants (as long as you stay out of their way). However, you would be counting grains of sugar from your sugar canister for a long, long time. Atoms and molecules are extremely small—far, far smaller than a grain of sugar. Counting atoms or molecules is not only unwise, it is absolutely impossible. One drop of water contains about \(10^{22}\) molecules of water. If you counted 10 molecules every second for 50 years without stopping, you would have counted only \(1.6 \times 10^{10}\) molecules. Put another way, at that counting rate, it would take you over 30 trillion years to count the water molecules in one tiny drop. Chemists of the past needed a name that stood for a very large number of items. Amadeo Avogadro (1776-1856), an Italian scientist, provided such a number. He is responsible for the counting unit of measure called the mole. A mole \(\left( \text{mol} \right)\) is the amount of a substance that contains \(6.02 \times 10^{23}\) representative particles of that substance. The mole is the SI unit for amount of a substance. Just like the dozen and the gross, it is a name that stands for a number. There are therefore \(6.02 \times 10^{23}\) water molecules in a mole of water molecules. There also would be \(6.02 \times 10^{23}\) bananas in a mole of bananas—if such a huge number of bananas ever existed. The number \(6.02 \times 10^{23}\) is called Avogadro's number , the number of representative particles in a mole. It is an experimentally determined number. A representative particle is the smallest unit in which a substance naturally exists. For the majority of elements, the representative particle is the atom. Iron, carbon, and helium consist of iron atoms, carbon atoms, and helium atoms, respectively. Seven elements exist in nature as diatomic molecules and they are \(\ce{H_2}\), \(\ce{N_2}\), \(\ce{O_2}\), \(\ce{F_2}\), \(\ce{Cl_2}\), \(\ce{Br_2}\), and \(\ce{I_2}\). The representative particle for these elements is the molecule. Likewise, all molecular compounds such as \(\ce{H_2O}\) and \(\ce{CO_2}\) exist as molecules, and so the molecule is their representative particle. For ionic compounds such as \(\ce{NaCl}\) and \(\ce{Ca(NO_3)_2}\), the representative particle is the formula unit. A mole of any substance contains Avogadro's number \(\left( 6.02 \times 10^{23} \right)\) of representative particles. Converting Between Number of Atoms to Moles and Vice Versa We can use Avogadro's number as a conversion factor, or ratio, in dimensional analysis problems. If we are given the number of atoms of an element X, we can convert it into moles by using the relationship \[\text{1 mol X} = 6.022 \times 10^{23} \text{ X atoms} \nonumber \] Example \(\PageIndex{1}\): Moles of Carbon The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is \(4.72 \times 10^{24}\) atoms of carbon? Solution Steps for Problem Solving The element carbon exists in two primary forms: graphite and diamond. How many moles of carbon atoms is \(4.72 \times 10^{24}\) atoms of carbon? Identify the "given" information and what the problem is asking you to "find." Given: \(4.72 \times 10^{24}\) C atoms Find: mol C List other known quantities. \(1\, mol = 6.022 \times 10^{23}\) C atoms Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. \[4.72 \times 10^{24} \: \cancel{\text{C} \: \ce{atoms}} \times \frac{1 \: \text{mol} \: \ce{C}}{6.02 \times 10^{23} \: \cancel{\text{C} \: \ce{atoms}}} = 7.84 \: \text{mol} \: \ce{C} \nonumber \] Think about your result. The given number of carbon atoms was greater than Avogadro's number, so the number of moles of \(\ce{C}\) atoms is greater than 1 mole. Since Avogadro's number is a measured quantity with three significant figures, the result of the calculation is rounded to three significant figures. Molar Mass The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. One mole of isotopically pure carbon-12 has a mass of 12 g. For an element, the molar mass is the mass of 1 mol of atoms of that element; for a covalent molecular compound, it is the mass of 1 mol of molecules of that compound; for an ionic compound, it is the mass of 1 mol of formula units. That is, the molar mass of a substance is the mass (in grams per mole) of 6.022 × 10 23 atoms, molecules, or formula units of that substance. In each case, the number of grams in 1 mol is the same as the number of atomic mass units that describe the atomic mass, the molecular mass, or the formula mass, respectively. The molar mass of any substance is its atomic mass, molecular mass, or formula mass in grams per mole. The periodic table lists the atomic mass of carbon as 12.011 amu; the average molar mass of carbon—the mass of 6.022 × 10 23 carbon atoms—is therefore 12.011 g/mol: Substance (formula) Basic Unit Atomic, Molecular, or Formula Mass (amu) Molar Mass (g/mol) carbon (C) atom 12.011 (atomic mass) 12.011 ethanol (C2H5OH) molecule 46.069 (molecular mass) 46.069 calcium phosphate [Ca3(PO4)2] formula unit 310.177 (formula mass) 310.177 Converting Between Grams and Moles The molar mass of any substance is the mass in grams of one mole of representative particles of that substance. The representative particles can be atoms, molecules, or formula units of ionic compounds. This relationship is frequently used in the laboratory. Suppose that for a certain experiment you need 3.00 moles of calcium chloride \(\left( \ce{CaCl_2} \right)\). Since calcium chloride is a solid, it would be convenient to use a balance to measure the mass that is needed. Dimensional analysis will allow you to calculate the mass of \(\ce{CaCl_2}\) that you should measure, as shown in Example \(\PageIndex{2}\). Example \(\PageIndex{2}\): Calcium Chloride Calculate the mass of 3.00 moles of calcium chloride (CaCl 2 ). Solution Steps for Problem Solving Calculate the mass of 3.00 moles of calcium chloride (CaCl2). Identify the "given" information and what the problem is asking you to "find." Given: 3.00 moles of CaCl2 Find: g CaCl2 List other known quantities. 1 ml CaCl2 = 110.98 g CaCl2 Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. \(3.00 \: \cancel{\text{mol} \: \ce{CaCl_2}} \times \dfrac{110.98 \: \text{g} \: \ce{CaCl_2}}{1 \: \cancel{\text{mol} \: \ce{CaCl_2}}} = 333 \: \text{g} \: \ce{CaCl_2}\) Think about your result. NaN Example \(\PageIndex{3}\): Water How many moles are present in 108 grams of water? Solution Steps for Problem Solving How many moles are present in 108 grams of water? Identify the "given" information and what the problem is asking you to "find." Given: 108 g H2O Find: mol H2O List other known quantities. \(1 \: \text{mol} \: \ce{H_2O} = 18.02 \: \text{g}\) H2O Prepare a concept map and use the proper conversion factor. NaN Cancel units and calculate. \(108 \: \cancel{\text{g} \: \ce{H_2O}} \times \dfrac{1 \: \text{mol} \: \ce{H_2O}}{18.02 \: \cancel{\text{g} \: \ce{H_2O}}} = 5.99 \: \text{mol} \: \ce{H_2O}\) Think about your result. NaN Exercise \(\PageIndex{1}\): Nitrogen Gas What is the mass of \(7.50 \: \text{mol}\) of Nitrogen gas \(\ce{N2}\)? Answer: 210 g Conversions Between Mass and Number of Particles In "Conversions Between Moles and Mass", you learned how to convert back and forth between moles and the number of representative particles. Now you have seen how to convert back and forth between moles and mass of a substance in grams. We can combine the two types of problems into one. Mass and number of particles are both related to moles. To convert from mass to number of particles or vice-versa, it will first require a conversion to moles, as shown in Figure \(\PageIndex{1}\) and Example \(\PageIndex{4}\). Example \(\PageIndex{4}\): Chlorine How many molecules is \(20.0 \: \text{g}\) of chlorine gas, \(\ce{Cl_2}\)? Solution Steps for Problem Solving How many molecules is \(20.0 \: \text{g}\) of chlorine gas, \(\ce{Cl_2}\)? Identify the "given" information and what the problem is asking you to "find." Given: 20.0 g Cl2 Find: # Cl2 molecules List other known quantities. 1 mol Cl2 = 70.90 g Cl2, 1mol Cl2 = 6.022 x 1023 Cl2 molecules Prepare a concept map and use the proper conversion factor. The conversion factors are 1 mole Cl2 over 70.90 grams Cl2, and 6.022 times 10^23 Cl2 molecules over 1 mole Cl2. Cancel units and calculate. \(20.0 \: \cancel{\text{g} \: \ce{Cl_2}} \times \dfrac{1 \: \cancel{\text{mol} \: \ce{Cl_2}}}{70.90 \: \cancel{\text{g} \: \ce{Cl_2}}} \times \dfrac{6.02 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}}{1 \: \cancel{\text{mol} \: \ce{Cl_2}}} = 1.70 \times 10^{23} \: \text{molecules} \: \ce{Cl_2}\) Think about your result. Since the given mass is less than half of the molar mass of chlorine, the resulting number of molecules is less than half of Avogadro's number. Exercise \(\PageIndex{2}\): Calcium Chloride How many formula units are in 25.0 g of CaCl 2 ? Answer: 1.36 x 10 23 CaCl 2 formula units Summary A mole of any substance contains Avogadro's number \(\left( 6.02 \times 10^{23} \right)\) of representative particles. The molar mass of a substance is defined as the mass in grams of 1 mole of that substance. Calculations involving conversions between moles of a material and the mass of that material are described. Calculations are illustrated for conversions between mass and number of particles. |
Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/11%3A_Nucleophilic_Acyl_Substitution_Reactions/11.06%3A_Acyl_Phosphates | Thioester formation Thioesters, which are themselves quite reactive in acyl substitution reactions (but less so than acyl phosphates), play a crucial role in the metabolism of fatty acids The ‘acyl X group’ in a thioester is a thiol. Coenzyme A is a thiol-containing coenzyme that plays a key role in metabolism. Coenzyme A is often abbreviated 'HSCoA' in order to emphasize the importance of the thiol functionality. Coenzyme A serves as a 'carrier' group in lipid biosynthesis, and is attached by a thioester linkage to growing fatty acid chains. Palmityl is shown below as an example of a typical fatty acyl-CoA thioester. As we look at reactions involving thioesters in this and future sections, we will frequently see Coenzyme A playing a key role. We will also see the formation and breaking of thioester linkages between an acyl group and other thiol-containing species, such as a cysteine residue on the enzyme: The term 'thioesterification' refers to the formation of a thioester functional group. In a typical biochemical thioesterification reaction, a carboxylate is first converted into an acyl phosphate (in other words, it is activated), then the acyl phosphate undergoes an acyl substitution reaction with a thiol nucleophile. Thioesterification reaction: Mechanism: 1. activation phase: 2. acyl substitution phase: Fatty acids such as palmitate , from fats and oils in your food, are converted to a coenzyme A thioester prior to being broken down by the fatty acid degradation pathway. A transthioesterification reaction is a thioester to thioester conversion - in other words, an acyl group is transferred from one thiol to another. Transthioesterification: Mechanism: For example, when your body synthesizes fatty acids, the two-carbon fatty acid 'building block' acetyl CoA is first converted to acetyl ACP (EC 2.3.1.38). ACP is an abbreviation for 'Acyl Carrier Protein', a modified protein with a thiol-containing prosthetic group attached to one of its serine side chains. Throughout the fatty acid chain elongation process, the growing hydrocarbon chain remains linked to ACP. The pyruvate dehydrogenase complex (EC 1.2.4.1) catalyzes one of the most central of all central metabolism reactions, the conversion of pyruvate to acetyl-CoA , which links the gycolytic pathway to the citric acid (Krebs) cycle. The reaction is quite complex, and we are not yet equipped to follow it through from start to finish (we will finally be ready to do this in section 17.3). The final step, however, we can understand: it is a transthioesterification , involving a dithiol coenzyme called dihydrolipoamide and coenzyme A. Given the information below, draw out a reasonable mechanism for the reaction. Ubiquitin is a protein which plays a key role in many cellular processes by reversibly attaching to other proteins, thus altering or regulating their function. Recently, a team of researchers uncovered details of the mechanism by which ubiquitin (abbreviated Ub) is transferred by the ubiquitin activating enzyme (abbreviated E1) to target proteins. In the first part of this process, the carboxy terminus of ubiquitin is linked to a cysteine side chain on E1, as shown in the incomplete reaction sequence below. Complete the figure by drawing the structures of species A and B. Formation of esters Esterification refers to the formation of a new ester functional group. In a typical biochemical esterification, a thioester is subjected to nucleophilic attack from an alcohol, leading to the formation of an ester and a thiol. Esterification reaction (from thioester): Mechanism: The reaction below is from the synthesis of triacylglycerol, the form in which fat is stored in our bodies. Phase 1 (transthioesterification): Phase 2 (esterification): The reaction, catalyzed by monoacylglycerol acyltransferase (EC 2.3.1.22), begins (phase 1 above) with a preliminary transthioesterification step in which the fatty acyl group is transferred from coenzyme A to a cysteine residue in the active site of the enzyme. Recall that it is a common strategy for enzymes to first form a covalent link to one substrate before catalyzing the 'main' chemical reaction. In phase 2 of the reaction, the fatty acyl group is now ready to be transferred to glycerol, trading its thioester linkage to the cysteine for a new ester linkage to one of the alcohol groups on glycerol. An esterification reaction has tremendous importance in the history of drug development, a story that we heard in the introduction to this chapter. The discovery of penicillin was arguably one of the most important events in the history of modern medicine. The key functional group in penicillin is the four-membered lactam (recall that a lactam is a cyclic amide). Penicillin, and later g enerations of antibiotic drugs, have saved countless lives from once-deadly bacterial infections. The elucidation of the chemical mechanism of penicillin action was also a milestone in our developing understanding of how drugs function on a molecular level. We now know that penicillin, and closely related drugs such as ampicillin and amoxycillin, work by inhibiting an enzyme that is involved in the construction of the peptide component bacterial cell walls. The details of the wall-building reaction itself are outside the scope of this discussion, but it is enough to know that the process involves the participation of a nucleophilic serine residue in the active site of the enzyme. The penicillin molecule is able to enter the active site, and once inside, the lactam group serves as an electrophilic 'bait' for the nucleophilic serine: Although you might expect that an amide-to-ester conversion such as what is shown above would be energetically unfavorable based on the reactivity trends we have learned, this lactam is in fact much more reactive than an ordinary amide group due to the effect of ring strain: recall from section 3.2 that four-membered rings are highly strained, and considerable energy is released when they are opened. Ring strain also accounts for why penicillin has a tendency to degrade: when in contact with water, the lactam will spontaneously hydrolyze over time, which opens the ring and forms a carboxylate group. Unfortunately, many strains of bacteria have acquired an enzyme called \(\beta \)-lactamase (EC 3.5.2.6), that catalyzes rapid hydrolysis of the lactam ring in penicillin-based drugs, rendering them inactive. These bacteria are consequently resistant to penicillin and related antibiotics. As you are probably aware, the evolution of drug resistance in bacteria is a major, world-wide health problem, and scientists are engaged in a constant battle to develop new antibiotics as the older ones become less and less effective. In a transesterification reaction, one ester is converted into another by an acyl substitution reaction. Mechanism for a transesterification reaction: If studying organic chemistry sometimes gives you a headache, you might want to turn to a transesterification reaction for help. Prostaglandins are a family of molecules that promote a wide range of biological processes, including inflammation. Acetylsalicylic acid, commonly known as aspirin, acts by transferring - through a transesterification reaction - an acetyl group to a serine residue on the enzyme responsible for the biosynthesis of prostaglandin H2 (one member of the prostaglandin family). Acetylation of this serine blocks a channel leading to the active site, effectively shutting down the enzyme, impeding prostaglandin production, and inhibiting the inflammation process that causes headaches. In section 11.8, we will see two laboratory acyl substitution reactions that lead to the formation of aspirin and ibuprofen. Discuss the key structural feature of aspirin that makes it so effective at transferring its acetyl group - in other words, why is the ester group in aspirin more reactive than a typical ester? Amide formation An activated carboxylate group (in other words, acyl phosphate or acyl-AMP) can be converted to an amide through nucleophilic attack by an amine. Mechanism for amide formation: The amino acid biosynthesis pathways provide examples of amide formation in biology. The amino acid glutamine is synthesized in most species by converting the carboxylate side chain of glutamate (another amino acid) to an amide, after first activating the carboxylate by monophosphorylation: (EC 6.3.1.2) A similar process takes place in the synthesis of asparagine from aspartate, except that the activated carboxylate in this case is an acyl-AMP: *In the asparagine synthesis reaction, the ammonia nucleophile actually comes from hydrolysis of a glutamine molecule. A enzyme in bacteria is thought to be responsible for resistance to a class of antibiotics that includes apramycin , ribostamycin and paromomycin . The enzyme catalyzes acetylation of the antibiotic compound with acetyl-CoA as an additional substrate. The structure of acetylated apramycin is shown below. a) Identify the acetyl group that has been transferred to apramycin, (and thus inactivating it). b) What functional group acts as an acetyl group donor? What functional group acts as an acetyl group? c) What is the coproduct of the reaction? |
Bookshelves/Introductory_Chemistry/Beginning_Chemistry_(Ball)/03%3A_Atoms_Molecules_and_Ions/3.01%3A_Prelude_to_Atoms_Molecules_and_Ions | Although not an SI unit, the angstrom (Å) is a useful unit of length. It is one ten-billionth of a meter, or 10 −10 m. Why is it a useful unit? The ultimate particles that compose all matter are about 10 −10 m in size, or about 1 Å. This makes the angstrom a natural—though not approved—unit for describing these particles. The angstrom unit is named after Anders Jonas Ångström, a nineteenth-century Swedish physicist. Ångström's research dealt with light being emitted by glowing objects, including the sun. Ångström studied the brightness of the different colors of light that the sun emitted and was able to deduce that the sun is composed of the same kinds of matter that are present on the earth. By extension, we now know that all matter throughout the universe is similar to the matter that exists on our own planet. Anders Jonas Ångstrom, a Swedish physicist, studied the light coming from the sun. His contributions to science were sufficient to have a tiny unit of length named after him, the angstrom, which is one ten-billionth of a meter. Source: Photo of the sun courtesy of NASA's Solar Dynamics Observatory. |
Ancillary_Materials/Reference/Reference_Tables/Equilibrium_Constants/E5%3A_Acid_Dissociation_Constants_of_Organics | The following table provides p K a and K a values for selected weak acids. All values are from Martell, A. E.; Smith, R. M. Critical Stability Constants , Vols. 1–4. Plenum Press: New York, 1976. Unless otherwise stated, values are for 25 o C and for zero ionic strength. Those values in brackets are considered less reliable. Weak acids are arranged alphabetically by the names of the neutral compounds from which they are derived. In some cases—such as acetic acid—the compound is the weak acid. In other cases—such as for the ammonium ion—the neutral compound is the conjugate base. Chemical formulas or structural formulas are shown for the fully protonated weak acid. Successive acid dissociation constants are provided for polyprotic weak acids; where there is ambiguity, the specific acidic proton is identified. To find the K b value for a conjugate weak base, recall that \[K_\text{a} \times K_\text{b} = K_\text{w} \nonumber\] for a conjugate weak acid, HA, and its conjugate weak base, A – . compound conjugate acid pKa Ka acetic acid \(\ce{CH3COOH}\) 4.757 \(1.75 \times 10^{-5}\) adipic acid NaN 4.42 5.42 \(3.8 \times 10^{-5}\) \(3.8 \times 10^{-6}\) alanine NaN 2.348 (\(\ce{COOH}\)) 9.867 (\(\ce{NH3}\)) \(4.49 \times 10^{-3}\) \(1.36 \times 10^{-10}\) aminobenzene NaN 4.601 \(2.51 \times 10^{-5}\) 4-aminobenzene sulfonic acid NaN 3.232 \(5.86 \times 10^{-4}\) 2-aminobenzoic acid NaN 2.08 (\(\ce{COOH}\)) 4.96 (\(\ce{NH3}\)) \(8.3 \times 10^{-3}\) \(1.1 \times 10^{-5}\) 2-aminophenol (\(T = 20 \text{°C}\)) NaN 4.78 (\(\ce{NH3}\)) 9.97 (OH) \(1.7 \times 10^{-5}\) \(1.05 \times 10^{-10}\) ammonia \(\ce{NH4+}\) 9.244 \(5.70 \times 10^{-10}\) arginine NaN 1.823 (COOH) 8.991 (\(\ce{NH3}\)) [12.48] (\(\ce{NH2}\)) \(1.50 \times 10^{-2}\) \(1.02 \times 10^{-9}\) [\(3.3 \times 10^{-13}\)] arsenic acid \(\ce{H3AsO4}\) 2.24 6.96 11.50 \(5.8 \times 10^{-3}\) \(1.1 \times 10^{-7}\) \(3.2 \times 10^{-12}\) asparagine (\(\mu = 0.1 \text{ M}\)) NaN 2.14 (COOH) 8.72 (\(\ce{NH3}\)) \(7.2 \times 10^{-3}\) \(1.9 \times 10^{-9}\) aspartic acid NaN 1.990 (\(\alpha\)-COOH) 3.900 (\(\beta\)-COOH) 10.002 (\(\ce{NH3}\)) \(1.02 \times 10^{-2}\) \(1.26 \times 10^{-4}\) \(9.95 \times 10^{-11}\) benzoic acid NaN 4.202 \(6.28 \times 10^{-5}\) benzylamine NaN 9.35 \(4.5 \times 10^{-10}\) boric acid (\(pK_\text{a2}, pK_\text{a3: } T = 20 \text{°C}\)) \(\ce{H3BO3}\) 9.236 [12.74] [13.80] \(5.81 \times 10^{-10}\) [\(1.82 \times 10^{-13}\)] [\(1.58 \times 10^{-14}\)] carbonic acid \(\ce{H2CO3}\) 6.352 10.329 \(4.45 \times 10^{-7}\) \(4.69 \times 10^{-11}\) catechol NaN 9.40 12.8 \(4.0 \times 10^{-10}\) \(1.6 \times 10^{-13}\) chloracetic acid \(\ce{ClCH2COOH}\) 2.865 \(1.36 \times 10^{-3}\) chromic acid (\(pK_\text{a1: } T = 20 \text{°C}\)) \(\ce{H2CrO4}\) –0.2 6.51 1.6 \(3.1 \times 10^{-7}\) citric acid NaN 3.128 (COOH) 4.761 (COOH) 6.396 (COOH) \(7.45 \times 10^{-4}\) \(1.73 \times 10^{-5}\) \(4.02 \times 10^{-7}\) cupferron (\(\mu = 0.1 \text{ M}\)) NaN 4.16 \(6.9 \times 10^{-5}\) cysteine NaN [1.71] (COOH) 8.36 (SH) 10.77 (\(\ce{NH3}\)) [\(1.9 \times 10^{-2}\)] \(4.4 \times 10^{-9}\) \(1.7 \times 10^{-11}\) dichloracetic acid \(\ce{Cl2CHCOOH}\) 1.30 \(5.0 \times 10^{-2}\) diethylamine \(\ce{(CH3CH2)2NH2+}\) 10.933 \(1.17 \times 10^{-11}\) dimethylamine \(\ce{(CH3)2NH2+}\) 10.774 \(1.68 \times 10^{-11}\) dimethylglyoxime NaN 10.66 12.0 \(2.2 \times 10^{-11}\) \(1. \times 10^{-12}\) ethylamine \(\ce{CH3CH2NH3+}\) 10.636 \(2.31 \times 10^{-11}\) ethylenediamine \(\ce{+H3NCH2CH2NH3+}\) 6.848 9.928 \(1.42 \times 10^{-7}\) \(1.18 \times 10^{-10}\) ethylenediaminetetracetic acid (EDTA) (\(\mu = 0.1 \text{ M}\)) NaN 0.0 (COOH) 1.5 (COOH) 2.0 (COOH) 2.66 (COOH) 6.16 (NH) 10.24 (NH) 1.0 \(3.2 \times 10^{-2}\) \(1.0 \times 10^{-2}\) \(2.2 \times 10^{-3}\) \(6.9 \times 10^{-7}\) \(5.8 \times 10^{-11}\) formic acid \(\ce{HCOOH}\) 3.745 \(1.80 \times 10^{-4}\) fumaric acid NaN 3.053 4.494 \(8.85 \times 10^{-4}\) \(3.21 \times 10^{-5}\) glutamic acid NaN 2.33 (\(\alpha\)-COOH) 4.42 (\(\lambda\)-COOH) 9.95 (\(\ce{NH3}\)) \(5.9 \times 10^{-3}\) \(3.8\times 10^{-5}\) \(1.12 \times 10^{-10}\) glutamine NaN 2.17 (COOH) 9.01 (\(\ce{NH3}\)) \(6.8 \times 10^{-3}\) \(9.8 \times 10^{-10}\) glycine NaN 2.350 (COOH) 9.778 (\(\ce{NH3}\)) \(4.47 \times 10^{-3}\) \(1.67 \times 10^{-10}\) glycolic acid \(\ce{HOOCH2COOH}\) 3.881 (COOH) \(1.48 \times 10^{-4}\) histidine (\(\mu = 0.1 \text{ M}\)) NaN 1.7 (COOH) 6.02 (NH) 9.08 (\(\ce{NH3}\)) \(2. \times 10^{-2}\) \(9.5 \times 10^{-7}\) \(8.3 \times 10^{-10}\) hydrogen cyanide \(\ce{HCN}\) 9.21 \(6.2 \times 10^{-10}\) hydrogen fluroride \(\ce{HF}\) 3.17 \(6.8 \times 10^{-4}\) hydrogen peroxide \(\ce{H2O2}\) 11.65 \(2.2 \times 10^{-12}\) hydrogen sulfide \(\ce{H2S}\) 7.02 13.9 \(9.5 \times 10^{-8}\) \(1.3 \times 10^{-14}\) hydrogen thiocyanate \(\ce{HSCN}\) 0.9 \(1.3 \times 10^{-1}\) 8-hydroxyquinoline NaN 4.9 (NH) 9.81 (OH) \(1.2 \times 10^{-5}\) \(1.6 \times 10^{-10}\) hydroxylamine \(\ce{HONH3+}\) 5.96 \(1.1 \times 10^{-6}\) hypobromous acid \(\ce{HOBr}\) 8.63 \(2.3 \times 10^{-9}\) hypochlorous acid \(\ce{HOCl}\) 7.53 \(3.0\times 10^{-8}\) hypoiodous acid \(\ce{HOI}\) 10.64 \(2.3 \times 10^{-11}\) iodic acid \(\ce{HIO3}\) 0.77 \(1.7 \times 10^{-1}\) isoleucine NaN 2.319 (COOH) 9.754 (\(\ce{NH3}\)) \(4.8 \times 10^{-3}\) \(1.76 \times 10^{-10}\) leucine NaN 2.329 (COOH) 9.747 (\(\ce{NH3}\)) \(4.69 \times 10^{-3}\) \(1.79 \times 10^{-10}\) lysine (\(\mu = 0.1 \text{ M}\)) NaN 2.04 (COOH) 9.08 (\(\alpha \text{-} \ce{NH3}\)) 10.69 (\(\epsilon \text{-} \ce{NH3}\)) \(9.1 \times 10^{-3}\) \(8.3 \times 10^{-10}\) \(2.0 \times 10^{-11}\) maleic acid NaN 1.910 6.332 \(1.23 \times 10^{-2}\) \(4.66 \times 10^{-7}\) malic acid NaN 3.459 (COOH) 5.097 (COOH) \(3.48 \times 10^{-4}\) \(8.00 \times 10^{-6}\) malonic acid \(\ce{HOOCCH2COOH}\) 2.847 5.696 \(1.42 \times 10^{-3}\) \(2.01 \times 10^{-6}\) methionine (\(\mu = 0.1 \text{ M}\)) NaN 2.20 (COOH) 9.05 (\(\ce{NH3}\)) \(6.3 \times 10^{-3}\) \(8.9 \times 10^{-10}\) methylamine \(\ce{CH3NH3+}\) 10.64 \(2.3 \times 10^{-11}\) 2-methylanaline NaN 4.447 \(3.57 \times 10^{-5}\) 4-methylanaline NaN 5.084 \(8.24 \times 10^{-6}\) 2-methylphenol NaN 10.28 \(5.2 \times 10^{-11}\) 4-methylphenol NaN 10.26 \(5.5 \times 10^{-11}\) nitrilotriacetic acid (\(T = 20 \text{°C}), pK_\text{a1: } \mu = 0.1 \text{ M}\)) NaN 1.1 (COOH) 1.650 (COOH) 2.940 (COOH) 10.334 (\(\ce{NH3}\)) \(8. \times 10^{-2}\) \(2.24 \times 10^{-2}\) \(1.15 \times 10^{-3}\) \(4.63 \times 10^{-11}\) 2-nitrobenzoic acid NaN 2.179 \(6.62 \times 10^{-3}\) 3-nitrobenzoic acid NaN 3.449 \(3.56 \times 10^{-4}\) 4-nitrobenzoic acid NaN 3.442 \(3.61 \times 10^{-4}\) 2-nitrophenol NaN 7.21 \(6.2 \times 10^{-8}\) 3-nitrophenol NaN 8.39 \(4.1 \times 10^{-9}\) 4-nitrophenol NaN 7.15 \(7.1 \times 10^{-8}\) nitrous acid \(\ce{HNO2}\) 3.15 \(7.1 \times 10^{-4}\) oxalic acid \(\ce{H2C2O4}\) 1.252 4.266 \(5.60 \times 10^{-2}\) \(5.42 \times 10^{-5}\) 1,10-phenanthroline NaN 4.86 \(1.38 \times 10^{-5}\) phenol NaN 9.98 \(1.05 \times 10^{-10}\) phenylalanine NaN 2.20 (COOH) 9.31 (\(\ce{NH3}\)) \(6.3 \times 10^{-3}\) \(4.9 \times 10^{-10}\) phosphoric acid \(\ce{H3PO4}\) 2.148 7.199 12.35 \(7.11 \times 10^{-3}\) \(6.32 \times 10^{-8}\) \(4.5 \times 10^{-13}\) phthalic acid NaN 2.950 5.408 \(1.12 \times 10^{-3}\) \(3.91 \times 10^{-6}\) piperdine NaN 11.123 \(7.53 \times 10^{-12}\) proline NaN 1.952 (COOH) 10.650 (NH) \(1.12 \times 10^{-2}\) \(2.29 \times 10^{-11}\) propanoic acid \(\ce{CH3CH2COOH}\) 4.874 \(1.34 \times 10^{-5}\) propylamine \(\ce{CH3CH2CH2NH3+}\) 10.566 \(2.72 \times 10^{-11}\) pyridine NaN 5.229 \(5.90 \times 10^{-6}\) resorcinol NaN 9.30 11.06 \(5.0 \times 10^{-10}\) \(8.7 \times 10^{-12}\) salicylic acid NaN 2.97 (COOH) 13.74 (OH) \(1.1 \times 10^{-3}\) \(1.8 \times 10^{-14}\) serine NaN 2.187 (COOH) 9.209 (\(\ce{NH3}\)) \(6.50 \times 10^{-3}\) \(6.18 \times 10^{-10}\) succinic acid NaN 4.207 5.636 \(6.21 \times 10^{-5}\) \(2.31 \times 10^{-6}\) sulfuric acid \(\ce{H2SO4}\) strong 1.99 — \(1.0 \times 10^{-2}\) sulfurous acid \(\ce{H2SO3}\) 1.91 7.18 \(1.2 \times 10^{-2}\) \(6.6 \times 10^{-8}\) D-tartaric acid NaN 3.036 (COOH) 4.366 (COOH) \(9.20 \times 10^{-4}\) \(4.31 \times 10^{-5}\) threonine NaN 2.088 (COOH) 9.100 (\(\ce{NH3}\)) \(8.17 \times 10^{-3}\) \(7.94 \times 10^{-10}\) thiosulfuric acid \(\ce{H2S2O3}\) 0.6 1.6 \(3. \times 10^{-1}\) \(3. \times 10^{-2}\) trichloracetic acid (\(\mu = 0.1 \text{ M}\)) \(\ce{Cl3CCOOH}\) 0.66 \(2.2 \times 10^{-1}\) triethanolamine \(\ce{(HOCH2CH2)3NH+}\) 7.762 \(1.73 \times 10^{-8}\) triethylamine \(\ce{(CH3CH2)3NH+}\) 10.715 \(1.93 \times 10^{-11}\) trimethylamine \(\ce{(CH3)3NH+}\) 9.800 \(1.58 \times 10^{-10}\) tris(hydroxymethyl)amino methane (TRIS or THAM) \(\ce{(HOCH2)3CNH3+}\) 8.075 \(8.41 \times 10^{-9}\) tryptophan (\(\mu = 0.1 \text{ M}\)) NaN 2.35 (COOH) 9.33 (\(\ce{NH3}\)) \(4.5 \times 10^{-3}\) \(4.7 \times 10^{-10}\) tyrosine (\(pK_\text{a1: } \mu = 0.1 \text{ M}\)) NaN 2.17 (COOH) 9.19 (\(\ce{NH3}\)) 10.47 (OH) \(6.8 \times 10^{-3}\) \(6.5 \times 10^{-10}\) \(3.4 \times 10^{-11}\) valine NaN 2.286 (COOH) 9.718 (\(\ce{NH3}\)) \(5.18 \times 10^{-3}\) \(1.91 \times 10^{-10}\) |
Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/10%3A_Equilibrium/10.05%3A_Disturbing_a_Reaction_at_Equilibrium-_Le_Ch%C3%A2telier%E2%80%99s_Principle/10.5.01%3A_The_Effect_of_a_Concentration_Change_on_Equilibrium | Consider the following system under equilibrium: \[ \underbrace{\ce{Fe^{3+}(aq)}}_{\text{colorless}} + \underbrace{ \ce{SCN^{-}(aq)}}_{\text{colorless}} \rightleftharpoons \underbrace{\ce{FeSCN^{2+}(aq)}}_{\text{red}} \nonumber \] If more \(Fe^{3+}\) is added to the reaction, what will happen? According to Le Chatelier's Principle, the system will react to minimize the stress. Since Fe 3 + is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right , producing more FeSCN 2 + . For this particular reaction, we will be able to see that this has happened, as the solution will become a darker red color. There are a few different ways to state what happens here when more Fe 3 + is added, all of which have the same meaning: equilibrium shifts to the right equilibrium shifts to the product side the forward reaction is favored What changes does this cause in the concentrations of the reaction participants? \(\ce{Fe^{3+}}\) Since this is what was added to cause the stress, the concentration of \(\ce{Fe^{3+}}\) will increase. (A shorthand way to indicate this: \(\ce{[Fe]^{3+}\: \uparrow}\) (Reminder: the square brackets represent "concentration") \(\ce{SCN^{-}(aq)}\) Equilibrium will shift to the right, which will use up the reactants. The concentration of \(\ce{SCN^{-}(aq)}\) will decrease \(\ce{[SCN]^{-}\: \downarrow}\) as the rate of the forward reaction increases. \(\ce{FeSCN^{2+}}\) When the forward reaction rate increases, more products are produced, and the concentration of \(\ce{FeSCN^{2+}}\) will increase. \(\ce{[FeSCN]^{2+}} \uparrow \) How about the value of K eq ? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of K eq will be unchanged. The value of K eq does not change when changes in concentration cause a shift in equilibrium. What if more FeSCN 2 + is added? Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favor the reverse reaction, since the reverse reaction will use up the additional FeSCN 2 + . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}(aq)}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts (in a side reaction) with something already in the reaction. Let's remove SCN - from the system (perhaps by adding some Pb 2 + ions—the lead(II) ions will form a precipitate with SCN - , removing them from the solution). What will happen now? Equilibrium will shift to replace SCN - —the reverse reaction will be favored because that is the direction that produces more SCN - . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored (but also ↓ because it was removed) \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added |
Courses/North_Dakota_State_University/CHEM_354L%3A_Majors_Organic_Chemistry_Lab/Organic_Chemistry_Reactions/Extra_Credit_19 | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/04%3A_Organic_Compounds-_Cycloalkanes_and_their_Stereochemistry/4.07%3A_Conformations_of_Monosubstituted_Cyclohexanes | Objectives After completing this section, you should be able to account for the greater stability of the equatorial conformers of monosubstituted cyclohexanes compared to their axial counterparts, using the concept of 1,3‑diaxial interaction. compare the gauche interactions in butane with the 1,3‑diaxial interactions in the axial conformer of methylcyclohexane. arrange a given list of substituents in increasing or decreasing order of 1,3‑diaxial interactions. Key Terms Make certain that you can define, and use in context, the key term below. 1,3‑diaxial interaction Study Notes 1,3-Diaxial interactions are steric interactions between an axial substituent located on carbon atom 1 of a cyclohexane ring and the hydrogen atoms (or other substituents) located on carbon atoms 3 and 5. Be prepared to draw Newman-type projections for cyclohexane derivatives as the one shown for methylcyclohexane. Note that this is similar to the Newman projections from chapter 3 such as n -butane. Newman projections of methylcyclohexane and n ‑butane Because axial bonds are parallel to each other, substituents larger than hydrogen generally suffer greater steric crowding when they are oriented axial rather than equatorial. Consequently, substituted cyclohexanes will preferentially adopt conformations in which the larger substituents assume equatorial orientation . When the methyl group in the structure above occupies an axial position it suffers steric crowding by the two axial hydrogens located on the same side of the ring. The conformation in which the methyl group is equatorial is more stable, and thus the equilibrium lies in this direction. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane may be determined by the conformational equilibrium of the compound. The corresponding equilibrium constant is related to the energy difference between the conformers, and collecting such data allows us to evaluate the relative tendency of substituents to exist in an equatorial or axial location.A table of these free energy values (sometimes referred to as A values) may be examined by clicking here. Looking at the energy values in this table, it is clear that the apparent "size" of a substituent (in terms of its preference for equatorial over axial orientation) is influenced by its width and bond length to cyclohexane, as evidenced by the fact that an axial vinyl group is less hindered than ethyl, and iodine slightly less than chlorine. We noted earlier that cycloalkanes having two or more substituents on different ring carbon atoms exist as a pair (sometimes more) of configurational stereoisomers. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable and do not easily interconvert, whereas, conformational isomers normally interconvert rapidly. In examining possible structures for substituted cyclohexanes, it is useful to follow two principles: Chair conformations are generally more stable than other possibilities. Substituents on chair conformers prefer to occupy equatorial positions due to the increased steric hindrance of axial locations. Strain values for other cyclohexane substituents can also be considered. The relative steric hindrance experienced by different substituent groups oriented in an axial versus equatorial location on cyclohexane determined the amount of strain generated. The strain generated can be used to evaluate the relative tendency of substituents to exist in an equatorial or axial location. Looking at the energy values in this table, it is clear that as the size of the substituent increases, the 1,3-diaxial energy tends to increase, also. Note that it is the size and not the molecular weight of the group that is important. Table 4.7.1 summarizes some of these strain values values. 0 1 2 3 Substituent -ΔG° (kJ/mol) Substituent -ΔG° (kJ/mol) \(\ce{CH_3\bond{-}}\) 7.4 \(\ce{HO\bond{-}}\) 3.9 \(\ce{CH_2H_5\bond{-}}\) 7.6 \(\ce{N#C\bond{-}}\) 0.8 \(\ce{(CH_3)_2CH\bond{-}}\) 9.2 \(\ce{CH_3O\bond{-}}\) 2.1 \(\ce{(CH_3)_3C\bond{-}}\) 20.0 \(\ce{HO_2C\bond{-}}\) 5.9 \(\ce{F\bond{-}}\) 1.4 \(\ce{H_2C=CH\bond{-}}\) 2.0 \(\ce{Cl\bond{-}}\) 2.4 \(\ce{C_6H_5\bond{-}}\) 12.0 \(\ce{Br\bond{-}}\) 2.4 NaN NaN \(\ce{I\bond{-}}\) 2.4 NaN NaN Exercises Exercise \(\PageIndex{1}\) In the molecule, cyclohexyl ethyne there is little steric strain, why? Answer The ethyne group is linear and therefore does not affect the hydrogens in the 1,3 positions to say to the extent as a bulkier or a bent group (e.g. ethene group) would. This leads to less of a strain on the molecule. Exercise \(\PageIndex{2}\) Calculate the energy difference between the axial and equatorial conformations of bromocyclohexane? Answer Bromocyclohexane will have two 1,3 diaxial interactions. The table above states that the total strain between axial and equatorial bromocyclohexane will be 2.4 kJ/mol. Exercise \(\PageIndex{3}\) Using your answer from the previous question estimate the percentages of axial and equatorial conformations of bromocyclohexane at 25 o C. Answer Remembering that the axial conformation is higher in energy, the energy difference between the two conformations is ΔE = (E equatorial - E axial) = (0 - 2.4 kJ/mol) = -2.4 kJ/mol. After converting o C to Kelvin and kJ/mol to J/mol we can use the equation ΔE = -RT lnK to find that -ΔE/RT = lnK or (2.4 x 10 3 J/mol) / (8.313 kJ/mol K • 298 K) = lnK. From this we calculate that K = 2.6. Because the ring flip reaction is an equilibrium we can say that K = [Equatorial] / [Axial]. If assumption is made that [Equatorial] = X then [Axial] must be 1-X. Plugging these values into the equilibrium expression produces K = [X] / [1-X]. After plugging in the calculated value for K, X can be solved algebraically. 2.6 = [X] / [1-X] → 2.6 - 2.6X = X → 2.6 = 3.6X → 2.6/3.6 = X = 0.72. This means that bromocyclohexane is in the equatorial position 72% of the time and in the axial position 28% of the time. Exercise \(\PageIndex{4}\) There very little in 1,3-diaxial strain when going from a methyl substituent (3.8 kJ/mol) to an ethyl substituent (4.0 kJ/mol), why? It may help to use molecular model to answer this question. Answer The fact that C-C sigma bonds can freely rotate allows the ethyl subsistent to obtain a conformation which places the bulky CH 3 group away from the cyclohexane ring. This forces the ethyl substituent to have only have 1,3- diaxial interactions between hydrogens, which only provides a slight difference to a methyl group. |
Bookshelves/Introductory_Chemistry/Beginning_Chemistry_(Ball)/00%3A_Front_Matter/07%3A_Preface | In 1977, chemists Theodore L. Brown and H. Eugene LeMay (joined in subsequent editions by Bruce Bursten and Julia Burdge) published a general chemistry textbook titled Chemistry: The Central Science. Since that time, the label the central science has become more and more associated with chemistry above all other sciences. Why? Follow along, if you will. Science is grounded, first and foremost, in mathematics. Math is the language of science. Any study of true science must use math as an inescapable tool. The most fundamental science is physics, the study of matter and energy. (For the sake of argument, I include astronomy as part of physics.) Then we progress to the study of the description of matter and how that description can change—that’s chemistry. As this point, however, several directions are possible. Do you want to study the chemistry of living things? That’s biology. The chemistry of the earth? That’s geology. The chemistry of how compounds work in our body? That’s pharmacology. The application of chemistry to better our lives? That’s engineering (chemical engineering, to be more specific, but we’ve just opened the door to the applied sciences). Granted, there are connections between more fundamental sciences and others—geophysics, astrobiology, and so forth—but a map of the sciences and their interconnections shows the most obvious branches after chemistry. This is why we consider chemistry the central science. This concept is reinforced by the fact that many science majors require a course or two of chemistry as part of their curriculum (indeed, perhaps this is the reason you are using this textbook). Do you want to study biology? You’ll need some chemistry courses. Are you a geology major? You’ll need to know some chemistry. Many engineering disciplines, not just chemical engineering, require some background in chemistry as well. The reason that chemistry is required by so many other disciplines is that it is, to overuse the word, central. Chemistry is not just central; it’s all around you. You participate in chemistry every day. This idea is one of the major themes in this book—Introductory Chemistry. Chemistry is all around you, and you practice it every day whether you know it or not. Throughout these chapters, I will attempt to convince you that you play with chemicals every day, perform chemistry every day, and depend on chemistry every day. This is what makes chemistry an integral part, and what should make chemistry an integral part, of the modern literate adult. The goal of this textbook is not to make you an expert. True expertise in any field is a years-long endeavor. Here I will survey some of the basic topics of chemistry. This survey should give you enough knowledge to appreciate the impact of chemistry in everyday life and, if necessary, prepare you for additional instruction in chemistry. The text starts with an introduction to chemistry. Some users might find this a throwaway chapter, but I urge you to look it over. Many people—even scientists—do not know what science really is, and we all can benefit if we learn what science is and, importantly, what science is not. Chemistry, like all sciences, is inherently quantitative, so Chapter 2 "Measurements" discusses measurements and the conventions for expressing them. Yes, chemistry has conventions and arbitrarily adopted, agreed-on standards against which everything is expressed. Students are sometimes dismayed to learn that a hard science like chemistry has arbitrary standards. But then, all fields have their arbitrary standards that experts in that field must master if they are to be considered “experts.” Chemistry, like other sciences, is no different. Chemistry is based on atoms, so that concept comes next. Atoms make molecules, another important topic in chemistry. But atoms and molecules can change—a fundamental concept in chemistry. Therefore, unlike some other competing texts, I introduce chemical change early. Chemistry is little without the concept of chemical change, so I deem it important to introduce the concept as early as possible. Quantity is also important in chemistry—I’m being repetitious. After chemical change comes a discussion of the unit of chemical change, the mole, and how it is used to relate chemicals to each other (a process known as stoichiometry). A discussion of the gas phase comes next—again earlier than in other texts. It is important for students to understand that we can model the physical properties of a phase of matter. Models are a crucial part of science, so reinforcing that idea earlier rather than later gives students a general understanding that they can apply to later material. Energy is also an important topic in chemistry, so now that atoms and molecules, chemical reactions, and stoichiometry have been introduced, I include energy as a quantitative property. With this, the basic topics of chemistry are introduced; the remaining chapters discuss either more applied topics or topics less crucial to their survey of knowledge even if they are fundamental to our understanding: electronic structure, bonding, phases, solutions, acids and bases, chemical equilibrium, oxidation and reduction, and nuclear chemistry. I finish the text with a quick introduction to organic chemistry, if only to whet the appetites of those who thirst to know more. Throughout each chapter, I present two features that reinforce the theme of the textbook—that chemistry is all around you. The first is a feature titled, appropriately, “Chemistry Is Everywhere.” These features examine a topic of the chapter and demonstrate how this topic shows up in everyday life. In Chapter 1 "What Is Chemistry?", “Chemistry Is Everywhere” focuses on the personal hygiene products that you may use every morning: toothpaste, soap, and shampoo, among others. These products are chemicals, aren’t they? Ever wonder about the chemical reactions that they undergo to give you clean and healthy teeth or shiny hair? I will explore some of these chemical reactions in future chapters. But this feature makes it clear that chemistry is, indeed, everywhere. The other feature focuses on chemistry that you likely indulge in every day: eating and drinking. In the “Food and Drink App,” I discuss how the chemistry of the chapter applies to things that you eat and drink every day. Carbonated beverages depend on the behavior of gases, foods contain acids and bases, and we actually eat certain rocks. (Can you guess which rocks without looking ahead?) Cooking, eating, drinking, and metabolism—we are involved with all these chemical processes all the time. These two features allow us to see the things we interact with every day in a new light—as chemistry. Each section starts with one or more Learning Objectives, which are the main points of the section. Key Takeaways, which review the main points, end each section. Each chapter is full of examples to illustrate the key points, and each example is followed by a similar Test Yourself exercise to see if a student understands the concept. Each section ends with its own set of paired exercises to practice the material from that section, and each chapter ends with Additional Exercises that are more challenging or require multiple steps or skills to answer. The mathematical problems in this text have been treated in one of two ways: either as a conversion-factor problem or as a formula problem. It is generally recognized that consistency in problem solving is a positive pedagogical tool. Students and instructors may have different ways to work problems mathematically, and if it is mathematically consistent, the same answer will result. However, I have found it better to approach mathematical exercises in a consistent fashion, without (horrors!) cutesy shortcuts. Such shortcuts may be useful for one type of problem, but if students do not do a problem correctly, they are clueless as to why they went wrong. Having two basic mathematical approaches (converting and formulas) allows the text to focus on the logic of the approach, not the tricks of a shortcut. Inundations of unnecessary data, such as the densities of materials, are minimized for two reasons. First, they contribute nothing to understanding the concepts. Second, as an introductory textbook, this book focuses on the concepts and does not serve as a reference of data. There are other well-known sources of endless data should students need them. Good luck, and good chemistry, to you all! |
Courses/University_of_British_Columbia/CHEM_100%3A_Foundations_of_Chemistry/15%3A_Chemical_Equilibrium/15.04%3A_The_Equilibrium_Constant_-_A_Measure_of_How_Far_a_Reaction_Goes | Learning Objectives Write equilibrium constant expressions. Use equilibrium constant expressions to solve for unknown concentrations. Use known concentrations to solve for the equilibrium constants. Explain what the value of \(K\) means in terms of relative concentrations of reactants and products. In the previous section, you learned about reactions that can reach a state of equilibrium, in which the concentration of reactants and products aren't changing. If these amounts are changing, we should be able to make a relationship between the amount of product and reactant when a reaction reaches equilibrium. The Equilibrium Constant Equilibrium reactions are those that do not go to completion, but are in a state where the reactants are reacting to yield products and the products are reacting to produce reactants. In a reaction at equilibrium, the equilibrium concentrations of all reactants and products can be measured. The equilibrium constant (\(K\)) is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants. Sometimes, subscripts are added to the equilibrium constant symbol \(K\), such as \(K_\text{eq}\), \(K_\text{c}\), \(K_\text{p}\), \(K_\text{a}\), \(K_\text{b}\), and \(K_\text{sp}\). These are all equilibrium constants and are subscripted to indicate special types of equilibrium reactions. There are some rules about writing equilibrium constant expressions that need to be learned: Concentrations of products are multiplied on the top of the expression. Concentrations of reactants are multiplied together on the bottom. Coefficients in the equation become exponents in the equilibrium constant expression. Solids, liquids, and solvents are assigned a value of 1, so their concentrations do not affect the value of K. Example \(\PageIndex{1}\) Write the equilibrium constant expression for: \[\ce{CO} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons \ce{CH_4} \left( g \right) + \ce{H_2O} \left( g \right) \nonumber \] Solution \[K = \dfrac{\left[ \ce{CH_4} \right] \left[ \ce{H_2O} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2} \right]^3} \nonumber \] *Note that the coefficients become exponents. Also, note that the concentrations of products in the numerator are multiplied . The same is true of the reactants in the denominator. Example \(\PageIndex{2}\) Write the equilibrium constant expression for: \[2 \ce{TiCl_3} \left( s \right) + 2 \ce{HCl} \left( g \right) \rightleftharpoons 2 \ce{TiCl_4} \left( s \right) + \ce{H_2} \left( g \right) \nonumber \] Solution \[K = \dfrac{\left[ \ce{H_2} \right]}{\left[ \ce{HCl} \right]^2} \nonumber \] *Note that the solids have a value of 1, and multiplying or dividing by 1 does not change the value of K. Example \(\PageIndex{2}\) Write the equilibrium constant expression for: \[\ce{P_4} \left( s \right) + 6 \ce{Cl_2} \left( g \right) \rightleftharpoons 4 \ce{PCl_3} \left( s \right) \nonumber \] Solution \[K = \dfrac{1}{\left[ \ce{Cl_2} \right]^6} \nonumber \] *Note that the only product is a solid, which is defined to have a value of 1. That leaves just 1 on top in the numerator. Example \(\PageIndex{3}\) Write the equilibrium constant expression for: \[\ce{H_2O} \left( l \right) \rightleftharpoons \ce{H^+} \left( aq \right) + \ce{OH^-} \left( aq \right) \nonumber \] Solution \[K = \left[ \ce{H^+} \right] \left[ \ce{OH^-} \right] \nonumber \] *Note that the water is the solvent, and thus has a value of 1. Dividing by 1 does not change the value of K. Equilibrium Constant Expressions The equilibrium constant value is the ratio of the concentrations of the products over the reactants. This means that we can use the value of \(K\) to predict whether there are more products or reactants at equilibrium for a given reaction. What can the value of K eq tell us about a reaction? If K eq is very large , the concentration of the products is much greater than the concentration of the reactants. The reaction essentially "goes to completion"; all, or most of, the reactants are used up to form the products. If K eq is very small , the concentration of the reactants is much greater than the concentration of the products. The reaction does not occur to any great extent—most of the reactants remain unchanged, and there are few products produced. When K eq is not very large or very small (close to a value of 1) then roughly equal amounts of reactants and products are present at equilibrium. Here are some examples to consider: Reaction Chemical Equations Equilibrium Constant the decomposition of ozone, \(\ce{O3}\) \(\ce{2O_3(g) \rightleftharpoons 3O_2(g)}\) \(\mathrm{K_{eq}=2.0 \times 10^{57}}\) the decomposition of ozone, \(\ce{O3}\) \(K_{eq}\) is very large, indicating that mostly \(\ce{O2}\) is present in an equilibrium system, with very little \(\ce{O3}\). \(K_{eq}\) is very large, indicating that mostly \(\ce{O2}\) is present in an equilibrium system, with very little \(\ce{O3}\). production of nitrogen monoxide \(\ce{N_2(g) + O_2(g) \rightleftharpoons 2NO(g)}\) \(\mathrm{K_{eq}=1.0 \times 10^{-25}}\) production of nitrogen monoxide Very little \(\ce{NO}\) is produced by this reaction; \(\ce{N2}\) and \(\ce{O2}\) do not react readily to produce \(\ce{NO}\) (this is lucky for us—otherwise we would have little oxygen to breath in our atmosphere!). Very little \(\ce{NO}\) is produced by this reaction; \(\ce{N2}\) and \(\ce{O2}\) do not react readily to produce \(\ce{NO}\) (this is lucky for us—otherwise we would have little oxygen to breath in our atmosphere!). reaction of carbon monoxide and water \(\ce{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}\) \(\mathrm{K_{eq}=5.09} \,(\text{at 700 K})\) reaction of carbon monoxide and water The concentrations of the reactants are very close to the concentrations of the products at equilibrium. The concentrations of the reactants are very close to the concentrations of the products at equilibrium. If the equilibrium constant is 1 or nearly 1, it indicates that the molarities of the reactants and products are about the same. If the equilibrium constant value is a large number, like 100, or a very large number, like \(1 \times 10^{15}\), it indicates that the products (numerator) are a great deal larger than the reactants. This means that at equilibrium, the great majority of the material is in the form of products and it is said that the "products are strongly favored". If the equilibrium constant is small, like 0.10, or very small, like \(1 \times 10^{-12}\), it indicates that the reactants are much larger than the products and the reactants are strongly favored. With large \(K\) values, most of the material at equilibrium is in the form of products and with small \(K\) values, most of the material at equilibrium is in the form of the reactants. The equilibrium constant expression is an equation that we can use to solve for \(K\) or for the concentration of a reactant or product. Example \(\PageIndex{4}\) Determine the value of \(K\) for the reaction \[\ce{SO_2} \left( g \right) + \ce{NO_2} \left( g \right) \rightleftharpoons \ce{SO_3} \left( g \right) + \ce{NO} \left( g \right) \nonumber \] when the equilibrium concentrations are: \(\left[ \ce{SO_2} \right] = 1.20 \: \text{M}\), \(\left[ \ce{NO_2} \right] = 0.60 \: \text{M}\), \(\left[ \ce{NO} \right] = 1.6 \: \text{M}\), and \(\left[ \ce{SO_3} \right] = 2.2 \: \text{M}\). Solution Step 1: Write the equilibrium constant expression: \[K = \dfrac{\left[ \ce{SO_3} \right] \left[ \ce{NO} \right]}{\left[ \ce{SO_2} \right] \left[ \ce{NO_2} \right]} \nonumber \] Step 2: Substitute in given values and solve: \[K = \dfrac{\left( 2.2 \right) \left( 1.6 \right)}{\left( 1.20 \right) \left( 0.60 \right)} = 4.9 \nonumber \] Example \(\PageIndex{5}\) Consider the following reaction: \[\ce{CO} \left( g \right) + \ce{H_2O} \left( g \right) \rightleftharpoons \ce{H_2} \left( g \right) + \ce{CO_2} \left( g \right) \nonumber \] with \(K = 1.34\). If the \(\left[ \ce{H_2O} \right] = 0.100 \: \text{M}\), \(\left[ \ce{H_2} \right] = 0.100 \: \text{M}\), and \(\left[ \ce{CO_2} \right] = 0.100 \: \text{M}\) at equilibrium, what is the equilibrium concentration of \(\ce{CO}\)? Solution Step 1: Write the equilibrium constant expression: \[K = \dfrac{\left[ \ce{H_2} \right] \left[ \ce{CO_2} \right]}{\left[ \ce{CO} \right] \left[ \ce{H_2O} \right]} \nonumber \] Step 2: Substitute in given values and solve: \[1.34 = \dfrac{\left( 0.100 \right) \left( 0.100 \right)}{\left[ \ce{CO} \right] \left( 0.100 \right)} \nonumber \] Solving for \(\left[ \ce{CO} \right]\), we get: \(\left[ \ce{CO} \right] = 0.0746 \: \text{M}\) Summary The equilibrium constant expression is a mathematical relationship that shows how the concentrations of the products vary with the concentration of the reactants. If the value of \(K\) is greater than 1, the products in the reaction are favored. If the value of \(K\) is less than 1, the reactants in the reaction are favored. If \(K\) is equal to 1, neither reactants nor products are favored. Vocabulary Equilibrium constant (\(K\)) - A mathematical ratio that shows the concentrations of the products divided by the concentrations of the reactants. Contributions & Attributions Modified by Tom Neils (Grand Rapids Community College) |
Courses/University_of_Illinois_Springfield/CHE_124%3A_General_Chemistry_for_the_Health_Professions_(Morsch_and_Andrews)/07%3A_Energy_and_Chemical_Processes/7.E%3A_Energy_and_Chemical_Processes_(Exercises) | 7.1: Energy and Its Units Concept Review Exercises What is the relationship between energy and heat? What units are used to express energy and heat? Answers Heat is the exchange of energy from one part of the universe to another. Heat and energy have the same units. Joules and calories are the units of energy and heat. Exercises Define energy . What is heat? What is the relationship between a calorie and a joule? Which unit is larger? What is the relationship between a calorie and a kilocalorie? Which unit is larger? Express 1,265 cal in kilocalories and in joules. Express 9,043.3 J in calories and in kilocalories. One kilocalorie equals how many kilojoules? One kilojoule equals how many kilocalories? Many nutrition experts say that an average person needs 2,000 Cal per day from his or her diet. How many joules is this? Baby formula typically has 20.0 Cal per ounce. How many ounces of formula should a baby drink per day if the RDI is 850 Cal? Answers Energy is the ability to do work. 2. Heat is a form of enery (thermal) that can be transferred from one object to another. 1 cal = 4.184 J; the calorie is larger. 4. 1 kilocalorie(1 Cal) = 1000 cal; the kcal is larger. 1.265 kcal; 5,293 J 6. 2161.4 cal; 2.1614 kcal 1 kcal = 4.184 kJ 8. 1 kJ = 0.239 kcal 9. 8.4 × 10 6 J 10. 42.5 oz 7.2: Heat and Temperature Concept Review Exercise 1. Describe the relationship between heat transfer and the temperature change of an object. 2. Describe what happens when two objects that have different temperatures come into contact with one another. Answer 1. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat. 2. The temperature of the hot object decreases and the temperature of the cold object increases as heat is transferred from the hot object to the cold object. The change in temperature of each depends on the identity and properties of each substance. Exercises 1. The melting point of mercury is −38.84 o C. Convert this value to degrees Fahrenheit and Kelvin. 2. A pot of water is set on a hot burner of a stove. What is the direction of heat flow? 3. Some uncooked macaroni is added to a pot of boiling water. What is the direction of heat flow? 4. How much energy in calories is required to heat 150 g of H 2 O from 0°C to 100°C? 5. How much energy in calories is required to heat 125 g of Fe from 25°C to 150°C? 6. If 250 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum? 7. If 195 cal of heat were added to 33.2 g of Hg at 56.2°C, what is the final temperature of the mercury? 8. A sample of copper absorbs 145 cal of energy, and its temperature rises from 37.8°C to 41.7°C. What is the mass of the copper? 9. A large, single crystal of sodium chloride absorbs 98.0 cal of heat. If its temperature rises from 22.0°C to 29.7°C, what is the mass of the NaCl crystal? 10. If 1.00 g of each substance in Table 7.2.1 were to absorb 100 cal of heat, which substance would experience the largest temperature change? 11. If 1.00 g of each substance in Table 7.2.1 were to absorb 100 cal of heat, which substance would experience the smallest temperature change? 12. Determine the heat capacity of a substance if 23.6 g of the substance gives off 199 cal of heat when its temperature changes from 37.9°C to 20.9°C. 13. What is the heat capacity of gold if a 250 g sample needs 133 cal of energy to increase its temperature from 23.0°C to 40.1°C? Answers 1. -37.91 0 F and 234.31 K 2. Heat flows into the pot of water. 3. Heat flows to the macaroni. 4. 15,000 cal 5. 1,690 cal 6. 49.0°C 7. 234°C 8. 404 g 9. 61 g 10. Mercury would experience the largest temperature change. 11. hydrogen (H 2 ) 12. 0.496 cal/g•°C 13. 0.031 cal/g•°C 7.3: Phase Changes Concept Review Exercises Explain what happens when heat flows into or out of a substance at its melting point or boiling point. How does the amount of heat required for a phase change relate to the mass of the substance? What is the direction of heat transfer in boiling water? What is the direction of heat transfer in freezing water? What is the direction of heat transfer in sweating? Answers 1. The energy goes into changing the phase, not the temperature. 2. The amount of heat is a constant per gram of substance. 3. Boiling. Heat is being added to the water to get it from the liquid state to the gas state. 4. Freezing. Heat is exiting the system in order to go from liquid to solid. Another way to look at it is to consider the opposite process of melting. Energy is consumed (endothermic) to melt ice (solid to liquid) so the opposite process (liquid to solid) must be exothermic. 5. Sweating. Heat is consumed to evaporate the moisture on your skin which lowers your temperature. Exercises How much energy is needed to melt 43.8 g of Au at its melting point of 1,064°C? How much energy is given off when 563.8 g of NaCl solidifies at its freezing point of 801°C? What mass of ice can be melted by 558 cal of energy? How much ethanol (C 2 H 5 OH) in grams can freeze at its freezing point if 1,225 cal of heat are removed? What is the heat of vaporization of a substance if 10,776 cal are required to vaporize 5.05 g? Express your final answer in joules per gram. If 1,650 cal of heat are required to vaporize a sample that has a heat of vaporization of 137 cal/g, what is the mass of the sample? What is the heat of fusion of water in calories per mole? What is the heat of vaporization of benzene (C 6 H 6 ) in calories per mole? What is the heat of vaporization of gold in calories per mole? What is the heat of fusion of iron in calories per mole? Answers 670 cal 2. 69,630 cal 3. 6.98 g 4. 27.10 g 8,930 J/g 6. 12.0 g 1,440 cal/mol 8. 7,350 cal/mol 9. 80,600 cal/mol 10. 3,530 cal/mol 7.4: Bond Energies and Chemical Reactions Concept Review Exercises What is the connection between energy and chemical bonds? Why does energy change during the course of a chemical reaction? Two different reactions are performed in two identical test tubes. In reaction A, the test tube becomes very warm as the reaction occurs. In reaction B, the test tube becomes cold. Which reaction is endothermic and which is exothermic? Explain. Classify "burning paper" as endothermic or exothermic processes. Answers Chemical bonds have a certain energy that is dependent on the elements in the bond and the number of bonds between the atoms. Energy changes because bonds rearrange to make new bonds with different energies. Reaction A is exothermic because heat is leaving the system making the test tube feel hot. Reaction B is endothermic because heat is being absorbed by the system making the test tube feel cold. "Burning paper" is exothermic because burning (also known as combustion) releases heat Exercises Using the data in Table 7.4.1, calculate the energy of one C–H bond (as opposed to 1 mol of C–H bonds). Recall that 1 mol = 6.022 x 10 23 C–H bonds Using the data in Table 7.4.1, calculate the energy of one C=C bond (as opposed to 1 mol of C=C bonds). Recall that 1 mol = 6.022 x 10 23 C=C bonds Is a bond-breaking process exothermic or endothermic? Is a bond-making process exothermic or endothermic? Is each chemical reaction exothermic or endothermic? 2SnCl 2 (s) + 33 kcal → Sn(s) + SnCl 4 (s) CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(ℓ) + 213 kcal Is each chemical reaction exothermic or endothermic? C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g) + 137 kJ C(s, graphite) + 1.9 kJ → C(s, diamond) Answers 1. 1.661 × 10 −19 cal 2. 2.408 x 10 -19 cal 3. endothermic 4. exothermic 5. a. endothermic b. exothermic 6. a. exothermic b. endothermic 7.5: The Energy of Biochemical Reactions Concept Review Exercise What is the energy content per gram of proteins, carbohydrates, and fats? Answer proteins and carbohydrates: 4 kcal/g; fats: 9 kcal/g Exercises An 8 oz serving of whole milk has 8.0 g of fat, 8.0 g of protein, and 13 g of carbohydrates. Approximately how many kilocalories does it contain? A serving of potato chips has 160 kcal. If the chips have 15 g of carbohydrates and 2.0 g of protein, about how many grams of fat are in a serving of potato chips? The average body temperature of a person is 37°C, while the average surrounding temperature is 22°C. Is overall human metabolism exothermic or endothermic? Cold-blooded animals absorb heat from the environment for part of the energy they need to survive. Is this an exothermic or an endothermic process? If the reaction ATP → ADP gives off 7.5 kcal/mol, then the reverse process, ADP → ATP requires 7.5 kcal/mol to proceed. How many moles of ADP can be converted to ATP using the energy from 1 serving of potato chips (see Exercise 2)? If the oxidation of glucose yields 670 kcal of energy per mole of glucose oxidized, how many servings of potato chips (see Exercise 2) are needed to provide the same amount of energy? Answers 156 kcal 2. 10.2 g 3. exothermic 4. endothermic 5. 21.3 mol 6. 4.2 servings Additional Exercises Sulfur dioxide (SO 2 ) is a pollutant gas that is one cause of acid rain. It is oxidized in the atmosphere to sulfur trioxide (SO 3 ), which then combines with water to make sulfuric acid (H 2 SO 4 ). Write the balanced reaction for the oxidation of SO 2 to make SO 3 . (The other reactant is diatomic oxygen.) When 1 mol of SO 2 reacts to make SO 3 , 23.6 kcal of energy are given off. If 100 lb (1 lb = 454 g) of SO 2 were converted to SO 3 , what would be the total energy change? Ammonia (NH 3 ) is made by the direct combination of H 2 and N 2 gases according to this reaction: N 2 (g) + 3H 2 (g) → 2NH 3 (g) + 22.0 kcal Is this reaction endothermic or exothermic? What is the overall energy change if 1,500 g of N 2 are reacted to make ammonia? A 5.69 g sample of iron metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H 2 O at 22.6°C. Assuming that the water gained all the heat lost by the iron, what is the final temperature of the H 2 O and Fe? A 5.69 g sample of copper metal was heated in boiling water to 99.8°C. Then it was dropped into a beaker containing 100.0 g of H 2 O at 22.6°C. Assuming that the water gained all the heat lost by the copper, what is the final temperature of the H 2 O and Cu? When 1 g of steam condenses, 540 cal of energy is released. How many grams of ice can be melted with 540 cal? When 1 g of water freezes, 79.9 cal of energy is released. How many grams of water can be boiled with 79.9 cal? The change in energy is +65.3 kJ for each mole of calcium hydroxide [Ca(OH) 2 ] according to the following reaction: Ca(OH) 2 (s) → CaO(s) + H 2 O(g) How many grams of Ca(OH) 2 could be reacted if 575 kJ of energy were available? The thermite reaction gives off so much energy that the elemental iron formed as a product is typically produced in the liquid state: 2Al(s) + Fe 2 O 3 (s) → Al 2 O 3 (s) + 2Fe(ℓ) + 204 kcal How much heat will be given off if 250 g of Fe are to be produced? A normal adult male requires 2,500 kcal per day to maintain his metabolism. Nutritionists recommend that no more than 30% of the calories in a person’s diet come from fat. At 9 kcal/g, what is the maximum mass of fat an adult male should consume daily? At 4 kcal/g each, how many grams of protein and carbohydrates should an adult male consume daily? A normal adult male requires 2,500 kcal per day to maintain his metabolism. At 9 kcal/g, what mass of fat would provide that many kilocalories if the diet was composed of nothing but fats? At 4 kcal/g each, what mass of protein and/or carbohydrates is needed to provide that many kilocalories? The volume of the world’s oceans is approximately 1.34 × 10 24 cm 3 . How much energy would be needed to increase the temperature of the world’s oceans by 1°C? Assume that the heat capacity of the oceans is the same as pure water. If Earth receives 6.0 × 10 22 J of energy per day from the sun, how many days would it take to warm the oceans by 1°C, assuming all the energy went into warming the water? Does a substance that has a small specific heat require a small or large amount of energy to change temperature? Explain. Some biology textbooks represent the conversion of adenosine triphosphate (ATP) to adenosine diphosphate (ADP) and phosphate ions as follows: ATP → ADP + phosphate + energy What is wrong with this reaction? Assuming that energy changes are additive, how much energy is required to change 15.0 g of ice at −15°C to 15.0 g of steam at 115°C? (Hint: you will have five processes to consider.) Answers 2SO 2 + O 2 → 2SO 3 16,700 kcal 2. exothermic 1177 kcal about 23.1°C 4. about 23.0°C 5. 6.76 g 6. 0.148 g 652 g 8. 457 kcal 83.3 g 438 g 10. a. 278 g b. 625 g 11. 1.34 × 10 24 cal 93 days 12. A substance with smaller specific heat requires less energy per unit of mass to raise its temperature, 13. A reactant is missing: H 2 O is missing. 14. Total energy = 11,019 cal |
Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/13%3A_The_Group_13_Elements/13.04%3A_The_Elements/13.4A%3A_Appearance | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Courses/Williams_School/Chemistry_I/06%3A_Chemical_Bonding_and_Molecular_Geometry/6.E%3A_Chemical_Bonding_and_Molecular_Geometry_(Exercises) | 7.1: Ionic Bonding Q7.1.1 Does a cation gain protons to form a positive charge or does it lose electrons? S7.1.1 The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost. Q7.1.2 Iron(III) sulfate [Fe 2 (SO 4 ) 3 ] is composed of Fe 3+ and \(\ce{SO4^2-}\) ions. Explain why a sample of iron(III) sulfate is uncharged. Q7.1.3 Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co? S7.1.3 P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals. Q7.1.4 Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd? Q7.1.5 Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: P Mg Al O Cl Cs S7.1.5 P 3– ; Mg 2+ ; Al 3+ ; O 2– ; Cl – ; Cs + Q7.1.6 Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: I Sr K N S In S7.1.6 I - Sr 2+ K + N 3- S 2- In 3+ Q7.1.7 Write the electron configuration for each of the following ions: As 3– I – Be 2+ Cd 2+ O 2– Ga 3+ Li + (h) N 3– (i) Sn 2+ (j) Co 2+ (k) Fe 2+ (l) As 3+ S7.1.7 [Ar]4 s 2 3 d 10 4 p 6 ; [Kr]4 d 10 5 s 2 5 p 6 1 s 2 [Kr]4 d 10 ; [He]2 s 2 2 p 6 ; [Ar]3 d 10 ; 1 s 2 (h) [He]2 s 2 2 p 6 (i) [Kr]4 d 10 5 s 2 (j) [Ar]3 d 7 (k) [Ar]3 d 6 , (l) [Ar]3 d 10 4 s 2 Q7.1.8 Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): Cl Na Mg Ca K Br Sr (h) F Q7.1.9 Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element: Al Br Sr Li As S S7.1.9 1 s 2 2 s 2 2 p 6 3 s 2 3 p 1 ; Al 3+ : 1 s 2 2 s 2 2 p 6 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 5 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 5 s 2 ; Sr 2+ : 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; 1 s 2 2 s 1 ; Li + : 1 s 2 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 3 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 10 4 s 2 4 p 6 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 4 ; 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 Q7.1.10 From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.) 7.3: Covalent Bonding Why is it incorrect to speak of a molecule of solid NaCl? NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules. What information can you use to predict whether a bond between two atoms is covalent or ionic? Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table: Cl 2 CO MnO NCl 3 CoBr 2 K 2 S CO CaF 2 (h) HI (i) CaO (j) IBr (k) CO 2 ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k) Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond. From its position in the periodic table, determine which atom in each pair is more electronegative: Br or Cl N or O S or O P or S Si or N Ba or P N or K Cl; O; O; S; N; P; N From its position in the periodic table, determine which atom in each pair is more electronegative: N or P N or Ge S or F Cl or S H or C Se or P C or Si From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: C, F, H, N, O Br, Cl, F, H, I F, H, O, P, S Al, H, Na, O, P Ba, H, N, O, As H, C, N, O, F; H, I, Br, Cl, F; H, P, S, O, F; Na, Al, H, P, O; Ba, H, As, N, O From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: As, H, N, P, Sb Cl, H, P, S, Si Br, Cl, Ge, H, Sr Ca, H, K, N, Si Cl, Cs, Ge, H, Sr Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom? N, O, F, and Cl Which is the most polar bond? C–C C–H N–H O–H Se–H Identify the more polar bond in each of the following pairs of bonds: HF or HCl NO or CO SH or OH PCl or SCl CH or NH SO or PO CN or NN HF; CO; OH; PCl; NH; PO; CN Which of the following molecules or ions contain polar bonds? O 3 S 8 \(\ce{O2^2-}\) \(\ce{NO3-}\) CO 2 H 2 S \(\ce{BH4-}\) 7.4: Lewis Symbols and Structures Q7.4.1 Write the Lewis symbols for each of the following ions: As 3– I – Be 2+ O 2– Ga 3+ Li + N 3– S7.4.1 eight electrons: eight electrons: no electrons Be 2+ ; eight electrons: no electrons Ga 3+ ; no electrons Li + ; eight electrons: Q7.4.2 Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements: Cl Na Mg Ca K Br Sr F Q7.4.3 Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: MgS Al 2 O 3 GaCl 3 K 2 O Li 3 N KF (a) ; (b) ; (c) ; (d) ; (e) ; (f) In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element: (a) (b) (c) (d) Write the Lewis structure for the diatomic molecule P 2 , an unstable form of phosphorus found in high-temperature phosphorus vapor. Write Lewis structures for the following: H 2 HBr PCl 3 SF 2 H 2 CCH 2 HNNH H 2 CNH (h) NO – (i) N 2 (j) CO (k) CN – Write Lewis structures for the following: O 2 H 2 CO AsF 3 ClNO SiCl 4 H 3 O + \(\ce{NH4+}\) (h) \(\ce{BF4-}\) (i) HCCH (j) ClCN (k) \(\ce{C2^2+}\) (a) In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule . (b) ; (c) ; (d) ; (e) ; (f) ; (g) ; (h) ; (i) ; (j) ; (k) Write Lewis structures for the following: ClF 3 PCl 5 BF 3 \(\ce{PF6-}\) Write Lewis structures for the following: SeF 6 XeF 4 \(\ce{SeCl3+}\) Cl 2 BBCl 2 (contains a B–B bond) SeF 6 : ; XeF 4 : ; \(\ce{SeCl3+}\): ; Cl 2 BBCl 2 : Write Lewis structures for: \(\ce{PO4^3-}\) \(\ce{ICl4-}\) \(\ce{SO3^2-}\) HONO Correct the following statement: “The bonds in solid PbCl 2 are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl 2 are located on the Cl – ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.” Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb 2+ ion has a 6 s 2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons. Write Lewis structures for the following molecules or ions: SbH 3 XeF 2 Se 8 (a cyclic molecule with a ring of eight Se atoms) Methanol, H 3 COH, is used as the fuel in some race cars. Ethanol, C 2 H 5 OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO 2 and H 2 O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas. Many planets in our solar system contain organic chemicals including methane (CH 4 ) and traces of ethylene (C 2 H 4 ), ethane (C 2 H 6 ), propyne (H 3 CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl 2 CO. Write the Lewis structures for carbon tetrachloride and phosgene. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: 1 s 2 2 s 2 2 p 5 1 s 2 2 s 2 2 p 6 3 s 2 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 4 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 10 4 p 1 The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms. the amino acid serine: urea: pyruvic acid: uracil: carbonic acid: (a) ; (b) ; (c) ; (d) ; (e) A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules. How are single, double, and triple bonds similar? How do they differ? Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond. 7.5: Formal Charges and Resonance Write resonance forms that describe the distribution of electrons in each of these molecules or ions. selenium dioxide, OSeO nitrate ion, \(\ce{NO3-}\) nitric acid, HNO 3 (N is bonded to an OH group and two O atoms) benzene, C 6 H 6 : the formate ion: Write resonance forms that describe the distribution of electrons in each of these molecules or ions. sulfur dioxide, SO 2 carbonate ion, \(\ce{CO3^2-}\) hydrogen carbonate ion, \(\ce{HCO3-}\) (C is bonded to an OH group and two O atoms) pyridine: the allyl ion: (a) ; (b) ; (c) ; (d) ; (e) Write the resonance forms of ozone, O 3 , the component of the upper atmosphere that protects the Earth from ultraviolet radiation. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, \(\ce{NO2-}\). In terms of the bonds present, explain why acetic acid, CH 3 CO 2 H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown: Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond. CO 2 CO (a) (b) CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO 2 has double bonds. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate. Determine the formal charge of each element in the following: HCl CF 4 PCl 3 PF 5 H: 0, Cl: 0; C: 0, F: 0; P: 0, Cl 0; P: 0, F: 0 Determine the formal charge of each element in the following: H 3 O + \(\ce{SO4^2-}\) NH 3 \(\ce{O2^2-}\) H 2 O 2 Calculate the formal charge of chlorine in the molecules Cl 2 , BeCl 2 , and ClF 5 . Cl in Cl 2 : 0; Cl in BeCl 2 : 0; Cl in ClF 5 : 0 Calculate the formal charge of each element in the following compounds and ions: F 2 CO NO – \(\ce{BF4-}\) \(\ce{SnCl3-}\) H 2 CCH 2 ClF 3 SeF 6 (h) \(\ce{PO4^3-}\) Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures: O 3 SO 2 \(\ce{NO2-}\) \(\ce{NO3-}\) ; (b) ; (c) ; (d) Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON? Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH? HOCl Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO? Draw the structure of hydroxylamine, H 3 NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges? The structure that gives zero formal charges is consistent with the actual structure: Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule: IF IF 3 IF 5 IF 7 Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound. NF 3 ; Which of the following structures would we expect for nitrous acid? Determine the formal charges: Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H 2 SO 4 , which has two oxygen atoms and two OH groups bonded to the sulfur. 7.6: Strengths of Ionic and Covalent Bonds Which bond in each of the following pairs of bonds is the strongest? C–C or \(\mathrm{C=C}\) C–N or \(\mathrm{C≡N}\) \(\mathrm{C≡O}\) or \(\mathrm{C=O}\) H–F or H–Cl C–H or O–H C–N or C–O Using the bond energies in Table , determine the approximate enthalpy change for each of the following reactions: \(\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)\) \(\ce{CH4}(g)+\ce{I2}(g)⟶\ce{CH3I}(g)+\ce{HI}(g)\) (c) \(\ce{C2H4}(g)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{2H2O}(g)\) −114 kJ; 30 kJ; (c) −1055 kJ Using the bond energies in Table , determine the approximate enthalpy change for each of the following reactions: \(\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)\) \(\mathrm{H_2C=CH_2}(g)+\ce{H2}(g)⟶\ce{H3CCH3}(g)\) (c) \(\ce{2C2H6}(g)+\ce{7O2}(g)⟶\ce{4CO2}(g)+\ce{6H2O}(g)\) When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: The greater bond energy is in the figure on the left. It is the more stable form. How does the bond energy of HCldiffer from the standard enthalpy of formation of HCl( g )? Using the standard enthalpy of formation data in Appendix G , show how the standard enthalpy of formation of HCl( g ) can be used to determine the bond energy. \(\ce{HCl}(g)⟶\dfrac{1}{2}\ce{H2}(g)+\dfrac{1}{2}\ce{Cl2}(g)\hspace{20px}ΔH^\circ_1=−ΔH^\circ_{\ce f[\ce{HCl}(g)]}\\ \dfrac{1}{2}\ce{H2}(g)⟶\ce{H}(g)\hspace{105px}ΔH^\circ_2=ΔH^\circ_{\ce f[\ce H(g)]}\\ \underline{\dfrac{1}{2}\ce{Cl2}(g)⟶\ce{Cl}(g)\hspace{99px}ΔH^\circ_3=ΔH^\circ_{\ce f[\ce{Cl}(g)]}}\\ \ce{HCl}(g)⟶\ce{H}(g)+\ce{Cl}(g)\hspace{58px}ΔH^\circ_{298}=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3\) \(\begin{align} D_\ce{HCl}=ΔH^\circ_{298}&=ΔH^\circ_{\ce f[\ce{HCl}(g)]}+ΔH^\circ_{\ce f[\ce H(g)]}+ΔH^\circ_{\ce f[\ce{Cl}(g)]}\\ &=\mathrm{−(−92.307\:kJ)+217.97\:kJ+121.3\:kJ}\\ &=\mathrm{431.6\:kJ} \end{align}\) Using the standard enthalpy of formation data in Appendix G , calculate the bond energy of the carbon-sulfur double bond in CS 2 . Using the standard enthalpy of formation data in Appendix G , determine which bond is stronger: the S–F bond in SF 4 ( g ) or in SF 6 ( g )? The S–F bond in SF 4 is stronger. Using the standard enthalpy of formation data in Appendix G , determine which bond is stronger: the P–Cl bond in PCl 3 ( g ) or in PCl 5 ( g )? Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond: The C–C single bonds are longest. Use the bond energy to calculate an approximate value of Δ H for the following reaction. Which is the more stable form of FNO 2 ? Use principles of atomic structure to answer each of the following: 1 The radius of the Ca atom is 197 pm; the radius of the Ca 2+ ion is 99 pm. Account for the difference. The lattice energy of CaO( s ) is –3460 kJ/mol; the lattice energy of K 2 O is –2240 kJ/mol. Account for the difference. (c) Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies. Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol) K 419 3050 Ca 590 1140 The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference. When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4 s electron in Ca requires more energy than removal of the 4 s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2 s electron. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice. For which of the following substances is the least energy required to convert one mole of the solid into separate ions? MgO SrO (c) KF CsF MgF 2 (d) The reaction of a metal, M, with a halogen, X 2 , proceeds by an exothermic reaction as indicated by this equation: \(\ce{M}(s)+\ce{X2}(g)⟶\ce{MX2}(s)\). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers. a large radius vs. a small radius for M +2 a high ionization energy vs. a low ionization energy for M (c) an increasing bond energy for the halogen a decreasing electron affinity for the halogen an increasing size of the anion formed by the halogen The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice. 4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy Which compound in each of the following pairs has the larger lattice energy? Note: Mg 2+ and Li + have similar radii; O 2– and F – have similar radii. Explain your choices. MgO or MgSe LiF or MgO (c) Li 2 O or LiCl Li 2 Se or MgO Which compound in each of the following pairs has the larger lattice energy? Note: Ba 2+ and K + have similar radii; S 2– and Cl – have similar radii. Explain your choices. K 2 O or Na 2 O K 2 S or BaS (c) KCl or BaS BaS or BaCl 2 Na 2 O; Na + has a smaller radius than K + ; BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; BaS; S has a larger charge Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? MgO SrO (c) KF CsF MgF 2 Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? K 2 S K 2 O (c) CaS Cs 2 S CaO (e) The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer. 7.7: Molecular Structure and Polarity Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear. What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical? Explain the difference between electron-pair geometry and molecular structure. Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry. Why is the H–N–H angle in NH 3 smaller than the H–C–H bond angle in CH 4 ? Why is the H–N–H angle in \(\ce{NH4+}\) identical to the H–C–H bond angle in CH 4 ? Explain how a molecule that contains polar bonds can be nonpolar. As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar. As a general rule, MX n molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH 3 (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they? Predict the electron pair geometry and the molecular structure of each of the following molecules or ions: SF 6 PCl 5 (c) BeH 2 \(\ce{CH3+}\) Both the electron geometry and the molecular structure are octahedral. Both the electron geometry and the molecular structure are trigonal bipyramid. (c) Both the electron geometry and the molecular structure are linear. Both the electron geometry and the molecular structure are trigonal planar. Identify the electron pair geometry and the molecular structure of each of the following molecules or ions: \(\ce{IF6+}\) CF 4 (c) BF 3 \(\ce{SiF5-}\) BeCl 2 What are the electron-pair geometry and the molecular structure of each of the following molecules or ions? ClF 5 \(\ce{ClO2-}\) (c) \(\ce{TeCl4^2-}\) PCl 3 SeF 4 \(\ce{PH2-}\) electron-pair geometry: octahedral, molecular structure: square pyramidal; electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; electron-pair geometry: tetrahedral, molecular structure: bent (109°) Predict the electron pair geometry and the molecular structure of each of the following ions: H 3 O + \(\ce{PCl4-}\) (c) \(\ce{SnCl3-}\) \(\ce{BrCl4-}\) ICl 3 XeF 4 (g) SF 2 Identify the electron pair geometry and the molecular structure of each of the following molecules: ClNO (N is the central atom) CS 2 (c) Cl 2 CO (C is the central atom) Cl 2 SO (S is the central atom) SO 2 F 2 (S is the central atom) XeO 2 F 2 (Xe is the central atom) (g) \(\ce{ClOF2+}\) (Cl is the central atom) electron-pair geometry: trigonal planar, molecular structure: bent (120°); electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: tetrahedral, molecular structure: tetrahedral; electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal Predict the electron pair geometry and the molecular structure of each of the following: IOF 5 (I is the central atom) POCl 3 (P is the central atom) (c) Cl 2 SeO (Se is the central atom) ClSO + (S is the central atom) F 2 SO (S is the central atom) \(\ce{NO2-}\) (g) \(\ce{SiO4^4-}\) Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments? ClF 5 \(\ce{ClO2-}\) (c) \(\ce{TeCl4^2-}\) PCl 3 SeF 4 \(\ce{PH2-}\) (g) XeF 2 All of these molecules and ions contain polar bonds. Only ClF 5 , \(\ce{ClO2-}\), PCl 3 , SeF 4 , and \(\ce{PH2-}\) have dipole moments. Which of the molecules and ions in Exercise contain polar bonds? Which of these molecules and ions have dipole moments? H 3 O + \(\ce{PCl4-}\) (c) \(\ce{SnCl3-}\) \(\ce{BrCl4-}\) ICl 3 XeF 4 (g) SF 2 Which of the following molecules have dipole moments? CS 2 SeS 2 (c) CCl 2 F 2 PCl 3 (P is the central atom) ClNO (N is the central atom) SeS 2 , CCl 2 F 2 , PCl 3 , and ClNO all have dipole moments. Identify the molecules with a dipole moment: SF 4 CF 4 (c) Cl 2 CCBr 2 CH 3 Cl H 2 CO The molecule XF 3 has a dipole moment. Is X boron or phosphorus? P The molecule XCl 2 has a dipole moment. Is X beryllium or sulfur? Is the Cl 2 BBCl 2 molecule polar or nonpolar? nonpolar There are three possible structures for PCl 2 F 3 with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them. Describe the molecular structure around the indicated atom or atoms: the sulfur atom in sulfuric acid, H 2 SO 4 [(HO) 2 SO 2 ] the chlorine atom in chloric acid, HClO 3 [HOClO 2 ] (c) the oxygen atom in hydrogen peroxide, HOOH the nitrogen atom in nitric acid, HNO 3 [HONO 2 ] the oxygen atom in the OH group in nitric acid, HNO 3 [HONO 2 ] the central oxygen atom in the ozone molecule, O 3 (g) each of the carbon atoms in propyne, CH 3 CCH (h) the carbon atom in Freon, CCl 2 F 2 (i) each of the carbon atoms in allene, H 2 CCCH 2 tetrahedral; trigonal pyramidal; (c) bent (109°); trigonal planar; bent (109°); bent (109°); (g) CH 3 CCH tetrahedral, CH 3 CCH linear; (h) tetrahedral; (i) H 2 CCCH 2 linear; H 2 CCCH 2 trigonal planar Draw the Lewis structures and predict the shape of each compound or ion: CO 2 \(\ce{NO2-}\) (c) SO 3 \(\ce{SO3^2-}\) A molecule with the formula AB 2 , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape. A molecule with the formula AB 3 , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate: \(\ce{CS3^2-}\) CS 2 (c) CS predict the molecular shapes for \(\ce{CS3^2-}\) and CS 2 and explain how you arrived at your predictions (a) ; (b) ; (c) ; \(\ce{CS3^2-}\) includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS 2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear What is the molecular structure of the stable form of FNO 2 ? (N is the central atom.) A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure? The Lewis structure is made from three units, but the atoms must be rearranged: Use the simulation to perform the following exercises for a two-atom molecule: Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A. With a partial positive charge on A, turn on the electric field and describe what happens. (c) With a small partial negative charge on A, turn on the electric field and describe what happens. Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens. Use the simulation to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles. Sketch the bond dipoles and molecular dipole (if any) for O 3. Explain your observations. Look at the bond dipoles for NH 3 . Use these dipoles to predict whether N or H is more electronegative. (c) Predict whether there should be a molecular dipole for NH 3 and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis. The molecular dipole points away from the hydrogen atoms. Use the Molecule Shape simulator to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers. Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select H 2 O. Switch between the “real” and “model” modes. Explain the difference observed. The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°. Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select “model” mode and S 2 O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle. |
Courses/Howard_University/General_Chemistry%3A_An_Atoms_First_Approach/Unit_4%3A__Thermochemistry/Chapter_10%3A_Gases/Chapter_10.2%3A_Gas_Pressure | 0 1 NaN Howard University General Chemistry: An Atoms First Approach Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Unit 1: Atomic Theory Unit 2: Molecular Structure Unit 3: Stoichiometry Unit 4: Thermochem & Gases Unit 5: States of Matter Unit 6: Kinetics & Equilibria Unit 7: Electro & Thermo Chemistry Unit 8: Materials Learning Objectives To describe and measure the pressure of a gas. At the macroscopic level, a complete physical description of a sample of a gas requires four quantities: temperature (expressed in kelvins), volume (expressed in liters), amount (expressed in moles), and pressure (in atmospheres). As we explain in this section and Section 10.3 , these variables are not independent. If we know the values of any three of these quantities, we can calculate the fourth and thereby obtain a full physical description of the gas. Temperature, volume, and amount have been discussed in previous chapters. We now discuss pressure and its units of measurement. Units of Pressure Any object, whether it is your computer, a person, or a sample of gas, exerts a force on any surface with which it comes in contact. The air in a balloon, for example, exerts a force against the interior surface of the balloon, and a liquid injected into a mold exerts a force against the interior surface of the mold, just as a chair exerts a force against the floor because of its mass and the effects of gravity. If the air in a balloon is heated, the increased kinetic energy of the gas eventually causes the balloon to burst because of the increased pressure( P ) The amount of force (F)(A)P = F/A of the gas, the force ( F ) per unit area ( A ) of surface: \( P=\dfrac{F}{A} \tag{10.2.1}\) Pressure is dependent on both the force exerted and the size of the area to which the force is applied. We know from Equation 10.2.1 that applying the same force to a smaller area produces a higher pressure. When we use a hose to wash a car, for example, we can increase the pressure of the water by reducing the size of the opening of the hose with a thumb. The units of pressure are derived from the units used to measure force and area. In the English system, the units of force are pounds and the units of area are square inches, so we often see pressure expressed in pounds per square inch (lb/in 2 , or psi), particularly among engineers. For scientific measurements, however, the SI units for force are preferred. The SI unit for pressure, derived from the SI units for force (newtons) and area (square meters), is the newton per square meter (N/m 2 ), which is called the pascal (Pa) The SI unit for pressure. The pascal is newtons per square meter: N/m 2 , after the French mathematician Blaise Pascal (1623–1662): \( 1\;Pa=1\;N/m^{2} \tag{10.2.1}\) To convert from pounds per square inch to pascals, multiply psi by 6894.757 [1 Pa = 1 psi (6894.757)]. Blaise Pascal (1623–1662) In addition to his talents in mathematics (he invented modern probability theory), Pascal did research in physics and was an author and a religious philosopher as well. His accomplishments include invention of the first syringe and the first digital calculator and development of the principle of hydraulic pressure transmission now used in brake systems and hydraulic lifts. Example 10.2.1 Assuming a paperback book has a mass of 2.00 kg, a length of 27.0 cm, a width of 21.0 cm, and a thickness of 4.5 cm, what pressure does it exert on a surface if it is lying flat? standing on edge in a bookcase? Given: mass and dimensions of object Asked for: pressure Strategy: A Calculate the force exerted by the book and then compute the area that is in contact with a surface. B Substitute these two values into Equation 10.2.1 to find the pressure exerted on the surface in each orientation. Solution: The force exerted by the book does not depend on its orientation. Recall from Section 9.1 that the force exerted by an object is F = ma , where m is its mass and a is its acceleration. In Earth’s gravitational field, the acceleration is due to gravity (9.8067 m/s 2 at Earth’s surface). In SI units, the force exerted by the book is therefore F = ma = (2.00 kg)(9.8067 m/s 2 ) = 19.6 (kg·m)/s 2 = 19.6 N A We calculated the force as 19.6 N. When the book is lying flat, the area is (0.270 m)(0.210 m) = 0.0567 m 2 . B The pressure exerted by the text lying flat is thus \( P= \dfrac{19.6 \; N}{0.0567 \; m^{2}} = 3.46\times 10^{2} \; Pa \) A If the book is standing on its end, the force remains the same, but the area decreases: \( \left (21.0 \;cm \right )\left (4.5 \;cm \right ) = \left (0.210 \;m \right )\left (0.045 \;m \right ) = 9.5 \times × 10^{-3} \; m^{2} \) B The pressure exerted by the book in this position is thus \( P= \dfrac{19.6 \; N}{9.5\times 10^{-3} \; m^{2}} = 2.1 \times 10^{3} \; Pa \) Thus the pressure exerted by the book varies by a factor of about six depending on its orientation, although the force exerted by the book does not vary. Exercise What pressure does a 60.0 kg student exert on the floor when standing flat-footed in the laboratory in a pair of tennis shoes (the surface area of the soles is approximately 180 cm 2 )? as she steps heel-first onto a dance floor wearing high-heeled shoes (the area of the heel = 1.0 cm 2 )? Answers: 3.27 × 10 4 Pa (4.74 lb/in. 2 ) 5.9 × 10 6 Pa (8.5 × 10 2 lb/in. 2 ) Atmospheric Pressure Just as we exert pressure on a surface because of gravity, so does our atmosphere. We live at the bottom of an ocean of gases that becomes progressively less dense with increasing altitude. Approximately 99% of the mass of the atmosphere lies within 30 km of Earth’s surface, and half of it is within the first 5.5 km ( Figure 9.3 ). Every point on Earth’s surface experiences a net pressure called atmospheric pressure . The pressure exerted by the atmosphere is considerable: a 1.0 m 2 column, measured from sea level to the top of the atmosphere, has a mass of about 10,000 kg, which gives a pressure of about 100 kPa: \( Pressure = \dfrac{\left ( 1.0\times 10^{4} \; kg \right )\left ( 9,807 \cancel{m}/s^{2} \right )}{1.0 \; m^{\cancel{2}}} = 0.98 \times 10^{5} \; Pa =98 \; kPa \tag{10.2.3}\) Figure 10.2.1 Atmospheric Pressure Each square meter of Earth’s surface supports a column of air that is more than 200 km high and weighs about 10,000 kg at Earth’s surface, resulting in a pressure at the surface of 1.01 × 10 5 N/m 2 . This corresponds to a pressure of 101 kPa = 760 mmHg = 1 atm. In English units, this is about 14 lb/in. 2 , but we are so accustomed to living under this pressure that we never notice it. Instead, what we notice are changes in the pressure, such as when our ears pop in fast elevators in skyscrapers or in airplanes during rapid changes in altitude. We make use of atmospheric pressure in many ways. We can use a drinking straw because sucking on it removes air and thereby reduces the pressure inside the straw. The atmospheric pressure pushing down on the liquid in the glass then forces the liquid up the straw. Atmospheric pressure can be measured using a barometer A device used to measure atmospheric pressure. , a device invented in 1643 by one of Galileo’s students, Evangelista Torricelli (1608–1647). A barometer may be constructed from a long glass tube that is closed at one end. It is filled with mercury and placed upside down in a dish of mercury without allowing any air to enter the tube. Some of the mercury will run out of the tube, but a relatively tall column remains inside ( Figure 10.2.1 ). Why doesn’t all the mercury run out? Gravity is certainly exerting a downward force on the mercury in the tube, but it is opposed by the pressure of the atmosphere pushing down on the surface of the mercury in the dish, which has the net effect of pushing the mercury up into the tube. Because there is no air above the mercury inside the tube in a properly filled barometer (it contains a vacuum ), there is no pressure pushing down on the column. Thus the mercury runs out of the tube until the pressure exerted by the mercury column itself exactly balances the pressure of the atmosphere. Under normal weather conditions at sea level, the two forces are balanced when the top of the mercury column is approximately 760 mm above the level of the mercury in the dish, as shown in Figure 10.2.2 This value varies with meteorological conditions and altitude. In Denver, Colorado, for example, at an elevation of about 1 mile, or 1609 m (5280 ft), the height of the mercury column is 630 mm rather than 760 mm. Figure 10.2.2 A Mercury Barometer The pressure exerted by the atmosphere on the surface of the pool of mercury supports a column of mercury in the tube that is about 760 mm tall. Because the boiling point of mercury is quite high (356.73°C), there is very little mercury vapor in the space above the mercury column. Mercury barometers have been used to measure atmospheric pressure for so long that they have their own unit for pressure: the millimeter of mercury (mmHg) A unit of pressure, often called the torr. , often called the torr A unit of pressure. One torr is the same as 1 mmHg. , after Torricelli. Standard atmospheric pressure The atmospheric pressure required to support a column of mercury exactly 760 mm tall, which is also referred to as 1 atmosphere (atm). is the atmospheric pressure required to support a column of mercury exactly 760 mm tall; this pressure is also referred to as 1 atmosphere (atm) Also referred to as standard atmospheric pressure, it is the atmospheric pressure required to support a column of mercury exactly 760 mm tall. . These units are also related to the pascal: \( 1 \; atm = 760 \; mmHg = 760 \; torr = 1.01325 \times × 10^{5} \;Pa = 101.325 \; kPa \tag{10.2.4}\) Thus a pressure of 1 atm equals 760 mmHg exactly and is approximately equal to 100 kPa. While mercury barometers were the workhorses for pressure measurement into the last quarter of the 20th century, they have been replaced by electronic gauges. One motivation was safety. Mercury and mercury vapor are heavy metal poisons. The oldest and simplest replacement for mercury barometers are aneroid gauges. The simplest version of an aneroid gauge (Figure 6.2.3) has a thin metal diaphram which expands and contracts. The diaphragm is connected to a dial by a mechanical linkage. Aneroid gauges can either be absolute or differential. Differential aneroid gauges compare the pressure in the gauge housing to that being measured. A common use for aneroid gauges was airplane altimeters. Figure 6.2.2 An Aneroid Barometer An Aneroid Barometer uses the expansion/contraction of a metal diaphragm to move the indicator dial. This drawing from the Wikipedia shows a simple aneroid barometer used for an airplane altimeter. Most modern pressure sensors are based on strain gauge which convert pressure into a strain on a semiconductor element. The strain is converted into an electrical signal by the piezoelectric effect as a change in resistance monitored by a resistance bridge. Alternatively pressure is related to the change in frequency of a quartz crystal oscillator. Modern balances are also based on this later principle. A very accurate type of pressure sensor uses a metal diaphragm as one part of a capacitor. As the pressure changes the diaphragm moves altering the capacitance. Handbooks are available from sensor manufacturers with details including OMEGA and WIKA . Below atmospheric pressures another type of gauge is used based on measuring temperature change of a hot wire as a function of pressure. This is best suited to pressures below atmospheric. Older types called thermocouple or Pirani gauges measure only up to a few Torr. Modern variations called convection gauges can measure up to atmospheric pressure A summary of the various types of pressure gauge in use today can be found in a technical note at the Kurt Lesker site Pressure gauges specify whether the measurement is absolute (relative to zero pressure) or gauge (relative to the standard atmosphere). One must be careful about which kind a gauge is when buying or using one.. Example 10.2.2 One of the authors visited Rocky Mountain National Park several years ago. After departing from an airport at sea level in the eastern United States, he arrived in Denver (altitude 5280 ft), rented a car, and drove to the top of the highway outside Estes Park (elevation 14,000 ft). He noticed that even slight exertion was very difficult at this altitude, where the atmospheric pressure is only 454 mmHg. Convert this pressure to atmospheres. kilopascals. Given: pressure in millimeters of mercury Asked for: pressure in atmospheres and kilopascals Strategy: Use the conversion factors in Equation 10.2.4 to convert from millimeters of mercury to atmospheres and kilopascals. Solution: From Equation 10.2,4 , we have 1 atm = 760 mmHg = 101.325 kPa. The pressure at 14,000 ft in atm is thus \( P=\left ( 454 \; \cancel{mmHg} \right )\left ( \dfrac{1 \; atm}{760 \; \cancel{mmHg}} \right )=0.597 \; atm\) The pressure in kPa is given by \( P=\left ( 0.597 \; \cancel{atm} \right )\left ( \dfrac{101.325 \; kPa}{1 \; \cancel{atm}} \right )=80.5 \; kPa\) Exercise Mt. Everest, at 29,028 ft above sea level, is the world’s tallest mountain. The normal atmospheric pressure at this altitude is about 0.308 atm. Convert this pressure to millimeters of mercury. kilopascals. Answer: a. 234 mmHg; b. 31.2 kPa Manometers Barometers measure atmospheric pressure, but manometers A device used to measure the pressures of samples of gases contained in an apparatus. measure the pressures of samples of gases contained in an apparatus. The key feature of a manometer is a U-shaped tube containing mercury (or occasionally another nonvolatile liquid). A closed-end manometer is shown schematically in part (a) in Figure 10.2.3 . When the bulb contains no gas (i.e., when its interior is a near vacuum), the heights of the two columns of mercury are the same because the space above the mercury on the left is a near vacuum (it contains only traces of mercury vapor). If a gas is released into the bulb on the right, it will exert a pressure on the mercury in the right column, and the two columns of mercury will no longer be the same height. The difference between the heights of the two columns is equal to the pressure of the gas. Figure 10.2.3 The Two Types of Manometer (a) In a closed-end manometer, the space above the mercury column on the left (the reference arm) is essentially a vacuum ( P ≈ 0), and the difference in the heights of the two columns gives the pressure of the gas contained in the bulb directly. (b) In an open-end manometer, the left (reference) arm is open to the atmosphere ( P ≈ 1 atm), and the difference in the heights of the two columns gives the difference between atmospheric pressure and the pressure of the gas in the bulb. If the tube is open to the atmosphere instead of closed, as in the open-end manometer shown in part (b) in Figure 10.2.3 , then the two columns of mercury have the same height only if the gas in the bulb has a pressure equal to the atmospheric pressure. If the gas in the bulb has a higher pressure, the mercury in the open tube will be forced up by the gas pushing down on the mercury in the other arm of the U-shaped tube. The pressure of the gas in the bulb is therefore the sum of the atmospheric pressure (measured with a barometer) and the difference in the heights of the two columns. If the gas in the bulb has a pressure less than that of the atmosphere, then the height of the mercury will be greater in the arm attached to the bulb. In this case, the pressure of the gas in the bulb is the atmospheric pressure minus the difference in the heights of the two columns. Example 10.2.3 Suppose you want to construct a closed-end manometer to measure gas pressures in the range 0.000–0.200 atm. Because of the toxicity of mercury, you decide to use water rather than mercury. How tall a column of water do you need? (At 25°C, the density of water is 0.9970 g/cm 3 ; the density of mercury is 13.53 g/cm 3 .) Given: pressure range and densities of water and mercury Asked for: column height Strategy: A Calculate the height of a column of mercury corresponding to 0.200 atm in millimeters of mercury. This is the height needed for a mercury-filled column. B From the given densities, use a proportion to compute the height needed for a water-filled column. Solution: A In millimeters of mercury, a gas pressure of 0.200 atm is \( P=\left ( 0.200 \; \cancel{atm} \right )\left ( \dfrac{760 \; mmHg}{1 \; \cancel{atm}} \right )=152 \; mmHg\) Using a mercury manometer, you would need a mercury column at least 152 mm high. B Because water is less dense than mercury, you need a taller column of water to achieve the same pressure as a given column of mercury. The height needed for a water-filled column corresponding to a pressure of 0.200 atm is proportional to the ratio of the density of mercury (d Hg )(d H 2 O ) \( \begin{matrix} \left ( height_{H_{2}O} \right )\left ( d_{H_{2}O} \right )=\left ( height_{Hg} \right )\left ( d_{Hg} \right ) \\ \\height_{H_{2}O}=\left ( height_{Hg} \right )\dfrac{d_{Hg}}{d_{H_{2}O}} \\ \\= \left ( 152 \; mmHg\right )\left ( \dfrac{13.53 \; \cancel{g/cm^{2}} }{0.9970 \; \cancel{g/cm^{2}}} \right ) \\ \\=2.06\times 10^{3} \; mm\; H_{2}O \end{matrix}\) This answer makes sense: it takes a taller column of a less dense liquid to achieve the same pressure. Exercise Suppose you want to design a barometer to measure atmospheric pressure in an environment that is always hotter than 30°C. To avoid using mercury, you decide to use gallium, which melts at 29.76°C; the density of liquid gallium at 25°C is 6.114 g/cm 3 . How tall a column of gallium do you need if P = 1.00 atm? Answer: 1.68 m The answer to Example 4 also tells us the maximum depth of a farmer’s well if a simple suction pump will be used to get the water out. If a column of water 2.06 m high corresponds to 0.200 atm, then 1.00 atm corresponds to a column height of \( \begin{matrix} \dfrac{h}{2.06 \; m} = \dfrac{1.00 \; \cancel{atm}}{0.200 \; \cancel{atm}} \\ \\h= 10.3 \; m \end{matrix}\) A suction pump is just a more sophisticated version of a straw: it creates a vacuum above a liquid and relies on atmospheric pressure to force the liquid up a tube. If 1 atm pressure corresponds to a 10.3 m (33.8 ft) column of water, then it is physically impossible for atmospheric pressure to raise the water in a well higher than this. Until electric pumps were invented to push water mechanically from greater depths, this factor greatly limited where people could live because obtaining water from wells deeper than about 33 ft was difficult. Summary Four quantities must be known for a complete physical description of a sample of a gas: temperature , volume , amount , and pressure . Pressure is force per unit area of surface; the SI unit for pressure is the pascal (Pa) , defined as 1 newton per square meter (N/m 2 ). The pressure exerted by an object is proportional to the force it exerts and inversely proportional to the area on which the force is exerted. The pressure exerted by Earth’s atmosphere, called atmospheric pressure , is about 101 kPa or 14.7 lb/in. 2 at sea level. Atmospheric pressure can be measured with a barometer , a closed, inverted tube filled with mercury. The height of the mercury column is proportional to atmospheric pressure, which is often reported in units of millimeters of mercury (mmHg) , also called torr . Standard atmospheric pressure , the pressure required to support a column of mercury 760 mm tall, is yet another unit of pressure: 1 atmosphere (atm) . A manometer is an apparatus used to measure the pressure of a sample of a gas. Key Takeaway Pressure is defined as the force exerted per unit area; it can be measured using a barometer or manometer. Key Equation Definition of pressure Equation 10.2.1 : P = F/A Conceptual Problems What four quantities must be known to completely describe a sample of a gas? What units are commonly used for each quantity? If the applied force is constant, how does the pressure exerted by an object change as the area on which the force is exerted decreases? In the real world, how does this relationship apply to the ease of driving a small nail versus a large nail? As the force on a fixed area increases, does the pressure increase or decrease? With this in mind, would you expect a heavy person to need smaller or larger snowshoes than a lighter person? Explain. What do we mean by atmospheric pressure ? Is the atmospheric pressure at the summit of Mt. Rainier greater than or less than the pressure in Miami, Florida? Why? Which has the highest atmospheric pressure—a cave in the Himalayas, a mine in South Africa, or a beach house in Florida? Which has the lowest? Mars has an average atmospheric pressure of 0.007 atm. Would it be easier or harder to drink liquid from a straw on Mars than on Earth? Explain your answer. Is the pressure exerted by a 1.0 kg mass on a 2.0 m 2 area greater than or less than the pressure exerted by a 1.0 kg mass on a 1.0 m 2 area? What is the difference, if any, between the pressure of the atmosphere exerted on a 1.0 m 2 piston and a 2.0 m 2 piston? If you used water in a barometer instead of mercury, what would be the major difference in the instrument? Answer Because pressure is defined as the force per unit area ( P = F/A ), increasing the force on a given area increases the pressure. A heavy person requires larger snowshoes than a lighter person. Spreading the force exerted on the heavier person by gravity (that is, their weight) over a larger area decreases the pressure exerted per unit of area, such as a square inch, and makes them less likely to sink into the snow. Numerical Problems Calculate the pressure in atmospheres and kilopascals exerted by a fish tank that is 2.0 ft long, 1.0 ft wide, and 2.5 ft high and contains 25.0 gal of water in a room that is at 20°C; the tank itself weighs 15 lb (d H 2 O = 1.00 g/cm 3 at 20°C). If the tank were 1 ft long, 1 ft wide, and 5 ft high, would it exert the same pressure? Explain your answer. Calculate the pressure in pascals and in atmospheres exerted by a carton of milk that weighs 1.5 kg and has a base of 7.0 cm × 7.0 cm. If the carton were lying on its side (height = 25 cm), would it exert more or less pressure? Explain your reasoning. If atmospheric pressure at sea level is 1.0 × 10 5 Pa, what is the mass of air in kilograms above a 1.0 cm 2 area of your skin as you lie on the beach? If atmospheric pressure is 8.2 × 10 4 Pa on a mountaintop, what is the mass of air in kilograms above a 4.0 cm 2 patch of skin? Complete the following table: atm kPa mmHg torr 1.4 NaN NaN NaN NaN NaN 723.0 NaN NaN 43.2 NaN NaN The SI unit of pressure is the pascal, which is equal to 1 N/m 2 . Show how the product of the mass of an object and the acceleration due to gravity result in a force that, when exerted on a given area, leads to a pressure in the correct SI units. What mass in kilograms applied to a 1.0 cm 2 area is required to produce a pressure of 1.0 atm? 1.0 torr? 1 mmHg? 1 kPa? If you constructed a manometer to measure gas pressures over the range 0.60–1.40 atm using the liquids given in the following table, how tall a column would you need for each liquid? The density of mercury is 13.5 g/cm 3 . Based on your results, explain why mercury is still used in barometers, despite its toxicity. Unnamed: 0 Liquid Density (20°C) Column Height (m) isopropanol 0.785 NaN coconut oil 0.924 NaN glycerine 1.259 NaN Answer 5.4 kPa or 5.3 × 10 −2 atm; 11 kPa, 1.1 × 10 −3 atm; the same force acting on a smaller area results in a greater pressure. Contributors Anonymous Modified by Joshua Halpern |
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