1044. Exercises - 1044.1. Energy Basics
(a) What is the specific heat of the substance?
(b) If it is one of the substances found in Table 9.1, 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 $19.3^{\circ} \mathrm{C}$ to $68.9^{\circ} \mathrm{C}$.
(a) What is the specific heat of the substance?
(b) If it is one of the substances found in Table 9.1, what is its likely identity?
12. An aluminum kettle weighs 1.05 kg .
(a) What is the heat capacity of the kettle?
(b) How much heat is required to increase the temperature of this kettle from $23.0^{\circ} \mathrm{C}$ to $99.0^{\circ} \mathrm{C}$ ?
(c) How much heat is required to heat this kettle from $23.0^{\circ} \mathrm{C}$ to $99.0^{\circ} \mathrm{C}$ if it contains 1.25 L of water (density of $0.997 \mathrm{~g} / \mathrm{mL}$ and a specific heat of $4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ )?
13. Most people find waterbeds uncomfortable unless the water temperature is maintained at about $85^{\circ} \mathrm{F}$. Unless it is heated, a waterbed that contains 892 L of water cools from $85^{\circ} \mathrm{F}$ to $72^{\circ} \mathrm{F}$ 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 $(\mathrm{kWh})=3.6 \times 10^{6} \mathrm{~J}$, and assume that the density of water is $1.0 \mathrm{~g} / \mathrm{mL}$ (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)?


1044. Exercises - 1044.2. Calorimetry
14. A $500-\mathrm{mL}$ bottle of water at room temperature and a $2-\mathrm{L}$ bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the $500-\mathrm{mL}$ 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 $500-\mathrm{mL}$ 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 9.5 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 9.6 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 9.6 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 $23^{\circ} \mathrm{C}$ with a density of $1.00 \mathrm{~g} / \mathrm{mL}$ must be mixed with 180 mL (about 6 oz ) of coffee at $95^{\circ} \mathrm{C}$ so that the resulting combination will have a temperature of $60^{\circ} \mathrm{C}$ ? Assume that coffee and water have the same density and the same specific heat.


1044. Exercises - 1044.2. Calorimetry
19. How much will the temperature of a cup ( 180 g ) of coffee at $95^{\circ} \mathrm{C}$ be reduced when a 45 g silver spoon (specific heat $0.24 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ ) at $25^{\circ} \mathrm{C}$ 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 $45-\mathrm{g}$ aluminum spoon (specific heat $0.88 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ ) at $24^{\circ} \mathrm{C}$ is placed in $180 \mathrm{~mL}(180 \mathrm{~g})$ of coffee at $85^{\circ} \mathrm{C}$ and the temperature of the two become equal.
(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water.
(b) The first time a student solved this problem she got an answer of $88^{\circ} \mathrm{C}$. 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 $240^{\circ} \mathrm{F}$. After it passes through the radiator it has a temperature of $175^{\circ} \mathrm{F}$. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of $4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$.
22. A $70.0-\mathrm{g}$ piece of metal at $80.0^{\circ} \mathrm{C}$ is placed in 100 g of water at $22.0^{\circ} \mathrm{C}$ contained in a calorimeter like that shown in Figure 9.12. The metal and water come to the same temperature at $24.6^{\circ} \mathrm{C}$. How much heat did the metal give up to the water? What is the specific heat of the metal?


1044. Exercises - 1044.2. Calorimetry
23. If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at $26.5^{\circ} \mathrm{C}$ in a calorimeter like that in Figure 9.12, what is the resulting temperature of the water?
24. A $0.500-\mathrm{g}$ sample of KCl is added to 50.0 g of water in a calorimeter (Figure 9.12). If the temperature decreases by $1.05^{\circ} \mathrm{C}$, what is the approximate amount of heat involved in the dissolution of the KCl , assuming the specific heat of the resulting solution is $4.18 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ ? Is the reaction exothermic or endothermic?
25. Dissolving 3.0 g of $\mathrm{CaCl}_{2}(s)$ in 150.0 g of water in a calorimeter (Figure 9.12) at $22.4^{\circ} \mathrm{C}$ causes the temperature to rise to $25.8^{\circ} \mathrm{C}$. What is the approximate amount of heat involved in the dissolution, assuming the specific heat of the resulting solution is $4.18 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ ? Is the reaction exothermic or endothermic?
26. When 50.0 g of $0.200 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})$ at $24.1^{\circ} \mathrm{C}$ is added to 100.0 g of $0.100 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})$ at $24.1^{\circ} \mathrm{C}$ in a calorimeter, the temperature increases to $25.2^{\circ} \mathrm{C}$ as $\mathrm{AgCl}(s)$ forms. Assuming the specific heat of the solution and products is $4.20 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$, calculate the approximate amount of heat in joules produced.


1044. Exercises - 1044.2. Calorimetry
27. The addition of 3.15 g of $\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}$ to a solution of 1.52 g of $\mathrm{NH}_{4} \mathrm{SCN}$ in 100 g of water in a calorimeter caused the temperature to fall by $3.1^{\circ} \mathrm{C}$. Assuming the specific heat of the solution and products is 4.20 $\mathrm{J} / \mathrm{g}{ }^{\circ} \mathrm{C}$, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation:
$\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{NH}_{4} \mathrm{SCN}(\mathrm{aq}) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(\mathrm{aq})+2 \mathrm{NH}_{3}(\mathrm{aq})+10 \mathrm{H}_{2} \mathrm{O}(\mathrm{I})$
28. The reaction of 50 mL of acid and 50 mL of base described in Example 9.5 increased the temperature of the solution by $6.9^{\circ} \mathrm{C}$. 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 $22.0^{\circ} \mathrm{C}$ ? Explain your answer.
29. If the 3.21 g of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in Example 9.6 were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer.


1044. Exercises - 1044.2. Calorimetry
30. When 1.0 g of fructose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)$, a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by $1.58^{\circ} \mathrm{C}$. If the heat capacity of the calorimeter and its contents is $9.90 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}$, what is $q$ for this combustion?
31. When a $0.740-\mathrm{g}$ sample of trinitrotoluene (TNT), $\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature increases from $23.4^{\circ} \mathrm{C}$ to $26.9^{\circ} \mathrm{C}$. The heat capacity of the calorimeter is $534 \mathrm{~J} /{ }^{\circ} \mathrm{C}$, 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 9.17), the temperature increases by $1.48{ }^{\circ} \mathrm{C}$. If the heat capacity of the calorimeter is $21.6 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}$, determine the heat produced by combustion of a ton of coal $\left(2.000 \times 10^{3}\right.$ 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?


1044. Exercises - 1044.2. Calorimetry
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 $1.1 \times 10^{3}$ 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 $/ \mathrm{g}$, 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, $0.6919 \mathrm{~g} / \mathrm{mL}$ ) 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.


1044. Exercises - 1044.3. Enthalpy
39. Explain how the heat measured in Example 9.5 differs from the enthalpy change for the exothermic reaction described by the following equation:
$\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$
40. Using the data in the check your learning section of Example 9.5, calculate $\Delta H$ in $\mathrm{kJ} / \mathrm{mol}$ of $\mathrm{AgNO}_{3}(\mathrm{aq})$ for the reaction: $\mathrm{NaCl}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{AgCl}(s)+\mathrm{NaNO}_{3}(a q)$
41. Calculate the enthalpy of solution ( $\Delta H$ for the dissolution) per mole of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ under the conditions described in Example 9.6.
42. Calculate $\Delta H$ 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.)
$\mathrm{Ba}(\mathrm{OH})_{2} \cdot 8 \mathrm{H}_{2} \mathrm{O}(s)+2 \mathrm{NH}_{4} \mathrm{SCN}(a q) \longrightarrow \mathrm{Ba}(\mathrm{SCN})_{2}(a q)+2 \mathrm{NH}_{3}(a q)+10 \mathrm{H}_{2} \mathrm{O}(l)$
43. Calculate the enthalpy of solution ( $\Delta H$ for the dissolution) per mole of $\mathrm{CaCl}_{2}$ (refer to Exercise 9.25).
44. Although the gas used in an oxyacetylene torch (Figure 9.7) 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 9.2. Considering the conditions for which the tabulated data are reported, suggest an explanation.


1044. Exercises - 1044.3. Enthalpy
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, $1.00 \mathrm{~g} / \mathrm{mL}$ ) and 200 mL of 0.150 M NaOH (density, $1.00 \mathrm{~g} / \mathrm{mL}$ ) are mixed?
$\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H^{\circ}=-58 \mathrm{~kJ}$
If both solutions are at the same temperature and the specific heat of the products is $4.19 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$, 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 $26.74^{\circ} \mathrm{C}$ to $27.93^{\circ} \mathrm{C}$. What is the heat capacity of the calorimeter and its contents?


1044. Exercises - 1044.3. Enthalpy
52. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, $6.00 \mathrm{kcal} / \mathrm{mol}$ ) was used in household refrigerators. What mass of $\mathrm{SO}_{2}$ must be evaporated to remove as much heat as evaporation of 1.00 kg of $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ (enthalpy of vaporization is $17.4 \mathrm{~kJ} / \mathrm{mol}$ )?
The vaporization reactions for $\mathrm{SO}_{2}$ and $\mathrm{CCl}_{2} \mathrm{~F}_{2}$ are $\mathrm{SO}_{2}(l) \longrightarrow \mathrm{SO}_{2}(g)$ and $\mathrm{CCl}_{2} \mathrm{~F}(l) \longrightarrow \mathrm{CCl}_{2} \mathrm{~F}_{2}(g)$, 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 $35.0^{\circ} \mathrm{C}$, as the heat provided when 100 g of steam is cooled from $110^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$.
54. Which of the enthalpies of combustion in Table 9.2 the table are also standard enthalpies of formation?
55. Does the standard enthalpy of formation of $\mathrm{H}_{2} \mathrm{O}(g)$ differ from $\Delta H^{\circ}$ for the reaction
$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) ?$
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 $\mathrm{HgO}(s)$ to $\mathrm{Hg}(\mathrm{I})$ and $\mathrm{O}_{2}(g)$ under standard conditions?


1044. Exercises - 1044.3. Enthalpy
57. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn , is burned to form $\mathrm{Mn}_{3} \mathrm{O}_{4}(s)$ at standard state conditions?
58. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe , is burned to form $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)$ at standard state conditions?
59. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
$4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-907 \mathrm{~kJ}$
$2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H=-113 \mathrm{~kJ}$
$3 \mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \quad \Delta H=-139 \mathrm{~kJ}$
Determine the total energy change for the production of one mole of aqueous nitric acid by this process.
60. Both graphite and diamond burn.
$\mathrm{C}(s$, diamond $)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$
For the conversion of graphite to diamond:
$\mathrm{C}(s$, graphite $) \longrightarrow \mathrm{C}(s$, diamond $) \quad \Delta H^{\circ}=1.90 \mathrm{~kJ}$
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: $\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)$
62. Which produces more heat?


1044. Exercises - 1044.3. Enthalpy
62. Which produces more heat?
$\mathrm{Os}(s) \longrightarrow 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{OsO}_{4}(s)$
or
$\mathrm{Os}(s) \longrightarrow 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{OsO}_{4}(g)$
for the phase change $\mathrm{OsO}_{4}(s) \longrightarrow \mathrm{OsO}_{4}(g) \quad \Delta H=56.4 \mathrm{~kJ}$
63. Calculate $\Delta \boldsymbol{H}^{\circ}$ for the process
$\mathrm{Sb}(s)+\frac{5}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{5}(s)$
from the following information:
$\mathrm{Sb}(s)+\frac{3}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{3}(s) \quad \Delta H^{\circ}=-314 \mathrm{~kJ}$
$\mathrm{SbCl}_{3}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SbCl}_{5}(s) \quad \Delta H^{\circ}=-80 \mathrm{~kJ}$
64. Calculate $\Delta H^{\circ}$ for the process $\mathrm{Zn}(s)+\mathrm{S}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s)$
from the following information:
$\mathrm{Zn}(s)+\mathrm{S}(s) \longrightarrow \mathrm{ZnS}(s) \quad \Delta H^{\circ}=-206.0 \mathrm{~kJ}$
$\mathrm{ZnS}(s)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{ZnSO}_{4}(s) \quad \Delta H^{\circ}=-776.8 \mathrm{~kJ}$
65. Calculate $\Delta H$ for the process $\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{Cl}_{2}(g)$ from the following information:


1044. Exercises - 1044.3. Enthalpy
$$
\begin{array}{lr}
\mathrm{Hg}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{HgCl}_{2}(s) & \Delta H=-224 \mathrm{~kJ} \\
\mathrm{Hg}(l)+\mathrm{HgCl}_{2}(s) \longrightarrow \mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) & \Delta H=-41.2 \mathrm{~kJ}
\end{array}
$$

66. Calculate $\Delta H^{\circ}$ for the process $\mathrm{Co}_{3} \mathrm{O}_{4}(s) \longrightarrow 3 \mathrm{Co}(s)+2 \mathrm{O}_{2}(g)$ from the following information:

$$
\begin{array}{lr}
\mathrm{Co}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CoO}(s) & \Delta H^{\circ}=-237.9 \mathrm{~kJ} \\
3 \mathrm{CoO}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Co}_{3} \mathrm{O}_{4}(s) & \Delta H^{\circ}=-177.5 \mathrm{~kJ}
\end{array}
$$


1044. Exercises - 1044.3. Enthalpy
67. Calculate the standard molar enthalpy of formation of $\mathrm{NO}(g)$ from the following data:
$\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=66.4 \mathrm{~kJ}$
$2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H^{\circ}=-114.1 \mathrm{~kJ}$
68. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:
(a) $\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)$
(b) $\mathrm{Si}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SiCl}_{4}(g)$
(c) $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$
(d) $2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(g)$
69. Using the data in Appendix G, calculate the standard enthalpy change for each of the following reactions:
(a) $\mathrm{Si}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{SiF}_{4}(g)$
(b) $2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(l)$
(c) $\mathrm{CH}_{4}(g)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{HCN}(g)+\mathrm{NH}_{3}(g)$;


1044. Exercises - 1044.3. Enthalpy
(c) $\mathrm{CH}_{4}(g)+\mathrm{N}_{2}(g) \longrightarrow \mathrm{HCN}(g)+\mathrm{NH}_{3}(g)$;
(d) $\mathrm{CS}_{2}(g)+3 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)+\mathrm{S}_{2} \mathrm{Cl}_{2}(g)$
70. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.
(a) $2 \mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow 4 \mathrm{Ag}(s)+\mathrm{O}_{2}(g)$
(b) $\mathrm{SnO}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Sn}(s)+\mathrm{CO}_{2}(g)$
(c) $\mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Cr}(s)+3 \mathrm{H}_{2} \mathrm{O}(l)$
(d) $2 \mathrm{Al}(s)+\mathrm{Fe}_{2} \mathrm{O}_{3}(s) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s)$
71. The decomposition of hydrogen peroxide, $\mathrm{H}_{2} \mathrm{O}_{2}$, 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 $\mathrm{H}_{2} \mathrm{O}_{2}$ under standard conditions.
$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)$


1044. Exercises - 1044.3. Enthalpy
$2 \mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{l}) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{O}_{2}(g)$
72. Calculate the enthalpy of combustion of propane, $\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$, for the formation of $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ and $\mathrm{CO}_{2}(\mathrm{~g})$. The enthalpy of formation of propane is $-104 \mathrm{~kJ} / \mathrm{mol}$.
73. Calculate the enthalpy of combustion of butane, $\mathrm{C}_{4} \mathrm{H}_{10}(g)$ for the formation of $\mathrm{H}_{2} \mathrm{O}(g)$ and $\mathrm{CO}_{2}(g)$. The enthalpy of formation of butane is $-126 \mathrm{~kJ} / \mathrm{mol}$.
74. Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned?
75. The white pigment $\mathrm{TiO}_{2}$ is prepared by the reaction of titanium tetrachloride, $\mathrm{TiCl}_{4}$, with water vapor in the gas phase: $\mathrm{TiCl}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{TiO}_{2}(s)+4 \mathrm{HCl}(g)$.
How much heat is evolved in the production of exactly 1 mole of $\mathrm{TiO}_{2}(s)$ under standard state conditions?
76. Water gas, a mixture of $\mathrm{H}_{2}$ and CO , is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: $\mathrm{C}(s)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{H}_{2}(g)$.
(a) Assuming that coke has the same enthalpy of formation as graphite, calculate $\Delta H^{\circ}$ for this reaction.


1044. Exercises - 1044.3. Enthalpy
(a) Assuming that coke has the same enthalpy of formation as graphite, calculate $\Delta H^{\circ}$ for this reaction.
(b) Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and additional hydrogen at high temperature and pressure in the presence of a suitable catalyst:
$2 \mathrm{H}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$.
Under the conditions of the reaction, methanol forms as a gas. Calculate $\Delta H^{\circ}$ 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 $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ and $\mathrm{CO}_{2}(\mathrm{~g})$.
77. In the early days of automobiles, illumination at night was provided by burning acetylene, $\mathrm{C}_{2} \mathrm{H}_{2}$. 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, $\mathrm{CaC}_{2}$ :
$\mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)$.
Calculate the standard enthalpy of the reaction. The $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{CaC}_{2}$ is $-15.14 \mathrm{kcal} / \mathrm{mol}$.
78. From the data in Table 9.2, determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: $\mathrm{CO}(g), \mathrm{CH}_{4}(g)$, or $\mathrm{C}_{2} \mathrm{H}_{2}(g)$.


1044. Exercises - 1044.3. Enthalpy
79. The enthalpy of combustion of hard coal averages $-35 \mathrm{~kJ} / \mathrm{g}$, that of gasoline, $1.28 \times 10^{5} \mathrm{~kJ} / \mathrm{gal}$. 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 $0.692 \mathrm{~g} / \mathrm{mL}$ (the same as the density of isooctane).
80. Ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, is used as a fuel for motor vehicles, particularly in Brazil.
(a) Write the balanced equation for the combustion of ethanol to $\mathrm{CO}_{2}(\mathrm{~g})$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$, and, using the data in Appendix G, calculate the enthalpy of combustion of 1 mole of ethanol.
(b) The density of ethanol is $0.7893 \mathrm{~g} / \mathrm{mL}$. Calculate the enthalpy of combustion of exactly 1 L of ethanol.
(c) 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, $\mathrm{C}_{8} \mathrm{H}_{18}$
$\left(\Delta H_{\mathrm{f}}^{\circ}=-208.4 \mathrm{~kJ} / \mathrm{mol}\right.$; density $\left.=0.7025 \mathrm{~g} / \mathrm{mL}\right)$.


1044. Exercises - 1044.3. Enthalpy
81. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane $\left[\mathrm{B}_{2} \mathrm{H}_{6}\right.$, produces $\mathrm{B}_{2} \mathrm{O}_{3}(\mathrm{~s})$ and $\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]$, methane $\left[\mathrm{CH}_{4}\right.$, produces $\mathrm{CO}_{2}(\mathrm{~g})$ and $\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]$, and hydrazine [ $\mathrm{N}_{2} \mathrm{H}_{4}$, produces $\mathrm{N}_{2}(\mathrm{~g})$ and $\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\right]$. 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 $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{B}_{2} \mathrm{H}_{6}(\mathrm{~g}), \mathrm{CH}_{4}(\mathrm{~g})$, and $\mathrm{N}_{2} \mathrm{H}_{4}(I)$ may be found in Appendix G .
82. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions?
83. Ethylene, $\mathrm{C}_{2} \mathrm{H}_{4}$, 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.
$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$
Using the data in the table in Appendix G, calculate $\Delta H^{\circ}$ for the reaction.


1044. Exercises - 1044.3. Enthalpy
Using the data in the table in Appendix G, calculate $\Delta H^{\circ}$ for the reaction.
84. The oxidation of the sugar glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, is described by the following equation:
$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-2816 \mathrm{~kJ}$
The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body.
(a) How much heat in kilojoules can be produced by the metabolism of 1.0 g of glucose?
(b) How many Calories can be produced by the metabolism of 1.0 g of glucose?
85. Propane, $\mathrm{C}_{3} \mathrm{H}_{8}$, is a hydrocarbon that is commonly used as a fuel.
(a) Write a balanced equation for the complete combustion of propane gas.
(b) Calculate the volume of air at $25^{\circ} \mathrm{C}$ and 1.00 atmosphere that is needed to completely combust 25.0 grams of propane. Assume that air is 21.0 percent $\mathrm{O}_{2}$ 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 $25^{\circ} \mathrm{C}$ and 1.00 atm contains 0.275 g of $\mathrm{O}_{2}$.)


1044. Exercises - 1044.3. Enthalpy
(c) The heat of combustion of propane is $-2,219.2 \mathrm{~kJ} / \mathrm{mol}$. Calculate the heat of formation, $\Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$ of propane given that $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{H}_{2} \mathrm{O}(\mathrm{I})=-285.8 \mathrm{~kJ} / \mathrm{mol}$ and $\Delta H_{\mathrm{f}}^{\circ}$ of $\mathrm{CO}_{2}(\mathrm{~g})=-393.5 \mathrm{~kJ} / \mathrm{mol}$.
(d) 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).
(a) Assume that natural gas is pure methane and determine the volume of natural gas in cubic feet that was required to heat the house. The average temperature of the natural gas was $56^{\circ} \mathrm{F}$; at this temperature and a pressure of 1 atm, natural gas has a density of $0.681 \mathrm{~g} / \mathrm{L}$.
(b) How many gallons of LPG (liquefied petroleum gas) would be required to replace the natural gas used? Assume the LPG is liquid propane [ $\mathrm{C}_{3} \mathrm{H}_{8}$ : density, $0.5318 \mathrm{~g} / \mathrm{mL}$; enthalpy of combustion, $2219 \mathrm{~kJ} / \mathrm{mol}$ for the formation of $\mathrm{CO}_{2}(g)$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{I}]$ and the furnace used to burn the LPG has the same efficiency as the gas furnace.


1044. Exercises - 1044.3. Enthalpy
(c) What mass of carbon dioxide is produced by combustion of the methane used to heat the house?
(d) What mass of water is produced by combustion of the methane used to heat the house?
(e) What volume of air is required to provide the oxygen for the combustion of the methane used to heat the house? Air contains $23 \%$ oxygen by mass. The average density of air during the month was $1.22 \mathrm{~g} / \mathrm{L}$. (f) How many kilowatt-hours ( $1 \mathrm{kWh}=3.6 \times 10^{6} \mathrm{~J}$ ) 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 $40 \%$ ?


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
87. Which bond in each of the following pairs of bonds is the strongest?
(a) $\mathrm{C}-\mathrm{C}$ or $\mathrm{C}=\mathrm{C}$
(b) $\mathrm{C}-\mathrm{N}$ or $\mathrm{C} \equiv \mathrm{N}$
(c) $\mathrm{C} \equiv \mathrm{O}$ or $\mathrm{C}=\mathrm{O}$
(d) $\mathrm{H}-\mathrm{F}$ or $\mathrm{H}-\mathrm{Cl}$
(e) $\mathrm{C}-\mathrm{H}$ or $\mathrm{O}-\mathrm{H}$
(f) $\mathrm{C}-\mathrm{N}$ or $\mathrm{C}-\mathrm{O}$
88. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:
(a) $\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \longrightarrow 2 \mathrm{HBr}(g)$
(b) $\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)$
(c) $\mathrm{C}_{2} \mathrm{H}_{4}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
89. Using the bond energies in Table 9.3, determine the approximate enthalpy change for each of the following reactions:
(a) $\mathrm{Cl}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{ClF}_{3}(g)$
(b) $\mathrm{H}_{2} \mathrm{C}=\mathrm{CH}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}_{3} \mathrm{CCH}_{3}(g)$


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
(c) $2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$
90. Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
(a) Use the bond energy found in Table 9.3 and Table 9.4 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
(b) Use the enthalpy of reaction and the bond energies for $\mathrm{H}_{2}$ and $\mathrm{Cl}_{2}$ to solve for the energy of one mole of HCl bonds.


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-184.7 \mathrm{~kJ} / \mathrm{mol}
$$

91. Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.
92. 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:

93. How does the bond energy of $\mathrm{HCl}(\mathrm{g})$ differ from the standard enthalpy of formation of $\mathrm{HCl}(\mathrm{g})$ ?
94. Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of $\mathrm{HCl}(\mathrm{g})$ can be used to determine the bond energy.
95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbonsulfur double bond in $\mathrm{CS}_{2}$.
96. Using the standard enthalpy of formation data in Appendix $G$, determine which bond is stronger: the $S-F$ bond in $\mathrm{SF}_{4}(g)$ or in $\mathrm{SF}_{6}(g)$ ?
97. Using the standard enthalpy of formation data in Appendix G , determine which bond is stronger: the $\mathrm{P}-\mathrm{Cl}$ bond in $\mathrm{PCl}_{3}(g)$ or in $\mathrm{PCl}_{5}(g)$ ?
98. Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:

$$
\begin{array}{llll}
\mathrm{H} \quad \mathrm{H} & \mathrm{H} & \mathrm{H}
\end{array}
$$


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
$$
\begin{array}{llll}
\mathrm{H} \quad \mathrm{H} & \mathrm{H} & \mathrm{H}
\end{array}
$$

$\mathrm{H} C \quad \mathrm{C} \quad \mathrm{C} \quad \mathrm{C} \quad \mathrm{C} \quad \mathrm{H}$
H H
99. Use the bond energy to calculate an approximate value of $\Delta H$ for the following reaction. Which is the more stable form of $\mathrm{FNO}_{2}$ ?

100. Use principles of atomic structure to answer each of the following: ${ }^{4}$
(a) The radius of the Ca atom is 197 pm ; the radius of the $\mathrm{Ca}^{2+}$ ion is 99 pm . Account for the difference.
(b) The lattice energy of $\mathrm{CaO}(s)$ is $-3460 \mathrm{~kJ} / \mathrm{mol}$; the lattice energy of $\mathrm{K}_{2} \mathrm{O}$ is $-2240 \mathrm{~kJ} / \mathrm{mol}$. 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 lonization Energy (kJ/mol) | Second lonization Energy (kJ/ <br> mol) |
| :---: | :---: | :---: |
| K | 419 | 3050 |
| Ca | 590 | 1140 |

(d) The first ionization energy of Mg is $738 \mathrm{~kJ} / \mathrm{mol}$ and that of Al is $578 \mathrm{~kJ} / \mathrm{mol}$. Account for this difference.


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
[^8]101. The lattice energy of LiF is $1023 \mathrm{~kJ} / \mathrm{mol}$, and the Li-F distance is 200.8 pm . NaF crystallizes in the same structure as LiF but with a $\mathrm{Na}-\mathrm{F}$ distance of 231 pm . Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175 , or $4090 \mathrm{~kJ} / \mathrm{mol}$ ? Explain your choice.
102. For which of the following substances is the least energy required to convert one mole of the solid into separate ions?
(a) MgO
(b) SrO
(c) KF
(d) CsF
(e) $\mathrm{MgF}_{2}$
103. The reaction of a metal, $M$, with a halogen, $X_{2}$, proceeds by an exothermic reaction as indicated by this equation: $\mathrm{M}(s)+\mathrm{X}_{2}(g) \longrightarrow \mathrm{MX}_{2}(s)$. For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.
(a) a large radius vs. a small radius for $\mathrm{M}^{+2}$
(b) a high ionization energy vs. a low ionization energy for $M$
(c) an increasing bond energy for the halogen
(d) a decreasing electron affinity for the halogen
(e) an increasing size of the anion formed by the halogen


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
(c) an increasing bond energy for the halogen
(d) a decreasing electron affinity for the halogen
(e) an increasing size of the anion formed by the halogen
104. The lattice energy of LiF is $1023 \mathrm{~kJ} / \mathrm{mol}$, and the Li-F distance is 201 pm . MgO crystallizes in the same structure as LiF but with a $\mathrm{Mg}-\mathrm{O}$ distance of 205 pm . Which of the following values most closely approximates the lattice energy of MgO: $256 \mathrm{~kJ} / \mathrm{mol}, 512 \mathrm{~kJ} / \mathrm{mol}, 1023 \mathrm{~kJ} / \mathrm{mol}, 2046 \mathrm{~kJ} / \mathrm{mol}$, or $4008 \mathrm{~kJ} /$ mol? Explain your choice.
105. Which compound in each of the following pairs has the larger lattice energy? Note: $\mathrm{Mg}^{2+}$ and $\mathrm{Li}^{+}$have similar radii; $\mathrm{O}^{2-}$ and $\mathrm{F}^{-}$have similar radii. Explain your choices.
(a) MgO or MgSe
(b) LiF or MgO
(c) $\mathrm{Li}_{2} \mathrm{O}$ or LiCl
(d) $\mathrm{Li}_{2} \mathrm{Se}$ or MgO
106. Which compound in each of the following pairs has the larger lattice energy? Note: $\mathrm{Ba}^{2+}$ and $\mathrm{K}^{+}$have similar radii; $\mathrm{S}^{2-}$ and $\mathrm{Cl}^{-}$have similar radii. Explain your choices.
(a) $\mathrm{K}_{2} \mathrm{O}$ or $\mathrm{Na}_{2} \mathrm{O}$
(b) $\mathrm{K}_{2} \mathrm{~S}$ or BaS
(c) KCl or BaS
(d) BaS or $\mathrm{BaCl}_{2}$
107. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
(a) MgO
(b) SrO
(c) KF
(d) CsF


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
107. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
(a) MgO
(b) SrO
(c) KF
(d) CsF
(e) $\mathrm{MgF}_{2}$
108. Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
(a) $\mathrm{K}_{2} \mathrm{~S}$
(b) $\mathrm{K}_{2} \mathrm{O}$
(c) CaS
(d) $\mathrm{Cs}_{2} \mathrm{~S}$
(e) CaO
109. The lattice energy of $K F$ is $794 \mathrm{~kJ} / \mathrm{mol}$, and the interionic distance is 269 pm . The $\mathrm{Na}-\mathrm{F}$ 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: $682 \mathrm{~kJ} / \mathrm{mol}, 794 \mathrm{~kJ} / \mathrm{mol}, 924 \mathrm{~kJ} / \mathrm{mol}, 1588 \mathrm{~kJ} / \mathrm{mol}$, or 3175 kJ/mol? Explain your answer.


1044. Exercises - 1044.4. Strengths of Ionic and Covalent Bonds
474 9•Exercises


1045. CHAPTER 10 <br> Liquids and Solids -
Figure 10.1 Solid carbon dioxide ("dry ice", left) sublimes vigorously when placed in a liquid (right), cooling the liquid and generating a dense mist of water above the cylinder. (credit: modification of work by Paul Flowers)


1046. CHAPTER OUTLINE - 1046.1. Intermolecular Forces
10.2 Properties of Liquids
10.3 Phase Transitions
10.4 Phase Diagrams
10.5 The Solid State of Matter
10.6 Lattice Structures in Crystalline Solids

INTRODUCTION Leprosy has been a devastating disease throughout much of human history. Aside from the symptoms and complications of the illness, its social stigma led sufferers to be cast out of communities and isolated in colonies; in some regions this practice lasted well into the twentieth century. At that time, the best potential treatment for leprosy was oil from the chaulmoogra tree, but the oil was extremely thick, causing blisters and making usage painful and ineffective. Healthcare professionals seeking a better application contacted Alice Ball, a young chemist at the University of Hawaii, who had focused her masters thesis on a similar plant. Ball initiated a sequence of procedures (repeated acidification and purification to change the characteristics of the oil and isolate the active substances (esters, discussed later in this text). The "Ball Method" as it later came to be called, became the standard treatment for leprosy for decades. In the liquid and solid states, atomic and molecular interactions are of considerable strength and play an important role in determining a number of physical properties of the substance. For example, the thickness, or viscosity, of the chaulmoogra oil was due to its intermolecular forces. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.


1047. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
- Identify the types of intermolecular forces experienced by specific molecules based on their structures
- Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state

As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase "intermolecular attraction" to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.

Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:

- Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
- Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.

The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles' KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. Figure 10.2 illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE , of a given substance.


1047. LEARNING OBJECTIVES -
FIGURE 10.2 Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)

As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between $\mathrm{H}_{2} \mathrm{O}$ molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid $\mathrm{H}_{2} \mathrm{O}$. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in Figure 10.3.

FIGURE 10.3 Condensation forms when water vapor in the air is cooled enough to form liquid water, such as (a) on the outside of a cold beverage glass or (b) in the form of fog. (credit a: modification of work by Jenny Downing; credit b: modification of work by Cory Zanker)

We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, $\mathrm{C}_{4} \mathrm{H}_{10}$, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter's fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in Figure 10.4.

FIGURE 10.4 Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state. (credit: modification of work by "Sam-Cat"/Flickr)

Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter.


1048. LINK TO LEARNING -
Access this interactive simulation (http://openstax.org/l/16phetvisual) on states of matter, phase transitions, and intermolecular forces. This simulation is useful for visualizing concepts introduced throughout this chapter.


1049. Forces between Molecules -
Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. Figure 10.5 illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy-430 kilojoules.

FIGURE 10.5 Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance's properties.

All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.


1050. Dispersion Forces -
One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron's location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species-a so-called dispersion force like that illustrated in Figure 10.6.

FIGURE 10.6 Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules.

Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close.

Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. $\mathrm{F}_{2}$ and $\mathrm{Cl}_{2}$ are gases at room temperature (reflecting weaker attractive forces); $\mathrm{Br}_{2}$ is a liquid, and $\mathrm{I}_{2}$ is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in Table 10.1.

Melting and Boiling Points of the Halogens


1050. Dispersion Forces -
Melting and Boiling Points of the Halogens

| Halogen | Molar Mass | Atomic Radius | Melting Point | Boiling Point |
| :---: | :---: | :---: | :---: | :---: |
| fluorine, $\mathrm{F}_{2}$ | $38 \mathrm{~g} / \mathrm{mol}$ | 72 pm | 53 K | 85 K |
| chlorine, $\mathrm{Cl}_{2}$ | $71 \mathrm{~g} / \mathrm{mol}$ | 99 pm | 172 K | 238 K |
| bromine, $\mathrm{Br}_{2}$ | $160 \mathrm{~g} / \mathrm{mol}$ | 114 pm | 266 K | 332 K |
| iodine, $\mathrm{I}_{2}$ | $254 \mathrm{~g} / \mathrm{mol}$ | 133 pm | 387 K | 457 K |
| astatine, $\mathrm{At}_{2}$ | $420 \mathrm{~g} / \mathrm{mol}$ | 150 pm | 575 K | 610 K |

TABLE 10.1

The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule's charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.


1052. London Forces and Their Effects -
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: $\mathrm{CH}_{4}$, $\mathrm{SiH}_{4}, \mathrm{GeH}_{4}$, and $\mathrm{SnH}_{4}$. Explain your reasoning.


1053. Solution -
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of $\mathrm{CH}_{4}, \mathrm{SiH}_{4}, \mathrm{GeH}_{4}$, and $\mathrm{SnH}_{4}$ are approximately $16 \mathrm{~g} / \mathrm{mol}, 32 \mathrm{~g} / \mathrm{mol}, 77 \mathrm{~g} /$ mol, and $123 \mathrm{~g} / \mathrm{mol}$, respectively. Therefore, $\mathrm{CH}_{4}$ is expected to have the lowest boiling point and $\mathrm{SnH}_{4}$ the highest boiling point. The ordering from lowest to highest boiling point is expected to be $\mathrm{CH}_{4}<\mathrm{SiH}_{4}<\mathrm{GeH}_{4}<$ $\mathrm{SnH}_{4}$.

A graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:


1054. Check Your Learning -
Order the following hydrocarbons from lowest to highest boiling point: $\mathrm{C}_{2} \mathrm{H}_{6}, \mathrm{C}_{3} \mathrm{H}_{8}$, and $\mathrm{C}_{4} \mathrm{H}_{10}$.


1055. Answer: -
$\mathrm{C}_{2} \mathrm{H}_{6}<\mathrm{C}_{3} \mathrm{H}_{8}<\mathrm{C}_{4} \mathrm{H}_{10}$. All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore $\mathrm{C}_{2} \mathrm{H}_{6}<\mathrm{C}_{3} \mathrm{H}_{8}<\mathrm{C}_{4} \mathrm{H}_{10}$.

The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers $n$-pentane, isopentane, and neopentane (shown in Figure 10.7) are $36^{\circ} \mathrm{C}, 27^{\circ} \mathrm{C}$, and $9.5^{\circ} \mathrm{C}$, respectively. Even though these compounds are composed of molecules with the same chemical formula, $\mathrm{C}_{5} \mathrm{H}_{12}$, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for $n$-pentane and least for neopentane. The elongated shape of $n$-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip's contact, the stronger the connection.

FIGURE 10.7 The strength of the dispersion forces increases with the contact area between molecules, as demonstrated by the boiling points of these pentane isomers.


1057. Geckos and Intermolecular Forces -
Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos' feet to behave this way.

Geckos' toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in Figure 10.8, with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces-weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko's weight.


1057. Geckos and Intermolecular Forces -
In 2014, two scientists developed a model to explain how geckos can rapidly transition from "sticky" to "non-sticky." Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos' feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Later research led by Alyssa Stark at University of Akron showed that geckos can maintain their hold on hydrophobic surfaces (similar to the leaves in their habitats) equally well whether the surfaces were wet or dry. Stark's experiment used a ribbon to gently pull the geckos until they slipped, so that the researchers could determine the geckos' ability to hold various surfaces under wet and dry conditions. Further investigations may eventually lead to the development of better adhesives and other applications.

FIGURE 10.8 Geckos' toes contain large numbers of tiny hairs (setae), which branch into many triangular tips (spatulae). Geckos adhere to surfaces because of van der Waals attractions between the surface and a gecko's millions of spatulae. By changing how the spatulae contact the surface, geckos can turn their stickiness "on" and "off." (credit photo: modification of work by "JC*+A!"/Flickr)


1058. LINK TO LEARNING -
Watch this video (http://openstax.org/l/16kellaraut) to learn more about Kellar Autumn's research that determined that van der Waals forces are responsible for a gecko's ability to cling and climb.


1059. Dipole-Dipole Attractions -
Recall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule-a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl . In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction-the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in Figure 10.9.

FIGURE 10.9 This image shows two arrangements of polar molecules, such as HCl , that allow an attraction between the partial negative end of one molecule and the partial positive end of another.

The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar $\mathrm{F}_{2}$ molecules. Both HCl and $\mathrm{F}_{2}$ consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K , molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to "stick together" to form a liquid, whereas the relatively weaker dispersion forces between nonpolar $\mathrm{F}_{2}$ molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl ( 188 K ) compared to $\mathrm{F}_{2}(85 \mathrm{~K})$ is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar $\mathrm{F}_{2}$ molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.


1060. Dipole-Dipole Forces and Their Effects -
Predict which will have the higher boiling point: $\mathrm{N}_{2}$ or CO. Explain your reasoning.


1061. Solution -
CO and $\mathrm{N}_{2}$ are both diatomic molecules with masses of about 28 amu , so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because $\mathrm{N}_{2}$ is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar $\mathrm{N}_{2}$ molecules, so CO is expected to have the higher boiling point.


1062. Check Your Learning -
Predict which will have the higher boiling point: ICl or $\mathrm{Br}_{2}$. Explain your reasoning.


1063. Answer: -
ICl. ICl and $\mathrm{Br}_{2}$ have similar masses ( $\sim 160 \mathrm{amu}$ ) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; $\mathrm{Br}_{2}$ is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.


1064. Hydrogen Bonding -
Nitrosyl fluoride (ONF, molecular mass 49 amu ) is a gas at room temperature. Water ( $\mathrm{H}_{2} \mathrm{O}$, molecular mass 18 amu ) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces. Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded ( 4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of $\mathrm{F}, \mathrm{O}$, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with $\mathrm{F}-\mathrm{H}, \mathrm{O}-\mathrm{H}$, or $\mathrm{N}-\mathrm{H}$ moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF $\cdots \mathrm{HF}, \mathrm{H}_{2} \mathrm{O} \cdots \mathrm{HOH}$, and $\mathrm{H}_{3} \mathrm{~N} \cdots \mathrm{HNH}_{2}$, in which the hydrogen bonds are denoted by dots. Figure 10.10 illustrates hydrogen bonding between water molecules.

FIGURE 10.10 Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules.


1064. Hydrogen Bonding -
FIGURE 10.10 Water molecules participate in multiple hydrogen-bonding interactions with nearby water molecules.

Despite use of the word "bond," keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to $10 \%$ as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.

Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group $15\left(\mathrm{NH}_{3}, \mathrm{PH}_{3}, \mathrm{AsH}_{3}\right.$, and $\left.\mathrm{SbH}_{3}\right)$, group 16 hydrides $\left(\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{~S}, \mathrm{H}_{2} \mathrm{Se}\right.$, and $\mathrm{H}_{2} \mathrm{Te}$ ), and group 17 hydrides ( $\mathrm{HF}, \mathrm{HCl}, \mathrm{HBr}$, and HI ). The boiling points of the heaviest three hydrides for each group are plotted in Figure 10.11. As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.

FIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3,4 , and 5 .


1064. Hydrogen Bonding -
FIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3,4 , and 5 .

If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect $\mathrm{NH}_{3}$ to boil at about $-120^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}$ to boil at about $-80^{\circ} \mathrm{C}$, and HF to boil at about $-110^{\circ} \mathrm{C}$. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in Figure 10.12. The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding.

FIGURE 10.12 In comparison to periods 3-5, the binary hydrides of period 2 elements in groups 17, 16 and 15 (F, O and N , respectively) exhibit anomalously high boiling points due to hydrogen bonding.


1066. Effect of Hydrogen Bonding on Boiling Points -
Consider the compounds dimethylether $\left(\mathrm{CH}_{3} \mathrm{OCH}_{3}\right)$, ethanol $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)$, and propane $\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}\right)$. Their boiling points, not necessarily in order, are $-42.1^{\circ} \mathrm{C},-24.8^{\circ} \mathrm{C}$, and $78.4^{\circ} \mathrm{C}$. Match each compound with its boiling point. Explain your reasoning.


1067. Solution -
The VSEPR-predicted shapes of $\mathrm{CH}_{3} \mathrm{OCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$, and $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ are similar, as are their molar masses ( $46 \mathrm{~g} / \mathrm{mol}, 46 \mathrm{~g} / \mathrm{mol}$, and $44 \mathrm{~g} / \mathrm{mol}$, respectively), so they will exhibit similar dispersion forces. Since $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}$ is nonpolar, it may exhibit only dispersion forces. Because $\mathrm{CH}_{3} \mathrm{OCH}_{3}$ is polar, it will also experience dipole-dipole attractions. Finally, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$ has an - OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{3}<\mathrm{CH}_{3} \mathrm{OCH}_{3}<\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$. The boiling point of propane is -42.1 ${ }^{\circ} \mathrm{C}$, the boiling point of dimethylether is $-24.8^{\circ} \mathrm{C}$, and the boiling point of ethanol is $78.5^{\circ} \mathrm{C}$.


1068. Check Your Learning -
Ethane $\left(\mathrm{CH}_{3} \mathrm{CH}_{3}\right)$ has a melting point of $-183^{\circ} \mathrm{C}$ and a boiling point of $-89^{\circ} \mathrm{C}$. Predict the melting and boiling points for methylamine $\left(\mathrm{CH}_{3} \mathrm{NH}_{2}\right)$. Explain your reasoning.


1069. Answer: -
The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. $\mathrm{CH}_{3} \mathrm{CH}_{3}$ and $\mathrm{CH}_{3} \mathrm{NH}_{2}$ are similar in size and mass, but methylamine possesses an - NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of $-93^{\circ} \mathrm{C}$ and a boiling point of $-6^{\circ} \mathrm{C}$.


1071. Hydrogen Bonding and DNA -
Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism's characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism's offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in Figure 10.13.

FIGURE 10.13 Two separate DNA molecules form a double-stranded helix in which the molecules are held together via hydrogen bonding. (credit: modification of work by Jerome Walker, Dennis Myts)

Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and $T$ share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure Figure 10.14.

FIGURE 10.14 The geometries of the base molecules result in maximum hydrogen bonding between adenine and thymine (AT) and between guanine and cytosine (GC), so-called "complementary base pairs."

The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily "unzip" down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication.


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- Distinguish between adhesive and cohesive forces
- Define viscosity, surface tension, and capillary rise
- Describe the roles of intermolecular attractive forces in each of these properties/phenomena

When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure 10.15, have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).

FIGURE 10.15 (a) Honey and (b) motor oil are examples of liquids with high viscosities; they flow slowly. (credit a: modification of work by Scott Bauer; credit b: modification of work by David Nagy)

The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table 10.2 shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases.

Viscosities of Common Substances at $25^{\circ} \mathrm{C}$


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
Viscosities of Common Substances at $25^{\circ} \mathrm{C}$

| Substance | Formula | Viscosity (mPa•s) |
| :---: | :---: | :---: |
| water | $\mathrm{H}_{2} \mathrm{O}$ | 0.890 |
| mercury | Hg | 1.526 |
| ethanol | $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ | 1.074 |
| octane | $\mathrm{C}_{8} \mathrm{H}_{18}$ | 0.508 |
| ethylene glycol | $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2}(\mathrm{OH})$ | 16.1 |
| honey | variable | $\sim 2,000-10,000$ |
| motor oil | variable | $\sim 50-500$ |

TABLE 10.2

The various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface-that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure 10.16, because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.

FIGURE 10.16 Attractive forces result in a spherical water drop that minimizes surface area; cohesive forces hold the sphere together; adhesive forces keep the drop attached to the web. (credit photo: modification of work by "OliBac"/Flickr)


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table 10.3. Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively "tough skin" that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure 10.17, even though they are denser than water, move on its surface because they are supported by the surface tension.

Surface Tensions of Common Substances at $25^{\circ} \mathrm{C}$

| Substance | Formula | Surface Tension (mN/m) |
| :---: | :---: | :---: |
| water | $\mathrm{H}_{2} \mathrm{O}$ | 71.99 |
| mercury | Hg | 458.48 |
| ethanol | $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ | 21.97 |
| octane | $\mathrm{C}_{8} \mathrm{H}_{18}$ | 21.14 |
| ethylene glycol | $\mathrm{CH}_{2}(\mathrm{OH}) \mathrm{CH}_{2}(\mathrm{OH})$ | 47.99 |

TABLE 10.3


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
TABLE 10.3

FIGURE 10.17 Surface tension (right) prevents this insect, a "water strider," from sinking into the water.
Surface tension is affected by a variety of variables, including the introduction of additional substances on the surface. In the late 1800s, Agnes Pockels, who was initially blocked from pursuing a scientific career but
studied on her own, began investigating the impact and characteristics of soapy and greasy films in water. Using homemade materials, she developed an instrument known as a trough for measuring surface contaminants and their effects. With the support of renowned scientist Lord Rayleigh, her 1891 paper showed that surface contamination significantly reduces surface tension, and also that changing the characteristics of the surface (compressing or expanding it) also affects surface tension. Decades later, Irving Langmuir and Katharine Blodgett built on Pockels' work in their own trough and important advances in surface chemistry. Langmuir pioneered methods for producing single-molecule layers of film; Blodgett applied these to the development of non-reflective glass (critical for film-making and other applications), and also studied methods related to cleaning surfaces, which are important in semiconductor fabrication.

The IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not "wet" the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure 10.18).


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
FIGURE 10.18 Differences in the relative strengths of cohesive and adhesive forces result in different meniscus shapes for mercury (left) and water (right) in glass tubes. (credit: Mark Ott)

If you place one end of a paper towel in spilled wine, as shown in Figure 10.19, the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action-when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.

FIGURE 10.19 Wine wicks up a paper towel (left) because of the strong attractions of water (and ethanol) molecules to the -OH groups on the towel's cellulose fibers and the strong attractions of water molecules to other water (and ethanol) molecules (right). (credit photo: modification of work by Mark Blaser)

Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many -OH groups. Water molecules are attracted to these -OH groups and form hydrogen bonds with them, which draws the $\mathrm{H}_{2} \mathrm{O}$ molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers.


1071. Hydrogen Bonding and DNA - 1071.1. Properties of Liquids
Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure 10.20. If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.

FIGURE 10.20 Depending upon the relative strengths of adhesive and cohesive forces, a liquid may rise (such as water) or fall (such as mercury) in a glass capillary tube. The extent of the rise (or fall) is directly proportional to the surface tension of the liquid and inversely proportional to the density of the liquid and the radius of the tube.

The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the
following equation:

$$
h=\frac{2 T \cos \theta}{r \rho g}
$$

In this equation, $h$ is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, $T$ is the surface tension of the liquid, $\theta$ is the contact angle between the liquid and the tube, $r$ is the radius of the tube, $\rho$ is the density of the liquid, and $g$ is the acceleration due to gravity, $9.8 \mathrm{~m} / \mathrm{s}^{2}$. When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of $0^{\circ}$. This is the situation for water rising in a glass tube.


1073. Capillary Rise -
At $25^{\circ} \mathrm{C}$, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm ?
For water, $T=71.99 \mathrm{mN} / \mathrm{m}$ and $\rho=1.0 \mathrm{~g} / \mathrm{cm}^{3}$.


1074. Solution -
The liquid will rise to a height $h$ given by: $h=\frac{2 T \cos \theta}{r \rho g}$
The Newton is defined as a $\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}$, and so the provided surface tension is equivalent to $0.07199 \mathrm{~kg} / \mathrm{s}^{2}$. The provided density must be converted into units that will cancel appropriately: $\rho=1000 \mathrm{~kg} / \mathrm{m}^{3}$. The diameter of the tube in meters is 0.00025 m , so the radius is 0.000125 m . For a glass tube immersed in water, the contact angle is $\theta=0^{\circ}$, so $\cos \theta=1$. Finally, acceleration due to gravity on the earth is $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$. Substituting these values into the equation, and cancelling units, we have:

$$
h=\frac{2\left(0.07199 \mathrm{~kg} / \mathrm{s}^{2}\right)}{(0.000125 \mathrm{~m})\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)}=0.12 \mathrm{~m}=12 \mathrm{~cm}
$$


1075. Check Your Learning -
Water rises in a glass capillary tube to a height of 8.4 cm . What is the diameter of the capillary tube?


1076. Answer: -
diameter $=0.36 \mathrm{~mm}$


1078. Biomedical Applications of Capillary Action -
Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure 10.21. When your finger is pricked, a drop of blood forms and holds together due to surface tension-the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrowdiameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising.

FIGURE 10.21 Blood is collected for medical analysis by capillary action, which draws blood into a small diameter glass tube. (credit: modification of work by Centers for Disease Control and Prevention)


1079. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define phase transitions and phase transition temperatures
- Explain the relation between phase transition temperatures and intermolecular attractive forces
- Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes

We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth's water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this module, the essential aspects of phase transitions are explored.


1080. Vaporization and Condensation -
When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid's vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 10.22 , and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.

(a)

(b)

(c)

FIGURE 10.22 In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from
the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.


1080. Vaporization and Condensation -
The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring "recapture" of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.


1082. Explaining Vapor Pressure in Terms of IMFs -
Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:


1083. Solution -
Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two -OH groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.


1084. Check Your Learning -
At $20^{\circ} \mathrm{C}$, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:

| Compound | methanol $\mathrm{CH}_{3} \mathrm{OH}$ | ethanol $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$ | propanol $\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}$ | butanol $\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}$ |
| :--- | :--- | :--- | :--- | :--- |
| Vapor Pressure at $20^{\circ} \mathrm{C}$ | 11.9 kPa | 5.95 kPa | 2.67 kPa | 0.56 kPa |


1085. Answer: -
All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:
$\mathrm{P}_{\text {methanol }}>\mathrm{P}_{\text {ethanol }}>\mathrm{P}_{\text {propanol }}>\mathrm{P}_{\text {butanol }}$.

As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic
energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 10.23. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.

FIGURE 10.23 Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, $K E$, to escape from the liquid into the gas phase.


1086. Boiling Points -
When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth's atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to $1 \mathrm{~atm}(101.3 \mathrm{kPa})$. Figure 10.24 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid's boiling point on surrounding pressure.

FIGURE 10.24 The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)


1088. A Boiling Point at Reduced Pressure -
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa . Use the graph in Figure 10.24 to determine the boiling point of water at this elevation.


1089. Solution -
The graph of the vapor pressure of water versus temperature in Figure 10.24 indicates that the vapor pressure of water is 68 kPa at about $90^{\circ} \mathrm{C}$. Thus, at about $90^{\circ} \mathrm{C}$, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.


1090. Check Your Learning -
The boiling point of ethyl ether was measured to be $10^{\circ} \mathrm{C}$ at a base camp on the slopes of Mount Everest. Use Figure 10.24 to determine the approximate atmospheric pressure at the camp.


1091. Answer: -
Approximately 40 kPa (0.4 atm)

The quantitative relation between a substance's vapor pressure and its temperature is described by the Clausius-Clapeyron equation:

$$
P=A e^{-\Delta H_{\mathrm{vap}} / R T}
$$

where $\Delta H_{\text {vap }}$ is the enthalpy of vaporization for the liquid, $R$ is the gas constant, and $A$ is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:

$$
\ln P=-\frac{\Delta H_{\mathrm{vap}}}{R T}+\ln A
$$

This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature $T_{1}$, the vapor pressure is $\mathrm{P}_{1}$, and at temperature $\mathrm{T}_{2}$, the vapor pressure is $\mathrm{P}_{2}$, the corresponding linear equations are:

$$
\ln P_{1}=-\frac{\Delta H_{\mathrm{vap}}}{R T_{1}}+\ln A \quad \text { and } \quad \ln P_{2}=-\frac{\Delta H_{\mathrm{vap}}}{R T_{2}}+\ln A
$$

Since the constant, $A$, is the same, these two equations may be rearranged to isolate $\ln A$ and then set them equal to one another:

$$
\ln P_{1}+\frac{\Delta H_{\mathrm{vap}}}{R T_{1}}=\ln P_{2}+\frac{\Delta H_{\mathrm{vap}}}{R T_{2}}
$$

which can be combined into:

$$
\ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)
$$


1093. Estimating Enthalpy of Vaporization -
Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At $34.0^{\circ} \mathrm{C}$, the vapor pressure of isooctane is 10.0 kPa , and at $98.8^{\circ} \mathrm{C}$, its vapor pressure is 100.0 kPa . Use this information to estimate the enthalpy of vaporization for isooctane.


1094. Solution -
The enthalpy of vaporization, $\Delta H_{\text {vap }}$, can be determined by using the Clausius-Clapeyron equation:

$$
\ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)
$$

Since we have two vapor pressure-temperature values ( $T_{1}=34.0^{\circ} \mathrm{C}=307.2 \mathrm{~K}, P_{1}=10.0 \mathrm{kPa}$ and $T_{2}=98.8^{\circ} \mathrm{C}=$ $372.0 \mathrm{~K}, P_{2}=100 \mathrm{kPa}$ ), we can substitute them into this equation and solve for $\Delta H_{\text {vap }}$. Rearranging the Clausius-Clapeyron equation and solving for $\Delta H_{\text {vap }}$ yields:

$$
\Delta H_{\mathrm{vap}}=\frac{R \cdot \ln \left(\frac{P_{2}}{P_{1}}\right)}{\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)}=\frac{(8.3145 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K}) \cdot \ln \left(\frac{100 \mathrm{kPa}}{10.0 \mathrm{kPa}}\right)}{\left(\frac{1}{307.2 \mathrm{~K}}-\frac{1}{372.0 \mathrm{~K}}\right)}=33,800 \mathrm{~J} / \mathrm{mol}=33.8 \mathrm{~kJ} / \mathrm{mol}
$$

Note that the pressure can be in any units, so long as they agree for both $P$ values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.


1095. Check Your Learning -
At $20.0^{\circ} \mathrm{C}$, the vapor pressure of ethanol is 5.95 kPa , and at $63.5^{\circ} \mathrm{C}$, its vapor pressure is 53.3 kPa . Use this information to estimate the enthalpy of vaporization for ethanol.


1096. Answer: -
$41,360 \mathrm{~J} / \mathrm{mol}$ or $41.4 \mathrm{~kJ} / \mathrm{mol}$


1098. Estimating Temperature (or Vapor Pressure) -
For benzene $\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)$, the normal boiling point is $80.1^{\circ} \mathrm{C}$ and the enthalpy of vaporization is $30.8 \mathrm{~kJ} / \mathrm{mol}$. What is the boiling point of benzene in Denver, where atmospheric pressure $=83.4 \mathrm{kPa}$ ?


1099. Solution -
If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, $\Delta H_{\text {vap, }}$, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:

$$
\ln \left(\frac{P_{2}}{P_{1}}\right)=\frac{\Delta H_{\mathrm{vap}}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)
$$

Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value ( $T_{1}=80.1^{\circ} \mathrm{C}=353.3 \mathrm{~K}, P_{1}=101.3 \mathrm{kPa}, \Delta H_{\text {vap }}=30.8$ $\mathrm{kJ} / \mathrm{mol}$ ) and want to find the temperature ( $T_{2}$ ) that corresponds to vapor pressure $P_{2}=83.4 \mathrm{kPa}$. We can substitute these values into the Clausius-Clapeyron equation and then solve for $T_{2}$. Rearranging the ClausiusClapeyron equation and solving for $T_{2}$ yields:
$T_{2}=\left(\frac{-R \cdot \ln \left(\frac{P_{2}}{P_{1}}\right)}{\Delta H_{\text {vap }}}+\frac{1}{T_{1}}\right)^{-1}=\left(\frac{-(8.3145 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}) \cdot \ln \left(\frac{83.4 \mathrm{kPa}}{101.3 \mathrm{kPa}}\right)}{30,800 \mathrm{~J} / \mathrm{mol}}+\frac{1}{353.3 \mathrm{~K}}\right)^{-1}=346.9 \mathrm{~K}$ or $73.8^{\circ} \mathrm{C}$


1100. Check Your Learning -
For acetone $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}$, the normal boiling point is $56.5^{\circ} \mathrm{C}$ and the enthalpy of vaporization is $31.3 \mathrm{~kJ} / \mathrm{mol}$. What is the vapor pressure of acetone at $25.0^{\circ} \mathrm{C}$ ?


1101. Answer: -
30.1 kPa