Problem 124
How is it possible for a reaction to be spontaneous yet endothermic?
Problem 128
Tell whether the free-energy changes, ΔG, for the processes listed in Problem 9.127 are likely to be positive, negative, or zero.
- When a bottle of perfume is opened, odorous molecules mix with air and slowly diffuse throughout the entire room. Is ΔG for the diffusion process positive, negative, or zero? What about ΔH and ΔS for the diffusion?
Problem 129
- Tell whether reactions with the following values of ΔH and ΔS are spontaneous or nonspontaneous and whether they are exothermic or endothermic. (a) ΔH = -128 kJ; ΔS = 35 J/K at 500 K (b) ΔH = +67 kJ; ΔS = -140 J/K at 250 K (c) ΔH = +75 kJ; ΔS = 95 J/K at 800 K
Problem 133
- 9.135 Suppose that a reaction has ΔH = + 41 kJ and ΔS = - 27 J/K. At what temperature, if any, will it change between spontaneous and nonspontaneous?
Problem 135
- Which of the reactions (a)–(d) in Problem 9.132 are spontaneous at all temperatures, which are nonspontaneous at all temperatures, and which have an equilibrium temperature?
Problem 136
- (d) What is the kinetic energy of an electron traveling at velocity (c)?
Problem 137
Problem 138
Ethyl alcohol has ΔHfusion = 5.02 kJ/mol and melts at - 114.1 °C. What is the value of ΔSfusion for ethyl alcohol?
Problem 141
What is the melting point of benzene in kelvin if ΔHfusion = 9.95 kJ/mol and ΔSfusion = 35.7 J/(K mol)?
- Metallic mercury is obtained by heating the mineral cinnabar (HgS) in air: HgS1s2 + O21g2 S Hg1l2 + SO21g2 (a) Use the data in Appendix B to calculate ΔH° in kilojoules for the reaction.
Problem 142
- Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4): Step 1. CO(g) + 2 H2(g) ⇌ CH3OH(l) ΔS° = -332 J/K Step 2. CH3OH(l) ⇌ CH4(g) + 1/2 O2(g) ΔS° = 162 J/K. (g) Calculate ΔG° for step 2. (d) Which term is more important, ΔH° or ΔS°? (i) Which term is more important, ΔH° or ΔS°?
Problem 143
Problem 143a
Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):
Step 1. CO(g) + 2 H2(g) → CH3OH(l) ΔS° = –332 J/K
Step 2. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K
(a) Calculate ΔH° in kilojoules for step 1.
Problem 143e
Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):
Step 1. CO(g) + 2 H2(g) S CH3OH(l) ΔS° = - 332 J/K
Step 2. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K
(e) In what temperature range is step 1 spontaneous?
Problem 143f
Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):
Step 1. CO(g) + 2 H2(g) → CH3OH(l) ΔS° = –332 J/K
Step 2. CH3OH(l) → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K
(f) Calculate ΔH° for step 2.
- Ethyl chloride 1C2H5Cl2, a substance used as a topical anes-thetic, is prepared by reaction of ethylene with hydrogen chloride: C2H41g2 + HCl1g2 ¡ C2H5Cl1g2 ΔH° = - 72.3 kJ How much PV work is done in kilojoules, and what is the value of ΔE in kilojoules if 89.5 g of ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is - 71.5 L?
Problem 144
- We said in Section 9.1 that the potential energy of water at the top of a dam or waterfall is converted into heat when the water dashes against rocks at the bottom. The potential energy of the water at the top is equal to EP = mgh, where m is the mass of the water, g is the acceleration of the falling water due to gravity 1g = 9.81 m>s22, and h is the height of the water. Assuming that all the energy is converted to heat, calculate the temperature rise of the water in degrees Celsius after falling over California's Yosemite Falls, a distance of 739 m. The specific heat of water is 4.18 J/(g·K).
Problem 145
- When a gaseous compound X containing only C, H, and O is burned in O2, 1 volume of the unknown gas reacts with 3 volumes of O2 to give 2 volumes of CO2 and 3 volumes of gaseous H2O. Assume all volumes are measured at the same temperature and pressure. (d) Combustion of 5.000 g of X releases 144.2 kJ heat. Look up ΔH°f values for CO21g2 and H2O1g2 in Appendix B, and calculate ΔH°f for compound X.
Problem 146
- Given 400.0 g of hot tea at 80.0 °C, what mass of ice at 0 °C must be added to obtain iced tea at 10.0 °C? The specific heat of the tea is 4.18 J>1g °C2 and ΔHfusion for ice is + 6.01 kJ>mol.
Problem 147
- Imagine that you dissolve 10.0 g of a mixture of NaNO3 and KF in 100.0 g of water and find that the temperature rises by 2.22 °C. Using the following data, calculate the mass of each compound in the original mixture. Assume that the specific heat of the solution is 4.18 J>1 g °C2 NaNO31s2 S NaNO31aq2 ΔH = + 20.4 kJ>mol KF1s2 S KF1aq2 ΔH = - 17.7 kJ>mol
Problem 148
- 9.149 Consider the reaction: 4 CO1g2 2 NO21g2 4 CO21g2 N21g2. Using the following information, determine ΔH° for the reaction at 25 °C. NO1g2 ΔH°f = + 91.3 kJ>mol CO21g2 ΔH°f = - 393.5 kJ>mol 2 NO1g2 + O21g2 S 2 NO21g2 ΔH° = - 116.2 kJ 2 CO1g2 + O21g2 S 2 CO21g2 ΔH° = - 566.0 kJ
Problem 149
- Combustion analysis of 0.1500 g of methyl tert-butyl ether, an octane booster used in gasoline, gave 0.3744 g of CO2 and 0.1838 g of H2O. When a flask having a volume of 1.00 L was evacuated and then filled with methyl tertbutyl ether vapor at a pressure of 100.0 kPa and a temperature of 54.8 °C, the mass of the flask increased by 3.233 g. (d) The enthalpy of combustion for methyl tert-butyl ether is ΔH° combustion = -3368.7 kJ>mol. What is its standard enthalpy of enthalpy of formation, ΔH°f?
Problem 150
Problem 151b
Phosgene, COCl2(g), is a toxic gas used as an agent of warfare in World War I. (b) Using the table of bond dissociation energies (Table 9.3) and the value ΔH°f = 716.7 kJ/mol for C(g), estimate ΔH°f for COCl2(g) at 25 °C. Compare your answer to the actual ΔH°f given in Appendix B, and explain why your calculation is only an estimate.
- Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are 112 CH3CO2H1l2 + Na2CO31s2 S CH3CO2Na1aq2 + CO21g2 + H2O1l2 122 CH3CO2H1l2 + NaHCO31s2 CH3CO2Na1aq2 + CO21g2 + H2O1l2 (c) How much heat in kilojoules is absorbed or liberated in each reaction? See Appendix B for standard heats of for- mation; ΔH°f = - 726.1 kJ>mol for CH3CO2 Na(aq).
Problem 152
Problem 152a
Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are
(1) CH3CO2H(l) + Na2CO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)
(2) CH3CO2H(l) + NaHCO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)
(a) Balance both equations.
Problem 152b
Acid spills are often neutralized with sodium carbonate or sodium hydrogen carbonate. For neutralization of acetic acid, the unbalanced equations are
(1) CH3CO2H(l) + Na2CO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)
(2) CH3CO2H(l) + NaHCO3(s) → CH3CO2Na(aq) + CO2(g) + H2O(l)
(b) How many kilograms of each substance is needed to neutralize a 1.000-gallon spill of pure acetic acid (density = 1.049 g/mL)?
Problem 153a
(a) Write a balanced equation for the reaction of potassium metal with water.
Problem 153b
(b) Use the data in Appendix B to calculate ΔH° for the reaction of potassium metal with water.
Problem 153c
(c) Assume that a chunk of potassium weighing 7.55 g is dropped into 400.0 g of water at 25.0 °C. What is the final temperature of the water if all the heat released is used to warm the water?
Problem 153d
(d) What is the molarity of the KOH solution prepared in part (c), and how many milliliters of 0.554 M H2SO4 are required to neutralize it?
- Hydrazine, a component of rocket fuel, undergoes combus- tion to yield N2 and H2O: N2H41l2 + O21g2 S N21g2 + 2 H2O1l2 (b) Use the following information to set up a Hess's law cycle, and then calculate ΔH° for the combustion reac- tion. You will need to use fractional coefficients for some equations. 2 NH31g2 + 3 N2O1g2 S 4 N21g2 + 3 H2O1l2 ΔH° = - 1011.2 kJ N2O1g2 + 3 H21g2 S N2H41l2 + H2O1l2 ΔH° = - 317.2 kJ 4 NH31g2 + O21g2 S 2 N2H41l2 + 2 H2O1l2 ΔH° = - 286.0 kJ H2O1l2 ΔH°f = - 285.8 kJ>mol
Problem 154
Ch.9 - Thermochemistry: Chemical Energy