CHEM 162: Thermodynamics Practice Problems Key 1. Consider the following reaction, NO(g) + O3(g) → NO2(g) + O2(g) ∆H°=-200.0 kJ Check all of the following statements that are true: ; a. This reaction is exothermic. ∆G° = ∆H° − T∆S°  b. This reaction is endothermic. − + ; c. If ∆S° is 71.3 J/K for this reaction, this reaction is always spontaneous.  d. Adding a catalyst to this reaction system will decrease the enthalpy change. ; e. This reaction will be even more spontaneous at higher temperatures if ∆S° is 71.3 J/K. ; f. When this reaction occurs, ∆Ssurr is positive.  g. Because O2(g) is the naturally occurring form of oxygen, S°=0 for O2(g) at 25°C. ; h. The activation energy for the forward reaction is lower than the activation energy for the reverse reaction. 2. Ammonium chloride dissolves readily in water. A student dissolves NH4Cl at 25°C, she notices that the beaker with the solution feels cold. Check all of the following statements that are true: ; a. This process is endothermic, so ∆H° is positive.  b. This process is exothermic, so ∆H° is negative. ∆G° = ∆H° − T∆S° ; c. The entropy change, ∆S°, for this process is positive. + +  d. The entropy change, ∆S°, for this process is negative.  e. The dissolving of ammonium chloride is spontaneous only at low temperatures. ; f. The dissolving of ammonium chloride is spontaneous only at high temperatures.  g. The dissolving of ammonium chloride is always spontaneous. 3. Consider the reaction at 25°C, 2 Cdiamond(s) + O2(g) → 2 CO(g), for which ∆S°=+186.0 J/K. Calculate S° for Cdiamond (in J/mol·K) given CO(g)’s S°=197.7KmolJ⋅ and O2(g)’s S°=205.0KmolJ⋅. ∆S° = [(1 mol)(197.7 KmolJ⋅)] − [(2 mol)(x) + (1 mol)(205.0 KmolJ⋅)] = 186.0 KJ = 395.4 KJ − [(2 mol)(x)] − 205.0 KJ = 186.0 KJ [(2 mol)(x)] = 395.4 − 205.0 − 186.0 KJ = 4.4 KJ → x = 2.2 KmolJ⋅ 4. Methane reacts with chlorine gas to produce carbon tetrachloride and hydrogen chloride gas, CH4(g) + 2 Cl2(g) → CCl4(l) + 4 HCl(g) ∆H°= −397.3 kJ CH4(g) CCl4(g) HCl(g) S° (J/mol·K) 186.2 309.7 186.9 a. If ∆S° for the reaction above is 425.1 J/K, calculate S° for chlorine gas. ∆S° = [(1 mol)(309.7KmolJ⋅) + (4 mol)(186.9KmolJ⋅)] − [(1 mol)(186.2KmolJ⋅) + (2 mol)(x)] = 425.1 KJ = 1057.3 KJ − 186.2 KJ − [(2 mol)(x)] = 425.1 KJ [(2 mol)(x)] = 1057.3 − 186.2 − 425.1 KJ = 446.0 KJ → x = 223.0 KmolJ⋅b. Explain if the reaction or process is always spontaneous, never spontaneous, or spontaneous only at high or low temperatures. Since ∆H° is negative and ∆S° is positive, the reaction is always spontaneous since both terms contribute to ∆G° being negative. c. Calculate the standard free energy change for the reaction. ∆G° = ∆H° − T∆S° = −397.3 kJ − (298.15 K)(425.1 KJ)J 1000kJ 1 = −524.0 kJ d. Calculate the equilibrium constant, K, for the reaction at standard state conditions. ∆G° = − RT ln K ln K = ()=×=∆⋅kJ1000J(298.15) 8.3145kJ) (-524.043- RTG- KmolJD211.374… K = e211.374 = 6×1091 e. Calculate the equilibrium constant, K, for the reaction at 175°C. =1212T1-T1 R∆H- KK lnD ×=×⋅298.15K1-448.15K1 8.3145J)10 -397.3- 106.288K lnKmolJ3912(=-53.643… K2 = 6.288x1091 (e-53.643) = 3x1068 f. Explain the difference between the equilibrium constants for d and e above. Since ∆H° is negative, the reaction is exothermic, so as temperature increases, the reaction shifts left, so the equilibrium constant is lower at higher temperature. 5. One source for hydrogen is the treatment of natural gas (primarily CH4) with steam, CH4(g) + H2O(g) → 3 H2(g) + CO(g). a. Calculate the standard enthalpy change (∆H°) and the standard entropy change (∆S°) for the reaction given the following: CH4(g) H2O(g) H2(g) CO(g) ∆H° (kJ/mol) -74.8 -241.8 − -110.5 S° (J/mol·K) 186 188.7 130.6 196.7 ∆H° = [(3 mol)(0 molkJ) + (1 mol)(-110.5 molkJ)] − [(1 mol)(-74.8 molkJ) + (1 mol)(-241.8 molkJ)] = -110.5 kJ – (-316.6 kJ) = +206.1 kJ ∆S° = [(3 mol)(130.6KmolJ⋅) + (1 mol)(196.7KmolJ⋅)] − [(1 mol)(186KmolJ⋅) + (1 mol)( 188.7KmolJ⋅)] = 213.8 KJ = +214 KJ b. Explain if the reaction or process is always spontaneous, never spontaneous, or spontaneous only at high or low temperatures. Since ∆H° and ∆S° are positive, ∆G° is negative only when the -T∆S° term dominates, so the reaction is only spontaneous at high temperatures.6. Consider that ∆S°=144.6 J/K for the sublimation of iodine, I2(s) → I2(g) ∆H°=62.4 kJ a. Calculate the standard free energy, ∆G°, for this phase change at 25°C. ∆G° = ∆H° − T∆S° = 62.4 kJ − (298.15 K)(144.6 KJ)J 1000kJ 1 = 19.3 kJ b. Is the reaction spontaneous at 25°C? Yes No c. Calculate the sublimation point (the temperature when sublimation occurs in °C) for iodine. Assume the values for ∆H° and ∆S° are valid over the range of temperatures involved. ∆G° = ∆H° − T∆S° = 0 ∆H° = T∆S° KJ3 144.6J1062.4∆S∆ΗT×==DD= 431.535 K T = 431.535 K - 273.15 = 158°C 7. Calculate ∆G° for the autoionization of water: H2O(l) → H+(aq) + OH−(aq). ∆G° = − RT ln Kw = − (8.3145 KmolJ⋅)(298.15 K) J 1000kJ 1 ln 1.0x10-14 = 79.9 kJ 8. The molar heat of vaporization for ethanol is 39.3 kJ/mol and the boiling point is 78.3°C. Calculate the entropy change for the vaporization of 50.0 g of ethanol. ∆G° = ∆H° − T∆S° = 0 ∆H° = T∆S° =×==K 351.451039.3T∆Η∆SmolJ3DD 111.822KmolJ⋅ 50.0 g C2H5OH×OHHC g 46.068OHHC mol 15252× =⋅KmolJ 111.822121 KJ 9. Explain whether the sublimation of dry ice, CO2(s), is always spontaneous, never spontaneous, or spontaneous only at high or low temperatures. CO2(s) → CO2(g) Sublimation is endothermic, and the entropy for the