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19 Chemical Thermodynamics Solutions to Exercises = - - CO(g) = -392.4 - (-166.23) - (-137.2) = In = K (8.314 10⁻³ = 35.922 = 35.9; K = 4 10¹⁵ (b) = - - CO(g) = -487.0 - (-238.6) - (-110.5) = -137.9 kJ The reaction is exothermic, so the value of K will decrease with increasing temperature, and the mole fraction of CH₃COOH will also decrease. Elevated temperatures must be used to increase the speed of the reaction. Thermodynamics cannot predict the rate at which a reaction reaches equilibrium. (c) = - TAS°; when = 0, = TAS° = S° - - = 159.8 - 126.8 197.9 = -164.9 = -0.1649 kJ/K The equilibrium favors products up to 836 K or 563°C, so the elevated temperatures to increase the rate of reaction can be safely employed. 19.101 (a) First calculate for each reaction: 1L + (A) = 6(-237.13) + 6(-394.4) - (-910.4) + 6(0) = -2878.8 kJ For C₆H₁₂O₆(s) + (B) = 2(-394.4) + 2(-174.8) - (-910.4) = -228.0 kJ For (A), In K = 2879 = 1162; K = 5 10⁵⁰⁴ For (B), In K = 228 = 92.026 = 92.0; K = 10³⁹ (b) Both these values for K are unimaginably large. However, K for reaction (A) is larger, because is more negative. The magnitude of the work that can be accomplished by coupling a reaction to its surroundings is measured by G. According to the calculations above, considerably more work can in principle be obtained from reaction (A), because is more negative. 19.102 (a) = -RT In K (Equation [19.20]); In K = Use = - TAS° to get at the two temperatures. Calculate and using data in Appendix C. 2CH₄(g) + = C₂H₆(g) + = -84.68 + 2(-74.8) = 64.92 = 603