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Homework 9 Key

# Homework 9 Key - CHEMISTRY 201X Homework#9 Equilibrium...

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CHEMISTRY 201X: Homework #9 Due : Monday, December 3, 2007 Equilibrium Applications (1) (6 pts) Snow and ice sublime spontaneously when the partial pressure of water vapor is below the equilibrium vapor pressure of ice. At 0 ° C the vapor pressure of ice is 0.0060 atm (the triple point of water). Taking the enthalpy of sublimation of ice to be 50.0 kJ/mol, calculate the partial pressure of water vapor below which ice will sublime spontaneously at 15 ° C. Use the Clausius-Clapeyron equation for equilibrium between H 2 O(s) and H 2 O(g): sub 1 1 2 2 1 H p p 1 1 50,000 1 1 ln = - ; ln = - p R T T 0.006 8.314 273.15 258.15 p 1 = 0.0017 atm (2) (14 pts) At 1200 K in the presence of solid carbon, an equilibrium mixture of CO(g) and CO 2 (g) (called “producer gas”) contains 98.3 mole percent CO and 1.69 mole percent CO 2 (g) when the total pressure is 1 atm. The reaction is CO 2 (g) + C(graphite) ZZX YZZ 2 CO(g) (a) (2 pts) Calculate p (CO) and p (CO 2 ) (in atm). p(CO) = x(CO)(1 atm) = 0.983 atm p(CO 2 ) = x(CO 2 )(1 atm) = 0.0169 atm (b) (2 pts) Calculate the equilibrium constant K p at 1200 K from these data. 2 2 p 2 p (CO) (0.983) K = = = 57.2 p(CO ) (0.0169) (c) (2 pts) Calculate G ° (in kJ) from these data. G ° = (8.314 J/K)(1200 K)ln(57.2) = 40370 J = 40.4 kJ (d) (4 pts) Use thermodynamic data to calculate G ° (in kJ) and K p at 1200 K. Compare with your results in parts (b) and (c). H ° = 2( 110.5 kJ) [( 393.5 kJ) + (0)] = +172.5 kJ S ° = 2(197.9 J/K) [(213.6 J/K) + (5.69 J/K)] = +176.5 J/K G ° = (+172.5 kJ) (1200 K)(+0.1765 kJ/K) = 39.3 kJ K p = exp[ ( 39300 J)/(8.314 J/K)(1200 K)] = exp(3.939) = 51.4 This estimate relies on assuming that H ° and S ° remain constant with temperature, which is strictly incorrect. The equilibrium constant at 1200 K is

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