Question
Substance
228.6 241.8 237.1 285.8 300.4 296.9 370.4 395.2 33.01 20.17 0 0 Gibbs Free Energy and Equilibrium The reaction SO2(g) + 2H2S(g) 3S(s) + 2H2O(g) is the basis of a suggested methodfor removal of SO2 from power-plant stack gases. The values below may be helpfulwhen answering questions about the process. Assume that the partial pressure of sulfurdioxide, Pso2 , is equal to the partial pressure of dihydrogensulfide, PH2s , and therefore Pso2 = PH2s . If the vapor pressure of water is24 torr , calculate the equilibrium partial pressure of SO2 ( PSO2 ) in the system at 298 K . Express the pressure in atmospheresusing two significant figures.
Explanation / Answer
We Know that : The givenReaction is : Greaction = Gproducts - Greactants = [ 3 (0) + 2 (-228.6) ] KJ - [(-300.4) + 2 (-33.01) ] KJ = - 90.78 KJ We Know that : Greaction = -2.303 R T logKp - 90.78 x 103 J = - 2.303 x 8.314 J / mol-K x 298 K xlog PH2O2 / PSO2 x PH2S2 PH2O2 / PSO2 x PH2S2 = 8.1 x1015 PSO2 = PH2S = X (24 torr )2 / 8.1 x 1015 = X3 X = 4.180 x 10-5 torr. The Pressure of SO2 gas is 4.180 x10-5 torr. Greaction = Gproducts - Greactants = [ 3 (0) + 2 (-228.6) ] KJ - [(-300.4) + 2 (-33.01) ] KJ = - 90.78 KJ We Know that : Greaction = -2.303 R T logKp - 90.78 x 103 J = - 2.303 x 8.314 J / mol-K x 298 K xlog PH2O2 / PSO2 x PH2S2 PH2O2 / PSO2 x PH2S2 = 8.1 x1015 PSO2 = PH2S = X (24 torr )2 / 8.1 x 1015 = X3 X = 4.180 x 10-5 torr. The Pressure of SO2 gas is 4.180 x10-5 torr. X = 4.180 x 10-5 torr. The Pressure of SO2 gas is 4.180 x10-5 torr.