Assume that the partial pressure of sulfur dioxide, PSO2, is equal to the partial pressure of...

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Assume that the partial pressure of sulfur dioxide, PSO2, is equal to the partial pressure of...

Assume that the partial pressure of sulfur dioxide, PSO2, is equal to the partial pressure of dihydrogen sulfide, PH2S, and therefore PSO2=PH2S. If the vapor pressure of water is 28 torr , calculate the equilibrium partial pressure of SO2 (PSO2) in the system at 298 K.

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Answer 1

Answer #1

SO2(g) + 2H2S (g)-------> 3S + 2H2O(g)

$\Delta$ G = $\Delta$ Gf products - $\Delta$ Gf reactants

= -2*228.59+0 -( -371.08 +2*-33.05)

= -20KJ/mole = -20000J/mole

$\Delta$ G = -RTlnKp

-20000 = -8.314*298*2.303logkP

logKp = 20000/5705.8483

= 3.5051

Kp = 10^3.5051 = 3199.6

PH2O = 28torr = 28/760 = 0.0368atm

Kp = P²H2O/PSo2 P²H2S

3199.6 = (0.0368)²/x*x²

x³ = 0.0368*0.0368/3199.6 = 4.23*10^-7

x = 7.5*10^-3 atm

PSo2 = x = 7.5*10^-3 atm

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Answer 2

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