Question

The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2from power-plant stack gases. The standard free energy of each substance are ΔG∘fS(s) = 0 kJ/mol, ΔG∘fH2O(g) = -228.57 kJ/mol, ΔG∘fSO2(g) = -300.4 kJ/mol, ΔG∘fH2S(g) = -33.01 kJ/mol.

If PSO2 = PH2S and the vapor pressure of water is 20 torr , calculate the equilibrium SO2 pressure in the system at 298 K.

Answer 1

ΔG∘fS(s) = 0 kJ/mol,

ΔG∘fH2O(g) = -228.57 kJ/mol,

ΔG∘fSO2(g) = -300.4 kJ/mol,

ΔG∘fH2S(g) = -33.01 kJ/mol.

SO2(g)+2H2S(g)←→3S(s)+2H2O(g)

ΔG∘rex    = ΔG∘f products - ΔG∘f reactants

               = (2*-228.57 +3*0) -(-300.4 +2*-33.01)

               = -90.72KJ/mole    = -90720J/mole

ΔG∘rex    = -RTlnKp

-90720    = -8.314*298*2.303logKp

-90720   = -5705.8483logKp

logKp    = -90720/-5705.8483

logKp    = 15.9

    Kp   = 10^15.9

   Kp    = 7.94*10^15

SO2(g)+2H2S(g)←→3S(s)+2H2O(g)

PSO2 = PH2S   = x

Kp   = P^2 H2O/PSo2*P^2 H2S

Kp   = (20)^2/x*x^2

7.94*10^15   = (20)^2/x*x^2

7.94*10^15   = 400/x^3

x^3               = 400/7.94*10^15

x^3            = 5.04*10^-14

x                = 3.64*10^-5

PSo2        = 3.64*10^-5 torr