Question

The reaction SO2(g)+2H2S(g)←−→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from 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.             

C) 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

delta G of reaction = deltaG for SO2 + 2*delta G for H2S - 3*deltaG of S - 2*delta G of H2) = -300.4 + (-33.01) *2

                                                                                                                           -3*(0) - 2* (-228.57) = 90.72 KJ/mole

   Now deltaG = nRT lnP2/P1

so       90.72   = 1* 8.314 * 298 *ln P2/20

ln P2/20 = 0.0366

2.303 log P2/20 =0.0366

so log P2/20 = 0.01589

P2/20 = 10^0.01589 =1.03727

P2 =   20*1.03727 = 20.745 torr