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...

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.

Part A What is the equilibrium constant for the reaction at 298 K?

Part B In principle, is this reaction a feasible method of removing SO2?

 -Yes -No

Part C If PSO2 = PH2S and the vapor pressure of water is 25 torr , calculate the equilibrium SO2 pressure in the system at 298 K.

Part D Would you expect the process to be more or less effective at higher temperatures?

 -More effective -Less effective

SO2 (g) + 2 H2S (g) <==> 3 S(s) + 2H2O (g)

ΔG°f S(s) = 0

ΔG°f H2O(g) = -228.57 kJ/mol

ΔG°f SO2(g) = -300.4 kJ/mol

ΔG°f H2S(g) = -33.01 kJ/mol

ΔG°r = 3 * ΔG°f S(s) + 2 * ΔG°f H2O(g) - ΔG°f SO2(g) – 2 * ΔG°f H2S(g)

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

= -90.72 kJ/mol

At T = 298 K

ΔGr = ΔG°r

a)

ΔGr = - R T ln(K)

R = 8.314 J/mol-K

90.72 = 8.314 x 10-3 * 298 * ln(K)

Ln(K) = 36.6

K = 8 x 1015

b)

Since ΔG°r is negative at these conditions, the reaction is feasible.

Yes

c)

K = p(H2O)2 / [p(SO2) * p(H2S)]

p(SO2) = p(H2S)

So,

K = p(H2O)2 / p(SO2)2

8 x 1015 = (25 torr / p(SO2))2

8.9 x 107 = 25 torr / p(SO2)

p(SO2) = 2.8 x 10-7 torr

d)

3 moles of gas are yielding 2 moles of gas in the reaction

Therefore ΔSr < 0

ΔGr = ΔHr – T ΔSr

ΔGr will become less negative at higher temperatures.

Process will be less effective at higher temperatures.