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.

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

Express your answer using one significant figure.

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.

Express your answer using one significant figure.

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

-More effective

-Less effective

Answer 1

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

ΔG_r = ΔG°_r

a)

ΔG_r = - 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 ΔS_r < 0

ΔG_r = ΔH_r – T ΔS_r

ΔGr will become less negative at higher temperatures.

Process will be less effective at higher temperatures.