Question

Using data from the Thermodynamic Data table, calculate ΔG for the reaction 2 H2S(g) + SO2(g) equilibrium reaction arrow 3 Srhombic(s) + 2 H2O(g) for the following conditions at 25°C. (Assume PA, PB, and PC are the partial pressures of H2S(g), SO2(g), and H2O(g), respectively.) PA = 1.3 ✕ 10−4 atm PB = 1.4 ✕ 10−2 atm PC = 2.9 ✕ 10−2 atm

Answer 1

The reaction is

2H₂S(g) + SO₂(g) equilibrium reaction arrow 3S_rhombic(s) + 2H₂O(g)

Absolute temperature T = (25+273) K = 298 K

We have ΔG = -2.303RTlogK_p

K_p = equilibrium constant in terms of partial pressure in case of gaseous reactions

ΔG = free energy of the reaction

R = Universal gas constant and its value is 8.314 J mol^-1 K^-1

T = absolute temperature in Kelvin scale

In the above reaction, as rhombic sulfur is a product that is solid at 25°C, so its partial pressure will be zero.

Thus the expression of equilibrium constant will be,

K_p = p_C² / (p_A² * p_B); where p_A, p_B and p_C are the partial pressures of H₂S(g), SO₂(g) and H₂O(g), respectively.

Thus putting the values of p_A, p_B and p_C in the expression of K_p, we get

K_p = (2.9*10^-2)² / {(1.3*10^-4)² * (1.4*10^-2)}

or, K_p = 3.55 * 10¹⁰

By using the values of K_p, R and T, from the expression of ΔG we get

ΔG = -2.303 * 8.314 * 298 * log(3.55 * 10¹⁰) J mol^-1

or ΔG = -60198 J mol^-1

Therefore the value of ΔG is -60198 J mol^-1