Calculate the equilibrium constant for the following reaction at 525 K: 2H₂S(g) → 2H₂(g) + S₂(g) Δ_rH° = 169.8 kJ mol^-1; Δ_rS°= 78.0 J K^-1 mol^-1

Question

Quizplus · Accepted Answer

The equilibrium constant can be calculated using the equation:
ΔG° = -RTlnK
where ΔG° is the standard Gibbs free energy change, R is the gas constant (8.314 J K-1 mol-1), T is the temperature in kelvin, and K is the equilibrium constant.

First, we need to convert the units of ΔS° to J K-1:
ΔS° = 78.0 J K-1 mol-1 / 1000 = 0.078 J K-1 mol-1

Next, we need to convert the units of ΔH° to J mol-1:
ΔH° = 169.8 kJ mol-1 * 1000 = 169800 J mol-1

Now, we can calculate ΔG°:
ΔG° = ΔH° - TΔS°
ΔG° = 169800 J mol-1 - (525 K * 0.078 J K-1 mol-1 * 2) = 168744 J mol-1

Finally, we can calculate K:
K = e^(-ΔG°/RT)
K = e^(-(168744 J mol-1) / (8.314 J K-1 mol-1 * 525 K))
K ≈ 1.51 × 10-13

Calculate the Equilibrium Constant for the Following Reaction at 525