Question

Calculate the reaction free energy of

H₂(g) + Cl₂(g)   2HCl(g)

when the concentrations are 0.026 mol L^-1 (H₂), 0.033 mol L^-1 (Cl₂), and 0.00184 mol L^-1 (HCl), and the temperature is 500K. For this reaction K_c = 85 at 500 K.

1. -5.3 kJ
2. -12.2 kJ
3. -22.2 kJ
4. -41.5 kJ

Answer 1

Reaction quotient

$Q_c = \frac{[HCl]^2}{[H_2][Cl_2]} \\ = \frac{0.00184^2}{(0.026 \times 0.033)}=0.003946 \\ K_c=85 \\ T= 500 \;K \\ R=8.314 \;J/mol \cdot K \\ ΔG=ΔG_o +Rtln(Q_c) \\ = -RT ln(K_c) + RT ln(Q_c) \\ = -8.314 \times 500 \times ln(85) + 8.314 \times 500 \times ln(0.003946) \\ = -41477 \;J/mol \\ ≈ -41.5 \;kJ/mol$

Answer: the last option -41.5 kJ/mol