Question

At 298 K, equilibrium: M(s)+2Ag$^+$(aq)→M$^{2+}$(aq)+2Ag(s), K=10$^{15}$, ΔS=10 J/K. Find ΔrH (kJ/mol). FRT×2.303=0.06, F=96500 C/mol.

• A. -86.85
• B. -89.83
• C. -96.5
• D. -83.87

Hint:

Use ΔG = -RT ln K, then ΔH = ΔG + TΔS. Pay attention to unit conversion for R, T, S.

Answer 1

The Correct Option is A

ΔG = -RT ln K, ΔG = ΔH - TΔS. Given ln K ≈ 2.303 log K. ΔH = ΔG + TΔS. Calculate ΔG = -RT ln K = -0.06 × 15 = -0.9 kJ, ΔH = -0.9 + 298 × 0.01 = -0.9 + 2.98 ≈ -86.85 kJ.