The value of log K for the reaction A ⇌ B at 298 K is ___________. (Nearest integer)
Given: Δ H°=-54.07 kJ mol⁻¹
Δ S°=10 J K⁻¹ mol⁻¹
(Take 2.303 × 8.314 × 298=5705 )

Solution

Chemistry ma’am · 3 steps

Get Δ G° from Δ H° - TΔ S° in joules, then turn it into log K.

1. Match the units. Δ H° is in kJ but Δ S° is in J, so convert first: Δ H° = -54070 J mol⁻¹.
2. Find ΔG°. Δ G° = Δ H° - TΔ S° = -54070 - (298 × 10) = -54070 - 2980 = -57050 J mol⁻¹.

   log K = (-Δ G°)/2.303RT = 57050/5705

3. So. log K = 10, so the answer is 10.

The common slip: 9. Adding TΔ S° instead of subtracting it gives Δ G° = -51090 J mol⁻¹ and log K ≈ 9, not 10.

The idea to keep. For K from enthalpy and entropy, go through Δ G° = Δ H° - TΔ S°, then use Δ G° = -2.303RT log K. Put everything in joules first.

Answer from the NTA final answer key. Solution by PrepWiser.