For the reaction A(g)⇌ B(g) at 495 K, ΔᵣG° = -9.478 kJ mol⁻¹.
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol⁻¹ K⁻¹; ln 10 = 2.303]

Solution

Chemistry ma’am · 4 steps

Turn Δᵣ G° into K first, and here it comes out as a neat 10.

1. Link ΔG° and K. The standard Gibbs energy change fixes the equilibrium constant. Keep Δᵣ G° in joules so the units match R.

   ln K = -(Δᵣ G°)/RT = 9478/(8.314 × 495) = 9478/4115.43 = 2.303

2. Find K. Since ln 10 = 2.303, ln K = ln 10, so K = 10. Here Δ n_g = 0, so the volume cancels and K is just the ratio of moles of B to moles of A.
3. Set up equilibrium. Let x millimoles of A change into B. At equilibrium, B is x and A is 22 - x, so x/(22 - x) = 10.
4. So. x = 220 - 10x, which gives 11x = 220 and x = 20. The equilibrium mixture has 20 millimoles of B.

The common slip: 2. Dropping the minus sign in Δᵣ G° = -RT ln K gives K = 0.1 and only 2 millimoles of B. A negative Δᵣ G° must give K > 1.

The idea to keep. Use Δᵣ G° = -RT ln K to get K, then build an equilibrium table. When Δ n_g = 0, plug moles straight into K because the volume cancels.

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