For the reaction :
A(l) → 2B(g)
Δ U = 2.1 kcal, Δ S = 20 cal K⁻¹ at 300 K
Hence ΔG in kcal is ________. (Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

A liquid turns into 2 mol of gas, so convert Δ U to Δ H before you touch Δ G.

1. Count gas moles. The liquid has no gas moles and the product has 2, so Δ n_g = 2 - 0 = 2. Because the gas moles change, Δ H is not equal to Δ U.
2. Find ΔH. Use Δ H = Δ U + Δ n_g RT with R ≈ 2 cal K⁻¹ mol⁻¹. Then Δ n_g RT = 2 × 2 × 300 = 1200 cal = 1.2 kcal, so Δ H = 2.1 + 1.2 = 3.3 kcal.

   Δ G = Δ H - TΔ S = 3.3 - (300 × 20)/1000 = 3.3 - 6 = -2.7 kcal

3. Match the units. TΔ S = 300 × 20 = 6000 cal, which is 6 kcal. Both terms must be in kcal before you subtract.
4. So. Δ G = -2.7 kcal, which rounds to −3. With R = 1.987 you get -2.71, which also rounds to −3.

The common slip: -4. Putting Δ U in place of Δ H gives 2.1 - 6 = -3.9, which rounds to −4. The Δ n_g RT term of 1.2 kcal is missing.

The idea to keep. Δ G needs Δ H, not Δ U. Whenever gas moles change, add Δ n_g RT first, and keep every term in the same energy unit.

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