For a dimerization reaction,
2A(g) → A₂(g)
at 298 K, ΔUᵒ = –20 kJ mol –1 , ΔSᵒ = –30 JK –1 mol –1 ,
then the ΔGᵒ will be _____ J. (Round off to the nearest integer)

Solution

Chemistry ma’am · 4 steps

You are given Δ U, not Δ H, so correct it with Δ n_g RT first.

1. Change in gas moles. Two moles of gas go in and one mole of gas comes out, so Δ n_g = 1 - 2 = -1.
2. Find enthalpy change. Use Δ H° = Δ U° + Δ n_g RT = -20000 + (-1)(8.314)(298). This gives -20000 - 2477.6 = -22477.6 J mol⁻¹.

   Δ G° = Δ H° - TΔ S° = -22477.6 - 298 × (-30)

3. Put in numbers. TΔ S° = 298 × (-30) = -8940 J mol⁻¹. Then Δ G° = -22477.6 + 8940 = -13537.6 J mol⁻¹.
4. So. Δ G° ≈ -13537.6 J, which rounds to -13538.

The common slip: -11060. Using Δ U in place of Δ H, with no Δ n_g RT correction, gives -11060, not -13538.

The idea to keep. Gibbs energy needs Δ H. If a question gives Δ U, first apply Δ H = Δ U + Δ n_g RT. Count only gas moles, and keep every term in joules.

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