Consider the following reaction at 298 K · 3/2 O_(2(g)) ⇌ O_(3(g)) · K_P=2.47 × 10⁻²⁹. Δᵣ G⊖ for the reaction is _________ kJ. (Given R=8.314 JK⁻¹ mol⁻¹)

Solution

Chemistry ma’am · 4 steps

A tiny K_P means a large positive Δᵣ G⊖: put it straight into -RT ln K.

1. The link. For a gas-phase reaction, the standard Gibbs energy change is tied to K_P by one equation.

   Δᵣ G⊖ = -RT ln K_P

2. Take the log. ln(2.47 × 10⁻²⁹) = ln 2.47 - 29 ln 10 = 0.904 - 66.775 = -65.871. The log is negative because K_P is far below 1.
3. Put in numbers. RT = 8.314 × 298 = 2477.6 J mol⁻¹. So Δᵣ G⊖ = -2477.6 × (-65.871) = 1.632 × 10⁵ J.
4. So. Δᵣ G⊖ ≈ 163.2 kJ, which rounds to 163.

The common slip: -163. Writing Δᵣ G⊖ = RT ln K_P without the minus sign gives -163, a negative value that cannot match such a tiny K_P.

The idea to keep. If K is less than 1, ln K is negative and Δᵣ G⊖ comes out positive. Check this sign before you write the answer.

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