JEE Main 2024 Chemistry question
Consider the following reaction at 298 K. (3/2)O₂(g) ⇌ O₃(g), = 2.47 × 10⁻²⁹. Δ_rG° for the reaction is kJ. (Given R = 8.314 JK⁻¹mol⁻¹)
Answer: 163
Solution
Δ_rG° = −2.303RT log = −2.303 × 8.314 × 298 × log(2.47 × 10⁻²⁹) = +163.2 × 10³ J ≈ 163 kJ.