In the decomposition equilibrium of a certain metallic carbonate, the partial pressure of CO₂ become — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

K_p is proportional to P_CO₂. So K₂/K₁ = 100. Using log(K₂/K₁) = (ΔH° / 2.303R) (1/T₁ - 1/T₂): log(100) = 2 = (ΔH° / (2.303 × 1.987)) (1/400 - 1/500). 2 = (ΔH° / 4.576) (100 / 200000). ΔH° = 2 × 4.576 × 2000 = 18304 cal/mol ≈ 18.3 kcal/mol. The closest option is 18.4 kcal/mol. Therefore, correct answer is A.