For the reaction: CuSO₄·3H₂O(s) ⇌ CuSO₄·H₂O(s) + 2H₂O(g); ΔH = 3360 cal. The dissociation pressure i — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

K_p1 = (P_H₂O)² = (7 × 10⁻³)² = 49 × 10⁻⁶ atm². Using ln(K_p2 / K_p1) = (ΔH / R) (1/T₁ - 1/T₂), we get ln(K_p2 / K_p1) = (3360 / 2) (1/300 - 1/400) = 1680 × (100 / 120000) = 1680 / 1200 = 1.4. Since ln2 = 0.7, 1.4 = 2 ln2 = ln(4). Thus K_p2 / K_p1 = 4. K_p2 = 4 × 49 × 10⁻⁶ = 196 × 10⁻⁶. New P_H₂O = √(196 × 10⁻⁶) = 14 × 10⁻³ = 1.4 × 10⁻² atm. Therefore, correct answer is B.