In equilibrium + – + H The equilibrium constant may change when — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

# Solution For the equilibrium: $CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)$ The equilibrium constant is: $$K_a = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]}$$ **What changes $K_a$?** $$K_a \text{ depends only on } \Delta G° = -RT\ln K_a$$ Since $\Delta G°$ is a state function dependent only on temperature: | Factor | Effect on $K_a$ | |--------|-----------------| | **Temperature** | ✓ **CHANGES** $K_a$ (shifts $\Delta G°$) | | **Pressure/Volume** | ✗ No effect (weak acid dissociation is not pressure-sensitive) | | **Concentration** | ✗ No effect (shifts equilibrium position, not $K_a$) | | **Catalyst** | ✗ No effect (speeds up forward & reverse equally) | **Why D is correct:** Only **temperature changes** affect the equilibrium constant. Increasing temperature shifts the equilibrium (for weak acids, typically increases $K_a$), while concentration, pressure, and catalysts only shift the equilibrium position without changing $K_a$ itself.