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

💡 Solution & Explanation

I need to see the options to explain why D is correct. Since they aren't provided, I'll cover the key principle: **For the equilibrium:** $CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+$ **The equilibrium constant $K_a$ changes only with:** - **Temperature** — This is the primary factor. $K_a$ is temperature-dependent through the Gibbs free energy relationship: $\Delta G° = -RT\ln K$ **The equilibrium constant does NOT change with:** - Concentration of reactants/products (shifts equilibrium position, not $K$) - Pressure/volume (for reactions in solution) - Presence of a catalyst (speeds up both forward and reverse reactions equally) - Amount of substance (intensive property) **If option D states "temperature,"** it's correct because: $$K_a = \frac{[CH_3COO^-][H_3O^+]}{[CH_3COOH]} \text{ depends on } T$$ The equilibrium constant is a function of temperature alone for a given reaction. Changing $T$ alters the ratio of forward to reverse reaction rate constants, thereby changing $K_a$.