Here are the solutions to the questions: e) Study the reaction below and answer the questions that follow. $$ \text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)} \quad \Delta H = +243 \text{ kJ mol}^{-1} $$ A: How does an increase in temperature affect the equilibrium position? The reaction is endothermic because $\Delta H$ is positive. According to Le Chatelier's principle, an increase in temperature will shift the equilibrium in the direction that absorbs heat. Therefore, the equilibrium position will shift to the right, favoring the formation of more $\text{NO}_2$. B: Equilibrium constant of the above reaction. Since the equilibrium shifts to the right (towards products) when the temperature is increased, the concentration of products ($\text{NO}_2$) will increase and the concentration of reactants ($\text{N}_2\text{O}_4$) will decrease at equilibrium. The equilibrium constant $K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}$. Therefore, the equilibrium constant will increase. f) Identify the acid/base conjugate pairs in the reaction below. $$ \text{C}_6\text{H}_5\text{COO}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{C}_6\text{H}_5\text{COOH(aq)} + \text{OH}^-\text{(aq)} $$ Acid/Conjugate Base Pair 1: $\text{H}_2\text{O}$ acts as an acid* by donating a proton ($\text{H}^+$). $\text{OH}^-$ is its conjugate base*. Base/Conjugate Acid Pair 2: $\text{C}_6\text{H}_5\text{COO}^-$ acts as a base* by accepting a proton ($\text{H}^+$). $\text{C}_6\text{H}_5\text{COOH}$ is its conjugate acid*. The conjugate pairs are: $\boxed{\text{H}_2\text{O} \text{ (acid)} / \text{OH}^- \text{ (conjugate base)}}$ $\boxed{\text{C}_6\text{H}_5\text{COO}^- \text{ (base)} / \text{C}_6\text{H}_5\text{COOH} \text{ (conjugate acid)}}$