Question

**Given:** Thermal decomposition of $\text{NaNO}_3$ in a closed vessel. **Equilibrium reaction:** $$2\text{NaNO}_3(s) \rightleftharpoons 2\text{NaNO}_2(s) + \text{O}_2(g)$$ **Equilibrium expression:** Since $\text{NaNO}_3$ and $\text{NaNO}_2$ are pure solids, their activities are unity and do not appear in the equilibrium expression: $$K_p = p_{\text{O}_2}$$ **Analysis of each option:** - **Option A:** $\text{NaNO}_2(s)$ is a pure solid (activity = 1). Adding it does **not** change $K_p$ or shift the equilibrium. ❌ Incorrect. - **Option B:** $\text{NaNO}_3(s)$ is a pure solid (activity = 1). Adding it does **not** change $K_p$ or shift the equilibrium. ❌ Incorrect. - **Option C:** The forward reaction is endothermic (bond breaking in $\text{NaNO}_3$). By Le Chatelier's principle, increasing temperature shifts equilibrium in the endothermic (forward) direction. ✅ Correct. - **Option D:** Since $K_p = p_{\text{O}_2}$ is constant at fixed temperature, increasing total pressure increases $p_{\text{O}_2}$. To restore equilibrium, the system shifts in the reverse direction (consuming $\text{O}_2$). ✅ Correct. **Answer: (C) and (D)**