For the gas phase reaction: 2(g) ⇌ (g) + (g); δ H = -43.5 kcal. Which one of the following is true f — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Relate the Enthalpy of the Forward and Reverse Reactions The given gas-phase chemical equation is: $$\ce{2NO(g) <=> N2(g) + O2(g)} \quad \Delta H = -43.5\text{ kcal}$$ The question asks us to evaluate the thermodynamic behaviour of the reverse reaction: $$\ce{N2(g) + O2(g) <=> 2NO(g)}$$ Since reversing a chemical reaction reverses the sign of its enthalpy change ($\Delta H$), we can find the enthalpy of the reverse reaction ($\Delta H_{\text{rev}}$) using the formula: $$\Delta H_{\text{rev}} = -\Delta H_{\text{fwd}}$$ $$\Delta H_{\text{rev}} = -(-43.5\text{ kcal}) = +43.5\text{ kcal}$$ Because $\Delta H_{\text{rev}} > 0$, the reaction $\ce{N2(g) + O2(g) <=> 2NO(g)}$ is endothermic. Step 2 - Analyze the Temperature Dependence of the Equilibrium Constant ($K$) The effect of temperature on the equilibrium constant is quantitatively described by the integrated form of the van 't Hoff equation: $$\log_{10}\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) = \frac{\Delta H^\circ}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)$$ For an endothermic reaction, $\Delta H^\circ > 0$. Let us assume we decrease the temperature from $T_1$ to $T_2$ (where $T_1 > T_2$, so $T_2 - T_1 < 0$): * Since $\Delta H^\circ > 0$, $R > 0$, and $(T_2 - T_1) < 0$, the right-hand side of the van 't Hoff equation becomes negative. * Therefore, $\log_{10}\left(\frac{K_2}{K_1}\right) < 0$, which mathematically means $\frac{K_2}{K_1} < 1$, or $K_2 < K_1$. Thus, for an endothermic reaction, a decrease in temperature ($T \downarrow$) causes the equilibrium constant ($K$) to decrease ($K \downarrow$). Step 3 - Evaluate the Options * **(A) K is independent of T**: Incorrect. The equilibrium constant $K$ depends on temperature for any reaction where $\Delta H^\circ \neq 0$. * **(B) K decreases as T decreases**: Correct. Since the synthesis of nitric oxide is an endothermic reaction ($\Delta H^\circ = +43.5\text{ kcal}$), the equilibrium constant decreases with a decrease in temperature. * **(C) K increases as T decreases**: Incorrect. This behaviour is characteristic of exothermic reactions ($\Delta H^\circ < 0$) rather than endothermic reactions. * **(D) K varies with addition of \ce{NO}**: Incorrect. According to Le Chatelier's principle, adding or removing reactants or products shifts the equilibrium position but does not change the value of the equilibrium constant $K$, which is a constant at a given temperature. $$\boxed{\text{B}}$$