The equilibrium constant for the reaction: (g) + (g) ⇌ 2(g) is 4.0 × 10^-4 at 2000 K. In the presenc — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Define the Equilibrium Constant For the given reversible gas-phase reaction: \[\ce{N2(g) + O2(g) <=> 2NO(g)}\] From a kinetic standpoint, the equilibrium constant (\(K_{\text{eq}}\)) for a reversible reaction is defined as the ratio of the rate constant of the forward reaction (\(k_{\text{f}}\)) to the rate constant of the backward reaction (\(k_{\text{b}}\)): \[K_{\text{eq}} = \frac{k_{\text{f}}}{k_{\text{b}}}\] At a constant absolute temperature of \(T = 2000\text{ K}\), the equilibrium constant is: \[K_{\text{eq}} = 4.0 \times 10^{-4}\] Step 2 - Analyze the Kinetic Role of a Catalyst A catalyst increases the speed of a chemical reaction by providing an alternative reaction pathway with a lower activation energy (\(E_{\text{a}}\)). Because the energy barrier is lowered by the exact same amount (\(\Delta E_{\text{a}}\)) for both the forward and reverse directions: * The rate constant of the forward reaction increases to \(k'_{\text{f}} = k_{\text{f}} \cdot e^{\Delta E_{\text{a}} / RT}\) * The rate constant of the backward reaction increases to \(k'_{\text{b}} = k_{\text{b}} \cdot e^{\Delta E_{\text{a}} / RT}\) Given that the equilibrium is attained \(10\) times faster in the presence of the catalyst, both the forward and backward rate constants are increased by a factor of \(10\): \[k'_{\text{f}} = 10 \cdot k_{\text{f}}\] \[k'_{\text{b}} = 10 \cdot k_{\text{b}}\] Step 3 - Determine the Catalyzed Equilibrium Constant The new equilibrium constant in the presence of the catalyst (\(K'_{\text{eq}}\)) is calculated as: \[K'_{\text{eq}} = \frac{k'_{\text{f}}}{k'_{\text{b}}}\] Substituting the catalyzed rate constants: \[K'_{eq} = \frac{10 \cdot k_{\text{f}}}{10 \cdot k_{\text{b}}} = \frac{k_{\text{f}}}{k_{\text{b}}} = K_{\text{eq}}\] The factor of \(10\) cancels out completely. This shows that the presence of a catalyst lowers the time needed to reach equilibrium but does not alter the equilibrium position or the value of the equilibrium constant. Furthermore, thermodynamically, the standard Gibbs free energy change (\(\Delta G^\circ\)) of the reaction is related to the equilibrium constant by: \[\Delta G^\circ = -RT \ln K_{\text{eq}}\] Since a catalyst does not alter the energy levels of the reactants or products, \(\Delta G^\circ\) remains unchanged. Since the temperature is kept constant at \(2000\text{ K}\), the equilibrium constant remains: \[K'_{\text{eq}} = \boxed{4.0 \times 10^{-4}}\] Step 4 - Evaluate the Options * **Option (A) \(4 \times 10^{-3}\)**: Incorrect. This assumes the equilibrium constant increases by a factor of 10, which represents a fundamental misunderstanding of the dynamic nature of catalysis. * **Option (B) \(4 \times 10^{-5}\)**: Incorrect. This assumes the equilibrium constant decreases by a factor of 10. * **Option (C) \(4 \times 10^{-4}\)**: Correct. The equilibrium constant is unaffected by a catalyst and remains constant at a constant temperature. * **Option (D) Unpredictable**: Incorrect. The physical behavior is fully governed by thermodynamic laws and is completely predictable.