At constant temperature, the equilibrium constant (Kp) for the decomposition reaction: ⇌ 2 is expres — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the Equilibrium Constant (\(K_p\)) The homogeneous gaseous decomposition of dinitrogen tetroxide (\(\ce{N2O4}\)) into nitrogen dioxide (\(\ce{NO2}\)) is represented by the following chemical equation: \[\ce{N2O4(g) <=> 2NO2(g)}\] Let us start with \(1\text{ mol}\) of pure \(\ce{N2O4(g)}\) initially. At equilibrium, let \(x\) be the extent of decomposition (or degree of dissociation): \begin{align*} \text{Moles of } \ce{N2O4} \text{ at equilibrium} &= 1 - x \\ \text{Moles of } \ce{NO2} \text{ at equilibrium} &= 2x \\ \text{Total moles of gas at equilibrium } (n_{\text{total}}) &= (1 - x) + 2x = 1 + x \end{align*} Using Dalton's law of partial pressures, the partial pressures of the gaseous components at a total equilibrium pressure \(P\) are: \[P_{\ce{N2O4}} = \left(\frac{1 - x}{1 + x}\right) P\] \[P_{\ce{NO2}} = \left(\frac{2x}{1 + x}\right) P\] The equilibrium constant in terms of partial pressures (\(K_p\)) is: \[K_p = \frac{\left(P_{\ce{NO2}}\right)^2}{P_{\ce{N2O4}}}\] Substituting the individual partial pressures into the expression: \[K_p = \frac{\left(\frac{2x}{1 + x} P\right)^2}{\left(\frac{1 - x}{1 + x}\right) P} = \frac{\frac{4x^2}{(1 + x)^2} P^2}{\left(\frac{1 - x}{1 + x}\right) P} = \frac{4x^2 P}{(1 + x)(1 - x)} = \frac{4x^2 P}{1 - x^2}\] Step 2 - Analyze the Temperature Dependence of \(K_p\) According to chemical thermodynamics, the equilibrium constant (\(K_p\)) is a state function that depends **only on temperature** for a given reaction. Its temperature dependence is described by the van 't Hoff equation: \[\frac{d \ln K_p}{dT} = \frac{\Delta H^\circ}{R T^2}\] Since the problem explicitly specifies that the reaction occurs at **constant temperature**, the value of the equilibrium constant \(K_p\) must remain strictly constant and invariant. Step 3 - Analyze the Relationship Between Pressure and the Extent of Decomposition Because \(K_p\) is a constant at a fixed temperature, any external change in the total pressure \(P\) will lead to a shift in the position of the equilibrium according to Le Chatelier's principle, causing a corresponding change in the extent of decomposition \(x\): * If the total pressure \(P\) is **increased**, the system shifts towards the side with fewer gaseous moles (the backward direction, \(\ce{2NO2 -> N2O4}\)), thereby **decreasing** the value of \(x\). * If the total pressure \(P\) is **decreased**, the system shifts towards the side with more gaseous moles (the forward direction, \(\ce{N2O4 -> 2NO2}\)), thereby **increasing** the value of \(x\). These adjustments in \(x\) occur in a precise manner such that the value of the mathematical ratio \(\frac{4x^2 P}{1 - x^2}\) remains perfectly constant and equal to \(K_p\). Step 4 - Evaluate the Options * **Option (A) "\(K_p\) increases with increase of \(P\)"**: Incorrect. As pressure \(P\) increases, the extent of decomposition \(x\) decreases to keep the ratio \(\frac{4x^2 P}{1 - x^2}\) constant. * **Option (B) "\(K_p\) increases with increase of \(x\)"**: Incorrect. Changing \(x\) by varying external factors like volume or pressure does not change the value of \(K_p\), which is a constant at a given temperature. * **Option (C) "\(K_p\) increases with decrease of \(x\)"**: Incorrect. The thermodynamic equilibrium constant is independent of the extent of decomposition. * **Option (D) "\(K_p\) remains constant with change in \(P\) and \(x\)"**: Correct. At a constant temperature, \(K_p\) is invariant. Any external changes in \(P\) are perfectly counterbalanced by shifts in \(x\) such that the value of \(K_p\) is preserved. \[\boxed{\text{D}}\]