If K1 and K2 are the equilibrium constants for a reversible reaction at T1 K and T2 K temperature, r — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the thermodynamic parameters of the reaction The problem describes a reversible reaction that occurs with "neither heat evolution nor absorption." In chemical thermodynamics, the exchange of heat at constant pressure is represented by the standard enthalpy change of the reaction (\(\Delta H^\circ\)). * A reaction with heat evolution (exothermic reaction) has a negative enthalpy change (\(\Delta H^\circ < 0\)). * A reaction with heat absorption (endothermic reaction) has a positive enthalpy change (\(\Delta H^\circ > 0\)). * A reaction with neither heat evolution nor absorption (thermoneutral reaction) has an enthalpy change of exactly zero: \[\Delta H^\circ = 0 \text{ J mol}^{-1}\] Step 2 - Apply the Van 't Hoff Equation The mathematical relationship showing how the equilibrium constant (\(K\)) of a reversible reaction varies with absolute temperature (\(T\)) is given by the integrated form of the Van 't Hoff equation: \[\log_{10}\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{2.303 R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] where: * \(K_1\) is the equilibrium constant at temperature \(T_1\text{ K}\). * \(K_2\) is the equilibrium constant at temperature \(T_2\text{ K}\). * \(\Delta H^\circ\) is the standard enthalpy of reaction. * \(R\) is the universal gas constant (\(8.314\text{ J K}^{-1}\text{ mol}^{-1}\)). Step 3 - Determine the relation between \(K_1\) and \(K_2\) Substituting the value \(\Delta H^\circ = 0\text{ J mol}^{-1}\) into the Van 't Hoff equation: \[\log_{10}\left(\frac{K_2}{K_1}\right) = \frac{0\text{ J mol}^{-1}}{2.303 \times 8.314\text{ J K}^{-1}\text{ mol}^{-1}} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)\] \[\log_{10}\left(\frac{K_2}{K_1}\right) = 0\] Taking the antilog (base 10) of both sides of the equation: \[\frac{K_2}{K_1} = 10^0 = 1\] \[K_2 = K_1\] Since this equality is derived without any restriction on the actual values of temperatures \(T_1\) and \(T_2\) (where \(T_1 < T_2\)), the equilibrium constant is entirely independent of temperature. Thus, \(K_1 = K_2\) at any temperature. Step 4 - Evaluate each option systematically * **Option (A) "\(K_1 > K_2\) at high temperature"**: Incorrect. This condition (where the equilibrium constant decreases with increasing temperature, as \(T_2 > T_1\)) holds only for exothermic reactions (\(\Delta H^\circ < 0\)). * **Option (B) "\(K_1 < K_2\) at high temperature"**: Incorrect. This condition (where the equilibrium constant increases with increasing temperature, as \(T_2 > T_1\)) holds only for endothermic reactions (\(\Delta H^\circ > 0\)). * **Option (C) "\(K_1 = K_2\) only at high temperature"**: Incorrect. The relationship \(K_1 = K_2\) is not restricted to high temperatures; it holds true across the entire temperature spectrum because the temperature coefficient of the equilibrium constant is zero. * **Option (D) "\(K_1 = K_2\) at any temperature"**: Correct. Because \(\Delta H^\circ = 0\), the equilibrium constant remains constant and invariant to temperature changes. Therefore, the correct option is D. \[\boxed{\text{D}}\]