Question

Consider a complex reaction taking place in three steps with rate constants \(k_1\), \(k_2\), and \(k_3\) respectively. The overall rate constant \(k\) is given by the expression \( k = \sqrt{\frac{k_1 k_3}{k_2}} \). If the activation energies of the three steps are 60, 30, and 10 kJ mol\(^{-1}\) respectively, then the overall energy of activation in kJ mol\(^{-1}\) is \(\_\_\_\_\)(Nearest integer).

Hint:

To find the overall activation energy in multi-step reactions, apply the Arrhenius equation to each step, then combine the activation energies weighted by their rate constants.

Answer 1

The expression for the overall rate constant is given as: \[ k = \sqrt{\frac{k_1 k_3}{k_2}}. \] Now, from the Arrhenius equation: \[ k = A e^{-\frac{E_a}{RT}}, \] where \(E_a\) is the activation energy, \(A\) is the pre-exponential factor, \(R\) is the gas constant, and \(T\) is the temperature. Taking the natural logarithm of the rate constants for each step: \[ \ln(k_1) = \ln(A_1) - \frac{E_1}{RT}, \quad \ln(k_2) = \ln(A_2) - \frac{E_2}{RT}, \quad \ln(k_3) = \ln(A_3) - \frac{E_3}{RT}. \] We need to determine the overall activation energy \(E_a\), which can be computed by adding the activation energies of the individual steps weighted by their respective rate constants. From the given activation energies \(E_1 = 60\), \(E_2 = 30\), and \(E_3 = 10\) kJ mol\(^{-1}\), the overall activation energy for the reaction is found to be approximately: \[ E_a = \frac{E_1 + E_3}{2} = \frac{60 + 10}{2} = 35 \, \text{kJ mol}^{-1}. \] However, by adjusting for the relative weightings of the rate constants, we find the overall activation energy to be closer to: \[ E_a \approx 25 \, \text{kJ mol}^{-1}. \]