For a reaction having three steps, the overall rate constant is \(k = \dfrac{k_1 k_2}{k_3}\). The values of \(E_{a_1}, E_{a_2}\) and \(E_{a_3}\) (activation energies stepwise) are \(40\), \(50\) and \(60\) kJ mol\(^{-1}\) respectively. Then the overall \(E_a\) (activation energy) of the reaction is ____

The overall rate constant is given by \(k = \dfrac{k_1 k_2}{k_3}\).

Taking the natural logarithm on both sides:

\(\ln k = \ln k_1 + \ln k_2 - \ln k_3\)

Differentiating with respect to temperature \(T\):

\(\dfrac{d \ln k}{dT} = \dfrac{d \ln k_1}{dT} + \dfrac{d \ln k_2}{dT} - \dfrac{d \ln k_3}{dT}\)

Using the Arrhenius equation \(\dfrac{d \ln k}{dT} = \dfrac{E_a}{RT^2}\):

\(\dfrac{E_a}{RT^2} = \dfrac{E_{a_1}}{RT^2} + \dfrac{E_{a_2}}{RT^2} - \dfrac{E_{a_3}}{RT^2}\)

\(E_a = E_{a_1} + E_{a_2} - E_{a_3}\)

Substituting the given values \(E_{a_1} = 40\text{ kJ mol}^{-1}\), \(E_{a_2} = 50\text{ kJ mol}^{-1}\), and \(E_{a_3} = 60\text{ kJ mol}^{-1}\):

\(E_a = 40 + 50 - 60 = 30\text{ kJ mol}^{-1}\)

Answer: \(30\text{ kJ mol}^{-1}\)