Consider a complex reaction taking place in three steps with rate constants $$k_1,\; k_2 \text{ and } k_3$$ respectively. The overall rate constant  $$k$$ is given by $$k=\sqrt{\frac{k_1k_3}{k_2}}$$. 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  $$\underline{\hspace{2cm}}.$$ (Nearest integer)

We need to find the overall activation energy given that the overall rate constant $$k = \sqrt{\frac{k_1 k_3}{k_2}}$$ and the activation energies of the three steps are 60, 30, and 10 kJ/mol respectively.

The Arrhenius equation states that each rate constant depends on its activation energy according to $$k_i = A_i \, e^{-E_{a_i}/(RT)}$$, and the overall rate constant also takes the form $$k = A \, e^{-E_a/(RT)}$$, where $$E_a$$ is the overall activation energy.

Since $$k = \sqrt{\frac{k_1 k_3}{k_2}} = \left(\frac{k_1 k_3}{k_2}\right)^{1/2}$$, taking the natural logarithm of both sides gives $$\ln k = \frac{1}{2}\bigl(\ln k_1 + \ln k_3 - \ln k_2\bigr).$$

Substituting $$\ln k_i = \ln A_i - \frac{E_{a_i}}{RT}$$ into this expression yields

$$\ln k = \frac{1}{2}\Bigl[\bigl(\ln A_1 - \frac{E_{a_1}}{RT}\bigr) + \bigl(\ln A_3 - \frac{E_{a_3}}{RT}\bigr) - \bigl(\ln A_2 - \frac{E_{a_2}}{RT}\bigr)\Bigr].$$

Rearranging then leads to $$\ln k = \frac{1}{2}\left(\ln A_1 + \ln A_3 - \ln A_2\right) - \frac{1}{2RT}\left(E_{a_1} + E_{a_3} - E_{a_2}\right).$$

By comparing this with $$\ln k = \ln A - \frac{E_a}{RT}$$, it follows that $$E_a = \frac{1}{2}\left(E_{a_1} + E_{a_3} - E_{a_2}\right).$$

Substituting the given values $$E_{a_1} = 60$$, $$E_{a_2} = 30$$, and $$E_{a_3} = 10$$ kJ/mol into this formula gives $$E_a = \frac{1}{2}(60 + 10 - 30) = \frac{1}{2}(40) = 20 \, \text{kJ mol}^{-1}.$$

The answer is 20 kJ mol$$^{-1}$$.