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The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

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From the Arrhenius equation,

\log \frac{k_2}{k_1}=\frac{E_a}{2.303 R}\left(\frac{T_2-T_1}{T_1 T_2}\right)...................................(i)

It is given that k_2=4 k_1
T1= 293 K

T2 = 313 K
Putting all these values in equation (i) we get,

\log 4=\frac{E_a}{2.303 \times 8.314}\left(\frac{313-293}{313 \times 293}\right)

\begin{aligned} & \Rightarrow 0.6021=\frac{20 \times E_a}{2.303 \times 8.314 \times 293 \times 313} \\ & \Rightarrow E_a=\frac{0.6021 \times 2.303 \times 8.314 \times 293 \times 313}{20}\end{aligned}

Activation Energy = 52.86 KJ/mol  
This is the required activation energy

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manish

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