The rate of a reaction triples when temperature changes from $20^{o} C$ to $50^{o} C$ . Calculate energy of activation for the reaction $(R = 8.314 {JK}^{- 1} {mol}^{- 1}, l o g_{10} 3 = 0.477) .$
(Give your answer to the nearest integer value)

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Solution

The Arrhenius equation is
$l o g_{10} \frac{k_{2}}{k_{1}} = \frac{E_{a}}{(R \times 2.303)} \times \frac{(T_{2} - T_{1})}{(T_{1} T_{2})}$
Given: $\frac{k_{2}}{k_{1}} = 3 : R = 8.314 {JK}^{- 1} {mol}^{- 1}; T_{1} = 20 + 273 = 293 K$
and $T_{2} = 50 + 273 = 323 K$
Substituting the given values in the Arrhenius equation,
$l o g_{10} 3 = \frac{E_{a}}{(8.314 \times 2.303)} \times \frac{(323 - 293)}{(323 \times 293)}$
$E_{a} = \frac{(2.303 \times 8.314 \times 323 \times 293 \times 0.477)}{30}$
$= 28811.8 {J mol}^{- 1}$
$= 28.81 {k J mol}^{- 1}$
$= 29 {k J mol}^{- 1}$

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