The rate constant of a reaction is $1.5' 10^{- 7} S^{- 1} at 50 ° C and 4.5' 10^{- 7} S^{- 1} at 100 ° C$ .The value of energy of activation is

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$2 . 2^{'} 10^{- 4} J {mol}^{- 1}$

$1.1' 10^{3} J {mol}^{- 1}$

$10^{3} J {mol}^{- 1}$

$3.3' 10^{4} J {mol}^{- 1}$

answer is A.

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Detailed Solution

$\log \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303} [\frac{T_{2} - T_{1}}{T_{1} T_{2}}]$

$Log \frac{4.5 \times 10^{- 7}}{1.5 \times 10^{- 7}} = \frac{E_{a}}{2.303 \times 8.314} [\frac{373 - 323}{323 \times 373}]$

$\Rightarrow \log 3 = \frac{E_{a}}{2.303 \times 8.314} [\frac{50}{323 \times 373}]$

$\Rightarrow 0.4771 = \frac{E_{a}}{2.303 \times 8.314} \times \frac{50}{323 \times 373}$

and then, $E_{a} = \frac{0.4771 \times 2.303 \times 8.314 \times 323 \times 373}{50}$ $∴ E_{a} = 2.2 \times 10^{- 4} J {mol}^{- 1}$

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