A first-order reaction is $50 %$ completed in $40$ minutes at $300 K$ and in $20$ minutes at $320 K$ . Calculate the activation energy of the reaction.

[Given: log $2 = 0.3010$ , log $4 = 0.6021$ , R= $8.314$ $J K^{- 1}$ $m o l^{- 1}$ ]

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Solution

When a first order reaction is $50 %$ completed, the time is equal to half life period.

A first order reaction is $50 %$ completed in $40$ minutes at $300$ K and in $20$ minutes at $320$ K.
$t_{1 / 2} = 40$ minutes
$t_{1 / 2}^{'} = 20$ minutes

The Arrhenius equation and temperature variation is given by the expression.
$l o g (\frac{k^{'}}{k}) = \frac{E_{a}}{2.303 R} [\frac{T^{'} - T}{T T^{'}}]$

Also, $t_{1 / 2} = \frac{0.693}{k}$ or $t_{1 / 2} \propto \frac{1}{k}$

Hence, $l o g (\frac{t_{1 / 2}}{t_{1 / 2}^{'}}) = \frac{E_{a}}{2.303 R} [\frac{T^{'} - T}{T T^{'}}]$

$l o g (\frac{40}{20}) = \frac{E_{a}}{2.303 \times 8.314 J / m o l / K} [\frac{320 K - 300 K}{300 K \times 320 K}]$

$0.3010 = \frac{E_{a}}{19.147 J / m o l / K} [0.0002083 / K]$

$E_{a} = 27664 J / m o l$

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