The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = `8.314 JK^(-1) mol^(-1)` and log2 = 0.301)A. `53.6 kJ mol^(-1)`B. 48.6 kJ `mol^(-1)`C. 58.5 kJ `mol^(-1)`D. 60.5 kJ `mol^(-1)`

The rate of a reaction doubles when its temperature changes from 300 K

1.	The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = `8.314 JK^(-1) mol^(-1)` and log2 = 0.301)A. `53.6 kJ mol^(-1)`B. 48.6 kJ `mol^(-1)`C. 58.5 kJ `mol^(-1)`D. 60.5 kJ `mol^(-1)`
Answer» Correct Answer - a log `(K_(2))/(K_(1)) = (-E_(a))/(2.030R) ((1)/(T_(2)) - (1)/(T_(1)))` `(K_(2))/(K_(1)) = 2 , T_(2) = 310 K " " T_(1) = 300` K `implies ` log `2 = (-E_(a))/(2.303 xx 8.314 ) ((1)/(310) - (1)/(300))` `implies E_(a) = 53598.6` J/mol = 53.6 KJ/mol

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The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = `8.314 JK^(-1) mol^(-1)` and log2 = 0.301)A. `53.6 kJ mol^(-1)`B. 48.6 kJ `mol^(-1)`C. 58.5 kJ `mol^(-1)`D. 60.5 kJ `mol^(-1)`

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