The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `48.6 kJ mol^(-1)`B. (b) `58.5 kJ mol^(-1)`C. `60.5 kJ mol^(-1)`D. `53.6 kJ mol^(-1)`

The rate of a reaction doubles when its temperature changes form `300

1.	The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `48.6 kJ mol^(-1)`B. (b) `58.5 kJ mol^(-1)`C. `60.5 kJ mol^(-1)`D. `53.6 kJ mol^(-1)`
Answer» Correct Answer - D As per Arrhenius equation: `log. (K_(2))/(K_(1)) = (E_(a))/(2.3R)[(T_(2)-T_(1))/(T_(2)T_(1))]` `2.3 log2 = (E_(a))/(8.314)[(10)/(300 xx 310)]` `:. E_(a) = 53.6 kJ mol^(-1)`Correct Answer - D As per Arrhenius equation: `log. (K_(2))/(K_(1)) = (E_(a))/(2.3R)[(T_(2)-T_(1))/(T_(2)T_(1))]` `2.3 log2 = (E_(a))/(8.314)[(10)/(300 xx 310)]` `:. E_(a) = 53.6 kJ mol^(-1)`

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The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `48.6 kJ mol^(-1)`B. (b) `58.5 kJ mol^(-1)`C. `60.5 kJ mol^(-1)`D. `53.6 kJ mol^(-1)`

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