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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. `53.6 JK^(-1) "mol"^(-1)`B. `48.6 JK^(-1) "mol"^(-1)`C. `58.5 JK^(-1) "mol"^(-1)`D. `60.5 JK^(-1) "mol"^(-1)` |
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Answer» Correct Answer - A Form Arrhenius equation, `log` `(k_(2))/(k_(1)) = (-E_(a))/(2.303R)((1)/(T_(2)) - (1)/(T_(1)))` Given, `(k_(2))/(k_(1)) = 2, T_(2) = 310 K` `T_(1) = 300 K` On putting values, `rArr log 2 = (-E_(a))/(2.303 xx 8.314) ((1)/(310) - (1)/(310))` `rArr E_(a) = 53603.93 J//mol = 53.6 kJ//mol` |
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