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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. `48.6kJ`B. `58.5 kJ mol^(-1)`C. `60.5 kJ`D. `53.6 kJ mol^(-1)` |
Answer» Correct Answer - D According to Arrhenius equation log `(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` Substituting the data, we have log `2=(E_(a))/(2.303(8.314JK^(-1)mol^(-1)))((310K-300K)/(300Kxx310K))` `E_(a)=53.6KJ//mole` |
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