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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. `53.6kJ" mol"^(-1)`B. `48.6kJ" mol"^(-1)`C. `58.5kJ" mol"^(-1)`D. `60.5kJ " mol"^(-1)` |
Answer» Correct Answer - A From 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)=310k,T_(1)=300k` On substituting the values, `"log"2=(-E_(a))/(2.303xx8.314)((1)/(310)-(1)/(300))` `orE_(a)=(log2xx2.303xx8.314xx310xx300)/(10)` `E_(a)=53598.6J//"mol"=53.6kJ//"mol"` |
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