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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. `58.5 kJ "mol"^(-1)`B. `60.5 kJ"mol"^(-1)`C. `53.6 kJ"mol"^(-1)`D. `48.6 kJ"mol"^(-1)` |
Answer» Correct Answer - C `log_(10)((k_(2))/(k_(1)))=(E_(a))/(2.303R)xx((1)/(T_(1))-(1)/(T_(2)))` `log2=(E_(a))/(2.303xx8.314)xx((1)/(300)-(1)/(310))` `E_(a)=(0.301xx2.303xx8.314xx300xx310)/(10)` `=53598 J "mol"^(-1)` `~~53.6kJ "mol"^(-1)` |
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