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The rate of a chemical reaction doubles ...

The rate of a chemical reaction doubles for an increases of 10 K in absolute temperature from 298 K. Calculate` E_e`.

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`log (k)_(2)(k)_(1)=(Ea)(2.303 (R)[(T)_(2)-(T)_(1)/(T)_ (1) (T)_(2)` `(T)_(1)=298 (K) (T)_(2)=298+10 =308 (K) (k)_(2)(k)_(1))=2` `log 2=(Ea)(2.303 * 8.314)(308-298)(298 *308)` `(Ea)=(2.303 * 8.314 * 298 * 308 * log 2)(10)`
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