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The rate of reaction triples when temper...

The rate of reaction triples when temperature changes form `20^(@)C` to `50^(@)C`. Calculate the energy of activation for the reaction `(R= 8.314JK^(-1)mol^(-1))`.

Text Solution

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The Arrhenius equation is,
`log.(k_(2))/(k_(1))=(E_(a))/(Rxx2.303)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Given : `(k_(2))/(k_(1))= 3, R = 8.314K^(-1)mol^(-1)`,
`T_(1)= 20+273= 293K, T_(2)= 50+ 273= 323K`
Substituting the given values in the Arrhenius eqaution.
`log3(E_(a))/(8.314xx2.303)[(323-293)/(323xx293)]`
`E_(a)=(2.303xx8.314xx323xx293xx0.477)/(30)`
`= 28811.8Jmol^(-1)`
`= 28.8118KJ mol^(-1)`
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