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The rate of a reaction quadruples when t...

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature

Text Solution

Verified by Experts

Given, `(k_(2))/(k_(1))=4`
`T_(2)=313, T_(1)=293`
Now, substituting the values in equation:
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R[((T_(2)-T_(1)))/(T_(1)T_(2))])`
`log4=(E_(a))/(2.303xx8.314JK^(-1)mol^(-1)[((313-293))/(313xx293)])`
`0.602 =(E_(a))/(19.1 JK^(-1)mol^(-1)[((20))/(91709)])`
`0.602=(E_(a))/(19.1JK^(-1)mol^(-1)(0.000218))`
`E_(a)=((0.602xx19.1))/(0.000218)`
`=52744.03J or 52.744`kJ/mol
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