Rate constant k of a reaction varies with temperature according to equation:
`log k=` constant ` - (E_a)/( 2.303 R).(1)/(T )`
What is the activation energy for the reaction. When a graph is plotted for log k versus `1/T` a straight line with a slope-6670 K is obtained. Calculate energy of activation for this reaction (R=8.314 `JK^(-1) mol^(-)` 1)

Answer

`127.711 KJ mol^(-1)`

Explanation

slope `= (-E_a)/(2.303 R) or - 6670 K = (-E_a)/(2.303 xx 8.314 JK^(-1) mol^(-1))`
or ` E_a = 6670 xx 2.303 xx 8.314 JK^(-1) mol^(-1)`
`= 127711 J mol^(-1) = 127 .711 KJ mol^(-1)`