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)

Explanation

Slope of graph for log k vs.
`(1)/(T ) =- (E_a)/(2.303 R)`
But slope =-4250k and R = 8.314 `JK^(-1) mol^(-1)`
` therefore E_a = + 4250 K xx 8.314 JK^(-1) mol^(-1) xx 2.303 `
` = 81373 J mol^(-1) = 81.375 KJ mol^(-1)`