The slope of the line in the graph of logk (k = rate constant) versus `1/T` for a reaction is - 5841 K. Calculate the energy of activation for this reaction. [R` = 8.314 JK^(-1) mol^(-)`]

Explanation

We know that slope `= -(E_a)/(2.303R)`
But slope =-5841 K and R= 8.314 `JK^(-) mol^(-)`
Substituting the values, we get :
`-5841 K =- (E_a)/(2.303 xx 8.314 JK^(-) mol^(-))`
` therefore E_a =2.303 xx 8.314 JK^(-1) mol^(-) xx 5841 K`

` ~= 111838 J mol^(-) = 111.838 KJ mol^(-)`