The slope of the line in the graph of log k (k = rate constant) versus `1/T `is - 5841. Calculate the activation energy of the reaction.

Explanation

we have that slope `=- (E_a)/(2.303 R)`
but slope `=- 5841 K and R= 8.314 JK^(-) mol^(-)`
substiting the values , we get :
` -5841 K =- ( E_a)/( 2.303 xx 8.314 JK^(-) mol^(-))`
` therefore E_a = 2.303 xx 8.314 JK^(-) mol ^(-) xx 5841 K`
` ~= 111838 J mol^(-) = 111.838 KJ mol^(-)`