The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. `log k = log A-(E_(a))/(2.303R)(1/T)`
Where `E_(a)` is the activation energy. When a graph is plotted for `log k vs 1//T`, a straight line with a slope of `-4250` K is obtained. Calcualte `E_(a)` for this reaction.

Explanation

slope of the graph with log k and along Y axis and `Y-1//T` along X axis`=(-Ea)/(2.303)R`.
Slope = `-4250 K` (Given)
`(-E_(a))/(2.303R) = (-4250K)`
Ea = `(4250 K) xx (2.303 ) xx (8.314 J K^(-1)mol^(-1))`
`=81375.35 J mol^(-1)=81.375 kJ mol^(-1)`