By plotting the graph between `log k vs 1/T` for first order reaction gives straight line having slope - 4670 K. The activation energy for this reaction is :

Options

A. `89.417J mol `
B. `89417.1 J mol^(-1)`
C. `89.417 KJ mol^(-1)`
D. `89417.1 KJ mol^(-1)`

Answer

C

Explanation

Hint : Slope `= - E_a//2.303 R `
`E_a =-(-4670 k) 2.303 xx 8.314 J mol^(-1)K^(-1)`
`= 89417.1 J mol^(-1) = 89.417 KJ mol^(-1)`