The decomposition of hydrocarbon follows the equation `k=(4.5xx10^(11)s^(-1))e^(-28000K//T)`
Calculate `E_(a)`.

Options

A. 669 kJ `mol^(-1)`
B. 232. 79 kJ `mol^(-1)`
C. `4.5 xx 10^(11) kJ mol^(-1)`
D. 28000 kJ `mol^(-1)`

Answer

B

Explanation

Arrhenius equation, `k = Ae^(-e_(a)//RT)`
Given equation is `k (4.5 xx 10^(11) s^(-1)) e^(-28000 K//T)`
Comparing both the equation, we get
`-(E_(a))/(RT) = (28000 K)/(T)`
`E_(a) = 28000 K xx R = 28000 K xx 8.314 K K^(-1) mol^(-1)`
`232.79 kJ mol^(-1)`