The slope of the line graph of `log k` versus `1//T` for the reaction `N_(2)O_(5) rarr2NO_(2) + 1//2O_(2)` is `-5000`.Calculate the energy of activation of the reaction (in `kJ K^(-1) mol^(-1)`).

Options

A. `95.7`
B. `9.57`
C. `957`
D. None

Answer

A

Explanation

Arrhenius equation :
`log k = log A - (2.303 E_(a))/(R ) xx (1)/(T)`
Slope `= (-E_(a))/(2.303 xx R) = - 5000`
`:. (-E_(a))/(2.303 xx 8.314 xx 10^(-3) kJ mol^(-1) K^(-1)) = - 5000`
`:. E_(a) = 95.7 kJ K^(-1) mol^(-1)`