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The slope of the line graph of log k ver...

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)`).

A

`95.7`

B

`9.57`

C

`957`

D

None

Text Solution

Verified by Experts

The correct Answer is:
A
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)`
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