The rate constant for the first order decomposition of `N_(2) O_(5)` is given by the following equation: `k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T)`
Calculate `E_a` for this reaction and rate constant if its half-life period be 300 minutes.

Explanation

`k=Ae^(-E_(a) //RT)" "…(i)`
Also `k=(2.5 xx 10^(14) s^(-1) ) e^((-25000K)//T)" "…(ii)`
From (i) and (ii), we have
`Ae^(-E_(a)//RT)=(2.5 xx 10^(-14) s^(-1) ) e^((-25000K)//T)`
Equating similar terms in the above equation, we have
`(-E_(a))/(RT) = (-25000K)/( T)`
or `E_(a) = R xx 25000 K`
or `E_(a) =8.314 J K^(-1) mol^(-1) xx 25000 K`
or `E_(a) = 207.85 kJ "mol"^(-1)`
For a first order reaction
`t_(1//2) = (0.693)/(k)` or `k=(0.693)/( t_(1//2))= (0.693)/(300 "minutes") = 2.31 xx 10^(-3) "minutes"^(-1)`
or `k=(0.693)/( 300 xx 60s) = (0.693)/(18000 s) = 3.85 xx 10^(-5) s^(-1)`.