If a first order reaction has activation energy of 25000 cal and a frequency factor of `5 xx 10^(12) sec^(-1)` , at what temperature will the reaction rate have a half-life of
1 minute

Answer

`277.3 K`

Explanation

`(i) E_a = 25000 cal = 25000 xx 4.187 J , t_(1//2) = 1 `minute `= 60 s`
` (ii) t_(1//2)= 80 days = 80 xx 24 xx 60 xx 60 = 6912000 s`
`t_(1//2) = 80 days = 6912000 s `
` k= ( 0.693 )/(6912000 ) = 1.002 xx 10^(-7) s^(-)`
Now log k ` = log A - (E_a)/(2.303 RT)`
` therefore log 1.002 xx 10^(-7) = log 5+ log 10^(12) - ([25000 xx 4.184 ])/(2.303 xx 8.314 T)`
` or 0.000 8677 -7-0.6989 - 12 = (-5462.9 )/( T)`
` or - 19.698 = (-5462.9 )/( T)`
` or T =(+5462.9 )/( 19.698 )= 277.3 K`