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

`373.2 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`
`k=(0.693 )/( t_(1//2)) = ( 0.693 )/(60 ) = 0.01155 s^(-)`
but log ` k= log A - (E_a)/( 2.303 RT)`
` therefore log 0.01155 = log 5 xx 10 ^(12) - ( 25000 xx 4.18 46 )/( 2.303 xx 8.314 xx T)`
` or - 1.9374 = 0.6989 + 12 - 5462 .9//T`
` or - 1-9374 -0.6989 - 12 =- (5462.9 )/( T)`
` or - 14.6363 =- (5462.9 )/(T)`
or ` T= (6462.9 )/( 14.6363 ) = 373 .2 K`