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If K (1) and K(2) are the rate constants...

If `K _(1) and K_(2)` are the rate constants at temperatures `T _(1) and T_(2)` respectively and `E _(a)` is the activatino energy, then :

A

`log "" (k _(1))/( k _(2)) =- (E _(a))/( 2.303R [(1)/(T _(1))- (1)/(T _(2))]`

B

`log "" (k _(2))/(k _(1)) = (E_(a))/(2.303R) [ (1)/(T _(2)) - (1)/(T _(2))]`

C

`log "" (k _(1))/(k _(2)) = (E _(a))/( 2.303) [ (1)/( T _(1) - (1)/(T _(2))]`

D

`log "" (k _(1))/( k _(2)) =- (E _(a))/( 2.303) [ (1)/(T _(2)) - (1)/(T _(1))]`

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The correct Answer is:
A
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