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How much faster would a reaction proceed...

How much faster would a reaction proceed at `25^(@)C` than at `0^(@)C` if the activation energy is `65 kJ`?

A

`2 "times"`

B

`16 "times"`

C

`11 "times"`

D

`6 "times"`

Text Solution

Verified by Experts

The correct Answer is:
c

`2.303"log"K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]`
`:. 2.303"log"K_(2)/K_(1)=(65xx10^(3))/(8.314)[(25)/(298xx273)]`
`:. K_(2)/K_(1)=11.05`
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