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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

`log. (K_2)/(k_1)= E_a/(2.303R)[(T_2-T_1)/(T_1T_2)]`
`:.log. (k_2)/k_1= (65xx 10^3)/(2.303 xx 8314)[(298-273)/(298xx 273)]`
`:.k_2/k_1= 11.05`
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