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The activation energy of a reaction is 5...

The activation energy of a reaction is 56.2 kJ/mol. The ratio of the rate constant at 300 K and 305 K is `(R = 8 J "mole"^(-1)K^(-1)`) about :

A

`1.25 `

B

`1.5 `

C

`1.10 `

D

`1.60 `

Text Solution

Verified by Experts

The correct Answer is:
B
Hint : `log k_2//k_1 = E_a//2.303 R [1//T_1 - 1//T_2]`
By putting values in this equation, we will get the answer after Solving `k_2//k_1 = 1.47 = 1.5`)
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