Question

The activation energy of a reaction is 5 kcal/mol.
The increase in the rate constant when its
temperature is raised from 300 to 305 K is
approximately
(A)
14%
(B) 50%
(C)
100%
(D)
25%​

Answer 1

i think b is the right answer for this question

Answer 2

The increase rate is 14%

Given:

Activation energy of a reaction = 5 kcal/mol.

Temperature raise = 300 to 305 K

To Find:

The increase in the rate constant

Solution:

As per Arrhenius equation -

log (k2)/k1 = Ea/2.303R ( T2- T1/T1T2)

= 500/ 2.303 x 2 ( 5/305 x 300)

= 0.0593

Now,

Antilog(0.0593) = 1.146

Hence,

k2/k1 = 1.146 or

k2/k1 - 1 = 1.146 - 1

= 0.146 0r

k2 - k1

= 0.146k1.

Which means k2 increases = 14.6%

Answer: The increase in the rate constant is 14%.

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