Question

The rate constant of a certain reaction is given by log K = A – $\frac{B}{T}$ + C log T. Then activation energy of reaction at 300 k is -

• A. $\frac{B}{T}$ + C log T
• B. [2.303 B + CT]R
• C. C log T
• D. B + CTR

Answer 1

Answer: (B)

$2.303 B + CT$R

Answer 2

log k = A – $\frac{B}{T}$ + C log T

OR ln k = 2.303 A – $\frac{B \times 2.303}{T}$ + C ln T

$\frac{d\mathcal{l}nk}{dT} = \left\lbrack \frac{2.303B}{T^{2}} + \frac{C}{T} \right\rbrack$

E_a = [2.303 B + CT] R