Question

For a first order reaction A ⟶ P, the temperature (T) dependent rate constant (k) was found to follow the equation log k = -2000(1/T) + 6.0. The pre-exponential factor A and the activation energy $E_a,$ respectively, are

(IIT-JEE, 2009)

• A. $1.0\times {10}^6{s}^{-1}and9.2kJmo{l}^{-1}$
• B. $6.0s^{-1}and16.6kJmo{l}^{-1}$
• C. $1.0\times {10}^6{s}^{-1}and16.6kJmo{l}^{-1}$
• D. $1.0\times {10}^6{s}^{-1}and38.3kJmo{l}^{-1}$

Answer 1

The correct option is D

$1.0\times {10}^6{s}^{-1}and38.3kJmo{l}^{-1}$

$k=A{e}^{-E_a/RT}$

or log k = log A - $\frac{E_a}{2.303RT}$

log k = -2000 $\left(\frac{1}{T}\right)$ + 6.0

On comparison, we get

log A = 6

or A = ${10}^6{s}^{-1}$

∴ $\frac{E_a}{2.303R}=2000$

or $E^a$ = 2.303 x 8.314 x 2000

$=38.3kJmo{l}^{-1}$