Q. For a first order reaction$\ A \rightarrow P$, the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation. $\log{k = - (2000)/T + 6.0}$ The pre-exponential factor $A$ and the activation energy $E_{a}$, respectively, are
A) $1.0 \times 10^{6}s^{- 1}and\ 9.2\ kJ\ mol^{- 1}$
B) $6.0\ s^{- 1}and\ 16.6\ kJ\ mol^{- 1}$
C) $1.0 \times 10^{6}s^{- 1}and\ 16.6\ kJ\ mol^{- 1}$
D) $1.0 \times 10^{6}s^{- 1}and\ 38.3\ kJ\ mol^{- 1}$
Explanation
(d)
Comparing the slope and intercept of the given equation with the following Arrhenius equation :
$\log{k = - \frac{E_{a}}{2.303RT} + \log A}$
Hence, $\log{A = 6\ i.e.,A = 10^{6}s^{- 1}}$
Comparing slope gives $E_{a} =$38.3 kJ/mol
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.