2003 AIPMT Chemical KineticsArrhenius equation Easy Question 7 of 7 in Arrhenius equation

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation, $k = A\,e^{-E^*/RT}$. Activation energy ($E^*$) of the reaction can be calculated by plotting

A k vs T
B k vs $\dfrac{1}{\log T}$
C log k vs $\dfrac{1}{T}$
D log k vs $\dfrac{1}{\log T}$

The answer is C.

Solution

Taking logarithms of the Arrhenius equation: $\ln k = \ln A - \dfrac{E^*}{RT}$
or $\log k = \log A - \dfrac{E^*}{2.303R}\cdot\dfrac{1}{T}$
This is a straight line (y = c + mx) when log k is plotted against $\dfrac{1}{T}$.
Slope $= -\dfrac{E^*}{2.303R}$, so $E^* = -2.303R \times$ slope.
So the activation energy is calculated by plotting log k vs $\dfrac{1}{T}$.

Concept Arrhenius plot

Arrhenius equation in past papers

7 questions from this chapter have appeared across 7 exam years.