Question

The activation energy (E$_a$) of a reaction can be determined from the slope of which of the following plots?

• A. ln $k$ vs. T
• B. $\dfrac{\ln k}{T}$ vs. T
• C. ln $k$ vs. $\dfrac{1}{T}$
• D. $\dfrac{T}{\ln k}$ vs. $\dfrac{1}{T}$

Hint:

Use the Arrhenius plot (ln $k$ vs. $\dfrac{1}{T}$) to find activation energy from the slope.

Answer 1

The Correct Option is C

The Arrhenius equation is given by:
k = A e$^{-E_a/RT}$
Taking natural log on both sides:
ln $k$ = ln $A$ − $\dfrac{E_a}{R} \cdot \dfrac{1}{T}$
This is in the form of a straight line $y = mx + c$, where the slope is $-E_a/R$.
Hence, a plot of ln $k$ vs. $\dfrac{1}{T}$ gives a straight line with slope equal to $-E_a/R$.