Question:
The activation energy (E$_a$) of a reaction can be determined from the slope of which of the following plots?
The activation energy (E$_a$) of a reaction can be determined from the slope of which of the following plots?
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Use the Arrhenius plot (ln $k$ vs. $\dfrac{1}{T}$) to find activation energy from the slope.
Updated On: Jul 1, 2025
- ln $k$ vs. T
- $\dfrac{\ln k}{T}$ vs. T
- ln $k$ vs. $\dfrac{1}{T}$
- $\dfrac{T}{\ln k}$ vs. $\dfrac{1}{T}$
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The Correct Option is C
Solution and Explanation
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 $-E_a/R$.
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 $-E_a/R$.
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