Question

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$, where $k$ is the rate constant of a reaction at temperature $T$ ?

• A. $\frac{-E_{a}}{2.303 R}$
• B. $\frac{-E_{a}}{R}$
• C. $\frac{E_{a}}{R}$
• D. $\frac{R}{E_{a}}$

Answer 1

Answer: (B)

Arrhenius equation is
$k=A e^{-E_{a} / R T}$
On taking log both sides,
$\ln k=\ln A-\frac{E_{a}}{R T}$
or $\ln k=-\frac{E_{a}}{R T}+\ln A$
Since, the reaction, $y=m x+ c$
when a graph is plotted between $\ln k$ and $\frac{1}{T}$ a straight line is obtained.
The slope of the straight line is equal to $-\frac{E_{a}}{R}$.