Question

According to Arrhenius equation rate constant $k$ is equal to $A{e}^{-E_a/RT}\cdot$ Which of the following options represents the graph of $\ln kvs$ $\frac{1}{T}$?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

According to Arrhenius equation, $k=A{e}^{-Ea/RT}$

Taking $\log$ on both side $\ln k=\ln \left(A.e\frac{-E_a}{RT}\right)$

$\ln k=\ln A-\frac{E_a}{RT}$

$\ln k=-\frac{E_a}{R}\times \frac{1}{T}+\ln A$

This equation can be related to equation of straight line as shown above.

From the graph, it is very clear that slope of the plot

$=-\frac{E_a}{R}$ and intercept $=\ln A.$