Question

According to Arrhenius equation rate constant $k$ is equal to $Ae^{-E_{a} /RT}\cdot$ Which of the following options represents the graph of $\ln \,k \,vs$ $\frac{1}{T}$?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

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

Taking $\log$ on both side $\ln \,k=\ln\, \left(A . e \frac{-E_{a}}{R T}\right)$

$\ln\, k=\ln\, A -\frac{E_{a}}{R T}$

$\ln \,k=-\frac{E_{a}}{R} \times \frac{1}{T}+\ln A$

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 $=\ln \,A .$