Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

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

Where,
$k=\text{Rate constant}$
$A=\text{Pre-Exponential Factor}$
$E_{\alpha }=\text{Activation Energy}$
$R=\text{Gas constant}$
$T=\text{Temperature}$

Taking log on both sides,

$\ln k=\ln \left(A{e}^{-\frac{Ea}{RT}}\right)$

$\ln k=\ln A+\ln e^{-\frac{Ea}{RT}}$

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

On comparing above equation with

$y=mx+c$

We get,

$m\left(\text{Slope}\right)=-\frac{Ea}{R}$

$c\left(\text{Intercept}\right)=\ln A$

Thus the graph $\ln kvs\frac{1}{T}$ will be linearly decreasing with the negative slope.


Hence Option (a) is correct.