The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value…
The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value $(R=$ gas constant, $E_\alpha=$ Activation energy)
$\frac{R}{E_a}$
$\frac{E_a}{R}$
$\frac{-E_a}{R}$
$\frac{-R}{E_a}$
Solution
Arrhenius equation,
$k=A e^{-E_a / \mathrm{RT}}$
On taking log on both sides, we get
$\ln k=\ln \mathrm{A}-\frac{E_a}{R T}$
A graph between in $k$ and $1 / \mathrm{T}$ is a straight line with $-\frac{E_a}{R}$ slope.