Question

The slope of the graph drawn between $\ln k$ and $\frac{1}{T}$ as per Arrhenius equation gives the value $(R=$ gas constant, $E_a=$ Activation energy)

• A. $\frac{R}{E_a}$
• B. $\frac{E_a}{R}$
• C. $\frac{-E_a}{R}$
• D. $\frac{-R}{E_a}$

Answer 1

Answer: (C)

We know that, Arrhenius equation,

$=A{e}^{-E_a/RT}$



On taking log on both sides, we get

$\log k=\log A-\frac{E_a}{2.303RT}$

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

$\therefore$ A graph between $\ln k$ and $1/T$ is a straight line (according to the equation of straight line, viz. $y=mx+c)$ with $-\frac{E_a}{R}$ slope (negative) and In A intercept.