Question

The temperature dependence of the rate constant $k$ is expressed as $k=A{e}^{-E_a/RT}$.
When a plot between $\log k$ and $1/T$ is plotted we get the graph as shown. What is the value of the slope in the graph?

• A. $\frac{E_a}{RT}$
• B. $-\frac{E_a}{2.303R}$
• C. $-\frac{E_a}{2.303RT}\log A$
• D. $-\frac{E_a}{2.303}\frac{R}{T}$

Answer 1

The correct option is B $-\frac{E_a}{2.303R}$

K=$A{e}^{{-E}_a/RT}\ln K=ln\left(A{e}^{{-E}_a/RT}\right)\ln K=\ln A+(-\frac{E_a}{RT})\ln e\ln K=\ln A-\frac{E_a}{RT}$ (1)

now changing $ln$ to $log$

$\log K=\log A+(-\frac{E_a}{2.303R})\frac{1}{T}$ (2)

this represent straight time as: $y=c+\left(m\right)x$

$\therefore$ slope=$-\frac{E_a}{2.303R}$