Question

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

What is the value of slope in the graph?

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

Answer 1

Answer: (B)

$K=A{e}^{-E_a/RT}$
$\ln K=\ln \left(A{e}^{-E_a/RT}\right)$
$\ln K=\ln A+\left(-\frac{E_a}{RT}\right)\ln e$
$\ln K=\ln A-\frac{E_a}{RT}$ (1)
now changing $\ln$ to $\log$
$\log K=\log A+\left(-\frac{E_a}{2.303R}\right)\frac{1}{T}$ (2)
this represent straight time as: $y=c+(m)x$
$\therefore$ slope $=-\frac{E_a}{2.303R}$