Question

The temperature dependence of the rate constant $k$ is expressed as $k=Ae^{-E_{a} /RT}\cdot$ 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. $-\frac{E_{a}}{2.303 R}$
• C. $-\frac{E_{a}}{2.303 RT} log\,A\quad$
• D. $-\frac{E_{a}}{2.303 T} \frac{R}{T}$

Answer 1

Answer: (B)

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