Question

$K$ represents the rate constant of a reaction when log $K$ is plotted against $1/T$ ($T=$ temperature) the graph obtained is a:

• A. curve
• B. a straight line with a constant positive slope
• C. a straight line with constant negative slope
• D. a straight line with no slope

Answer 1

The correct option is C a straight line with constant negative slope

$K\propto e^{-E_a/RT}$

$\therefore logK=\frac{-E_a}{RT}+constant$

$E_a$ is always positive. Therefore graph of $logK$ vs $\frac{1}{T}$ is a straight line and has a constant negative slope.

Hence, option C is correct.