Question

When a graph is plotted between log K and $\left( \frac{1}{T} \right),$ the slope of line obtained represents: (K = rate constant, T = temperature)

• A. $\frac{{{E}_{a}}}{2.303R}$
• B. $-\frac{{{E}_{a}}}{R}$
• C. $-\frac{{{E}_{a}}}{2.303R}$
• D. $\log A$

Answer 1

Answer: (C)

$-\frac{{{E}_{a}}}{2.303R}$

Answer 2

The Arrhenius equation can be written as $\log \,K=\log A-\frac{E}{2.303RT}$ On comparing this equation with general equation of a straight line, $y=mx+c$ We get, $y=\log \,K,$ $x=\frac{1}{T},$ $m=-\frac{E}{2.303R},$ $c=\log A$ i.e., if we plot a graph between log K (at Y-axis) and $\frac{1}{T}$ (at X-axis), then the slope of the line obtained will be equal to $-\frac{e}{2.303\,R}.$