Question

The plot of $logk$ vs $\frac{1}{T}$ yields a straight line. The slope of the line would be equal to:

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

Answer 1

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

The Arrhenius equation is $k=A{e}^{-E_a/RT}$ or $logk=logA-\frac{E_a}{2.303RT}.$
Thus, when $logk$ is plotted against $\frac{1}{T}$, the slope is $tan\theta =-\frac{E_a}{2.303R}$.