Q. When a graph is plotted between $\ln k\ and\ 1/T$ for a first order reaction, a straight line is obtained. The slope of the line is equal to
A) $- \frac{E_{a}}{2.303}$
B) $- \frac{E_{a}}{2.303R}$
C) $- \frac{2.303}{E_{a}R}$
D) $- \frac{E_{a}}{R}$
Explanation
(d)
Arrhenius equation is:
$k = Ae^{- E_{a}/RT}$
$\ln k = \ln A - \frac{E_{a}}{RT}$
$\ln k = \ln A - \frac{E_{a}}{RT}$
$Slope = - \frac{E_{a}}{R}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.