The graph of ln(rate) against 1/T for a chemical reaction gives a straight line with a negative slope. This slope is equal to
A$-\frac{E_a}{R}$CORRECT
B$\frac{E_a}{R}$
C$-\frac{R}{E_a}$
D$\frac{R}{E_a}$
GCEChemistryEnergetics and Kinetics2022
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Why the answer is A, and why the others tempt you.
From the Arrhenius equation, $k = Ae^{-E_a/RT}$, taking natural logarithms gives $\ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T}$. A plot of $\ln k$ (or $\ln$ rate) against $\frac{1}{T}$ is a straight line with slope $-\frac{E_a}{R}$. The negative sign confirms the negative slope observed experimentally.
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