In the Arrhenius plot of ln k versus 1/T, what does the slope of the straight line equal?

1. Taking ln of the Arrhenius equation: $\ln k = -\dfrac{E_a}{R}\cdot\dfrac{1}{T} + \ln A$. 2. Comparing with $y = mx + c$, slope $= -\dfrac{E_a}{R}$. 3. The intercept is $\ln A$, so option B is the intercept, not the slope. 4. Option C is the slope of the log (base 10) plot, not the ln plot. _Source: syllabus Chemistry Ch 3 "Chemical Kinetics", p.19_