The figure below represents the graph between ln k vs 1/T. The slope of the straight line will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle -\frac{E_a}{R} \)
The effect of temperature on the rate constant is given by the Arrhenius equation, where A is the frequency (pre-exponential) factor and Ea the activation energy:
\( \displaystyle k=A\,e^{-E_a/RT} \)
Taking the natural logarithm of both sides turns it into the equation of a straight line (y = mx + c):
\( \displaystyle \ln k=\ln A-\frac{E_a}{R}\left(\frac{1}{T}\right) \)
With y = ln k and x = 1/T, the slope is −Ea/R and the intercept is ln A. The slope is negative, which is why the line in the graph falls as 1/T increases.
Why the other options are incorrect
Option (2) \( \frac{E_a}{R} \) — Incorrect. The sign is missing. A positive slope would mean k decreases as temperature rises.
Option (3) \( \frac{E_a}{2.303R} \) — Incorrect. The sign is missing, and the factor 2.303 belongs to a plot of log k, not ln k.
Option (4) \( -\frac{E_a}{2.303R} \) — Incorrect. −Ea/2.303R is the slope of log k vs 1/T, obtained by dividing the ln form by 2.303. The graph here uses ln k.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – …
