According to the Arrhenius rate equation, rate constant \(k\) is equal to \(A.e^{-E_a/RT}\). Which of the following options represents the graph of \(ln k \) vs \(\frac{1}{T}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → a straight line falling from a positive intercept on the ln k axis (negative slope)
Taking the natural logarithm of the Arrhenius equation turns it into the equation of a straight line:
\( k = A\,e^{-E_a/RT} \)
\( \displaystyle \ln k = \ln A - \frac{E_a}{R}\left(\frac{1}{T}\right) \)
This has the form y = c + mx with y = ln k and x = 1/T. So the plot of ln k against 1/T is a straight line with
\( \text{slope} = -\frac{E_a}{R} \) (negative, since Ea is positive)
\( \text{intercept} = \ln A \) (on the ln k axis)
A line that starts at a point on the ln k axis and slopes downwards is the graph in Option (1). The slope gives Ea = −R × slope.
Why the other options are incorrect
Option (2) rising line with a positive intercept — Incorrect. A positive slope would mean −Ea/R > 0, i.e. a negative activation energy, with k falling as T rises.
Option (3) rising line through the origin — Incorrect. The slope has the wrong sign, and a line through the origin would need ln A = 0.
Option (4) rising line cutting the 1/T axis — Incorrect. Again a positive slope, which would mean a negative activation energy.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – …
