The activation energy of a reaction can be determined from the slope of which of the following graphs?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → ln k vs. 1/T
The temperature dependence of the rate constant is given by the Arrhenius equation:
\( \displaystyle k=Ae^{-E_a/RT} \)
In its logarithmic form the equation is a straight line, y = mx + c, with y = ln k, x = 1/T, slope m = −Ea/R and intercept c = ln A.
So a plot of ln k against 1/T is a straight line with slope −Ea/R, and
\( \displaystyle E_a=-R\times\text{slope} \)
Why the other options are incorrect
Option (1) ln k vs. T — Incorrect. Ln k depends on 1/T, not on T, so this plot is a curve and has no constant slope.
Option (2) ln k/T vs. T — Incorrect. This combination does not arise from the Arrhenius equation and is not linear.
Option (4) ln k vs. T2 — Incorrect. Again a curve; no term in T2 appears in the equation.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – …
