The temperature dependence of the rate constant (k) of a chemical reaction is written in terms of the Arrhenius equation, k = Ae−Ea/RT. The activation energy (Ea) of the reaction can be calculated by plotting:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → log k vs 1/T
Taking the logarithm of the Arrhenius equation turns it into a straight-line form in 1/T:
\( \displaystyle \ln k=-\frac{E_a}{RT}+\ln A \)
Since ln k = 2.303 log k, the same line in base-10 logarithms is:
\( \displaystyle \log k=-\frac{E_a}{2.303R}\cdot\frac{1}{T}+\log A \)
Comparing with y = mx + c, a plot of log k against 1/T is a straight line with slope −Ea/2.303R, so:
\( \displaystyle E_a=-2.303R\times\text{slope} \)
Why the other options are incorrect
Option (2) log k vs T — Incorrect. log k varies with 1/T, not with T, so this plot is a curve with no constant slope.
Option (3) k vs T — Incorrect. k rises exponentially with temperature, so the plot is a curve, and Ea cannot be read from a single slope.
Option (4) k vs 1/T — Incorrect. Without the logarithm, k = Ae−Ea/RT is an exponential function of 1/T, so this plot is a curve too.
NCERT (2026-27 edition):
3.4 Temperature Dependence of the Rate of a Reaction – “…
\( \displaystyle \ln k=-\frac{E_a}{RT}+\ln A \qquad (3.19) \)
The plot of ln k vs 1/T gives a straight line according to the equation (3.19) as shown in Fig. 3.10. … In Fig. 3.10, slope = – Ea/R and intercept = ln A. So we can calculate Ea and A using these values.”
Fig. 3.10: A plot between ln k and 1/T