The plot of ln k against 1/T is given in the figure. What is its slope?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → -Ea/R
The Arrhenius equation relates the rate constant to temperature:
\( \displaystyle k = A\,e^{-E_a/RT} \)
Taking natural logarithms:
\( \displaystyle \ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A \)
This has the form y = mx + c with y = ln k and x = 1/T. So the plot of ln k against 1/T is a straight line of slope = −Ea/R and intercept ln A, falling to the right as in the figure.
Why the other options are incorrect
Option (2) Ea/R — Incorrect. A positive slope would mean ln k falls as T rises; the line in the figure goes down, and the sign in the equation is negative.
Option (3) Ea/2.303R — Incorrect. The factor 2.303 appears only when log10 k is plotted, and even then the slope is negative.
Option (4) -Ea/2.303R — Incorrect. This is the slope of log k against 1/T; the graph here is of ln k.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – “… In the Arrhenius equation (3.18) the factor e–Ea/RT corresponds to the fraction of molecules that have kinetic energy greater than Ea. Taking natural logarithm of both sides of equation (3.18) \(\ln k = -\frac{E_a}{RT} + \ln A\) (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.”
