A graph is plotted between \(ln k\) and \(\frac{1}{T}\), what is the value of slope?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → slope \(= -\frac{E_a}{R}\)
The Arrhenius equation is k = A e−Ea/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 is a straight line with slope = −Ea/R and intercept = ln A. The slope is negative because k increases as 1/T decreases (as T rises).
Why the other options are incorrect
Option (1) slope = Ea/R — Incorrect. It misses the minus sign; ln k falls as 1/T rises.
Option (3) slope = k/R — Incorrect. k is the variable plotted on the y-axis, not part of the slope.
Option (4) slope = −k/R — Incorrect. The slope contains the activation energy, not k.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – “… Taking natural logarithm of both sides of equation (3.18)
ln k = − Ea/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. Thus, it has been found from Arrhenius equation (3.18) that increasing the temperature or decreasing the activation energy will result in an increase in the rate of the reaction and an exponential increase in the rate constant. In Fig. 3.10, slope = − Ea/R and intercept = ln A.”
