According to Arrhenius equation \(k = Ae^{-\frac{E_a}{RT}}\). If \(ln k\) vs \(\frac{1}{T}\) graph plotted, what will be the intercept of this plot?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(ln A\)
The Arrhenius equation is
\( k = A\,e^{-E_a/RT} \)
Taking natural logarithms of both sides,
\( \ln k = -\displaystyle\frac{E_a}{R}\cdot\displaystyle\frac{1}{T} + \ln A \)
This has the form y = mx + c with y = ln k and x = 1/T, so the constant term, ln A, is the intercept of the straight line.
Why the other options are incorrect
Option (1) \(-\frac{E_a}{RT}\) — Incorrect. This is the term (slope × 1/T) that changes with temperature, not the intercept; the slope itself is −Ea/R.
Option (2) \(\frac{1}{T}\) — Incorrect. 1/T is the variable on the x-axis, not the intercept.
Option (4) \(ln K\) — Incorrect. Ln k is the quantity on the y-axis; it changes with temperature, so it cannot be the fixed intercept.
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. … In Fig. 3.10, slope = − Ea/R and intercept = ln A.”
