With reference to Arrhenius equation \( k = Ae^{-E_a/RT} \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The rate constant increases exponentially with decreasing activation energy and increasing temperature.
The Arrhenius equation relates the rate constant to the activation energy and the temperature:
\( \displaystyle k = A\,e^{-E_a/RT} \)
The factor \( e^{-E_a/RT} \) is the fraction of molecules with energy equal to or greater than \( E_a \). Taking logarithms:
\( \displaystyle \ln k = \ln A - \frac{E_a}{RT} \)
The exponent \( -E_a/RT \) becomes less negative in two ways:
when \( E_a \) decreases (a smaller numerator), as happens when a catalyst is added;
when T increases (a larger denominator); for many reactions a rise of 10 K nearly doubles the rate constant.
Since k depends on this exponent through \( e^{x} \), the rate constant increases exponentially with decreasing activation energy and increasing temperature.
Why the other options are incorrect
Option (1) The rate constant increases exponentially with decreasing activation energy and decreasing temperature. — Incorrect. Lowering T makes \( E_a/RT \) larger and so lowers k.
Option (3) The rate constant increases exponentially with increasing activation energy and decreasing temperature. — Incorrect. Both changes make the exponent more negative, so k decreases.
Option (4) The rate constant decreases exponentially with decreasing activation energy and increasing temperature. — Incorrect. These are the right changes, but they increase k; they do not decrease it.
NCERT (2026-27 edition):
Temperature Dependence of the Rate of a Reaction – …
