The rate constant of a reaction is given by k = PZe−E/RT. Which factor should decrease for the reaction to proceed faster?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → E
In collision theory, Z is the collision frequency, P is the probability (steric) factor for proper orientation, and \( e^{-E/RT} \) is the fraction of molecules whose energy is equal to or greater than the activation energy E.
\( \displaystyle k=PZ\,e^{-E/RT} \)
E stands in the negative exponent. When E decreases, \( -\frac{E}{RT} \) becomes less negative and \( e^{-E/RT} \) becomes larger: more molecules can cross the energy barrier, so k increases and the reaction proceeds faster.
Why the other options are incorrect
Option (1) T — Incorrect. A lower T makes \( \frac{E}{RT} \) larger, so \( e^{-E/RT} \) becomes smaller and the reaction slows down.
Option (2) Z — Incorrect. k is proportional to Z; fewer collisions per second give a slower reaction.
Option (4) P — Incorrect. k is proportional to P; a smaller P means fewer properly oriented (effective) collisions and a slower reaction.
NCERT (2026-27 edition):
3.5 Collision Theory of Chemical Reactions – “… Another factor which affects the rate of chemical reactions is activation energy (as we have already studied). For a bimolecular elementary reaction A + B → Products rate of reaction can be expressed as
\( \displaystyle \mathrm{Rate}=Z_{AB}\,e^{-E_a/RT} \qquad (3.23) \)
where ZAB represents the collision frequency of reactants, A and B and e−Ea/RT represents the fraction of molecules with energies equal to or greater than Ea. … To account for effective collisions, another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented i.e., \( \mathrm{Rate}=PZ_{AB}\,e^{-E_a/RT} \)”