For the first order reaction:
(A) The degree of dissociation is equal to \((1-e^{-kt})\)
(B) A plot of reciprocal concentration of the reactant v/s time gives a straight line
(C) The time taken for the completion of 75% reaction is thrice the half-life of the reaction
(D) 1st order reaction never completes
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (D)
For a first order reaction the integrated rate law is [R] = [R]0e−kt and t1/2 = 0.693/k.
(A) Degree of dissociation = (1 − e−kt) — True. Fraction reacted α = ([R]0 − [R])/[R]0 = 1 − e−kt.
(B) Plot of 1/[R] versus time is a straight line — False. That is the behaviour of a second order reaction; for first order, ln[R] (or log [R]0/[R]) versus t is linear.
(C) t75% is thrice the half-life — False. 75% completion means two successive half-lives (100 → 50 → 25%), so t75% = (2.303/k) log 4 = 2t1/2.
(D) A first order reaction never completes — True. [R] = [R]0e−kt becomes zero only as t → ∞; in every half-life only half of what is left reacts.
NCERT (2026-27 edition):
First Order Reactions – “… Equation (3.9) can also be written as ln([R]/[R]0) = −kt. Taking antilog of both sides
[R] = [R]0 e–kt (3.14)”
Half-Life of a Reaction – “… Half-life for a first order reaction is t1/2 = 0.693/k”