If we plot graph for the first order reaction involving reactant R, then it will be a straight line if plotted for:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(log\frac{[R]}{[R]_0}\) against time, t
For a first order reaction the integrated rate equation is:
\( \ln[R] = \ln[R]_0 - kt \)
Rearranging and changing to common logarithms:
\( \displaystyle \log\frac{[R]}{[R]_0} = -\frac{k}{2.303}\,t \)
This has the form y = mx. A plot of log([R]/[R]0) against t is a straight line through the origin with slope −k/2.303 (equivalently, log([R]0/[R]) against t has slope +k/2.303).
Why the other options are incorrect
Option (1) [R]/[R]0 against time — Incorrect. [R]/[R]0 = e–kt, an exponential decay curve, not a straight line.
Option (2) [R]0 against time — Incorrect. [R]0 is the initial concentration, a fixed number; it does not change with time, so this plot does not follow the reaction at all.
Option (4) [R] against time — Incorrect. A straight line of [R] against t (slope −k) is the test for a zero order reaction; for first order this plot is a curve.
NCERT (2026-27 edition):
First Order Reactions – …
