The integrated rate equation for a first-order reaction is \(\frac{kt}{2.303} = logR_o − logR_t\). The straight line graph is obtained by plotting:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → time vs \(log R_t\)
A plot gives a straight line when the equation can be written in the form y = mx + c. Rearranging the integrated rate equation of a first order reaction:
\( \displaystyle \frac{kt}{2.303} = \log R_0 - \log R_t \)
\( \displaystyle \log R_t = -\frac{k}{2.303}\,t + \log R_0 \)
This is y = mx + c with y = log Rt and x = t. So log Rt against time is a straight line with slope = −k/2.303 and intercept = log R0.
Why the other options are incorrect
Option (1) time vs \(R_t\) — Incorrect. For a first order reaction Rt = R0e−kt, an exponential curve; [R] against t is a straight line only for a zero order reaction.
Option (3) time vs \(\frac{1}{R_t}\) — Incorrect. 1/[R] against t is a straight line for a second order reaction (1/[R] = kt + 1/[R]0).
Option (4) time vs \(R_o\) — Incorrect. R0 is the initial concentration, a constant; this would be a horizontal line that says nothing about the rate.
NCERT (2026-27 edition):
First Order Reactions – …
