The figure given below represents the graph between log([R]_0)/([R])…

The correct answer is Option (1) → \( \frac{k}{2.303} \)

For a first order reaction, the integrated rate equation is:

    \( \displaystyle k=\frac{2.303}{t}\log\frac{[R]_0}{[R]} \)

Rearranging it to show the dependence on time:

    \( \displaystyle \log\frac{[R]_0}{[R]}=\frac{k}{2.303}\,t \)

This has the form y = mx, a straight line through the origin. With log([R]₀/[R]) on the y-axis and time on the x-axis, the slope is k/2.303. The slope is positive, as the rising line in the figure shows.

Why the other options are incorrect

• Option (2) \( \frac{-k}{2.303} \) — Incorrect. This is the slope of the plot of log [R] against time, since log [R] = −kt/2.303 + log [R]₀. That line falls with time.

• Option (3) \( \frac{-2.303}{k} \) — Incorrect. This is the negative reciprocal of the slope. It has the unit of time, while the slope has the unit of k (s⁻¹).

• Option (4) \( \frac{2.303}{k} \) — Incorrect. This is the reciprocal of the slope; it has the unit of time, while the slope has the unit of k.

NCERT (2026-27 edition):
First Order Reactions – …

Note: The same/similar question has been asked in different years by NTA as per the details given below: