For the graphical representation of the first-order reaction The slop…

The correct answer is Option (1) → k

The integrated rate equation of a first order reaction is

    \( \displaystyle k = \frac{1}{t}\ln\frac{[R]_0}{[R]} \quad\Rightarrow\quad \ln\frac{[R]_0}{[R]} = kt \)

This has the form y = mx with y = ln([R]₀/[R]) and x = t. The graph in the question plots ln([R]₀/[R]) against time, so it is a straight line through the origin with slope = k.

Why the other options are incorrect

• Option (2) −k — Incorrect. This is the slope of the plot of ln[R] against t, which falls with time.

• Option (3) k/2.303 — Incorrect. This would be the slope if log([R]₀/[R]) (base 10) were plotted against t; the graph here uses the natural logarithm.

• Option (4) −k/2.303 — Incorrect. This is the slope of the plot of log[R] against t.

NCERT (2026-27 edition):
First Order Reactions – “… Comparing equation (3.9) with y = mx + c, if we plot ln [R] against t (Fig. 3.4) we get a straight line with slope = –k and intercept equal to ln [R]₀. The first order rate equation (3.10) can also be written in the form

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

… If we plot a graph between log [R]₀/[R] vs t, (Fig. 3.5), the slope = k/2.303”