The figure shows how the concentration of the reactant changes with time for a reaction R → P.
As the reaction proceeds and the concentration of R falls, the rate of the reaction:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → stays the same
The rate at any moment is −d[R]/dt, the negative of the slope of the concentration–time plot.
The figure is a straight line, so its slope is the same at every point. The rate is therefore constant, equal to k (k = −slope), even as [R] falls.
A rate that does not depend on the concentration of the reactant is the mark of a zero order reaction:
\( \displaystyle [\ce{R}] = -kt + [\ce{R}]_0 \)
Why the other options are incorrect
Option (2) falls in proportion to [R] — Incorrect. A falling rate would make the plot curve and flatten out with time, as for a first order reaction; this line does not flatten.
Option (3) rises — Incorrect. A rising rate would make the line steeper with time; its slope does not change.
Option (4) falls in proportion to [R]2 — Incorrect. This too would give a curve whose slope becomes smaller with time.
NCERT (2026-27 edition):
3.2.3 Order of a Reaction – “A zero order reaction means that the rate of reaction is independent of the concentration of reactants.”
3.3.1 Zero Order Reactions – “Comparing (3.6) with equation of a straight line, y = mx + c, if we plot [R] against t, we get a straight line (Fig. 3.3) with slope = –k and intercept equal to [R]0.”
Fig. 3.3: Variation in the concentration vs time plot for a zero order reaction