The figure shows how the concentration of the reactant (a) and of the product (b) change with time for a reaction R → P.
According to the curves, how does the instantaneous rate of the reaction change as time passes?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → It decreases, because the tangents to the curves become less steep with time.
The figure gives the instantaneous rate at time t as the slope of the tangent drawn to the curve at t.
Rate from the tangent — rinst = −d[R]/dt = −slope in (a), and rinst = d[P]/dt = slope in (b).
Shape of the curves — Both curves are steep at the start and flatten out as time passes, so a tangent drawn at a later time is less steep. The rate is therefore largest at the start and decreases with time.
Why the other options are incorrect
Option (1) It stays the same throughout the reaction. — Incorrect. A constant rate would give a straight line; both curves bend and flatten, so the slope of the tangent changes with time.
Option (2) It increases, because the concentration of the product increases. — Incorrect. The rate is the slope of the tangent, not the concentration; [P] keeps rising, but curve (b) rises less and less steeply.
Option (4) It increases at first and then decreases. — Incorrect. Both curves are steepest at t = 0; there is no early part where they become steeper.
NCERT (2026-27 edition):
3.1 Rate of a Chemical Reaction – “It can be determined graphically by drawing a tangent at time t on either of the curves for concentration of R and P vs time t and calculating its slope (Fig. 3.1).”
Fig. 3.1: Instantaneous and average rate of a reaction