JEE Challenger
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ChemMedAdv 2026 • Paper 2

Graphical Representations for First Order Reaction Kinetics

For a first-order reaction R→P\text{R} \rightarrow \text{P} at a given temperature, kk is the rate constant. For this reaction, at the given temperature, the concentrations of R\text{R} and P\text{P} at a time tt are [R][\text{R}] and [P][\text{P}], respectively. The correct graphical representation(s) for this reaction is(are)

Options

A
Option A
B
Option B
C
Option C
Correct
D
Option D
Correct

Topics & Concepts

Chemical KineticsOrder of ReactionRate of Reaction

Step-by-Step Solution

For a first-order elementary reaction R→P\text{R} \rightarrow \text{P} with rate constant kk:

  1. Integrated Rate Law: The rate of disappearance of reactant R\text{R} is given by: −d[R]dt=k[R]-\frac{\text{d}[\text{R}]}{\text{d}t} = k[\text{R}] Integrating this differential equation yields: [R]=[R]0e−kt[\text{R}] = [\text{R}]_0 e^{-kt} where [R]0[\text{R}]_0 is the initial concentration of R\text{R} at t=0t = 0.

  2. Concentration of Product [P][\text{P}]: Since [P]=[R]0−[R][\text{P}] = [\text{R}]_0 - [\text{R}], we have: [P]=[R]0(1−e−kt)[\text{P}] = [\text{R}]_0 \left(1 - e^{-kt}\right)

Analysis of Options:

  • Option (A): Taking the first and second derivatives of [P][\text{P}] with respect to time tt: d[P]dt=k[R]0e−kt>0\frac{\text{d}[\text{P}]}{\text{d}t} = k [\text{R}]_0 e^{-kt} > 0 d2[P]dt2=−k2[R]0e−kt<0\frac{\text{d}^2[\text{P}]}{\text{d}t^2} = -k^2 [\text{R}]_0 e^{-kt} < 0 Since the second derivative is negative (d2[P]dt2<0\frac{\text{d}^2[\text{P}]}{\text{d}t^2} < 0), the graph of [P][\text{P}] vs tt must be concave downwards (it increases initially and then levels off asymptotically to [R]0[\text{R}]_0). Graph (A) shows a concave upwards curve, which is incorrect.

  • Option (B): The rate of change of concentration of R\text{R} is: d[R]dt=−k[R]\frac{\text{d}[\text{R}]}{\text{d}t} = -k[\text{R}] A plot of d[R]dt\frac{\text{d}[\text{R}]}{\text{d}t} versus [R][\text{R}] is a straight line passing through the origin with a negative slope (−k-k). Graph (B) shows a positive slope, which is incorrect.

  • Option (C): The rate of formation of product P\text{P} is: d[P]dt=k[R]=k[R]0e−kt\frac{\text{d}[\text{P}]}{\text{d}t} = k[\text{R}] = k[\text{R}]_0 e^{-kt} This shows that d[P]dt\frac{\text{d}[\text{P}]}{\text{d}t} decreases exponentially with time tt from an initial maximum value of k[R]0k[\text{R}]_0 at t=0t = 0 towards zero as t→∞t \to \infty. Graph (C) correctly represents this exponential decay, so option (C) is correct.

  • Option (D): The rate constant kk depends only on temperature and is independent of time tt or concentration. Thus, at a given temperature, a plot of kk versus tt is a horizontal straight line parallel to the time axis. Graph (D) correctly represents this, so option (D) is correct.

Correct Answer: The correct options are C and D.

Graphical Representations for First Order Reaction Kinetics | Chemistry PYQ Solution - JEE Challenger