JEE Challenger
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ChemMedAdv 2026 • Paper 1

Concentration versus Time Graph for Reversible First Order Reaction

For a reversible reaction R⇌P\mathbf{R} \rightleftharpoons \mathbf{P}, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants kfk_f and kbk_b, respectively. At time zero, the concentration of R\mathbf{R} is [R]0[\mathbf{R}]_0 and the concentration of P\mathbf{P} is zero. At any given time, [R][\mathbf{R}] and [P][\mathbf{P}] are the concentrations of R\mathbf{R} and P\mathbf{P}, respectively. If kb=4kfk_b = 4k_f, the correct graphical representation of the reaction is

Options

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

Topics & Concepts

Chemical KineticsOrder of ReactionRate of Reaction

Step-by-Step Solution

To determine the correct graphical representation for the given reversible reaction R⇌kfkbP\mathbf{R} \underset{k_b}{\overset{k_f}{\rightleftharpoons}} \mathbf{P}, we can analyze the concentration ratios at initial state (t=0t = 0) and at equilibrium (t→∞t \to \infty).

1. Initial Conditions (t=0t = 0)

At time t=0t = 0: [R]=[R]0  ⟹  [R][R]0=1[\mathbf{R}] = [\mathbf{R}]_0 \implies \frac{[\mathbf{R}]}{[\mathbf{R}]_0} = 1 [P]=0  ⟹  [P][R]0=0[\mathbf{P}] = 0 \implies \frac{[\mathbf{P}]}{[\mathbf{R}]_0} = 0

2. Conservation of Mass

By the law of conservation of mass, at any time tt: [R]+[P]=[R]0[\mathbf{R}] + [\mathbf{P}] = [\mathbf{R}]_0

Dividing throughout by [R]0[\mathbf{R}]_0: [R][R]0+[P][R]0=1\frac{[\mathbf{R}]}{[\mathbf{R}]_0} + \frac{[\mathbf{P}]}{[\mathbf{R}]_0} = 1

3. Equilibrium Conditions (t→∞t \to \infty)

At dynamic equilibrium, the rate of the forward reaction equals the rate of the backward reaction: rf=rbr_f = r_b kf[R]eq=kb[P]eqk_f [\mathbf{R}]_{eq} = k_b [\mathbf{P}]_{eq}

Given that kb=4kfk_b = 4k_f, we substitute this into the equilibrium rate expression: kf[R]eq=4kf[P]eqk_f [\mathbf{R}]_{eq} = 4k_f [\mathbf{P}]_{eq} [R]eq=4[P]eq[\mathbf{R}]_{eq} = 4[\mathbf{P}]_{eq}

Using the mass conservation relation at equilibrium: [R]eq+[P]eq=[R]0[\mathbf{R}]_{eq} + [\mathbf{P}]_{eq} = [\mathbf{R}]_0 4[P]eq+[P]eq=[R]04[\mathbf{P}]_{eq} + [\mathbf{P}]_{eq} = [\mathbf{R}]_0 5[P]eq=[R]05[\mathbf{P}]_{eq} = [\mathbf{R}]_0

Therefore, the equilibrium concentration ratios are: [P]eq[R]0=15=0.2\frac{[\mathbf{P}]_{eq}}{[\mathbf{R}]_0} = \frac{1}{5} = 0.2 [R]eq[R]0=45=0.8\frac{[\mathbf{R}]_{eq}}{[\mathbf{R}]_0} = \frac{4}{5} = 0.8

4. Kinetics and Graph Characteristics

For a first-order reversible reaction, the concentration ratios approach their equilibrium values exponentially with time:

  • The ratio [R][R]0\frac{[\mathbf{R}]}{[\mathbf{R}]_0} starts at 1.01.0 and exponentially decays towards 0.80.8.
  • The ratio [P][R]0\frac{[\mathbf{P}]}{[\mathbf{R}]_0} starts at 0.00.0 and exponentially rises towards 0.20.2.

Comparing this behavior with the given options:

  • Option (C) correctly shows [R][R]0\frac{[\mathbf{R}]}{[\mathbf{R}]_0} approaching 0.80.8 and [P][R]0\frac{[\mathbf{P}]}{[\mathbf{R}]_0} approaching 0.20.2 as time increases.

Correct Option: (C)

Concentration versus Time Graph for Reversible First Order Reaction | Chemistry PYQ Solution - JEE Challenger