For a reversible reaction \(\mathbf{R} \rightleftharpoons \mathbf{P}\), at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constant \(k_{f}\) and \(k_{b}\), respectively. At time zero, the concentration of \(\mathbf{R}\) is \([\mathrm{R}]_{0}\) and the concentration of \(\mathbf{P}\) is zero. At any given time \([\mathbf{R}]\) and \([\mathbf{P}]\) are the concentration of \(\mathbf{R}\) and \(\mathbf{P}\), respectively. If \(k_{b}=4 k_{\mathrm{f}}\), the correct graphical representation of the reaction is

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Chemical Kinetics

For a reversible reaction

R ⇌ P

, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constant

k_{f}

and

k_{b}

, respectively. At time zero, the concentration of

R

[R]_{0}

and the concentration of

P

is zero. At any given time

[R]

and

[P]

are the concentration of

R

and

P

, respectively. If

k_{b} = 4 k_{f}

, the correct graphical representation of the reaction is

Answer: (C)

For the reversible reaction

R ⇌ P

, the equilibrium constant is:

K_{c} = \frac{k _{f}}{k _{b}} = \frac{k _{f}}{4 k _{f}} = \frac{1}{4} = \frac{[ P ] _{e q}}{[ R ] _{e q}}

This means at equilibrium:

[R]_{e q} = 4 [P]_{e q}

Since we start with

[R]_{0}

and

[P] = 0

, and the total amount is conserved:

[R]_{e q} + [P]_{e q} = [R]_{0}

4 [P]_{e q} + [P]_{e q} = [R]_{0}

[P]_{e q} = 0.2 [R]_{0} and [R]_{e q} = 0.8 [R]_{0}

Option (C) shows this equilibrium ratio correctly, with

[P]_{e q} / [R]_{0} = 0.2

and

[R]_{e q} / [R]_{0} = 0.8

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option