JEE (Advanced) 2026, Paper 1, Chemistry, Section 1 — single correct option, +3 for the correct option, 0 if unanswered, −1 otherwise.
Q.2 For a reversible reaction R ⇌ P, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants kf and kb, respectively. At time zero, the concentration of R is [R]0 and the concentration of P is zero. At any given time, [R] and [P] are the concentrations of R and P, respectively. If kb = 4kf, the correct graphical representation of the reaction is
A vessel of pure R. Start the reaction and watch every molecule that turns into P, and every P that turns back. The two rate meters show why the reaction slows down, the graph draws itself from the same numbers, and the chamber tells you when it has stopped changing — even though the molecules never stop.
100 molecules of R and not a single P. Press START — the question's values are already filled in.
The question fixes only kb = 4kf. kf = 1 s⁻¹ and [R]0 = 1 M are convenient starting numbers — change them and see what moves and what does not.
Every change restarts the reaction from pure R and keeps the last run as a ghost. Pick a ratio, or type your own kf and kb in the panel — nothing is required.
The plateau is set by the ratio of the rate constants; how quickly the reaction gets there is set by their sum. Take d[P]/dt = kf[R] − kb[P] with [R] = [R]0 − [P]:
So the distance still to go, [P]eq − [P], shrinks as e−(kf+kb)t: a straight line on a log scale, and its slope measures kf + kb — which is how you would get both rate constants from one experiment.
Four tests that any correct graph of this reaction has to pass, applied to each printed option. The measurements are read off the redrawn graphs; the verdict column fills in once the experiment has told you what the right answer looks like.
A first-order approach to equilibrium always bends over (d²[P]/dt² < 0), keeps [R] + [P] = [R]0, and levels off where kf[R] = kb[P].
What the chamber has shown so far, written down only when it has actually happened, and a log of every run you complete.
Nothing observed yet — start the reaction.
No completed runs yet. A run is logged when the chamber reaches equilibrium.
Each card runs the rate law with one mistaken idea, on the values in the panel, and shows which printed graph the mistake leads to.
Run the experiment and pick the matching graph — the derivation opens with the answer.
The chamber has not found the matching graph yet. Run the reaction, reach equilibrium and pass the graph challenge — or use Reveal anyway in the header if you are revising.
Both directions are elementary first-order steps, so each rate is its rate constant times the concentration it uses up. Equilibrium is the moment the two become equal.
The backward step is four times faster per molecule, so at balance there must be four times as much R as P for the two rates to match.
Every R lost becomes a P, so the total is always the starting amount.
Substituting [R] = [R]0 − [P] into the rate law and integrating from [P] = 0:
So [R]/[R]0 starts at 1 and [P]/[R]0 at 0, both change fastest at the start, both bend over smoothly, their sum is always 1, and they level off at 0.8 and 0.2 after a few multiples of 1/(5kf).
(A) levels off at 0.5 and 0.5 — that needs kb = kf. (B) levels off with P at 0.8 — that needs kf = 4kb, the rate constants the wrong way round. (D) never levels off, its [P] curve bends upward, and where its curves cross [R] + [P] is about 0.6 of [R]0, not 1. (C) starts at 1 and 0, bends over, and settles at 0.8 and 0.2.
The one idea this question is built on, and where it sits in the syllabus.
At equilibrium: forward rate = backward rate.
NOT forward reaction = 0, and NOT backward reaction = 0. Both reactions continue, molecule by molecule. Only the amounts stop changing, because every R that becomes P is matched by a P that becomes R.
Live: the molecules keep swapping while the graph line stays flat.
The plateau depends only on kb/kf. Double both rate constants and it settles in the same place, twice as fast.
The distance from equilibrium shrinks as e−(kf+kb)t — faster than kf alone would give, because both directions pull towards the plateau.
Starting values right? Sum conserved? Bends the right way? Levels off where the rates balance? A graph that fails any one of them cannot be right.
For a one-step reversible reaction, K is the forward rate constant over the backward one — the bridge between kinetics and equilibrium.
| Exam | JEE Advanced 2026, Paper 1 |
|---|---|
| Question | Chemistry, Section 1, Q.2 |
| Type | Single correct option (+3 / 0 / −1) |
| Chapter | Chemical Kinetics |
| Topic | Reversible (opposing) first-order reactions |
| Also uses | Dynamic equilibrium, K = kf/kb, reading kinetics graphs |
| Difficulty | Moderate |
| Answer | — run the experiment |