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Graphical representation of a reversible first-order reaction R ⇌ P with kb = 4kf
JEE Advanced 2026
Paper 1 · Question 2
Chemistry
Reversible (opposing) first-order reactions
Moderate
A glass vessel of pure R. Start the reaction and watch R turn into P and P turn back, molecule by molecule, while two rate meters show why it slows: kf[R] falls as R is used up and kb[P] rises as P builds up. The clock stops to make you predict where it will settle, the chamber reaches dynamic equilibrium at 0.8 and 0.2 with the molecules still swapping, an equation board derives it, and only then do you pick the printed graph that matches your own experiment.
Open the interactive simulationSub-topics covered
- Rate law for opposing elementary first-order steps
- Dynamic equilibrium: forward rate equals backward rate
- Equilibrium constant as a ratio of rate constants, K = kf/kb
- Mass balance [R] + [P] = [R]0
- Integrated rate law for a reversible first-order reaction
- Observed relaxation constant kf + kb
- Reading and interpreting kinetics graphs
What it teaches
- Both directions are elementary first-order steps, so the forward rate is kf[R] and the backward rate is kb[P]; at t = 0 there is no P and the backward rate is exactly zero
- Equilibrium is the moment the two rates become equal, kf[R]eq = kb[P]eq - not the moment either reaction stops. Both keep running, molecule by molecule, and only the amounts stop changing
- With kb = 4kf, balancing the rates needs four times as much R as P, so [R]eq = 4[P]eq
- Every R lost becomes a P, so [R] + [P] = [R]0 at every instant; with [R]eq = 4[P]eq this gives 5[P]eq = [R]0, so [P]eq/[R]0 = 0.2 and [R]eq/[R]0 = 0.8
- The equilibrium constant is the ratio of the rate constants: K = [P]eq/[R]eq = kf/kb = 1/4
- The plateau depends only on kb/kf; [R]0 cancels in the ratios and the absolute size of kf sets only the time scale
- The approach to equilibrium is exponential with the observed constant kf + kb = 5kf, faster than kf alone, because both directions pull towards the balance point
- A correct graph must start at 1 and 0, keep [R] + [P] = [R]0, bend over (a first-order approach is always concave down) and level off where the rates balance - option D fails all three of the last tests, option A levels off at kb = kf, option B at kf = 4kb
Simulation ID
- ADV-2026-P1-CHE-Q02 · revision 1
See this simulation in the library → · open the simulation file on its own