PYQ Vault

JEE Mains Chemistry · Chemical Kinetics

Rate of Reaction and Stoichiometry

The average rate is a change in concentration over a time interval; one rate of reaction is shared by every species once each species' rate is divided by its coefficient.

All Chemical Kinetics notesJEE Mains Chemistry strategy

Why this matters

Nine PYQs, four of them multiple choice, and two from 2026. Five convert the rate of one species into the rate of another through the coefficients of the balanced equation; four measure a rate — an average rate read from a plot or from a change in concentration, or the iodide–hydrogen peroxide clock experiment. Two ideas cover the page.

Concept 1 of 2: Average rate from a change in concentration

A rate says how fast a concentration changes. Over an interval, divide the change in concentration by the time taken. A reactant's concentration falls, so a minus sign keeps the rate positive. On a concentration–time plot, read both curves over the same interval: for A→nBA \rightarrow nB, the rise in [B][B] is nn times the fall in [A][A].

Definition

  • Average rate =−Δ[R]Δt=+Δ[P]Δt= -\dfrac{\Delta[\text{R}]}{\Delta t} = +\dfrac{\Delta[\text{P}]}{\Delta t} for a 1 : 1 reaction.
  • Instantaneous rate =−d[R]dt= -\dfrac{d[\text{R}]}{dt}, the slope of the tangent to the curve at that time.
  • Unit: mol L−1^{-1} s−1^{-1}. A rate per minute is 60 times a rate per second; a rate per hour is 60 times a rate per minute.
  • For A→nBA \rightarrow nB: n=Δ[B]−Δ[A]n = \dfrac{\Delta[B]}{-\Delta[A]} over the same interval.
  • The iodide–H2O2\mathrm{H_2O_2} clock: H2O2+2I−+2H+→I2+2H2O\mathrm{H_2O_2 + 2I^- + 2H^+ \rightarrow I_2 + 2H_2O}. A little thiosulphate turns the I2\mathrm{I_2} back to I−\mathrm{I^-} as it forms. When the thiosulphate is used up, I2\mathrm{I_2} builds up and turns starch blue. Use fresh starch, keep thiosulphate LESS than KI, and record the time the instant the blue appears.

Average rate

rav=−Δ[R]Δt=+Δ[P]Δtr_{\text{av}} = -\frac{\Delta[\text{R}]}{\Delta t} = +\frac{\Delta[\text{P}]}{\Delta t}

Worked example

For A→nBA \rightarrow nB, [A][A] falls from 0.080 M to 0.060 M in the first 5 min while [B][B] rises from 0 to 0.040 M. Find nn, and the average rate of disappearance of AA in mol L−1^{-1} h−1^{-1}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q26Moderate

Example 1 · Chemical Kinetics · Rate of Reaction and Stoichiometry

Given above is the concentration vs time plot for a dissociation reaction : A→nBA \rightarrow nB. Based on the data of the initial phase of the reaction (initial 10 min ), the value of n is ____\_\_\_\_ .

Clock reaction: the time and the thiosulphate

The time is recorded the instant the blue colour appears, because that is when the thiosulphate runs out. Thiosulphate must be LESS than KI; with more thiosulphate, iodine never builds up and the blue never comes.

Minutes divided as if they were hours

A change of 0.1 mol L−1^{-1} in 20 minutes is 0.11/3=0.3\dfrac{0.1}{1/3} = 0.3 mol L−1^{-1} h−1^{-1}. Convert the time into the unit the question asks for before dividing.
Drill 3 more on average rate from a change in concentration

Concept 2 of 2: Rates of different species through the coefficients

In aA+bB→cC+dDaA + bB \rightarrow cC + dD, each species changes at its own speed: AA is used up aa times as fast as the reaction runs, DD forms dd times as fast. Divide each species' rate by its coefficient and all of them give one number, the rate of reaction.

Definition

  • r=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dtr = -\dfrac{1}{a}\dfrac{d[A]}{dt} = -\dfrac{1}{b}\dfrac{d[B]}{dt} = \dfrac{1}{c}\dfrac{d[C]}{dt} = \dfrac{1}{d}\dfrac{d[D]}{dt}.
  • From species X to species Y: rate of Y =yx×= \dfrac{y}{x} \times rate of X, where xx and yy are their coefficients.
  • The rate of reaction is the per-coefficient value, not the rate of any one species.
  • Check the unit asked for: 1 mmol dm−3^{-3} s−1^{-1} =10−3= 10^{-3} mol dm−3^{-3} s−1^{-1}.

Rate of reaction

r=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dtr = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Worked example

For 4NH3+5O2→4NO+6H2O\mathrm{4NH_3 + 5O_2 \rightarrow 4NO + 6H_2O}, oxygen is used up at 0.025 mol L−1^{-1} s−1^{-1}. Find the rate of reaction and the rate of formation of water.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q36Moderate

Example 2 · Chemical Kinetics · Rate of Reaction and Stoichiometry

Observe the following reactions at T(K)T(K) (I) A → products. (II) 5Br−(aq)+BrO3−(aq)+6H+(aq)→3Br2(aq)+3H2O(l)5{Br}^{-}(aq) + {BrO}_{3}^{-}(aq) + 6H^{+}(aq) \rightarrow 3{Br}_{2}(aq) + 3H_{2}O(l) Both the reactions are started at 10.00 am . The rates of these reactions at 10.10 am are same. The value of −Δ[Br−]Δt-\frac{\Delta\left\lbrack Br^{-} \right\rbrack}{\Delta t} at 10.10 am is 2×10−4 mol L−1Min−12 \times10^{- 4}\text{ }mol{\text{ }L}^{- 1}Min^{- 1}. The concentration of A at 10.10 am is 10−2 mol L−110^{- 2}\text{ }mol{\text{ }L}^{- 1}. What is the first order rate constant (in min⁡−1\min^{- 1} ) of reaction I?

Two species' rates taken as equal

In 2N2O5→4NO2+O2\mathrm{2N_2O_5 \rightarrow 4NO_2 + O_2}, NO2\mathrm{NO_2} forms twice as fast as N2O5\mathrm{N_2O_5} is used up, not at the same rate. Divide each species' rate by its own coefficient before comparing.

A species' rate reported as the rate of reaction

The rate of reaction is the species' rate divided by its coefficient. If Br−\mathrm{Br^-} (coefficient 5) is used up at 5×10−45 \times 10^{-4} mol L−1^{-1} s−1^{-1}, the rate of reaction is 1×10−41 \times 10^{-4}, not 5×10−45 \times 10^{-4}.
Drill 4 more on rates of different species through the coefficients

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Average rate from a change in concentration

    Average rate

    rav=−Δ[R]Δt=+Δ[P]Δtr_{\text{av}} = -\frac{\Delta[\text{R}]}{\Delta t} = +\frac{\Delta[\text{P}]}{\Delta t}
  • Rates of different species through the coefficients

    Rate of reaction

    r=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dtr = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = \frac{1}{c}\frac{d[C]}{dt} = \frac{1}{d}\frac{d[D]}{dt}

Watch out for (4)

Test yourself on Chemical Kinetics

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Chemical Kinetics chapter testDrill all 9 questions