PYQ Vault
EbooksTutor

MHT-CET Chemistry · Chemical Kinetics

Rate of Reaction, Stoichiometry and Average Rate

For aA + bB → cC + dD the single rate is −(1/a)d[A]/dt = −(1/b)d[B]/dt = (1/c)d[C]/dt = (1/d)d[D]/dt; one species' rate converts to another's through the ratio of coefficients, and the average rate is Δ[X]/Δt.

All Chemical Kinetics notesMHT-CET Chemistry strategy

Why this matters

31 PYQs, none HARD — the chapter's entry page and its most repetitive: given the rate at which one species appears or disappears, find another's (N₂ + 3H₂ → 2NH₃ and 2N₂O₅ → 4NO₂ + O₂ recur every year), write the rate expression with the right signs and reciprocals, or read a balanced equation back off a rate expression. The only errors are a missed coefficient ratio or a sign.

Concept 1 of 2: Converting One Species' Rate to Another's: Multiply by the Coefficient Ratio

Each mole of reaction consumes aa of A and makes cc of C, so C appears ca\dfrac{c}{a} times as fast as A disappears. Divide every rate by its coefficient to get the ONE rate of reaction, then multiply by the target's coefficient.

Definition

  • N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3: NH3_3 forms twice as fast as N2_2 disappears (2.22×10−3→4.44×10−32.22 \times 10^{-3} \to 4.44 \times 10^{-3}); N2_2 disappears at half the NH3_3 rate (0.088→0.0440.088 \to 0.044).
  • 2N2O5→4NO2+O22\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2: O2_2 forms at half the N2_2O5_5 rate (0.02→0.010.02 \to 0.01); NO2_2 forms at twice it (0.06→0.120.06 \to 0.12, x→2xx \to 2x); the rate of reaction from NO2_2 rising 5.2×10−35.2 \times 10^{-3} M in 100100 s is 145.2×10−3100=1.3×10−5\dfrac{1}{4}\dfrac{5.2 \times 10^{-3}}{100} = 1.3 \times 10^{-5}.
  • 3I−+S2O82−→I3−+2SO42−3\text{I}^- + \text{S}_2\text{O}_8^{2-} \to \text{I}_3^- + 2\text{SO}_4^{2-}: with SO42−_4^{2-} at 2.2×10−22.2 \times 10^{-2}, S2_2O82−_8^{2-} goes at 1.1×10−21.1 \times 10^{-2} and I−^- at 3.3×10−23.3 \times 10^{-2}; SO42−_4^{2-} at 0.0440.044 means I−^- at 0.0660.066.
  • 2A+B→3C2\text{A} + \text{B} \to 3\text{C}: C at 1.3×10−41.3 \times 10^{-4} means A at 23×1.3×10−4=8.66×10−5\dfrac23 \times 1.3 \times 10^{-4} = 8.66 \times 10^{-5}. A+3B→2C\text{A} + 3\text{B} \to 2\text{C}: A at 1.41.4 means C at 2.82.8. 2NH3→N2+3H22\text{NH}_3 \to \text{N}_2 + 3\text{H}_2 with rate 2.5×10−62.5 \times 10^{-6}: H2_2 at 7.5×10−67.5 \times 10^{-6}.
  • Equal coefficients, equal rates: CH3_3Br + OH−^- → CH3_3OH + Br−^-, NO2_2 + CO → NO + CO2_2. Average rate =Δ[product]Δt= \dfrac{\Delta[\text{product}]}{\Delta t}: 0.0520=0.0025\dfrac{0.05}{20} = 0.0025 M s−1^{-1}.

Rate of reaction

rate=−1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt\text{rate} = -\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+6H2O4\text{NH}_3 + 5\text{O}_2 \to 4\text{NO} + 6\text{H}_2\text{O}, NH3_3 disappears at 0.240.24 mol dm−3^{-3} s−1^{-1}. Find the rates of disappearance of O2_2 and of formation of H2_2O.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q51Easy

Example 1 · Chemical Kinetics · Rate of Reaction, Stoichiometry and Average Rate

For the reaction N2(g)+3H2(g)→2NH3(g)\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g}), rate of disappearance of N2(g)\text{N}_2(\text{g}) is 2.22×10−32.22 \times 10^{-3} mol dm−3^{-3}. What is the rate of appearance of NH3(g)\text{NH}_3(\text{g})?

Multiplying when you should divide

N2_2 at 0.0440.044 means NH3_3 at 0.0880.088; NH3_3 at 0.0880.088 means N2_2 at 0.0440.044. Set up the equal-rates chain and read the direction off the coefficients rather than guessing which way the factor goes.
Drill 15 more on converting one species' rate to another's: multiply by the coefficient ratio

Concept 2 of 2: Writing the Rate Expression, and Reading the Equation Back From It

Reactants carry a minus sign (their concentration falls), products a plus; each term is divided by its own coefficient. Reverse the reading: a negative term with 12\dfrac12 in front is a reactant with coefficient 22.

Definition

  • N2+3H2⇌2NH3\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3: −d[N2]dt=−13d[H2]dt=+12d[NH3]dt-\dfrac{d[\text{N}_2]}{dt} = -\dfrac13\dfrac{d[\text{H}_2]}{dt} = +\dfrac12\dfrac{d[\text{NH}_3]}{dt}; cross-multiplied, 3d[NH3]dt=−2d[H2]dt3\dfrac{d[\text{NH}_3]}{dt} = -2\dfrac{d[\text{H}_2]}{dt}.
  • 2NO+2H2→N2+2H2O2\text{NO} + 2\text{H}_2 \to \text{N}_2 + 2\text{H}_2\text{O}: −d[NO]dt=d[H2O]dt-\dfrac{d[\text{NO}]}{dt} = \dfrac{d[\text{H}_2\text{O}]}{dt} (same coefficient), d[N2]dt=−12d[H2]dt=12d[H2O]dt\dfrac{d[\text{N}_2]}{dt} = -\dfrac12\dfrac{d[\text{H}_2]}{dt} = \dfrac12\dfrac{d[\text{H}_2\text{O}]}{dt}.
  • Reading back: −12d[x]dt=−d[y]dt=12d[z]dt-\dfrac12\dfrac{d[x]}{dt} = -\dfrac{d[y]}{dt} = \dfrac12\dfrac{d[z]}{dt} is 2x+y→2z2x + y \to 2z; 13d[x]dt=−12d[y]dt=−d[Z]dt\dfrac13\dfrac{d[x]}{dt} = -\dfrac12\dfrac{d[y]}{dt} = -\dfrac{d[Z]}{dt} is 2y+Z→3x2y + Z \to 3x; −1ad[A]dt=−1bd[B]dt=1cd[C]dt=1dd[D]dt-\dfrac1a\dfrac{d[A]}{dt} = -\dfrac1b\dfrac{d[B]}{dt} = \dfrac1c\dfrac{d[C]}{dt} = \dfrac1d\dfrac{d[D]}{dt} is aA+bB→cC+dDaA + bB \to cC + dD.
  • Options that put a coefficient IN FRONT (−3d[H2]dt-3\dfrac{d[\text{H}_2]}{dt}) instead of as a reciprocal, or drop the minus on a reactant, are the standard distractors.

Signs and reciprocals

reactant: −1coeffd[⋅]dtproduct: +1coeffd[⋅]dt\text{reactant: } -\frac{1}{\text{coeff}}\frac{d[\cdot]}{dt} \qquad \text{product: } +\frac{1}{\text{coeff}}\frac{d[\cdot]}{dt}

Worked example

Write the rate expression for 2H2O2→2H2O+O22\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2 and relate the O2_2 rate to the H2_2O2_2 rate.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 19 April Shift II · Q58Moderate

Example 2 · Chemical Kinetics · Rate of Reaction, Stoichiometry and Average Rate

Which from following is a correct representation of reaction rate for reaction stated below?
N2( g)+3H2( g)⇌2NH3( g)N_{2(\text{ }g)}+ 3H_{2(\text{ }g)}\rightleftharpoons 2NH_{3(\text{ }g)}

Coefficient in front instead of as a reciprocal

The term for H2_2 is −13d[H2]dt-\dfrac13\dfrac{d[\text{H}_2]}{dt}, not −3d[H2]dt-3\dfrac{d[\text{H}_2]}{dt}. Option (A) on every N2_2/H2_2/NH3_3 stem is the inverted version.
Drill 8 more on writing the rate expression, and reading the equation back from it

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)

Watch out for (2)

Test yourself on Chemical Kinetics

20 past MHT-CET questions from this chapter, timed at 18 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 31 questions