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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.

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Why this matters

26 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

Intuition

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.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical KineticsEASY
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})?

[Q51 · Shift 1 · 2022]

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 16 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

Intuition

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.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical KineticsMODERATE
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)}

[Q58 · 19 April Shift II · 2025]

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

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