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MHT-CET Chemistry · Chemical Kinetics

Rate Law, Order, Molecularity and Rate Expression

rate = k[A]ˣ[B]ʸ with x + y the (experimental) order; k is the rate at unit concentrations, depends only on temperature, and its unit depends on the order; molecularity is the number of species in an elementary step and equals the order only for elementary reactions.

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Why this matters

38 PYQs, none HARD — the chapter's second-largest page. A third are k from a rate and concentrations (or the reverse), a third are the order read off a rate law or a concentration experiment, and the rest are the change in rate when concentrations are doubled or halved, the order-versus-molecularity contrast, and three named examples (H₂O₂ decomposition first order, H₂ + I₂ second, H₂ + Br₂ order 3/2). Nothing here needs more than substitution; the traps are conceptual — order is experimental, k is concentration-independent.

Concept 1 of 4

The Rate Law and the Order: Exponents Come From Experiment, Not From the Equation

Intuition

The rate law r=k[A]x[B]yr = k[A]^x[B]^y is measured, and x+yx + y is the order. The exponents need not match the stoichiometric coefficients, can be fractional or zero, and a species with exponent zero does not affect the rate at all.

Definition

  • 2NO+2H2→N2+2H2O2\text{NO} + 2\text{H}_2 \to \text{N}_2 + 2\text{H}_2\text{O} with r=k[NO]2[H2]r = k[\text{NO}]^2[\text{H}_2]: first order in H2_2, overall order 33. r=k[H2][I2]r = k[\text{H}_2][\text{I}_2]: overall 22, first in each. r=k[A]1.5[B]2.5r = k[A]^{1.5}[B]^{2.5}: order 44.
  • Rate proportional to [NO2]2[\text{NO}_2]^2 and independent of [CO]: r=k[NO2]2[CO]0r = k[\text{NO}_2]^2[\text{CO}]^0. Second order in NO, first in Cl2_2: r=k[NO]2[Cl2]r = k[\text{NO}]^2[\text{Cl}_2]. First in CHCl3_3, half in Cl2_2: r=k[CHCl3][Cl2]1/2r = k[\text{CHCl}_3][\text{Cl}_2]^{1/2}, order 32\tfrac32.
  • From an experiment: tenfold [A][A] giving 100×100\times the rate means 10x=10010^x = 100, x=2x = 2.
  • Order is an experimental quantity; it may be an integer, a fraction or zero; it is NOT a theoretical quantity.
  • Named examples: 2H2O2→2H2O+O22\text{H}_2\text{O}_2 \to 2\text{H}_2\text{O} + \text{O}_2 and CH3_3CHO → CH4_4 + CO are first order; H2+I2→2HI\text{H}_2 + \text{I}_2 \to 2\text{HI} and 2NO2→2NO+O22\text{NO}_2 \to 2\text{NO} + \text{O}_2 are second; 2NO+2H22\text{NO} + 2\text{H}_2 and 2NO2+F22\text{NO}_2 + \text{F}_2 are third.

Rate law

r=k[A]x[B]y,order=x+y (experimental; may be 0, fractional)r = k[A]^x[B]^y,\qquad \text{order} = x + y \ (\text{experimental; may be } 0, \text{ fractional})

Worked example

Doubling [A][A] doubles the rate; doubling [B][B] quadruples it. Write the rate law and the overall order.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Chemical KineticsEASY
For the reaction 2NO(g)+2H2(g)⟶N2(g)+2H2O(s)2\text{NO}(g) + 2\text{H}_{2}(g) \longrightarrow \text{N}_{2}(g) + 2\text{H}_{2}\text{O}(s) rate =k[NO]2[H2]= k[\text{NO}]^{2}[\text{H}_{2}]. What is the order of reaction with respect to H2\text{H}_{2} and overall order of reaction respectively?

[Q53 · 15th May Shift 1 · 2023]

Reading the order off the balanced equation

2N2O5→4NO2+O22\text{N}_2\text{O}_5 \to 4\text{NO}_2 + \text{O}_2 is first order, and H2_2 + Br2_2 is order 32\tfrac32. The coefficients give molecularity for an elementary step; the order is measured.
Drill 12 more on the rate law and the order: exponents come from experiment, not from the equation

Concept 2 of 4

The Rate Constant: k = rate/([A]ˣ[B]ʸ), Its Units and Its Properties

Intuition

Substitute the given concentrations into the rate law and divide the rate by the concentration product. The unit of kk is whatever makes the equation balance — (mol dm−3^{-3})1−n^{1 - n} s−1^{-1} for order nn — so a unit of s−1^{-1} means first order.

Definition

  • r=k[A][B]2r = k[A][B]^2: k=7.2×10−20.4×0.01=18k = \dfrac{7.2 \times 10^{-2}}{0.4 \times 0.01} = 18, 3.6×10−20.2×0.01=18\dfrac{3.6 \times 10^{-2}}{0.2 \times 0.01} = 18 mol−2^{-2} dm6^6 s−1^{-1}; r=k[A]2[B]r = k[A]^2[B]: 0.221×0.25=0.88\dfrac{0.22}{1 \times 0.25} = 0.88, 1.8×10−20.04×0.1=4.5\dfrac{1.8 \times 10^{-2}}{0.04 \times 0.1} = 4.5, 0.240.25×0.2=4.8\dfrac{0.24}{0.25 \times 0.2} = 4.8.
  • Rate from kk: 6.25×12×0.2=1.256.25 \times 1^2 \times 0.2 = 1.25 (r=k[A]2[B]r = k[A]^2[B]); 6.25×1×0.22=0.256.25 \times 1 \times 0.2^2 = 0.25 (r=k[A][B]2r = k[A][B]^2). Concentration from a rate: [A]=0.256.25×0.25=0.16[A] = \dfrac{0.25}{6.25 \times 0.25} = 0.16.
  • Units: zero order mol dm−3^{-3} s−1^{-1}; first s−1^{-1}; second mol−1^{-1} dm3^3 s−1^{-1}; third mol−2^{-2} dm6^6 s−1^{-1}. A kk in hour−1^{-1} or s−1^{-1} is first order.
  • kk is independent of concentration, varies with temperature, equals the rate at unit concentrations; its unit DEPENDS on the order. For a first-order reaction the slope of rate against concentration is kk.

Rate constant

k=r[A]x[B]y,[k]=(mol dm−3)1−n s−1k = \frac{r}{[A]^x[B]^y},\qquad [k] = (\text{mol dm}^{-3})^{1-n}\,\text{s}^{-1}

Worked example

For r=k[A]2[B]r = k[A]^2[B], the rate is 3.2×10−33.2 \times 10^{-3} mol dm−3^{-3} s−1^{-1} when [A]=0.4[A] = 0.4 M and [B]=0.05[B] = 0.05 M. Find kk with its unit.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Chemical KineticsEASY
Rate of reaction, A+B→A + B \to product, is 7.2×10−2 mol dm−3s−17.2 \times 10^{-2}\,\text{mol dm}^{-3}\text{s}^{-1} at [A]=0.4 mol dm−3[A] = 0.4\,\text{mol dm}^{-3} and [B]=0.1 mol dm−3[B] = 0.1\,\text{mol dm}^{-3}. The reaction is first order in A and second order in B. Calculate rate constant.

[Q75 · 16th May Shift 1 · 2023]

Squaring the wrong concentration

r=k[A][B]2r = k[A][B]^2 squares [B][B]. With [A]=1[A] = 1 and [B]=0.2[B] = 0.2, squaring the wrong one turns 0.250.25 into 1.251.25 — and both are on the list, because the two rate laws appear on twin stems.
Drill 13 more on the rate constant: k = rate/([a]ˣ[b]ʸ), its units and its properties

Concept 3 of 4

How the Rate Changes When Concentrations Change: Multiply the Factors

Intuition

Scale each concentration by its factor raised to its order and multiply. Doubling a first-order reactant doubles the rate; doubling a second-order one quadruples it; halving one and doubling another with equal orders leaves it unchanged.

Definition

  • r=k[A][B]2r = k[A][B]^2: doubling [A][A] → ×2\times 2; doubling both → ×8\times 8; doubling A, B and a zero-order C → still ×8\times 8.
  • r=k[A][B]r = k[A][B]: doubling both → ×4\times 4; doubling [A][A] alone → ×2\times 2; doubling [B][B] and halving [A][A] → ×1\times 1; halving [B][B] alone → ×12\times \tfrac12.
  • 'Which change does NOT affect the rate' is the one whose factors multiply to 11.

Rate factor

r′r=([A]′[A])x([B]′[B])y\frac{r'}{r} = \left(\frac{[A]'}{[A]}\right)^x\left(\frac{[B]'}{[B]}\right)^y

Worked example

For r=k[A]2[B]r = k[A]^2[B], by what factor does the rate change if [A][A] is tripled and [B][B] is halved?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Chemical KineticsEASY
For the reaction, CH3Br(aq)+OH−(aq)→CH3OH(aq)+Br−(aq)\text{CH}_3\text{Br(aq)}+\text{OH}^-\text{(aq)}\to\text{CH}_3\text{OH(aq)}+\text{Br}^-\text{(aq)}, The rate law is rate =k[CH3Br][OH−]=k[\text{CH}_3\text{Br}][\text{OH}^-]. What is change in rate of reaction if concentration of both reactants is doubled?

[Q61 · 13th May Shift 2 · 2024]

Adding the factors

Doubling A and B in r=k[A][B]2r = k[A][B]^2 gives 2×4=82 \times 4 = 8, not 2+4=62 + 4 = 6. Factors multiply because the rate law is a product.
Drill 5 more on how the rate changes when concentrations change: multiply the factors

Concept 4 of 4

Order Versus Molecularity

Intuition

Molecularity counts the species that collide in ONE elementary step — always a positive whole number, never zero or fractional — and is a theoretical idea. Order is measured from the rate law and can be anything. For an elementary reaction the two coincide.

Definition

  • O3+O→2O2\text{O}_3 + \text{O} \to 2\text{O}_2, elementary, r=k[O3][O]r = k[\text{O}_3][\text{O}]: order 22, molecularity 22. NO2+NO2→2NO+O2\text{NO}_2 + \text{NO}_2 \to 2\text{NO} + \text{O}_2: order 22, molecularity 22.
  • C2H5I→C2H4+HI\text{C}_2\text{H}_5\text{I} \to \text{C}_2\text{H}_4 + \text{HI}, r=k[C2H5I]r = k[\text{C}_2\text{H}_5\text{I}]: order 11, molecularity 11 (unimolecular).
  • H2+Br2→2HBr\text{H}_2 + \text{Br}_2 \to 2\text{HBr}, r=k[H2][Br2]1/2r = k[\text{H}_2][\text{Br}_2]^{1/2}: bimolecular, order 32\tfrac32 — the standard example of the two differing.
  • For a complex (multistep) reaction, molecularity is defined per step and the overall order comes from the slow step.

The contrast

molecularity∈{1,2,3} (theoretical, per step);order∈R≥0 (experimental)\text{molecularity} \in \{1, 2, 3\} \text{ (theoretical, per step)};\qquad \text{order} \in \mathbb{R}_{\ge 0} \text{ (experimental)}

Worked example

For the elementary reaction 2NO+O2→2NO22\text{NO} + \text{O}_2 \to 2\text{NO}_2, state the molecularity, the rate law and the order.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Chemical KineticsEASY
What is the order and molecularity respectively for the elementary reaction given below? O3( g)+O(g)⟶2O2( g)O_{3(\text{ }g)}+O_{(g)}\longrightarrow 2O_{2(\text{ }g)} if r=k[O3][O]r = k\left\lbrack O_{3} \right\rbrack\lbrack O\rbrack

[Q57 · 21 April Shift I · 2025]

Calling H₂ + Br₂ 'monomolecular' because the order is fractional

Two molecules collide, so it is bimolecular whatever the order. Options pairing 'monomolecular' with 32\tfrac32 are the planted confusion of the two ideas.
Drill 4 more on order versus molecularity

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