Order of Reaction
from Rate Data

Four experiments with varying [A] and [B] and their initial rates

The rate law

Compare pairs where only one concentration changes.

Rate₂/Rate₁ = ([X]₂/[X]₁)^order

1. Order in B — compare II and III, where [A] is fixed at 0.3. [B] doubles from 0.2 to 0.4.
2. Rate goes 7.2 × 10⁻² → 2.88 × 10⁻¹, a factor of 4. Doubling gave quadrupling, so order in B = 2.
3. Order in A — compare I and IV, where [B] is fixed at 0.1. [A] goes 0.1 → 0.4, a factor of 4.
4. Rate goes 6.0 × 10⁻³ → 2.40 × 10⁻², also a factor of 4. So order in A = 1.
5. Rate law: Rate = k[A][B]², overall order 3.

Rate = k[A][B]², and experiment I: [A]=0.1, [B]=0.1, rate=6.0×10⁻³

k with units

Substitute one full row into the established rate law.

k = Rate/([A][B]²) ; units = (mol L⁻¹)^(1−n)·time⁻¹

1. k = 6.0 × 10⁻³ / (0.1 × 0.1²).
2. Denominator = 0.1 × 0.01 = 1.0 × 10⁻³.
3. k = 6.0 × 10⁻³ / 1.0 × 10⁻³ = 6.0.
4. Overall order n = 3, so units = (mol L⁻¹)^(1−3) min⁻¹ = mol⁻² L² min⁻¹.

Three experiments with the stated concentrations and initial rates

Orders in A and B

First identify any reactant whose change does nothing — that one is zero order.

Rate₂/Rate₁ = ([X]₂/[X]₁)^order

1. Compare experiments 1 and 2: [A] is fixed at 0.20 while [B] falls from 0.30 to 0.10, yet the rate is identical.
2. A change in [B] producing no change in rate means order in B = 0.
3. Now compare experiments 2 and 3: [A] doubles from 0.20 to 0.40. [B] also changes, but since B is zero order that is irrelevant.
4. Rate goes 5.07 × 10⁻⁵ → 1.43 × 10⁻⁴, a factor of 2.82.
5. 2^x = 2.82 gives x = 1.5. So order in A = 1.5, order in B = 0, overall order 1.5.

Rate = k[A]; experiment I gives k; experiment II rate = 4.0×10⁻²

[A] in experiment II

Since B is zero order, its concentration is irrelevant — the rate depends on [A] alone.

Rate = k[A]

1. From experiment I: k = Rate/[A] = 2.0 × 10⁻²/0.1 = 0.2 min⁻¹.
2. For experiment II: [A] = Rate/k = 4.0 × 10⁻²/0.2.
3. [A] = 0.2 mol L⁻¹.
4. The change in [B] from 0.1 to 0.2 is a deliberate distraction and plays no part.

k = 0.2 min⁻¹, [A] = 0.4, [B] = 0.4, Rate = k[A]

initial rate

Substitute into the rate law, ignoring the zero-order species.

Rate = k[A]

1. Rate = 0.2 × 0.4.
2. = 0.08 = 8.0 × 10⁻² mol L⁻¹ min⁻¹.
3. Option (b) is what you get by wrongly including [B] as first order — the intended trap.

Rate = k[A][B]², k = 2.0 × 10⁻⁶, [A] = 0.1, [B] = 0.2

initial rate

Straight substitution — the only care needed is squaring the right term.

Rate = k[A][B]²

1. [B]² = (0.2)² = 0.04.
2. Rate = 2.0 × 10⁻⁶ × 0.1 × 0.04.
3. = 2.0 × 10⁻⁶ × 4.0 × 10⁻³.
4. = 8.0 × 10⁻⁹ mol L⁻¹ s⁻¹.

2A + B → A₂B, initially [A]=0.1, [B]=0.2; now [A]=0.06

new rate

This is the stoichiometry trap — as A is consumed, B is consumed too, in the ratio given by the equation.

Rate = k[A][B]² ; ΔB = ΔA/2 from the coefficients

1. A has fallen by 0.1 − 0.06 = 0.04 mol L⁻¹.
2. The equation 2A + B → A₂B says two A are used per one B, so B falls by half as much: 0.04/2 = 0.02.
3. New [B] = 0.2 − 0.02 = 0.18 mol L⁻¹.
4. Rate = 2.0 × 10⁻⁶ × 0.06 × (0.18)² = 2.0 × 10⁻⁶ × 0.06 × 0.0324.
5. = 3.89 × 10⁻⁹ mol L⁻¹ s⁻¹.

Rate = k[A][B]², [B] tripled

Rate multiplier

The multiplier is the factor raised to that reactant's order.

Rate multiplier = (factor)^order

1. Order in B is 2 and the factor is 3.
2. Multiplier = 3² = 9.
3. So the rate increases nine-fold.

Rate = k[A][B]², both doubled

Rate multiplier

Contributions from separate reactants multiply.

multiplier = (f_A)^x × (f_B)^y

1. From A: 2¹ = 2.
2. From B: 2² = 4.
3. Total multiplier = 2 × 4 = 8.
4. The rate increases eight-fold.

Rate = k[X]², [X] tripled

Effect on rate of formation of Y

Second order means the rate goes as the square of the concentration.

Rate = k[X]²

1. Multiplier = 3² = 9.
2. So the rate of formation of Y increases nine times.
3. Note that 'second order kinetics' with a single reactant means second order in that reactant.

r = k[A]^½[B]²

Overall order

Overall order is simply the sum of the exponents, fractions included.

order = x + y

1. Exponent of A = ½ = 0.5.
2. Exponent of B = 2.
3. Sum = 0.5 + 2 = 2.5.

Rate = k[A]^{3/2}[B]⁻¹

Overall order

Negative exponents are subtracted, not ignored.

order = x + y

1. Exponent of A = 3/2.
2. Exponent of B = −1.
3. Sum = 3/2 + (−1) = 3/2 − 1 = 1/2.
4. So the reaction is half order overall — B actually inhibits it.

Rate = k[NO]² for 3NO → N₂O

Order

Read the exponent from the rate law, not the coefficient from the equation.

order = exponent in the experimental rate law

1. The coefficient of NO in the balanced equation is 3, which is a distraction.
2. The experimentally determined exponent is 2.
3. Therefore the order is 2, not 3.
4. This is a direct illustration that order and stoichiometry need not agree.

Rate = k[CHCl₃][Cl₂]^½

Overall order

Sum the exponents; note this is NCERT's own example of stoichiometry failing to predict order.

order = x + y

1. Exponent of CHCl₃ = 1.
2. Exponent of Cl₂ = ½.
3. Overall order = 1 + 0.5 = 1.5.
4. The balanced equation would have suggested 2, so this reaction proves the rule that order must be measured.

Rate = k[ester]¹[H₂O]⁰

Overall order

A zero exponent contributes nothing to the sum.

order = 1 + 0

1. Exponent of the ester = 1.
2. Exponent of water = 0, so water contributes nothing.
3. Overall order = 1.
4. This is a pseudo first order reaction: truly second order, but water's vast excess makes its concentration effectively constant.

Concentration ×2, rate ×2.83

Order

2.83 is not a whole power of 2 — it is 2^1.5.

rate ratio = (conc ratio)^order

1. Set 2^n = 2.83.
2. Take logs: n = log 2.83/log 2 = 0.4518/0.3010.
3. n = 1.5.
4. Confirm: 2^1.5 = 2 × √2 = 2 × 1.414 = 2.83. Correct.

Concentration ×3, rate ×5.2

Order

Recognise 5.2 as 3^1.5.

3^n = 5.2

1. n = log 5.2/log 3 = 0.716/0.477.
2. n = 1.5.
3. Check: 3^1.5 = 3√3 = 3 × 1.732 = 5.196 ≈ 5.2.

Rate = k[A]²[B], volume doubled

New rate as a fraction of the old

Doubling the volume halves every concentration simultaneously.

multiplier = (½)^x × (½)^y = (½)^(x+y)

1. Doubling the volume halves both [A] and [B].
2. Contribution from A: (½)² = ¼.
3. Contribution from B: (½)¹ = ½.
4. Total multiplier = ¼ × ½ = 1/8.
5. Equivalently, (½)^(overall order 3) = 1/8.

Second order, concentration halved

Rate multiplier

The multiplier is the factor raised to the order, including factors below one.

multiplier = (½)²

1. Factor = ½ and order = 2.
2. Multiplier = (½)² = ¼.
3. So the rate falls to one-fourth.

Rate = k[A][B], [A] ×2 and [B] ×½

Effect on rate

Opposing changes multiply and can cancel.

multiplier = 2¹ × (½)¹

1. From A: 2¹ = 2.
2. From B: (½)¹ = 0.5.
3. Total = 2 × 0.5 = 1.
4. So the rate is unchanged.

[A] = 0.31 at 30 s and 0.17 at 60 s

Average rate over that interval

Average rate over an interval is the concentration drop divided by the time taken.

r_av = −Δ[A]/Δt

1. Δ[A] = 0.17 − 0.31 = −0.14 mol L⁻¹.
2. Δt = 60 − 30 = 30 s.
3. r_av = −(−0.14)/30 = 0.14/30.
4. = 4.67 × 10⁻³ mol L⁻¹ s⁻¹.

Four quantities

Which requires experiment

Order is empirical; the others are read off or defined by the equation itself.

Rate = k[A]ˣ[B]ʸ with x, y measured

1. Stoichiometric coefficients and atom counts are read directly from the balanced equation.
2. Molecularity of an elementary step is defined by that step's own equation once the mechanism is known.
3. Order requires measuring how the rate responds to concentration changes.
4. So only order is genuinely experimental.

Rate = k[A]⁰

The correct interpretation

Zero order means constant rate, not zero rate and not constant concentration.

[A]⁰ = 1, so Rate = k

1. Anything to the power zero equals 1, so Rate = k × 1 = k.
2. The rate is constant, which is not the same as being zero — option (b) is a misreading.
3. Since the rate does not depend on [A], option (c) is wrong.
4. The concentration of A still falls steadily with time — it falls linearly. So (d) is wrong too.

Rate ∝ [A], independent of [B]

Rate law

Independence means a zero exponent, regardless of the coefficient 2.

Rate = k[A]¹[B]⁰

1. Proportional to [A] means the exponent of A is 1.
2. Independent of [B] means the exponent of B is 0.
3. Rate = k[A]¹[B]⁰ = k[A].
4. The coefficient 2 in the equation is deliberately misleading and must be ignored.

Overall order = 2

Units of k

Apply the general unit rule.

units of k = (mol L⁻¹)^(1−n)·time⁻¹

1. Put n = 2: (mol L⁻¹)^(1−2) = (mol L⁻¹)⁻¹.
2. (mol L⁻¹)⁻¹ = mol⁻¹ L.
3. So the units are mol⁻¹ L s⁻¹, equivalently L mol⁻¹ s⁻¹.

Effects of doubling each of A, B and C in turn

Overall order

Read each order from its own doubling result, then add.

order = x + y + z

1. Doubling A doubles the rate: 2¹ = 2, so x = 1.
2. Doubling B does nothing: 2⁰ = 1, so y = 0.
3. Doubling C gives eight times: 2³ = 8, so z = 3.
4. Overall order = 1 + 0 + 3 = 4.

Concentration ×2, rate ×¼

Order

A falling rate on rising concentration means a negative order.

2^n = 1/4

1. Set 2^n = 1/4.
2. 1/4 = 2⁻², so n = −2.
3. A negative order means the species inhibits the reaction — commonly by blocking a catalyst or by shifting an equilibrium in a pre-step.

Concentration ×4, rate ×2

Order

A rate change smaller than the concentration change means a fractional order below 1.

4^n = 2

1. Set 4^n = 2.
2. Since 4 = 2², this becomes 2^(2n) = 2¹.
3. Equating exponents: 2n = 1, so n = 0.5.
4. This is a half order reaction.

Order 3/2, pressure in bar, time in minutes

Units of k

The same unit rule applies with bar replacing mol L⁻¹.

units of k = (pressure)^(1−n)·time⁻¹

1. The rate itself has units bar min⁻¹.
2. k = Rate/(pressure)^{3/2}, so units = bar min⁻¹/bar^{3/2}.
3. = bar^(1 − 3/2) min⁻¹ = bar^{−1/2} min⁻¹.

Rate = k[H₂O₂][I⁻]

Overall order

Only species appearing in the rate law contribute; H⁺ does not appear at all.

order = sum of exponents

1. Exponent of H₂O₂ = 1.
2. Exponent of I⁻ = 1.
3. H⁺ does not appear, so its exponent is 0.
4. Overall order = 1 + 1 + 0 = 2, despite the equation summing to six reactant particles.

Order = 3/2

Units of k

Apply the general rule with a fractional n.

units of k = (mol L⁻¹)^(1−n)·s⁻¹

1. 1 − n = 1 − 3/2 = −1/2.
2. (mol L⁻¹)^{−1/2} = mol^{−1/2} L^{1/2}.
3. So the units are mol^{−1/2} L^{1/2} s⁻¹.

Rate = k[C₂H₅Cl]

Order and units of k

Read the exponent, then apply the unit rule.

order = 1 ; units = (mol L⁻¹)^0 s⁻¹

1. The exponent is 1, so the reaction is first order.
2. 1 − n = 1 − 1 = 0.
3. (mol L⁻¹)⁰ = 1, so the units reduce to s⁻¹.

Rate independent of every reactant concentration

Overall order

Every exponent must be zero, so the overall order is zero.

Rate = k[A]⁰[B]⁰ = k

1. If no concentration affects the rate, every exponent must be zero.
2. The overall order is the sum of those exponents, which is zero.
3. The rate then equals k, a constant.
4. Such reactions are real — NCERT cites ammonia decomposition on hot platinum at high pressure.

[A]: 0.1 → 0.4 (×4); rate: 1.0 × 10⁻³ → 1.6 × 10⁻² (×16)

Order in A

Express both ratios and match the power.

4^n = 16

1. Concentration ratio = 0.4/0.1 = 4.
2. Rate ratio = 1.6 × 10⁻²/1.0 × 10⁻³ = 16.
3. 4^n = 16 = 4², so n = 2.

Rate = k[A]²[B]^0.5, [A] ×½, [B] ×4

Rate multiplier

Compute each contribution and multiply.

multiplier = (½)² × 4^0.5

1. From A: (½)² = 0.25.
2. From B: 4^0.5 = √4 = 2.
3. Multiplier = 0.25 × 2 = 0.5.
4. So the rate is halved.

Candidate order values

The impossible one

Order is experimental and unrestricted; it is molecularity that is restricted to 1, 2 or 3.

—

1. Zero order is well documented, for instance ammonia on platinum.
2. Fractional orders such as 1.5 arise, as in acetaldehyde decomposition.
3. Negative orders occur when a species inhibits the reaction.
4. So all the listed values are possible, and option (a) is correct. The restriction to 1, 2 or 3 belongs to molecularity, not order.

Rate = 2 × 10⁻³ mol L⁻¹ s⁻¹, [A] = [B] = 0.1 M, first order in each

k with units

Substitute and derive the units from the overall order.

k = Rate/([A][B])

1. [A][B] = 0.1 × 0.1 = 0.01.
2. k = 2 × 10⁻³/0.01 = 0.2.
3. Overall order = 2, so units = (mol L⁻¹)⁻¹ s⁻¹ = L mol⁻¹ s⁻¹.
4. k = 0.2 L mol⁻¹ s⁻¹.

Doubling concentration quadruples rate, single reactant

Applicable integrated rate law

First find the order, then recall the matching integrated form.

2^n = 4 ⟹ n = 2 ; second order integrated law

1. Doubling gave quadrupling, so 2^n = 4 and n = 2.
2. The reaction is second order.
3. The second order integrated law is 1/[A] − 1/[A]₀ = kt.
4. Options (b) and (d) are first order forms, and (c) is the zero order form.

Four proposed changes

Which leaves k unchanged

k depends on temperature and catalyst, never on concentration.

k = A e^(−Ea/RT)

1. Doubling concentrations changes the rate but leaves k untouched, since k has no concentration dependence.
2. Raising or lowering the temperature changes k through the Arrhenius relation.
3. A catalyst lowers Ea and so raises k.
4. So only option (a) leaves k unaltered.

Slope = 0.5, intercept = −1 on a log–log plot

k

The intercept of a log–log plot is log k.

log(Rate) = log k + x·log[A]

1. Comparing with y = mx + c, the intercept c equals log k.
2. So log k = −1.
3. k = 10⁻¹ = 0.1.
4. The slope 0.5 separately tells us the order is half.

Rate = k[A], four candidate data sets

The consistent data set

First order means the rate must track concentration in exact proportion.

Rate ∝ [A]¹

1. In each option the concentration doubles at each step.
2. First order requires the rate to double too.
3. Option (a) shows 1 → 2 → 4, exactly doubling. Correct.
4. Option (b) quadruples (second order), (c) is constant (zero order), (d) multiplies by 8 (third order).

Order = 3/2, concentration ×4

Rate multiplier

Raise the factor to the fractional order.

multiplier = 4^{3/2}

1. 4^{3/2} = (√4)³.
2. √4 = 2, and 2³ = 8.
3. So the rate increases eight-fold.

A reactant shows zero order

Valid conclusion

Zero order means the rate does not respond to that concentration — usually because of excess or saturation.

Rate = k[A]ˣ[B]⁰

1. The reactant is still consumed; it simply does not influence the rate. So (b) is wrong.
2. It does appear in the balanced equation, so (c) is wrong.
3. The reaction proceeds normally, so (d) is wrong.
4. The usual physical causes are a vast excess (as with water in ester hydrolysis) or a saturated catalyst surface (as with ammonia on platinum). So (a) is correct.

Rate = k[A]²[B], [A] ×3, [B] ×2

Rate multiplier

Multiply each reactant's contribution.

multiplier = 3² × 2¹

1. From A: 3² = 9.
2. From B: 2¹ = 2.
3. Total = 9 × 2 = 18 times.