Chemical Kinetics — Rate Laws & Integrated Forms

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Thermodynamics tells you whether a reaction can happen. Kinetics tells you how fast it happens, and almost every exam numerical in this chapter comes from four equations: the rate law, the first-order integrated law, the half-life, and the Arrhenius equation.

What you will be able to do

• Write a rate law, find order from initial-rate data, and give the units of k for any order
• Use the integrated laws and half-life expressions for zero- and first-order reactions
• Solve first-order numericals: percentage completion, t½, gas-phase pressure data
• Use the Arrhenius equation to find Ea or k at a new temperature
• Explain collision theory, the role of a catalyst and the rate-determining step

The big picture

Four equations answer almost every question in this chapter.

In one line: The rate law says how rate depends on concentration; the integrated law turns that into concentration against time; the Arrhenius equation says how the rate constant depends on temperature.

1. Rate law: rate = k[A]ˣ[B]ʸ — found by experiment, not from the equation.
2. Integrated law: how much reactant is left after time t.
3. Half-life: the time for half the reactant to go.
4. Arrhenius: k rises steeply with temperature because more molecules have enough energy.

Think of it like this: Picture a crowd leaving a stadium. If one gate lets out a fixed number per minute whatever the crowd size, that is zero order. If a fixed fraction of whoever is left leaves each minute, that is first order — the crowd halves in the same time again and again, however big it was.

Rate and rate law

How fast, measured how.

For aA + bB → cC + dD the rate is written so that it is the same whichever species you watch:

Rate = −(1/a) d[A]/dt = −(1/b) d[B]/dt = +(1/c) d[C]/dt = +(1/d) d[D]/dt

Minus sign for reactants (their concentration falls); units mol L⁻¹ s⁻¹

Solved example 01 — Relating rates

For 2N₂O₅ → 4NO₂ + O₂, the rate of formation of NO₂ is 0.04 mol L⁻¹ s⁻¹. Find the rate of disappearance of N₂O₅ and the rate of reaction.

1. Rate of reaction = (1/4) d[NO₂]/dt = 0.04/4 = 0.01 mol L⁻¹ s⁻¹.
2. −d[N₂O₅]/dt = 2 × rate = 0.02 mol L⁻¹ s⁻¹.

Answer: N₂O₅ disappears at 0.02 mol L⁻¹ s⁻¹; rate of reaction 0.01 mol L⁻¹ s⁻¹.

Rate = k [A]ˣ [B]ʸ
Order = x + y

x and y come from experiment; they need not equal a and b

Key idea: Units of k depend on order: (mol L⁻¹)⁽¹⁻ⁿ⁾ s⁻¹. Zero order: mol L⁻¹ s⁻¹. First order: s⁻¹. Second order: L mol⁻¹ s⁻¹. Third order: L² mol⁻² s⁻¹. Read the units and you know the order.

Finding order from initial rates

Solved example 02 — Method of initial rates

For A + B → products: (1) [A] = 0.1, [B] = 0.1, rate = 2.0 × 10⁻³; (2) [A] = 0.2, [B] = 0.1, rate = 8.0 × 10⁻³; (3) [A] = 0.1, [B] = 0.2, rate = 4.0 × 10⁻³ (mol L⁻¹ s⁻¹). Find the rate law and k.

1. Experiments 1 → 2: [A] doubles, rate × 4 = 2² → order 2 in A.
2. Experiments 1 → 3: [B] doubles, rate × 2 → order 1 in B.
3. k = 2.0 × 10⁻³ / (0.1² × 0.1) = 2.0.

Answer: Rate = k[A]²[B]; overall order 3; k = 2.0 L² mol⁻² s⁻¹.

Order and molecularity

Two numbers students mix up every year.

	Order	Molecularity
Definition	Sum of powers in the experimental rate law	Number of particles colliding in one elementary step
Found by	Experiment	Proposed mechanism
Values	0, 1, 2, 3 or a fraction	Whole number 1, 2 or 3 (never 0)
Applies to	Overall reaction	Each elementary step only

• Pseudo-first-order: a second-order reaction looks first order when one reactant (often water) is in huge excess. Examples: acid hydrolysis of an ester (CH₃COOC₂H₅ + H₂O), inversion of cane sugar.
• Rate-determining step: in a multi-step mechanism the slowest step fixes the rate law.
• Molecularity above 3 is almost never seen: four particles colliding at once is too unlikely.

Zero-order reactions

A constant rate, independent of concentration.

[A] = [A]₀ − kt
t½ = [A]₀ / 2k

Rate = k; units of k: mol L⁻¹ s⁻¹

• Examples: decomposition of NH₃ on a hot platinum surface (surface saturated), the H₂ + Cl₂ photochemical reaction, some enzyme reactions at high substrate concentration.
• The reaction finishes completely at t = [A]₀/k.
• Half-life is proportional to the starting concentration.

First-order reactions

The most examined section of the chapter.

k = (2.303 / t) log ( [A]₀ / [A] )

Equivalently ln[A] = ln[A]₀ − kt, or [A] = [A]₀ e⁻ᵏᵗ

t½ = 0.693 / k

Independent of starting concentration

Key idea: Shortcuts worth memorising: after n half-lives, the fraction left is (1/2)ⁿ. 75% complete = 2 half-lives; 87.5% = 3; 93.75% = 4. t(99%) = 2 × t(90%), and t(99.9%) ≈ 10 × t½.

• Examples: radioactive decay, decomposition of N₂O₅, H₂O₂, and many gas-phase decompositions.
• Gas-phase pressure data for A(g) → B(g) + C(g): with initial pressure p₀ and total pressure pₜ, k = (2.303/t) log [p₀ / (2p₀ − pₜ)].

Solved example 03 — Half-life from k

A first-order reaction has k = 1.155 × 10⁻³ s⁻¹. Find t½ and the time for 75% completion.

1. t½ = 0.693 / (1.155 × 10⁻³) = 600 s.
2. 75% complete means two half-lives.

Answer: t½ = 600 s (10 min); 75% completion takes 1200 s (20 min).

Solved example 04 — Percentage completion

A first-order reaction is 20% complete in 10 min. How long for 75% completion? (log 1.25 = 0.0969, log 4 = 0.602)

1. 20% complete: [A]₀/[A] = 100/80 = 1.25. k = (2.303/10) × 0.0969 = 0.0223 min⁻¹.
2. 75% complete: [A]₀/[A] = 4. t = (2.303/0.0223) × 0.602.

Answer: About 62 min.

Solved example 05 — Gas-phase decomposition

For A(g) → B(g) + C(g), first order, p₀ = 0.40 atm and after 100 s the total pressure is 0.60 atm. Find k. (log 2 = 0.301)

1. p(A) left = 2p₀ − pₜ = 0.80 − 0.60 = 0.20 atm.
2. k = (2.303/100) log (0.40/0.20) = 0.02303 × 0.301.

Answer: k ≈ 6.93 × 10⁻³ s⁻¹ (so t½ = 100 s — the pressure of A has exactly halved).

Check yourself

• C1. The rate constant of a reaction is 3 × 10⁻⁴ s⁻¹. What is its order?
• C2. A first-order reaction has t½ = 20 min. What fraction remains after 1 hour?

Answers at the end.

Reading the graphs

Plot the right thing and get a straight line.

Order	Straight-line plot	Slope	Half-life vs [A]₀
Zero	[A] against t	−k	t½ ∝ [A]₀
First	ln[A] against t	−k	Independent
First	log[A] against t	−k / 2.303	Independent
First	log([A]₀/[A]) against t	+k / 2.303	—
Second (one reactant)	1/[A] against t	+k	t½ ∝ 1/[A]₀

Key idea: Rate against concentration: zero order gives a horizontal line; first order gives a straight line through the origin with slope k.

Temperature and the Arrhenius equation

Why a 10 K rise can double the rate.

k = A · exp(−Ea / RT)
log k = log A − Ea / (2.303 RT)

A = frequency (pre-exponential) factor; Ea = activation energy; plot log k vs 1/T: slope = −Ea / 2.303R

log (k₂ / k₁) = [Ea / (2.303 R)] × [(T₂ − T₁) / (T₁ T₂)]

The two-temperature form used in nearly every numerical

• Temperature coefficient: k(T + 10)/k(T) is usually about 2 to 3 near room temperature.
• The factor exp(−Ea/RT) is the fraction of collisions with energy ≥ Ea. A small rise in T greatly increases this fraction — that, not collision frequency, is why rate rises so fast.
• Higher Ea → rate more sensitive to temperature.

Solved example 06 — Activation energy from two temperatures

The rate constant doubles when the temperature rises from 298 K to 308 K. Find Ea. (R = 8.314 J K⁻¹ mol⁻¹, log 2 = 0.301)

1. log 2 = [Ea / (2.303 × 8.314)] × [10 / (298 × 308)].
2. Ea = 0.301 × 2.303 × 8.314 × 298 × 308 / 10.
3. = 0.301 × 19.147 × 9178.4 J mol⁻¹.

Answer: Ea ≈ 52.9 kJ mol⁻¹.

Collision theory and catalysts

What has to happen for molecules to react.

• A reaction happens only when molecules collide with energy at least Ea and in the right orientation.
• Rate = P × Z × exp(−Ea/RT), where Z is collision frequency and P the steric (orientation) factor.
• Effective collisions are a tiny fraction of all collisions.

Key idea: A catalyst provides a new path with lower Ea. It speeds the forward and backward reactions equally, so it does not change ΔH, ΔG or the equilibrium constant K — it only helps equilibrium arrive sooner. It is unchanged in mass and composition at the end.

Check yourself

• C3. Does a catalyst change the value of the equilibrium constant?
• C4. Why does a 10 K rise often double the rate though collision frequency rises only about 2%?

Answers at the end.

Exam trap gallery

Where marks are lost in kinetics.

#	What students write	What is actually correct	Why
1	Order = stoichiometric coefficients.	Order comes from experiment.	The balanced equation does not show the mechanism.
2	Molecularity can be zero or fractional.	Molecularity is a whole number 1, 2 or 3; order can be zero or fractional.	Molecularity counts colliding particles.
3	t½ of a first-order reaction depends on starting concentration.	It is independent: t½ = 0.693/k.	Only zero- and second-order half-lives depend on [A]₀.
4	A catalyst shifts the equilibrium to the right.	A catalyst does not change K.	It lowers Ea for both directions equally.
5	Rate rises with T mainly because molecules collide more often.	Mainly because more collisions have energy ≥ Ea.	The exp(−Ea/RT) term changes far faster than Z.
6	Using ln with 2.303.	k = (2.303/t) log(…) or k = (1/t) ln(…).	Use 2.303 only with log (base 10).
7	75% completion takes 1.5 half-lives.	It takes 2 half-lives (50% then half of the remaining 50%).	First-order decay is geometric, not linear.
8	Pseudo-first-order reactions are truly first order.	They are higher order; one reactant is in large excess.	Its concentration hardly changes, so it hides in k.

Memory hook: “Units tell order: s⁻¹ is one, mol L⁻¹ s⁻¹ is zero, L mol⁻¹ s⁻¹ is two.” · “Point six nine three over k” — first-order half-life. · “Ninety-nine is twice ninety” — t(99%) = 2 t(90%). · “Catalyst: faster, not further.”

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The unit of k for a first-order reaction is: (Easy)

• (a) mol L⁻¹ s⁻¹
• (b) s⁻¹
• (c) L mol⁻¹ s⁻¹
• (d) L² mol⁻² s⁻¹

Q02. The half-life of a first-order reaction is: (Easy)

• (a) proportional to [A]₀
• (b) inversely proportional to [A]₀
• (c) independent of [A]₀
• (d) proportional to [A]₀²

Q03. Molecularity of a reaction cannot be: (Easy)

• (a) 1
• (b) 2
• (c) 3
• (d) 0

Q04. A catalyst increases the rate by: (Easy)

• (a) increasing Ea
• (b) decreasing Ea
• (c) increasing ΔH
• (d) changing K

Q05. For a zero-order reaction, the plot that is a straight line is: (Easy)

• (a) [A] vs t
• (b) ln[A] vs t
• (c) 1/[A] vs t
• (d) log k vs 1/T

Q06. For 2A → B, if −d[A]/dt = 0.2 mol L⁻¹ s⁻¹, then d[B]/dt is: (Moderate)

• (a) 0.4
• (b) 0.2
• (c) 0.1
• (d) 0.05

Q07. A first-order reaction has t½ = 30 min. The fraction remaining after 90 min is: (Moderate)

• (a) 1/3
• (b) 1/4
• (c) 1/8
• (d) 1/9

Q08. Doubling [A] quadruples the rate. The order in A is: (Moderate)

• (a) 0
• (b) 1
• (c) 2
• (d) 4

Q09. The slope of log k against 1/T is: (Moderate)

• (a) −Ea/R
• (b) −Ea/2.303R
• (c) Ea/2.303R
• (d) −2.303Ea/R

Q10. Acid hydrolysis of ethyl acetate is: (Moderate)

• (a) zero order
• (b) pseudo-first order
• (c) true second order
• (d) third order

Q11. For a zero-order reaction, t½ is: (Moderate)

• (a) 0.693/k
• (b) [A]₀/2k
• (c) 1/k[A]₀
• (d) 2k/[A]₀

Q12. If t(90%) of a first-order reaction is 20 min, t(99%) is: (Moderate)

• (a) 22 min
• (b) 30 min
• (c) 40 min
• (d) 200 min

Q13. Total pressure for A(g) → B(g) + C(g) after one half-life, starting from p₀, is: (Tough)

• (a) p₀
• (b) 1.5 p₀
• (c) 2 p₀
• (d) 0.5 p₀

Q14. Ea of a reaction is zero. Its rate constant: (Tough)

• (a) increases with T
• (b) decreases with T
• (c) is independent of T
• (d) is zero

Q15. Rate = k[A]⁰·⁵[B]¹·⁵ — the overall order is: (Tough)

• (a) 1
• (b) 1.5
• (c) 2
• (d) 3

Answer key with explanations

1. (b) s⁻¹ — (mol L⁻¹)⁰ s⁻¹.
2. (c) independent of [A]₀ — t½ = 0.693/k.
3. (d) 0 — At least one particle must take part.
4. (b) decreasing Ea — Lower-energy pathway.
5. (a) [A] vs t — [A] = [A]₀ − kt.
6. (c) 0.1 — Half as fast as A disappears.
7. (c) 1/8 — Three half-lives: (1/2)³.
8. (c) 2 — 2ˣ = 4.
9. (b) −Ea/2.303R — log k = log A − Ea/2.303RT.
10. (b) pseudo-first order — Water is in large excess.
11. (b) [A]₀/2k — Half of [A]₀ used at constant rate k.
12. (c) 40 min — t(99%) = 2 × t(90%).
13. (b) 1.5 p₀ — p(A) = p₀/2; p(B) + p(C) = p₀ → total 1.5p₀.
14. (c) is independent of T — k = A when Ea = 0.
15. (c) 2 — 0.5 + 1.5 = 2.

Advanced problems

Each needs two ideas from the chapter.

Solved example 07 — k at a new temperature

For a reaction Ea = 60 kJ mol⁻¹ and k = 2.0 × 10⁻³ s⁻¹ at 300 K. Estimate k at 310 K. (R = 8.314; antilog 0.337 ≈ 2.17)

1. log(k₂/k₁) = [60000 / (2.303 × 8.314)] × [10 / (300 × 310)].
2. = 3133.6 × 1.075 × 10⁻⁴ ≈ 0.337.
3. k₂/k₁ ≈ 2.17.

Answer: k ≈ 4.3 × 10⁻³ s⁻¹.

Solved example 08 — Radioactive decay

The half-life of ¹⁴C is 5730 years. How old is a wooden sample with 25% of the ¹⁴C of living wood?

1. Radioactive decay is first order.
2. 25% left = (1/2)² → two half-lives.

Answer: About 11 460 years.

Check-yourself answers

• C1. First order (units s⁻¹).
• C2. Three half-lives → 1/8 remains.
• C3. No — it speeds both directions equally; K is unchanged.
• C4. Because the fraction of collisions with energy ≥ Ea roughly doubles; collision frequency barely changes.

One-page revision

Read this page the night before the exam.

Rate and rate law

• Rate = −(1/a)d[A]/dt = +(1/c)d[C]/dt
• Rate = k[A]ˣ[B]ʸ; order = x + y
• Order from experiment only
• k units: (mol L⁻¹)⁽¹⁻ⁿ⁾ s⁻¹

Order vs molecularity

• Order: 0, fraction, any; overall
• Molecularity: 1, 2, 3; elementary step
• Pseudo-1st: ester hydrolysis, sucrose
• Slowest step decides the rate law

Zero order

• [A] = [A]₀ − kt
• t½ = [A]₀/2k ∝ [A]₀
• NH₃ on hot Pt
• [A] vs t: slope −k

First order

• k = (2.303/t) log([A]₀/[A])
• t½ = 0.693/k, independent of [A]₀
• t99 = 2 t90; t99.9 ≈ 10 t½
• Gas: k = (2.303/t) log[p₀/(2p₀ − pₜ)]

Arrhenius

• k = A·exp(−Ea/RT)
• log(k₂/k₁) = Ea/2.303R × (T₂−T₁)/T₁T₂
• log k vs 1/T: slope −Ea/2.303R
• 10 K rise → k × 2 to 3

Collisions and catalysts

• Need E ≥ Ea and right orientation
• Rate = PZ·exp(−Ea/RT)
• Catalyst lowers Ea both ways
• No change in ΔH, ΔG or K