Kinetics

Kinetics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Chemical Kinetics quantifies how fast reactants convert into products and identifies the molecular factors controlling that speed. The rate is defined as the decrease in reactant concentration (or increase in product concentration) per unit time, expressed in mol L⁻¹ s⁻¹.

The governing equation for most exam questions is the rate law: for aA + bB → products, Rate = k[A]ᵐ[B]ⁿ, where the order (m + n) is an experimental quantity, not the stoichiometric coefficients. The Arrhenius equation k = A·exp(−Ea/RT) links the rate constant to temperature and activation energy.

Must-know facts

• Zero-order t₁/₂ = [A]₀/(2k) — depends on initial concentration.
• First-order t₁/₂ = 0.693/k — independent of [A]₀ (most-tested case).
• Second-order (equal conc.) t₁/₂ = 1/(k[A]₀).
• A catalyst lowers Ea without changing K_eq or ΔH.
• Molecularity applies only to elementary steps and is always a positive integer; order can be zero, fractional, or negative.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Chemical Kinetics is the experimental and theoretical study of reaction rates and mechanisms. It separates how fast a reaction proceeds (kinetics) from how far it proceeds at equilibrium (thermodynamics).

Rate, Rate Law, and Order

Rate is always reported as a positive quantity: Rate = −d[R]/dt = +d[P]/dt. The rate law has the general form Rate = k[A]ᵐ[B]ⁿ, where k is the rate constant (units depend on overall order) and m + n is the overall order. Order is determined from data — initial rates, graphical plots of integrated forms, or the half-life method — never read off the balanced equation.

Integrated Rate Equations

Order	Integrated form	Half-life t₁/₂	Units of k
0	[A] = [A]₀ − kt	[A]₀/(2k)	mol L⁻¹ s⁻¹
1	ln[A] = ln[A]₀ − kt	0.693/k	s⁻¹
2	1/[A] = 1/[A]₀ + kt	1/(k[A]₀)	L mol⁻¹ s⁻¹

The first-order equation is most heavily tested: t₁/₂ depends only on k, which is why radioactive decay and many decomposition reactions show concentration-independent half-lives.

Arrhenius and Temperature Dependence

The Arrhenius equation, k = A·exp(−Ea/RT), gives a straight line when ln k is plotted against 1/T. The slope (−Ea/R) lets you extract Ea from two-temperature data:

$$\log\frac{k_2}{k_1} = \frac{E_a}{2.303,R}\left(\frac{T_2 - T_1}{T_1 T_2}\right)$$

with R = 8.314 J K⁻¹ mol⁻¹.

JEE trap: A 10 K rise near 300 K roughly doubles k because of the exponential term — students often miss that the factor, not the absolute value, of rate increase matters.

Pseudo-order and Catalysts

When one reactant (usually water) is in vast excess, its concentration stays effectively constant and the reaction becomes pseudo-first-order; the ester hydrolysis in aqueous medium is the textbook example. A catalyst provides an alternative pathway with a lower Ea, raising k at a given T without shifting ΔG or K_eq.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

The Extended tier treats Kinetics at the depth needed for JEE Advanced-style numericals, where distinguishing order from molecularity and applying the rate-determining step (RDS) rule decides full marks.

Order vs Molecularity — The Critical Distinction

Molecularity is a theoretical concept: the number of reactant molecules colliding in a single elementary step. It is always a positive integer (1, 2, or rarely 3) and never applies to a complex, multi-step reaction. Order is empirical, derived from the experimentally measured rate law, and may be zero, fractional, or even negative (when a product inhibits the reaction). For an elementary step, order equals molecularity; for a complex reaction, only the RDS appears in the overall rate law — fast steps equilibrate rapidly and contribute through equilibrium constants.

Worked Example — Arrhenius Slope

A reaction has k = 2.5 × 10⁻⁴ s⁻¹ at 300 K and k = 4.0 × 10⁻³ s⁻¹ at 320 K. Find Ea.

Plugging into the two-temperature form:

• log(4.0×10⁻³ / 2.5×10⁻⁴) = log(16) = 1.204
• (T₂ − T₁)/(T₁T₂) = 20/(300·320) = 2.083×10⁻⁴ K⁻¹
• Ea = 1.204 × 2.303 × 8.314 / 2.083×10⁻⁴ ≈ 88.3 kJ mol⁻¹

Mnemonic: “Slope = −Ea/R” — a steeper negative slope means a larger Ea.

Common Mistakes

• Writing rate as +d[R]/dt instead of −d[R]/dt — the integrated forms assume the negative sign, so sign errors propagate.
• Conflating Ea in cal/mol with R in J K⁻¹ mol⁻¹ inside one calculation. Standardise units first.
• Treating a catalyst as if it changes ΔH or K_eq. It lowers Ea only.
• Assuming t₁/₂ of a zero-order reaction is constant — it scales linearly with [A]₀.

Collision Theory in Brief

Rate = Z·ρ·f, where Z is the collision frequency, ρ the orientation (steric) factor, and f the fraction of collisions exceeding Ea. Raising temperature increases Z modestly but boosts f exponentially — the dominant reason reactions accelerate with heat.

Two Practice Prompts

1. For the reaction 2N₂O₅ → 4NO₂ + O₂, the rate of O₂ formation is 1.5 × 10⁻⁴ mol L⁻¹ s⁻¹. Express the rate in terms of (i) d[N₂O₅]/dt and (ii) d[NO₂]/dt, and determine the order if halving [N₂O₅] doubles t₁/₂.
2. At 600 K a first-order reaction is 50% complete in 25 minutes. How long will it take to reach 90% completion at the same temperature?

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