Chemical Kinetics

Chemical Kinetics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Chemical kinetics is the branch of physical chemistry that quantifies how fast reactants convert into products and identifies the molecular pathway that makes that speed possible. The core measurable quantity is the rate of reaction, defined as the change in concentration of a species per unit time. For a generic reaction A → products:

Rate = −d[A]/dt = k[A]ⁿ

where k is the rate constant and n is the order of reaction. The half-life (t₁/₂) for a first-order process is the universal 0.693/k — independent of initial concentration. The temperature dependence of k is governed by the Arrhenius equation: k = A·e^(−Eₐ/RT), where Eₐ is the activation energy. ECAT expects you to distinguish order (experimentally determined) from molecularity (theoretical count of molecules colliding in an elementary step). Three high-yield pointers: (1) rate is never written from the balanced equation, (2) units of k reveal the order, and (3) a catalyst lowers Eₐ without shifting equilibrium.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definition of Rate and Average vs Instantaneous Rate

The rate is the time derivative of concentration. The average rate over interval Δt uses the magnitude of Δ[conc]/Δt, while the instantaneous rate is the limit as Δt → 0, obtained graphically from the slope of the concentration–time curve. For multiple species A, B, C with stoichiometric coefficients a, b, c, rates are linked by the relation:

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

This factor ensures the rate is independent of which species is monitored.

Rate Law, Order, and Molecularity

The rate law has the form Rate = k[A]ᵐ[B]ⁿ, where exponents m and n are not read from the balanced equation — they are found from experiments (initial-rates method or graphical fitting of integrated laws). Order = m + n (overall) and applies to the overall reaction. Molecularity refers only to an elementary step and equals the number of molecules colliding simultaneously; unimolecular, bimolecular, and termolecular being the only realistic cases.

Integrated Rate Laws and Half-Life

For a first-order reaction A → products:

ln[A]ₜ = ln[A]₀ − kt → [A]ₜ = [A]₀·e^(−kt)

The corresponding t₁/₂ = 0.693/k — concentration-independent. A plot of ln[A] vs t is linear with slope −k.

For a second-order reaction (single reactant):

1/[A]ₜ = 1/[A]₀ + kt, and t₁/₂ = 1/(k[A]₀)

A plot of 1/[A] vs t is linear; here t₁/₂ depends on initial concentration.

Arrhenius Equation

The temperature dependence is given by:

k = A·e^(−Eₐ/RT), or ln k = ln A − Eₐ/RT

A plot of ln k versus 1/T is linear with slope −Eₐ/R, allowing Eₐ to be extracted from two temperatures:

ln(k₂/k₁) = (Eₐ/R)·(1/T₁ − 1/T₂)

Order	Integrated form (linear plot)	Slope	t₁/₂
0	[A]ₜ = [A]₀ − kt	−k	[A]₀/(2k)
1	ln[A]ₜ = ln[A]₀ − kt	−k	0.693/k
2	1/[A]ₜ = 1/[A]₀ + kt	+k	1/(k[A]₀)

Catalysts and Collision Theory

A catalyst provides an alternative pathway with lower activation energy, increasing k for both forward and reverse directions equally — ΔH and K_eq remain unchanged. Collision theory requires (i) sufficient kinetic energy (≥ Eₐ) and (ii) proper geometric orientation for a productive collision.

ECAT Question Patterns

Expect 3–5 MCQs covering: order identification from tabulated concentration data, unit analysis of k (s⁻¹ for 1st order, M⁻¹s⁻¹ for 2nd), Eₐ calculation from a 2-temperature dataset, and selecting the correct integrated-law plot.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases and Pseudo-Order Reactions

When one reactant is present in huge excess (e.g., hydrolysis in water), its concentration stays effectively constant and gets absorbed into k. The reaction then behaves as a lower-order pseudo-order reaction. For instance, the acid-catalyzed inversion of sucrose is pseudo-first-order in sucrose because [H⁺] is buffered. Recognizing this lets you correctly apply the first-order integrated law.

Mechanism and the Rate-Determining Step

A complex reaction proceeds via a mechanism — a sequence of elementary steps. The overall rate equals the rate of the slowest (rate-determining) step. Intermediates formed in fast prior steps are consumed before they accumulate and must not appear in the final rate law. Constructing the mechanism from kinetic data: if Rate = k[A][B] and a stoichiometric excess of B shows first-order-in-A only, then the RDS involves A alone, with B participating in a rapid pre-equilibrium — a classic ECAT trap.

Worked Example (Eₐ Calculation)

Given k₁ = 3.5 × 10⁻³ s⁻¹ at T₁ = 300 K and k₂ = 1.2 × 10⁻² s⁻¹ at T₂ = 320 K, find Eₐ.

Using ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂):

ln(1.2 × 10⁻² / 3.5 × 10⁻³) = ln(3.43) ≈ 1.233 (1/300 − 1/320) = (320 − 300)/(300·320) = 20/96000 = 2.083 × 10⁻⁴ K⁻¹ Eₐ = 1.233 × 8.314 / 2.083 × 10⁻⁴ ≈ 49.2 kJ/mol

Common Mistakes

• Using log₁₀ instead of ln without the 2.303 conversion factor in Arrhenius problems.
• Reading the order directly from coefficients in the balanced equation — valid only for elementary steps.
• Forgetting that a catalyst lowers Eₐ for both forward and reverse directions, so the equilibrium constant is unchanged.
• Confusing molecularity (≤3, integer) with order (any real number including fractional).

Connections to Other ECAT Topics

Kinetics bridges directly into chemical equilibrium (K_eq = k_f/k_r), thermodynamics (ΔG‡ from Eyring equation, related to Eₐ), and electrochemistry (current–overpotential relationships). Mastery here makes the equilibrium constant problems much easier.

Practice Prompts

1. The decomposition of N₂O₅ in CCl₄ gives a straight line when ln[conc] is plotted against time. What is the order, and what does the slope represent?
2. A reaction doubles in rate when temperature rises from 300 K to 310 K. Estimate Eₐ using the rule of thumb that a 10 K rise near 300 K roughly doubles k.

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