Reaction Kinetics

Reaction Kinetics

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

Reaction kinetics quantifies how fast reactants are consumed and products form, governed by the rate law: Rate = k[A]ᵐ[B]ⁿ, where k is the rate constant, m and n are experimental orders. Order is determined from data, never from balanced coefficients. Activation energy (Eₐ) is the minimum energy barrier; catalysts lower Eₐ without changing ΔH or K_eq. The Arrhenius equation, k = A·e^(−Eₐ/RT), shows k rises exponentially with T — a 10 °C jump often doubles the rate. Half-life for first-order reactions (t₁/₂ = 0.693/k) is independent of [A]₀, a fact MCQs love to test.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Rate, Rate Law, and Order

The instantaneous rate is the derivative: Rate = −d[Reactant]/dt = d[Product]/dt. The negative sign keeps the rate positive as reactant concentration falls. The differential rate law expresses rate in terms of concentrations; the integrated rate law expresses concentration as a function of time. The overall order (m + n) is found by running experiments at different initial concentrations — not by reading the balanced equation.

Integrated Forms and Half-Lives

Order	Integrated Law	Half-life (t₁/₂)
Zero	[A]ₜ = [A]₀ − kt	[A]₀ / (2k)
First	ln[A]ₜ = ln[A]₀ − kt	0.693 / k
Second	1/[A]ₜ = 1/[A]₀ + kt	1 / (k[A]₀)

Plotting [A] vs t (zero), ln[A] vs t (first), and 1/[A] vs t (second) gives a straight line — the classical way to identify order experimentally.

Arrhenius Behaviour

From k = A·e^(−Eₐ/RT), the two-temperature form is: ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂). This lets MDCAT numericals solve for Eₐ when two k–T pairs are given.

Molecularity vs Mechanism

Molecularity counts colliding particles in one elementary step (1, 2, or rarely 3). It must be an integer and applies only to single steps. The rate-determining step — the slowest step in the mechanism — dictates the overall rate law. A catalyst provides an alternative pathway with lower Eₐ; it never alters ΔH, K_eq, or overall stoichiometry.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Collision and Transition-State Theory

Collision theory requires that colliding molecules carry energy ≥ Eₐ and correct orientation; otherwise the collision is ineffective. Transition state theory refines this by defining the activated complex — a short-lived, high-energy configuration at the peak of the reaction coordinate diagram. Catalysts stabilise this complex, lowering Eₐ forward and reverse equally, so K_eq stays unchanged.

Worked Example

For a first-order decomposition, k = 3.2 × 10⁻⁴ s⁻¹ at 35 °C and 9.8 × 10⁻⁴ s⁻¹ at 50 °C. Find Eₐ. ln(9.8/3.2) = (Eₐ/8.314)[1/308 − 1/323] → 1.119 = (Eₐ/8.314)(1.507 × 10⁻⁴) → Eₐ ≈ 61.7 kJ mol⁻¹.

Common Traps

• Treating the stoichiometric coefficient as the order — only true for elementary steps.
• Writing rate without the negative sign on [Reactant].
• Applying t₁/₂ = 0.693/k to zero- or second-order reactions, where the formula depends on [A]₀.
• Assuming higher [A] always speeds the rate proportionally — only true when order = 1.

Adjacent Links

Mastery here feeds directly into chemical equilibrium (catalysts shift rates, not K), electrochemical kinetics (Butler–Volmer extends Arrhenius), and enzyme kinetics (Michaelis–Menten is a saturation form of rate laws).

Practice Prompts

1. A reaction doubles in rate when [A] doubles and [B] is tripled. Write the rate law and state the overall order.
2. Using k values at 400 K and 500 K, calculate Eₐ via the two-point Arrhenius form and predict k at 450 K.

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