Chemical Kinetics

The rate of a chemical reaction is the change in concentration of a reactant or product per unit time. As a reaction proceeds, reactant concentration falls and product concentration rises, so the rate can be tracked using either species.

The average rate over a time interval Δt is Rate = −Δ[R]/Δt = Δ[P]/Δt. A negative sign is placed before the reactant term (since its concentration decreases) so that the calculated rate comes out positive. As Δt is made smaller and smaller (Δt → 0), the average rate becomes the instantaneous rate, −d[R]/dt or d[P]/dt — the slope of the tangent to the concentration-versus-time curve at that instant.

For a general reaction aA + bB → cC + dD, the different species may appear/disappear at different numerical rates if the stoichiometric coefficients differ, so the true rate of reaction is obtained by dividing each species' rate of change by its own coefficient:

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

The units of rate are always concentration/time, i.e. mol L⁻¹ s⁻¹ (or mol L⁻¹ min⁻¹, mol L⁻¹ h⁻¹), irrespective of the order of the reaction.

Worked example: For 2N₂O₅(g) → 4NO₂(g) + O₂(g), the concentration of N₂O₅ falls from 1.6×10⁻² mol L⁻¹ to 1.4×10⁻² mol L⁻¹ in 10 s. Find the rate of reaction and the rate of formation of NO₂.

1. Rate of disappearance of N₂O₅ = −Δ[N₂O₅]/Δt = (1.6×10⁻² − 1.4×10⁻²)/10 = 2.0×10⁻⁴ mol L⁻¹ s⁻¹
2. Rate of reaction = −(1/2) d[N₂O₅]/dt = (1/2) × 2.0×10⁻⁴ = 1.0×10⁻⁴ mol L⁻¹ s⁻¹
3. Rate of formation of NO₂ = 4 × Rate of reaction = 4.0×10⁻⁴ mol L⁻¹ s⁻¹
4. Rate of formation of O₂ = 1 × Rate of reaction = 1.0×10⁻⁴ mol L⁻¹ s⁻¹

The rate of a reaction depends chiefly on the concentration of reactants, and also on temperature and the presence of a catalyst. The dependence on concentration is expressed through the rate law (rate expression), which relates rate to the concentrations of reactants, each raised to some power.

For a reaction, Rate = k[A]ˣ[B]ʸ, where k is the rate constant (specific reaction rate) and x, y are the powers to which [A] and [B] must be raised to match experimental data. These powers are not necessarily the stoichiometric coefficients of the balanced equation — they must be determined experimentally, usually by the initial rate method.

The order of reaction with respect to a reactant is the power of its concentration term in the rate law; the overall order is the sum of all such powers (x + y here). Order can be zero, a whole number, or even a fraction, and is always found experimentally.

When every reactant concentration is 1 mol L⁻¹, Rate = k, so the rate constant equals the rate at unit concentration. The units of k depend on the overall order n: k has units mol^(1−n) L^(n−1) s⁻¹. A zero order reaction has k in mol L⁻¹ s⁻¹, a first order reaction has k in s⁻¹, and a second order reaction has k in mol⁻¹ L s⁻¹.

Worked example: For A + B → Product: Experiment 1: [A]=0.1 mol L⁻¹, [B]=0.1 mol L⁻¹, rate=2.0×10⁻² mol L⁻¹s⁻¹. Experiment 2: [A]=0.2 mol L⁻¹, [B]=0.1 mol L⁻¹, rate=4.0×10⁻² mol L⁻¹s⁻¹. Experiment 3: [A]=0.1 mol L⁻¹, [B]=0.2 mol L⁻¹, rate=2.0×10⁻² mol L⁻¹s⁻¹.

Comparing 1 and 2 ([B] fixed): [A] doubles, rate doubles → first order in A. Comparing 1 and 3 ([A] fixed): [B] doubles, rate unchanged → zero order in B. So Rate = k[A]¹[B]⁰ = k[A], overall order = 1, and k = rate/[A] = 2.0×10⁻²/0.1 = 0.2 s⁻¹.

Molecularity of a reaction is the number of reacting species (atoms, ions or molecules) that must collide simultaneously in a single step for the reaction to occur. It is a theoretical concept decided purely by the mechanism, and applies only to elementary reactions — reactions occurring in a single step. Molecularity is always a whole number (1, 2, or rarely 3); it can never be zero or fractional, and values above 3 are essentially never observed since simultaneous collision of more than three species is extremely improbable.

Order, by contrast, is an experimental quantity read from the rate law and can be zero, fractional, or whole-number. For a single-step (elementary) reaction, order and molecularity usually match. But most laboratory reactions are complex (multi-step) reactions proceeding through a sequence of elementary steps called the reaction mechanism. For such reactions, the order is decided only by the slowest step, the rate-determining step (RDS), and can differ completely from what the overall balanced equation suggests.

Worked example: NO₂(g) + CO(g) → NO(g) + CO₂(g) is experimentally found to be second order in NO₂ and zero order in CO below 500 K, even though the balanced equation might suggest first order in each. A two-step mechanism explains this:

1. Step 1 (slow, rate-determining): NO₂ + NO₂ → NO₃ + NO
2. Step 2 (fast): NO₃ + CO → NO₂ + CO₂

Since the overall reaction can never be faster than its slowest step, the rate law is fixed entirely by Step 1: Rate = k[NO₂]², matching experiment and showing why a rate law need not follow the overall stoichiometric equation.

The rate law (differential form) gives instantaneous rate but not how concentration changes over time. Integrating it with respect to time gives the integrated rate equation, which relates concentration directly to time and is far more useful for extracting k from experimental data or predicting concentration after a given time.

Zero order reaction: Rate = −d[R]/dt = k[R]⁰ = k. Separating variables and integrating from [R]₀ at t = 0 to [R] at time t gives [R] = [R]₀ − kt. A plot of [R] against t is a straight line of slope −k and intercept [R]₀; k = ([R]₀ − [R])/t. Zero order kinetics occurs, for example, in some catalytic reactions where the catalytic surface is saturated with reactant.

First order reaction: Rate = −d[R]/dt = k[R]. Separating variables, −d[R]/[R] = k dt, and integrating from [R]₀ to [R] gives ln[R] = ln[R]₀ − kt, i.e. ln([R]₀/[R]) = kt, or in base-10 form k = (2.303/t) log([R]₀/[R]). A plot of log[R] against t is a straight line of slope −k/2.303.

Half-life (t½) is the time for reactant concentration to fall to half its initial value, [R] = [R]₀/2. For zero order: t½ = [R]₀/2k (depends on initial concentration). For first order, substituting [R] = [R]₀/2 gives t½ = (2.303/k) log 2 = 0.693/k — notably independent of initial concentration, a signature test for first order kinetics.

Worked example: A first order reaction has k = 4.6×10⁻² min⁻¹. Find the time needed for the reactant concentration to fall to 1/4 of its initial value.

t = (2.303/k) log([R]₀/[R]) = (2.303/4.6×10⁻²) × log 4 = 50.07 × 0.602 ≈ 30.1 min. (Equivalently, falling to 1/4 takes exactly two half-lives: t½ = 0.693/0.046 = 15.07 min, so 2 × t½ = 30.1 min — the two methods agree.)

A pseudo first order reaction is a reaction whose true rate law is second (or higher) order, but which behaves experimentally as first order because one reactant is present in such large excess that its concentration remains effectively constant throughout.

Acid hydrolysis of ethyl acetate: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (H⁺ catalysed). The true rate law is Rate = k[CH₃COOC₂H₅][H₂O]. Since water is the solvent and present in huge excess compared to the ester, its concentration barely changes during the reaction and can be treated as constant. The rate law simplifies to Rate = k′[CH₃COOC₂H₅], where k′ = k[H₂O] is the pseudo first order rate constant — the reaction now obeys first order kinetics even though it is truly second order.

Inversion (hydrolysis) of cane sugar: C₁₂H₂₂O₁₁ + H₂O → C₆H₁₂O₆ (glucose) + C₆H₁₂O₆ (fructose), H⁺ catalysed. Water is again in large excess, so Rate = k[C₁₂H₂₂O₁₁][H₂O] ≈ k′[C₁₂H₂₂O₁₁], another pseudo first order reaction.

Worked example: A bimolecular reaction has true rate constant k = 1.0×10⁻³ mol⁻¹ L s⁻¹ and is carried out in water where [H₂O] ≈ 55.5 mol L⁻¹ remains essentially constant. The pseudo first order rate constant is:

k′ = k[H₂O] = 1.0×10⁻³ × 55.5 = 5.55×10⁻² s⁻¹

This k′ can then be used directly in the first order formulas (k′ = (2.303/t) log([R]₀/[R]), t½ = 0.693/k′) even though the underlying mechanism is second order.

For most reactions, the rate constant (and hence the rate) roughly doubles or triples for every 10°C rise in temperature. This strong dependence is captured by the Arrhenius equation: k = A e^(−Ea/RT), where A is the pre-exponential (frequency) factor, Ea is the activation energy, R is the gas constant, and T is the absolute temperature.

Reacting molecules must possess a certain minimum energy — the threshold energy — before they can react on colliding. Activation energy Ea is the extra energy, above the average energy of the reactants, that must be supplied for molecules to cross this barrier and form the intermediate activated complex (transition state) before turning into products. Only a fraction of molecules, e^(−Ea/RT), have enough energy to react at a given temperature; raising T sharply increases this fraction, explaining the steep rise of rate with temperature.

Taking the natural log of the Arrhenius equation gives its linear form: ln k = ln A − Ea/RT, or in base-10 form: log k = log A − Ea/(2.303RT). A plot of log k against 1/T is a straight line with slope = −Ea/2.303R and intercept = log A; this graphical method is the standard way to determine Ea experimentally.

For two temperatures T₁ and T₂ with rate constants k₁ and k₂, subtracting the linear forms eliminates A: log(k₂/k₁) = (Ea/2.303R) × [(T₂ − T₁)/(T₁T₂)]. This two-point form lets Ea be calculated from just two (T, k) measurements.

A catalyst speeds up a reaction by providing an alternative reaction pathway with a lower activation energy, without itself being consumed. A lower Ea means a much larger fraction of molecules can now react, so the rate constant rises for both the forward and reverse reaction equally; the enthalpy change (ΔH) and the equilibrium position of the reaction are unaffected — the catalyst only helps the system reach equilibrium faster.

Worked example: The rate constant of a reaction increases from 4×10⁻³ s⁻¹ at 300 K to 8×10⁻³ s⁻¹ at 320 K. Calculate the activation energy. (log 2 = 0.301, R = 8.314 J K⁻¹ mol⁻¹)

log(k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
log(8×10⁻³/4×10⁻³) = log 2 = 0.301
0.301 = [Ea/(2.303 × 8.314)] × (20/96000)
0.301 = [Ea/19.147] × 2.083×10⁻⁴
Ea = 0.301 × 19.147/2.083×10⁻⁴ ≈ 2.766×10⁴ J mol⁻¹ ≈ 27.7 kJ mol⁻¹