Chemical Kinetics, Reaction Dynamics, Photophysical Kinetics and Surfaces
1. Rate laws, integrated forms and half-lives
The rate law r = k[A]^m[B]^n is found by experiment; the order m + n need not match the stoichiometry. Only for an elementary reaction, a single molecular event, does the order equal the molecularity (uni-, bi- or termolecular). A reaction with one reactant in large excess behaves as pseudo-first-order, k′ = k[B]₀. The integrated forms identify the order from which plot is straight:
| Order | Integrated form | Straight-line plot | Half-life |
|---|---|---|---|
| 0 | [A] = [A]₀ − kt | [A] vs t | [A]₀/(2k) |
| 1 | ln[A] = ln[A]₀ − k·t | ln[A] vs t | ln 2/k, independent of [A]₀ |
| 2 | 1/[A] = 1/[A]₀ + kt | 1/[A] vs t | 1/(k[A]₀) |
| n (≠ 1) | 1/[A]ⁿ⁻¹ = 1/[A]₀ⁿ⁻¹ + (n − 1)kt | 1/[A]ⁿ⁻¹ vs t | ∝ 1/[A]₀ⁿ⁻¹ |
The temperature dependence is Arrhenius: k = A e^(−Ea/RT), so ln(k₂/k₁) = −(Ea/R)(1/T₂ − 1/T₁) and the slope of ln k against 1/T is −Ea/R. Near room temperature a reaction with Ea ≈ 50 kJ/mol roughly doubles its rate for a 10 K rise.
2. Parallel, opposing and consecutive reactions; steady state; unimolecular reactions
Parallel first-order reactions A → B (k₁) and A → C (k₂) consume A with k₁ + k₂, and the product ratio [B]/[C] = k₁/k₂ stays constant throughout. Opposing reactions A ⇌ B (k_f, k_b) approach equilibrium exponentially with the observed constant k_f + k_b, and at equilibrium K = k_f/k_b. Consecutive reactions A → B → C (k₁, k₂) give [B] = [A]₀k₁/(k₂ − k₁)(e^(−k₁t) − e^(−k₂t)), which passes through a maximum at t_max = ln(k₁/k₂)/(k₁ − k₂); if k₂ ≫ k₁, B never builds up and the first step is rate-determining.
The steady-state approximation sets d[I]/dt ≈ 0 for a reactive intermediate present at low concentration, turning a set of differential equations into algebra. It explains the rate laws of chain reactions: for H₂ + Br₂ → 2HBr, with initiation Br₂ → 2Br, propagation Br + H₂ → HBr + H and H + Br₂ → HBr + Br, inhibition H + HBr → H₂ + Br and termination, it gives r = k[H₂][Br₂]^½/(1 + k′[HBr]/[Br₂]) — the half order in Br₂ reveals atom initiation, and the denominator the inhibition by product.
Unimolecular reactions in the gas phase (isomerisation, decomposition) follow the Lindemann–Hinshelwood mechanism: A + M ⇌ A* + M (k₁, k₋₁), A* → P (k₂). Steady state in A* gives k_uni = k₁k₂[M]/(k₋₁[M] + k₂). At high pressure k₋₁[M] ≫ k₂ and k_uni = k₁k₂/k₋₁, first order; at low pressure activation becomes rate-limiting and k_uni = k₁[M], second order overall — the characteristic fall-off as pressure drops.
3. Potential-energy surfaces, transition-state theory, isotope effects and fast reactions
For A + BC → AB + C the Born–Oppenheimer energy as a function of the two bond lengths is a potential-energy surface. Reactant and product valleys are joined by a minimum-energy path, the reaction coordinate, whose highest point is a saddle point — a maximum along the path and a minimum in every perpendicular direction. That point is the transition state. Classical trajectories run on the surface show how energy is used: for an early (reactant-like) barrier translational energy is most effective, for a late (product-like) barrier vibrational energy of the reactant is (Polanyi’s rules).
Transition-state theory assumes a quasi-equilibrium between reactants and activated complexes and gives the Eyring equation k = κ(k_BT/h)K‡, with κ the transmission coefficient (usually taken as 1). In thermodynamic form k = (k_BT/h) e^(ΔS‡/R) e^(−ΔH‡/RT), so a plot of ln(k/T) against 1/T gives ΔH‡ from the slope and ΔS‡ from the intercept. For a reaction in solution Ea = ΔH‡ + RT. A negative ΔS‡ indicates a more ordered activated complex — typical of associative bimolecular steps and cycloadditions. At 298 K, k_BT/h = 6.21 × 10¹² s⁻¹.
A primary kinetic isotope effect arises when the bond to the isotope breaks in the rate-determining step. The C–H stretch has more zero-point energy than C–D, and if that vibration becomes the reaction coordinate the difference is lost at the transition state, so the C–H bond has a smaller effective barrier: kH/kD = exp[Δ(ZPE)/kT], about 7 at 298 K for a full C–H stretch (2900 vs 2100 cm⁻¹). Values near 1 (secondary effects, 0.7–1.4) mean the C–H bond is not broken; values far above 7 point to tunnelling.
Fast reactions. In flow methods reactants are mixed in milliseconds and observed downstream (continuous flow) or after the flow is halted (stopped flow, down to about 1 ms). Relaxation methods perturb an equilibrium suddenly — a temperature jump of a few kelvin in microseconds, or a pressure jump — and follow the return with relaxation time τ: for A ⇌ B, 1/τ = k_f + k_b; for A + B ⇌ C, 1/τ = k_f([A]ₑ + [B]ₑ) + k_b. Flash photolysis and pump–probe lasers reach femtoseconds. Diffusion-controlled reactions react at every encounter; their rate constant is set by the solvent viscosity, k_d ≈ 8RT/(3η), about 7 × 10⁹ L mol⁻¹ s⁻¹ in water at 298 K.
4. Catalysis, enzyme kinetics and polymerisation kinetics
A catalyst provides a lower-energy pathway, raising forward and reverse rates by the same factor, so it does not change K or ΔG. Acid–base catalysis may be specific (rate ∝ [H₃O⁺]) or general (rate depends on every acid present, following the Brønsted catalysis law). Enzyme catalysis follows the Michaelis–Menten scheme E + S ⇌ ES → E + P: with a steady state in ES, v = V_max[S]/(K_M + [S]), where V_max = k_cat[E]₀ and K_M = (k₋₁ + k₂)/k₁ is the substrate concentration at which v = V_max/2. At low [S] the rate is first order in S, at high [S] zero order. The Lineweaver–Burk plot 1/v = (K_M/V_max)(1/[S]) + 1/V_max linearises the data; k_cat/K_M measures catalytic efficiency, with an upper limit set by diffusion. A competitive inhibitor raises the apparent K_M and leaves V_max unchanged; a non-competitive one lowers V_max and leaves K_M unchanged.
Polymerisation. In step-growth (condensation) polymerisation any two chains can join, and the Carothers equation gives the number-average degree of polymerisation X̄ₙ = 1/(1 − p), where p is the extent of reaction of the functional groups — p = 0.99 is needed for X̄ₙ = 100, which is why step-growth polymers need very pure monomers and near-complete conversion. In free-radical chain-growth polymerisation, with initiation, propagation and termination and a steady state in radicals, the rate of polymerisation is R_p = k_p[M](f k_d[I]/k_t)^½ — first order in monomer and half order in initiator — and the kinetic chain length is R_p divided by the rate of initiation.
5. Photophysical kinetics, quantum yield and quenching
The quantum yield Φ of a process is the number of events divided by the number of photons absorbed; for a primary photochemical step Φ ≤ 1, while chain reactions (H₂ + Cl₂) can give Φ of 10⁴ or more. Photons are counted in einsteins (moles of photons): one einstein at wavelength λ carries N_Ahc/λ, 239 kJ at 500 nm. After excitation, S₁ decays by unimolecular processes — fluorescence k_f, internal conversion k_ic and intersystem crossing k_isc — so its lifetime is τ₀ = 1/(k_f + k_ic + k_isc) and the fluorescence quantum yield Φ_F = k_fτ₀.
A quencher Q adds a bimolecular decay channel. In dynamic (collisional) quenching Q meets the excited molecule during its lifetime, and the Stern–Volmer equation follows: I₀/I = τ₀/τ = 1 + k_qτ₀[Q] = 1 + K_SV[Q], with k_q typically near the diffusion limit. In static quenching Q forms a non-emitting complex with the ground-state molecule; I₀/I still rises linearly with [Q], but the lifetime of the molecules that do emit is unchanged. So the two are distinguished by lifetimes, and by temperature: dynamic quenching grows with T (faster diffusion), static quenching usually falls (the complex dissociates).
6. Adsorption, surface catalysis, surface tension, colloids and macromolecules
| Feature | Physisorption | Chemisorption |
|---|---|---|
| Forces | van der Waals | chemical bonds |
| Enthalpy | about −20 to −40 kJ/mol | about −80 to −400 kJ/mol |
| Layers | multilayer possible | monolayer only |
| Specificity | low | high |
| Activation energy | none; fast and reversible | often activated; may be irreversible |
The Langmuir isotherm assumes a monolayer of equivalent, independent sites: θ = KP/(1 + KP), so θ is proportional to P at low pressure and saturates at 1; the linear form P/V = P/V_m + 1/(KV_m) gives the monolayer volume V_m. For dissociative adsorption θ = (KP)^½/(1 + (KP)^½). The empirical Freundlich isotherm x/m = kP^(1/n) (n > 1) fits heterogeneous surfaces at intermediate coverage. The BET isotherm extends Langmuir to multilayers, P/[V(P₀ − P)] = 1/(V_m c) + [(c − 1)/(V_m c)](P/P₀), and is the standard way to measure surface area from N₂ adsorption at 77 K.
In the Langmuir–Hinshelwood mechanism both reactants adsorb and react on the surface: r = kθ_Aθ_B = kK_AP_AK_BP_B/(1 + K_AP_A + K_BP_B)². At fixed P_B the rate passes through a maximum as P_A rises, because too much A crowds B off the surface. In the Eley–Rideal mechanism a gas-phase molecule strikes an adsorbed one, r = kθ_AP_B, with no such maximum.
Surface tension γ is the work to create unit area of surface; it produces the Laplace pressure ΔP = 2γ/r across a curved surface and the capillary rise h = 2γ cos θ/(ρgr). Viscosity η is measured by flow through a capillary (Poiseuille) or a falling sphere (Stokes, F = 6πηrv). Colloids (1–1000 nm) are lyophilic (solvent-loving, stable, reversible) or lyophobic (stabilised by charge); they scatter light (Tyndall effect), move in a field (electrophoresis, governed by the zeta potential) and are coagulated by counter-ions more strongly the higher their charge (Hardy–Schulze rule). Surfactants self-assemble into micelles above the critical micelle concentration, where surface tension stops falling and conductivity changes slope; micelles form only above the Krafft temperature. For macromolecules the number-average molar mass M̄ₙ = ΣNᵢMᵢ/ΣNᵢ (from osmometry) and the weight-average M̄_w = ΣNᵢMᵢ²/ΣNᵢMᵢ (from light scattering) differ, and M̄_w/M̄ₙ ≥ 1 is the polydispersity index.
Key takeaways
- Zero, first and second order give straight lines of [A], ln[A] and 1/[A] against t; the first-order half-life is ln 2/k whatever [A]₀.
- Opposing reactions relax with k_f + k_b; consecutive A → B → C peaks at t_max = ln(k₁/k₂)/(k₁ − k₂); Lindemann gives first order at high pressure and second order at low.
- Eyring k = (k_BT/h)e^(ΔS‡/R)e^(−ΔH‡/RT); primary kH/kD ≈ 7 when C–H breaks; diffusion limit 8RT/3η ≈ 7 × 10⁹ L mol⁻¹ s⁻¹ in water.
- Michaelis–Menten v = V_max[S]/(K_M + [S]); Carothers X̄ₙ = 1/(1 − p); radical R_p ∝ [M][I]^½. Stern–Volmer I₀/I = 1 + k_qτ₀[Q]; only dynamic quenching shortens the lifetime.
- Langmuir θ = KP/(1 + KP), Freundlich kP^(1/n), BET for surface area; the Langmuir–Hinshelwood rate has a maximum in P_A; micelles form above the CMC.
Practice questions (18)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A first-order reaction has k = 1.0 × 10⁻³ s⁻¹. What is its half-life, in s, to the nearest whole number?
Numerical answer — type the value.
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Answer: 693
t½ = ln 2/k = 0.6931/1.0 × 10⁻³ = 693 s, the same whatever the starting concentration. 1/k = 1000 s is the mean lifetime, not the half-life.The rate constant of a reaction doubles between 300 K and 310 K. What is its activation energy, in kJ/mol, to one decimal place? (R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 53.6
ln 2 = (Ea/R)(1/300 − 1/310) = (Ea/8.314)(1.0753 × 10⁻⁴), so Ea = 8.314 × 0.6931/1.0753 × 10⁻⁴ = 53 590 J/mol = 53.6 kJ/mol. Using log₁₀ 2 = 0.301 without 2.303 gives 23.3 kJ/mol.What fraction of the reactant remains after three half-lives of a first-order reaction? Give the answer to three decimal places.
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Answer: 0.125
Each half-life halves what is left: (½)³ = 0.125, i.e. 12.5%. A tempting error is 1 − 3 × 0.5 or 0.25, which treats the decay as linear or counts two half-lives.For consecutive first-order reactions A → B → C with k₁ = 0.20 min⁻¹ and k₂ = 0.10 min⁻¹, at what time, in minutes, does [B] reach its maximum? Give the answer to two decimal places.
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Answer: 6.93
t_max = ln(k₁/k₂)/(k₁ − k₂) = ln 2/0.10 = 6.93 min. The formula is symmetric in k₁ and k₂, so swapping them gives the same time, but the height of the maximum does depend on which step is slower.For the opposing first-order reactions A ⇌ B with k_f = 0.030 s⁻¹ and k_b = 0.010 s⁻¹, what is the relaxation time τ after a small perturbation, in s?
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Answer: 25
1/τ = k_f + k_b = 0.040 s⁻¹, so τ = 25 s. The equilibrium constant K = k_f/k_b = 3 is a separate quantity; taking 1/τ = k_f − k_b gives 50 s.A reaction in solution has ΔG‡ = 80.0 kJ/mol at 298 K. Using the Eyring equation with κ = 1, what is its rate constant, in s⁻¹, to three decimal places? (k_BT/h = 6.21 × 10¹² s⁻¹ at 298 K, R = 8.314 J mol⁻¹ K⁻¹)
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Answer: 0.059
k = (k_BT/h)e^(−ΔG‡/RT) = 6.21 × 10¹² × exp(−80 000/(8.314 × 298)) = 6.21 × 10¹² × e^(−32.29) = 6.21 × 10¹² × 9.48 × 10⁻¹⁵ = 0.0589 s⁻¹, i.e. 0.059. Each 5.7 kJ/mol on ΔG‡ changes k tenfold at this temperature.A C–H bond (2900 cm⁻¹) is broken in the rate-determining step, and the C–D analogue stretches at 2100 cm⁻¹. If all the stretching zero-point energy is lost at the transition state, estimate kH/kD at 298 K, to one decimal place. (kT/hc = 208.6 cm⁻¹ at 298 K is close enough.)
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Answer: 6.8
Δ(ZPE) = ½(2900 − 2100) = 400 cm⁻¹, and kH/kD = exp(Δ(ZPE)hc/kT) = exp(400/208.6) = exp(1.918) = 6.8. Forgetting the ½ in the zero-point energy gives exp(3.84) = 46, far above the observed range.In the Lindemann–Hinshelwood mechanism for a unimolecular gas-phase reaction, how does the observed order change as the pressure is lowered?
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Answer: A — From first order at high pressure to second order at low pressure
k_uni = k₁k₂[M]/(k₋₁[M] + k₂). At high [M] deactivation outpaces reaction and k_uni = k₁k₂/k₋₁, first order; at low [M] every activated molecule reacts, k_uni = k₁[M], so the rate k₁[A][M] is second order. This is the fall-off region.Estimate the diffusion-controlled rate constant in water at 298 K from k_d = 8RT/(3η), with η = 8.9 × 10⁻⁴ Pa s. Give the answer in units of 10⁹ L mol⁻¹ s⁻¹, to one decimal place. (R = 8.314 J mol⁻¹ K⁻¹)
Numerical answer — type the value.
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Answer: 7.4
k_d = 8 × 8.314 × 298/(3 × 8.9 × 10⁻⁴) = 7.42 × 10⁶ m³ mol⁻¹ s⁻¹ = 7.42 × 10⁹ L mol⁻¹ s⁻¹, i.e. 7.4. Leaving the result in m³ gives a number 1000 times smaller in these units.An enzyme obeys Michaelis–Menten kinetics with V_max = 100 μmol L⁻¹ min⁻¹. What is the rate, in μmol L⁻¹ min⁻¹, when [S] = 3K_M?
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Answer: 75
v = V_max[S]/(K_M + [S]) = 100 × 3K_M/(4K_M) = 75. At [S] = K_M the rate is half of V_max (50), and it approaches 100 only as [S] ≫ K_M.A competitive inhibitor is added to an enzyme obeying Michaelis–Menten kinetics. What happens to the apparent kinetic constants?
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Answer: A — K_M increases and V_max is unchanged
A competitive inhibitor binds the free enzyme at the active site, so enough substrate can outcompete it: V_max is reached eventually, but more substrate is needed, K_M(app) = K_M(1 + [I]/K_I). Lower V_max with unchanged K_M is non-competitive inhibition; both falling is uncompetitive.In a step-growth polymerisation, the extent of reaction of the functional groups is p = 0.99. What is the number-average degree of polymerisation by the Carothers equation?
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Answer: 100
X̄ₙ = 1/(1 − p) = 1/0.01 = 100. At p = 0.95 it would be only 20: step-growth chains grow long only at conversions very close to completion.A fluorophore has τ₀ = 10 ns and is quenched dynamically with k_q = 1.0 × 10¹⁰ L mol⁻¹ s⁻¹. What is I₀/I at a quencher concentration of 0.010 mol/L, to one decimal place?
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Answer: 2.0
K_SV = k_qτ₀ = 1.0 × 10¹⁰ × 1.0 × 10⁻⁸ = 100 L/mol, and I₀/I = 1 + K_SV[Q] = 1 + 100 × 0.010 = 2.0: half the emission is quenched, and the lifetime falls to 5 ns. Omitting the leading 1 gives 1.0.What is the energy of one einstein (one mole of photons) of 500 nm light, in kJ/mol, to one decimal place? (h = 6.626 × 10⁻³⁴ J s, c = 2.998 × 10⁸ m/s, N_A = 6.022 × 10²³ mol⁻¹)
Numerical answer — type the value.
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Answer: 239.3
E = N_Ahc/λ = 6.022 × 10²³ × 6.626 × 10⁻³⁴ × 2.998 × 10⁸/(500 × 10⁻⁹) = 2.393 × 10⁵ J/mol = 239.3 kJ/mol. Without N_A the answer is the energy of one photon, 3.97 × 10⁻¹⁹ J.Which observations indicate that fluorescence quenching is static rather than dynamic?
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Answer: A — The fluorescence lifetime is unchanged as the quencher is added; B — Quenching becomes less efficient as the temperature is raised
Static quenching removes molecules into a dark ground-state complex, so the survivors emit with the normal lifetime, and heating dissociates the complex, weakening quenching. A falling lifetime (τ₀/τ linear in [Q]) and faster quenching in less viscous solvents are the signatures of collisional, dynamic quenching.A gas adsorbs on a surface following the Langmuir isotherm with K = 0.50 bar⁻¹. What is the fractional coverage θ at 2.0 bar, to two decimal places?
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Answer: 0.5
θ = KP/(1 + KP) = (0.50 × 2.0)/(1 + 1.0) = 0.50. Using θ = KP gives 1.0, which is the low-pressure form stretched past its range; coverage can never exceed one monolayer in this model.In a Langmuir–Hinshelwood surface reaction A + B → P, the partial pressure of B is held fixed and that of A is increased steadily. The rate
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Answer: A — rises, passes through a maximum and then falls
r = kK_AP_AK_BP_B/(1 + K_AP_A + K_BP_B)²: the numerator grows as P_A but the squared denominator eventually grows faster, because A displaces B from the sites. Levelling off without a maximum is the Eley–Rideal or single-reactant Langmuir behaviour.Which statements about adsorption and colloids are correct?
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Answer: A — Physisorption can form multilayers, chemisorption only a monolayer; B — The BET isotherm is used to measure the surface area of solids
Weak van der Waals forces allow further layers, while chemical bonds need direct contact with the surface; BET fitted to N₂ adsorption at 77 K gives V_m and the area. Micelles appear above the CMC, and M̄_w ≥ M̄ₙ always, since the weight average favours heavy chains.