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26. Reaction kinetics

Syllabus
9701–2028–2029
Section
26
Level
A2

26.1 Rate equations, orders and rate constants

Syllabus
9701–2028–2029
Topic
26.1
Level
A2

Use the core language of experimental kinetics

Term Precise meaning
rate equation experimentally determined relationship between rate and reactant concentrations
order with respect to A power of [A] in the rate equation
overall order sum of all concentration powers
rate constant, k proportionality constant for a fixed reaction at a stated temperature
half-life, t₁/₂ time for a reactant concentration to fall to half its value
rate-determining step slow step that controls the observed rate
intermediate species made in one mechanism step and consumed in a later step

rate=k[A]2[B]⇒second order in A, first order in B, third order overall\mathrm{rate}=k[A]^2[B]\quad\Rightarrow\quad\text{second order in A, first order in B, third order overall}

Orders come from rate evidence, not coefficients in the overall equation. An intermediate cancels from the summed mechanism, and the rate-determining step is not automatically the first step.

Deduce and use a rate equation from data and graphs

Evidence when [A] changes Order in A Rate–[A] graph Concentration–time clue
rate unchanged 0 horizontal straight decrease while zero-order conditions hold
rate changes by same factor 1 straight through origin constant successive half-lives
rate changes by square of factor 2 upward curve successive half-lives increase
Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.100 0.100 2.00 × 10⁻³
2 0.200 0.100 8.00 × 10⁻³
3 0.100 0.200 2.00 × 10⁻³

Comparing 1→2, doubling [A] quadruples rate, so m = 2. Comparing 1→3, doubling [B] leaves rate unchanged, so n = 0.

rate=k[A]2[B]0=k[A]2\mathrm{rate}=k[A]^2[B]^0=k[A]^2

k=2.00×10−3(0.100)2=0.200 dm3 mol−1 s−1;[A]=0.150⇒rate=4.50×10−3 mol dm−3 s−1k=\frac{2.00\times10^{-3}}{(0.100)^2}=0.200\ \mathrm{dm^3\ mol^{-1}\ s^{-1}};\quad [A]=0.150\Rightarrow\mathrm{rate}=4.50\times10^{-3}\ \mathrm{mol\ dm^{-3}\ s^{-1}}

Compare experiments where only one concentration changes. A concentration–time gradient gives instantaneous rate; its curved shape alone does not identify order without the appropriate half-life or rate–concentration evidence.

A first-order reaction has a concentration-independent half-life

t1/2=0.693kt_{1/2}=\frac{0.693}{k}

At fixed temperature, k is constant, so every halving takes the same time even though the concentration and instantaneous rate both decrease. Constant successive half-lives are therefore evidence of first-order behaviour.

Time elapsed [A] from 0.800 mol dm⁻³ when t₁/₂ = 10.0 s
0 s 0.800
10 s 0.400
20 s 0.200
30 s 0.100

[A]t=[A]0(12)t/t1/2[A]_t=[A]_0\left(\frac12\right)^{t/t_{1/2}}

Constant half-life does not mean constant rate, and concentration-independent half-life is not a general rule for zero- or second-order reactions.

Calculate k from initial-rate data or first-order half-life

Evidence supplied Calculation route
initial rate and established orders k = rate/([A]ᵐ[B]ⁿ)
first-order half-life k = 0.693/t₁/₂

rate=k[A][B];k=4.38×10−6(0.0250)(0.0125)=1.40×10−2 dm3 mol−1 s−1\mathrm{rate}=k[A][B];\quad k=\frac{4.38\times10^{-6}}{(0.0250)(0.0125)}=1.40\times10^{-2}\ \mathrm{dm^3\ mol^{-1}\ s^{-1}}

t1/2=10.0 min=600 s;k=0.693600=1.16×10−3 s−1t_{1/2}=10.0\ \mathrm{min}=600\ \mathrm{s};\quad k=\frac{0.693}{600}=1.16\times10^{-3}\ \mathrm{s^{-1}}

Choose k units so that the rate equation yields mol dm⁻³ s⁻¹. Zero-, first-, second- and third-overall-order equations therefore have different k units.

Determine order before calculating k. The 0.693/t₁/₂ relationship applies only to first-order reactions, and time units set the reciprocal-time unit of k.

Test a mechanism against both the overall reaction and rate equation

For 2NO(g) + O₂(g) → 2NO₂(g), consider a fast equilibrium followed by a slow step.

2 NO⇌NX2OX2fast equilibrium\ce{2NO <=> N2O2}\quad\text{fast equilibrium}

NX2OX2+OX2→2 NOX2slow rate-determining step\ce{N2O2 + O2 -> 2NO2}\quad\text{slow rate-determining step}

The slow step gives rate ∝ [N₂O₂][O₂]. The preceding equilibrium makes [N₂O₂] proportional to [NO]², so substitution gives rate = k[NO]²[O₂]. Adding the two steps cancels N₂O₂ and reproduces the overall equation.

Species pattern across steps Identity
formed then consumed; absent overall intermediate (N₂O₂ here)
consumed then regenerated; absent overall catalyst

A proposed mechanism is consistent only if its summed steps give the overall reaction and its rate-determining logic gives the observed rate equation. The same checks can identify which listed step is rate determining.

Do not leave an intermediate in the observable rate equation without a supplied relationship that eliminates it, and do not infer mechanism solely from overall stoichiometric coefficients.

Raising temperature increases k and therefore increases rate

Raising temperature changes the energy distribution → a larger fraction of particles has energy at least equal to the activation energy → a larger fraction of collisions can react → the rate constant k increases → rate increases at the same concentrations.

The fraction beyond the activation-energy threshold can grow substantially even for a modest temperature rise, so k and rate may increase much more than collision frequency alone would suggest.

For one reaction, k is constant only at a specified temperature. Concentration appears separately in the rate equation; changing concentration changes rate without changing k when temperature and catalyst are unchanged.

Temperature does more than make particles move faster: the crucial kinetic effect is the increased fraction able to overcome Ea. A catalyst changes the available pathway and Ea rather than acting as a temperature increase.

26.2 Homogeneous and heterogeneous catalysts

Syllabus
9701–2028–2029
Topic
26.2
Level
A2

Classify catalysts by their phase relative to reactants

Catalyst type Phase relationship Example
homogeneous same phase as reactants Fe²⁺/Fe³⁺ ions with aqueous I⁻ and S₂O₈²⁻
heterogeneous different phase from reactants solid Fe with gaseous N₂ and H₂ in the Haber process

Both types provide an alternative route with lower activation energy and are regenerated overall. The classification depends on physical phases during reaction, not on whether the catalyst is a metal.

A catalyst may change chemically in individual steps even though it is reformed overall. 'Homogeneous' does not mean uniform appearance alone; the catalyst and reactants must share a phase.

Heterogeneous catalysis uses adsorption, bond weakening and desorption

Stage What happens at the solid surface Why it speeds reaction
1 adsorption reactants diffuse to and bond at active sites holds reacting species close and suitably oriented
2 activation/reaction adsorption weakens bonds within reactants; new bonds form provides a lower-energy surface pathway
3 desorption product–surface bonds break and products leave frees active sites for another cycle

In the Haber process, N₂ and H₂ adsorb on iron. Their bonds weaken, adsorbed atoms form N–H bonds, and NH₃ desorbs, leaving the iron surface available again.

In a catalytic converter, palladium, platinum and rhodium surfaces adsorb carbon monoxide and oxides of nitrogen. Surface reaction forms products including CO₂ and N₂, which desorb from the metal.

Adsorption is attachment to the surface, not absorption into the bulk or dissolution. Products must desorb; otherwise occupied active sites would stop further catalytic cycles.

A homogeneous catalyst is used in one step and reformed later

A homogeneous catalyst reacts in one elementary step to form a temporary different species, then is regenerated in a later step. Adding the steps cancels the catalyst cycle species and gives the overall reaction.

2 FeX3+(aq)+2 IX−(aq)→2 FeX2+(aq)+IX2(aq)\ce{2Fe^{3+}(aq) + 2I-(aq) -> 2Fe^{2+}(aq) + I2(aq)}

2 FeX2+(aq)+SX2OX8X2−(aq)→2 FeX3+(aq)+2 SOX4X2−(aq)\ce{2Fe^{2+}(aq) + S2O8^{2-}(aq) -> 2Fe^{3+}(aq) + 2SO4^{2-}(aq)}

Fe³⁺ is used in the first step and reformed in the second; equivalently the cycle may start from Fe²⁺. The two steps avoid the difficult direct collision between two negatively charged reactants, I⁻ and S₂O₈²⁻.

NOX2(g)+SOX2(g)→SOX3(g)+NO(g)\ce{NO2(g) + SO2(g) -> SO3(g) + NO(g)}

NO(g)+12 OX2(g)→NOX2(g)\ce{NO(g) + 1/2O2(g) -> NO2(g)}

NO₂ is used to oxidise SO₂ and is reformed when NO reacts with O₂. Cancelling NO/NO₂ across the cycle leaves the overall oxidation of SO₂ by oxygen.

Regenerated overall does not mean chemically unchanged at every stage. Identify the catalyst by consumed-then-reformed order; a formed-then-consumed species is an intermediate.