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26.1 Rate equations, orders and rate constants

Syllabus
9701–2028–2029
Topic
26.1
Level
A2

Learning objectives

26.1.1And use the terms rate equation, order• Explain and use the terms rate equation, order of reaction, overall order of reaction, rate constant, half-life, rate-determining step and intermediate26.1.2Rate equations and reaction order• The initial rates method and half-life method- (a) understand and use rate equations of the form rate = k [A]m[B]n (for which m and n are 0, 1 or 2)- (b) deduce the order of a reaction from concentration–time graphs or from experimental data relating to- (c) interpret experimental data in graphical form, including concentration–time and rate–concentration graphs- (d) calculate an initial rate using concentration data- (e) construct a rate equation26.1.3Show understanding that the half-life• (a) show understanding that the half-life of a first-order reaction is independent of concentration: (b) use the half-life of a first-order reaction in calculations26.1.4The numerical value of a rate constant, e.g• Calculate the numerical value of a rate constant, e.g. by:- (a) using the initial rates and the rate equation- (b) using the half-life, t- 2- 1 , and the equation k = 0.693 / t- 2- 126.1.5Mechanisms and rate-determining step• For a multi-step reaction:- (a) suggest a reaction mechanism that is consistent with the rate equation and the equation for the overall reaction- (b) predict the order that would result from a given reaction mechanism and rate-determining step- (c) deduce a rate equation using a given reaction mechanism and rate-determining step for a given reaction- (d) identify an intermediate or catalyst from a given reaction mechanism- (e) identify the rate determining step from a rate equation and a given reaction mechanism26.1.6Effect of temperature change on the rate• Describe qualitatively the effect of temperature change on the rate constant and hence the rate of a reaction

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.