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What controls how fast a reaction goes, and how can the rate equation reveal the mechanism?

Explain rate of reaction using collision theory and the Boltzmann distribution, deduce rate equations and orders from experimental data, define the rate constant and half-life, and use the rate-determining step to propose a reaction mechanism, including the action of catalysts

A focused answer to the H2 Chemistry learning outcome on kinetics. Collision theory and the Boltzmann distribution, deducing order and the rate constant from data, half-life of a first-order reaction, the rate-determining step, and how catalysts (including enzymes) lower activation energy.

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to explain reaction rate using collision theory and the Boltzmann distribution, deduce rate equations and orders from experimental data, define and find the rate constant and half-life, and use the rate-determining step to propose mechanisms and explain catalysis. Order determination from data and the Boltzmann-distribution explanation are recurring high-mark questions.

The answer

Collision theory and the Boltzmann distribution

A reaction occurs when molecules collide with sufficient energy (at least the activation energy EaE_a) and the correct orientation. At any temperature, molecular energies follow the Boltzmann distribution, a skewed curve where the area under the curve to the right of EaE_a represents the fraction of molecules able to react.

Factors that increase rate:

  • Higher concentration or pressure: more frequent collisions.
  • Higher temperature: shifts the distribution to higher energies, so a much larger fraction exceeds EaE_a. This is the dominant reason a 1010 degree rise can roughly double the rate.
  • Catalyst: provides an alternative pathway with lower EaE_a, so a larger fraction of molecules can react.
  • Larger surface area: more contact for heterogeneous reactions.

Rate equations and order

The rate equation is determined experimentally, not from the stoichiometric equation:

rate=k[A]m[B]n\text{rate} = k[\text{A}]^m[\text{B}]^n

Here mm and nn are the orders with respect to A and B, and m+nm + n is the overall order. Deduce each order by changing one concentration at a time:

  • rate unchanged when concentration doubles: zero order
  • rate doubles when concentration doubles: first order
  • rate quadruples when concentration doubles: second order

The rate constant

The rate constant kk is found by substituting one experimental run into the rate equation. Its units depend on the overall order:

  • first order: s−1^{-1}
  • second order: mol−1^{-1} dm3^3 s−1^{-1}
  • zero order: mol dm−3^{-3} s−1^{-1}

kk increases with temperature (more molecules above EaE_a). It is independent of concentration.

Half-life and first-order reactions

For a first-order reaction, the half-life (time for the concentration to halve) is constant, independent of starting concentration. A constant half-life read from a concentration-time graph is the signature of first order:

t1/2=ln⁡2kt_{1/2} = \frac{\ln 2}{k}

The rate-determining step and mechanism

A multi-step reaction proceeds at the pace of its slowest step, the rate-determining step (RDS). Only species involved up to and including the RDS appear in the rate equation. So the rate equation is direct evidence for the mechanism: a species that is second order is involved twice in or before the slow step, and a species absent from the rate equation enters after the slow step.

Catalysis

A catalyst speeds a reaction by providing a route of lower EaE_a and is regenerated.

  • Homogeneous catalysts are in the same phase (e.g. aqueous ions catalysing aqueous reactions).
  • Heterogeneous catalysts are in a different phase (e.g. solid iron in the Haber process), working by adsorption of reactants onto the surface.
  • Enzymes are biological catalysts with active sites that bind substrates.

Examples in context

Example 1. Iodine clock in the lab. A JC practical times how long a fixed amount of product takes to appear as concentrations are varied. Because the time is inversely proportional to the initial rate, plotting 1/time1/\text{time} against concentration reveals the order. This is the standard way Paper 4 and Paper 2 test order determination without continuous monitoring.

Example 2. The role of the RDS in nucleophilic substitution. In organic chemistry, a tertiary halogenoalkane reacts by an SN1 mechanism whose slow step is the formation of a carbocation from the halogenoalkane alone. The rate equation is therefore first order in the halogenoalkane and zero order in the nucleophile, which is direct kinetic evidence for the two-step SN1 mechanism met later in Organic Chemistry.

Try this

Q1. State two factors that increase the rate of a reaction and explain each using collision theory. [3 marks]

  • Cue. Higher concentration (more frequent collisions); higher temperature (larger fraction above EaE_a). Either pair acceptable.

Q2. A first-order reaction has a half-life of 4040 s. Calculate the rate constant. [2 marks]

  • Cue. k=ln⁡2/t1/2=0.693/40=1.73×10−2k = \ln 2 / t_{1/2} = 0.693/40 = 1.73 \times 10^{-2} s−1^{-1}.

Q3. For rate=k[X]2[Y]\text{rate} = k[\text{X}]^2[\text{Y}], state the overall order and the units of kk. [2 marks]

  • Cue. Overall order 3; units mol−2^{-2} dm6^6 s−1^{-1}.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Specimen (9729)5 marksThe reaction between substances P and Q was studied. When [P] was doubled at constant [Q], the rate doubled. When [Q] was doubled at constant [P], the rate quadrupled. Deduce the rate equation, state the overall order, and explain what these results suggest about the species present in the rate-determining step.
Show worked answer →

Determine the order with respect to each reactant.

Doubling [P] doubles the rate, so rate is proportional to [P] to the power 1: first order in P.

Doubling [Q] quadruples the rate (factor of 4 = 2 squared), so rate is proportional to [Q] to the power 2: second order in Q.

Rate equation: rate=k[P][Q]2\text{rate} = k[\text{P}][\text{Q}]^2.

Overall order = 1 + 2 = 3 (third order).

The rate-determining step (the slowest step) involves the species in the rate equation: one molecule of P and two molecules of Q (or species derived from them) are involved up to and including the slow step.

Markers reward each order with justification, the full rate equation, the overall order, and the link to the rate-determining step.

2022 (style)4 marksExplain, using the Boltzmann distribution, why a small increase in temperature produces a large increase in the rate of a reaction, and why a catalyst increases the rate without being consumed.
Show worked answer →

The Boltzmann distribution shows the spread of molecular energies at a given temperature, with only molecules above the activation energy Ea able to react.

Raising the temperature shifts the distribution to higher energies, so a much larger fraction of molecules now exceeds Ea. Because the fraction above Ea rises steeply (exponentially), even a small temperature rise gives a large rate increase. (More frequent collisions also contribute, but the energy effect dominates.)

A catalyst provides an alternative reaction pathway with a lower activation energy. A larger fraction of molecules now exceeds the new, lower Ea, so the rate rises. The catalyst is regenerated, so it is not consumed overall.

Markers reward the link between Ea and the area under the curve, the steep rise in fraction above Ea, the alternative-pathway definition of a catalyst, and the regeneration point.

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