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How do rate equations, orders of reaction and the rate-determining step reveal a reaction mechanism?

Rate equations and orders of reaction, the rate constant, determining orders from concentration-time and rate-concentration data, the rate-determining step, and the Arrhenius relationship with activation energy.

An Eduqas A-Level Chemistry PI3 answer on rate equations and orders of reaction, the rate constant, determining orders from data, the rate-determining step and the Arrhenius equation.

Generated by Claude Opus 4.812 min answer

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  1. What this topic is asking
  2. Rate equations and orders
  3. Determining orders from data
  4. The rate-determining step
  5. The Arrhenius equation
  6. Examples in context
  7. Try this

What this topic is asking

Eduqas topic PI3 makes kinetics quantitative: rate equations, orders of reaction, the rate constant, finding orders from concentration-time and rate-concentration data, the rate-determining step and how the rate equation reveals the mechanism, and the Arrhenius relationship between rate constant and activation energy. It builds directly on the qualitative collision theory of topic C2.3.

Rate equations and orders

The rate constant kk is the proportionality constant in the rate equation; its units depend on the overall order and it increases with temperature.

Determining orders from data

There are two standard methods. From concentration-time data: a zero-order reactant gives a straight-line fall, a first-order reactant gives a curve with a constant half-life, and a second-order reactant gives a steeper curve with an increasing half-life. From initial-rate (rate-concentration) data: compare how the rate changes when one concentration is varied and the others held constant.

The rate-determining step

The Arrhenius equation

The rate constant rises sharply with temperature because more molecules exceed the activation energy. The Arrhenius equation expresses this:

Examples in context

Example 1. The iodine clock. Timing how long a fixed amount of product takes to form at different starting concentrations gives initial rates, from which orders are deduced; it is a classic Eduqas practical for finding a rate equation.

Example 2. Catalysed mechanisms. A catalyst can change the rate-determining step (and hence the rate equation) by opening a lower-energy route, which is why catalysed and uncatalysed versions of the same reaction can show different kinetics.

Try this

Q1. State what a constant half-life on a concentration-time graph tells you about the order. [1 mark]

  • Cue. The reaction is first order with respect to that reactant.

Q2. A reaction has rate =k[A]2= k[\text{A}]^2. Deduce the units of kk if rate is in mol dm−3s−1\text{mol dm}^{-3}\text{s}^{-1}. [1 mark]

  • Cue. k=rate[A]2k = \dfrac{\text{rate}}{[\text{A}]^2}, so units are mol dm−3s−1mol2dm−6=mol−1dm3s−1\dfrac{\text{mol dm}^{-3}\text{s}^{-1}}{\text{mol}^2\text{dm}^{-6}} = \text{mol}^{-1}\text{dm}^3\text{s}^{-1}.

Exam-style practice questions

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

Eduqas 20195 marksFor the reaction A+B→products\text{A} + \text{B} \rightarrow \text{products}, doubling [A][\text{A}] (with [B][\text{B}] constant) doubles the rate, and doubling [B][\text{B}] (with [A][\text{A}] constant) quadruples the rate. (a) Determine the order with respect to A and to B. (b) Write the rate equation and give the overall order. (c) State the units of the rate constant.
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(a) First order in A (rate proportional to [A]1[\text{A}]^1); second order in B (doubling [B][\text{B}] gives 22=42^2 = 4 times the rate) (2).

(b) Rate =k[A][B]2= k[\text{A}][\text{B}]^2; overall order =1+2=3= 1 + 2 = 3 (2).

(c) Rate has units mol dm−3s−1\text{mol dm}^{-3}\text{s}^{-1} and concentration cubed has units mol3dm−9\text{mol}^3\text{dm}^{-9}, so kk has units mol−2dm6s−1\text{mol}^{-2}\text{dm}^{6}\text{s}^{-1} (1).

Eduqas 20214 marksThe reaction NO2+CO→NO+CO2\text{NO}_2 + \text{CO} \rightarrow \text{NO} + \text{CO}_2 has the rate equation rate =k[NO2]2= k[\text{NO}_2]^2. (a) Deduce the order with respect to CO. (b) What does the rate equation suggest about the rate-determining step?
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(a) The order with respect to CO is zero, because CO does not appear in the rate equation, so changing [CO][\text{CO}] has no effect on the rate (1).

(b) The rate-determining step (the slowest step) involves two NO2\text{NO}_2 molecules and no CO (1). A possible slow step is NO2+NO2→NO3+NO\text{NO}_2 + \text{NO}_2 \rightarrow \text{NO}_3 + \text{NO} (1), with CO reacting in a later, fast step (1).

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