Skip to main content
ScotlandChemistrySyllabus dot point

How do rate equations reveal the mechanism of a reaction?

Rate equations of the form rate = k[A]^m[B]^n, the order of reaction with respect to each reactant and overall, the rate constant and its units, and the link between the rate equation, the rate-determining step and a reaction mechanism.

An SQA Advanced Higher Chemistry answer on kinetics, covering rate equations of the form rate = k[A]^m[B]^n, finding the order of reaction with respect to each reactant and overall, calculating the rate constant and its units, and using the rate equation to identify the rate-determining step and propose a reaction mechanism.

Generated by Claude Opus 4.814 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this key area is asking
  2. Rate equations and order
  3. Finding orders from initial-rate data
  4. The rate-determining step and mechanisms
  5. Examples in context
  6. Try this

What this key area is asking

The SQA wants you to use rate equations of the form rate=k[A]m[B]n\text{rate} = k[A]^m[B]^n, to determine the order of reaction from initial-rate data, to calculate the rate constant and its units, and to link the rate equation to the rate-determining step and a reaction mechanism. Finding orders from data, working out the units of kk, and reasoning about mechanisms are reliable exam earners.

Rate equations and order

Orders cannot be read off the balanced equation; they must be found by experiment. A reaction is zero order in a reactant if its rate is unaffected by that concentration, first order if the rate is proportional to the concentration, and second order if the rate is proportional to the concentration squared.

Finding orders from initial-rate data

The standard method compares experiments in which one concentration is changed while the others are held constant:

  • If doubling a concentration leaves the rate unchanged, the order is zero.
  • If doubling a concentration doubles the rate, the order is one (21=22^1 = 2).
  • If doubling a concentration multiplies the rate by four, the order is two (22=42^2 = 4).

The rate-determining step and mechanisms

For example, if a reaction A+2BA + 2B \rightarrow products has the rate equation rate=k[A][B]\text{rate} = k[A][B], the rate-determining step must involve one molecule of A and one of B; the second B reacts in a later, faster step that does not affect the rate. Reasoning between rate equation and mechanism is a key Advanced Higher skill.

Examples in context

Kinetics underpins how chemists control and understand reactions. The hydrolysis of a tertiary haloalkane is first order overall (rate=k[haloalkane]\text{rate} = k[\text{haloalkane}]), which is evidence for an SN1S_N1 mechanism with a slow first step forming a carbocation; a primary haloalkane is second order, evidence for a one-step SN2S_N2 mechanism. Enzyme-catalysed reactions show how concentration affects rate up to a saturation point. Industrially, knowing the order with respect to each reactant lets a chemist choose concentrations that maximise rate efficiently, and the temperature dependence of kk explains why warming a reaction (raising kk) speeds it up far more than the increase in collision frequency alone would suggest.

Try this

Q1. Define the term overall order of reaction. [1 mark]

  • Cue. The sum of the powers of the concentration terms in the rate equation.

Q2. A reaction is first order in A and zero order in B. Write the rate equation. [1 mark]

  • Cue. rate=k[A]\text{rate} = k[A].

Q3. State what the rate equation tells you about the rate-determining step. [1 mark]

  • Cue. Only species in or before the rate-determining step appear in the rate equation.

Exam-style practice questions

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

SQA AH 20194 marksFor the reaction A+BA + B \rightarrow products, the following initial-rate data were obtained. Experiment 1: [A]=0.10[A] = 0.10, [B]=0.10[B] = 0.10, rate =2.0×103= 2.0 \times 10^{-3}. Experiment 2: [A]=0.20[A] = 0.20, [B]=0.10[B] = 0.10, rate =4.0×103= 4.0 \times 10^{-3}. Experiment 3: [A]=0.10[A] = 0.10, [B]=0.20[B] = 0.20, rate =8.0×103= 8.0 \times 10^{-3}. (a) Find the order with respect to A and to B. (b) Write the rate equation. (c) Calculate the rate constant k.
Show worked answer →

Markers reward each order, the rate equation and the rate constant with units.

(a) From experiments 1 and 2, doubling [A][A] at constant [B][B] doubles the rate, so the reaction is first order in A. From experiments 1 and 3, doubling [B][B] at constant [A][A] multiplies the rate by four (×4=22\times 4 = 2^2), so the reaction is second order in B.

(b) The rate equation is rate=k[A][B]2\text{rate} = k[A][B]^2.

(c) Using experiment 1:

k=rate[A][B]2=2.0×1030.10×(0.10)2=2.0×1031.0×103=2.0 mol2l2s1k = \frac{\text{rate}}{[A][B]^2} = \frac{2.0 \times 10^{-3}}{0.10 \times (0.10)^2} = \frac{2.0 \times 10^{-3}}{1.0 \times 10^{-3}} = 2.0 \text{ mol}^{-2}\text{l}^2\text{s}^{-1}

SQA AH specimen2 marksA reaction has the rate equation rate=k[X]\text{rate} = k[X]. The proposed two-step mechanism is: Step 1 (slow) XYX \rightarrow Y; Step 2 (fast) Y+ZY + Z \rightarrow products. Explain why this mechanism is consistent with the rate equation.
Show worked answer →

The answer must link the slow step to the rate equation.

The slowest step of a mechanism is the rate-determining step, so only the species involved in or before it appear in the rate equation. Here step 1 (the slow step) involves only X, so the rate depends only on [X][X], giving rate=k[X]\text{rate} = k[X].

Z appears only in the fast step 2, which is not rate-determining, so [Z][Z] does not appear in the rate equation. The mechanism is therefore consistent with the observed first-order rate equation.

Related dot points

Sources & how we know this