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How can we express reaction rate as an equation and find the mechanism?

Rate equations, orders of reaction and the rate constant k, determining orders from initial-rate and concentration-time data, the Arrhenius equation, and using the rate-determining step to deduce a mechanism.

A focused answer to AQA A-Level Chemistry 3.1.9, covering rate equations and orders of reaction, the rate constant k, determining orders from data, the Arrhenius equation, and the rate-determining step in mechanisms.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The rate equation
  3. Finding orders from data
  4. The rate constant and Arrhenius equation
  5. The rate-determining step
  6. Try this

What this dot point is asking

AQA wants you to write rate equations and define order and the rate constant, determine orders from initial-rate and concentration-time data, use the Arrhenius equation, and link orders to the rate-determining step to propose a mechanism.

The rate equation

A reaction is zero order in a species if its concentration does not affect the rate, first order if rate is proportional to concentration, and second order if rate is proportional to concentration squared. The orders reflect how many particles of each species are involved up to and including the slowest step, which is why they cannot be predicted from the stoichiometric equation: the balanced equation describes the overall change, not the step-by-step mechanism. A species can even appear in the rate equation without appearing in the overall equation (a catalyst), or appear in the overall equation but be zero order (it reacts only after the rate-determining step).

Finding orders from data

Initial-rate method: change one reactant's concentration while keeping others constant and see how the initial rate changes. If doubling concentration doubles the rate, the order is 1; if it quadruples the rate, the order is 2; if the rate is unchanged, the order is 0.

Concentration-time graphs: a zero-order reactant gives a straight-line decrease (constant rate, so the gradient is constant); a first-order reactant gives an exponential decay with a constant half-life (the time for the concentration to halve is the same at every point, which is the diagnostic test for first order). A rate-concentration graph (rate plotted against concentration) is also used: a horizontal line is zero order, a straight line through the origin is first order, and an upward curve is second order. The units of the rate constant kk depend on the overall order: s−1\text{s}^{-1} for first order, mol−1 dm3 s−1\text{mol}^{-1}\,\text{dm}^3\,\text{s}^{-1} for second order, and mol dm−3 s−1\text{mol dm}^{-3}\,\text{s}^{-1} for zero order, which is why deriving kk's units from the rate equation is a routine exam step.

The rate constant and Arrhenius equation

The rate constant kk increases sharply with temperature, described by the Arrhenius equation:

The rate-determining step

Try this

Q1. State what is meant by the order of reaction with respect to a reactant. [1 mark]

  • Cue. The power to which its concentration is raised in the rate equation.

Q2. A reactant is first order. Describe its concentration-time graph. [1 mark]

  • Cue. Exponential decay with a constant half-life.

Exam-style practice questions

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

AQA 20194 marksWhen the concentration of A is doubled the rate doubles, and when the concentration of B is doubled the rate quadruples. Deduce the rate equation and the overall order.
Show worked answer →

Doubling [A][A] doubles the rate (21=22^1 = 2), so the reaction is first order with respect to A.

Doubling [B][B] quadruples the rate (22=42^2 = 4), so the reaction is second order with respect to B.

The rate equation is therefore rate=k[A][B]2\text{rate} = k[A][B]^2, and the overall order is 1+2=31 + 2 = 3 (third order).

Markers reward each order deduced from the factor change and the combined rate equation with overall order.

AQA 20224 marksA reaction is first order in X\text{X} and zero order in Y\text{Y}. When [X]=0.20 mol dm−3[\text{X}] = 0.20\ \text{mol dm}^{-3} and [Y]=0.10 mol dm−3[\text{Y}] = 0.10\ \text{mol dm}^{-3}, the rate is 4.0×10−3 mol dm−3 s−14.0 \times 10^{-3}\ \text{mol dm}^{-3}\,\text{s}^{-1}. Write the rate equation and calculate the rate constant, giving its units.
Show worked answer →

Because the reaction is first order in X\text{X} and zero order in Y\text{Y}, the rate equation is rate=k[X]\text{rate} = k[\text{X}] (the [Y]0[\text{Y}]^0 term equals 1 and is omitted).

Rearrange: k=rate[X]=4.0×10−30.20=2.0×10−2k = \dfrac{\text{rate}}{[\text{X}]} = \dfrac{4.0 \times 10^{-3}}{0.20} = 2.0 \times 10^{-2}.

Units: mol dm−3 s−1mol dm−3=s−1\dfrac{\text{mol dm}^{-3}\,\text{s}^{-1}}{\text{mol dm}^{-3}} = \text{s}^{-1}, so k=2.0×10−2 s−1k = 2.0 \times 10^{-2}\ \text{s}^{-1}.

Markers reward the rate equation omitting the zero-order term, the rearrangement, the value, and the units of s−1\text{s}^{-1} (characteristic of an overall first-order reaction).

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