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How do we find the rate equation of a reaction, and what does it tell us about the mechanism?

Orders of reaction and the rate equation, the rate constant and its units, concentration-time and rate-concentration graphs, half-life, the rate-determining step, and the Arrhenius equation.

An OCR H432 module 5 answer on rates: orders and the rate equation, the rate constant and its units, concentration-time and rate-concentration graphs, 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. Orders and the rate equation
  3. The rate constant and its units
  4. Graphs and half-life
  5. The rate-determining step
  6. The Arrhenius equation
  7. Examples in context
  8. Try this

What this topic is asking

OCR specification point 5.1.1 wants you to define orders of reaction, write and use the rate equation rate=k[A]m[B]n\text{rate} = k[A]^m[B]^n, work out the rate constant and its units, interpret concentration-time and rate-concentration graphs (including constant half-life for first order), link the rate equation to the rate-determining step, and use the Arrhenius equation. This is the quantitative version of the collision theory met in Module 3.

Orders and the rate equation

Orders are usually found by the initial rates method: change one reactant concentration at a time and see how the initial rate responds. If doubling a concentration doubles the rate, it is first order; if it quadruples the rate, second order; if there is no change, zero order.

The rate constant and its units

Graphs and half-life

On a concentration-time graph, a first-order reactant has a constant half-life (the time for the concentration to halve is the same throughout), which is a quick test for first order. The gradient at any point is the rate at that instant. On a rate-concentration graph: zero order gives a horizontal line, first order a straight line through the origin, and second order an upward curve.

The rate-determining step

The Arrhenius equation

A plot of ln⁡k\ln k against 1/T1/T is a straight line with gradient −Ea/R-E_a/R (so Ea=−R×gradientE_a = -R \times \text{gradient}) and intercept ln⁡A\ln A.

Examples in context

Example 1. Iodine clock reactions. A clock reaction releases a sudden colour change after a fixed amount of product forms, letting the initial rate be timed for different starting concentrations, the standard practical for finding orders.

Example 2. Why food spoils faster when warm. The Arrhenius equation explains why a modest temperature rise sharply increases the rate constant of the reactions that spoil food, which is why refrigeration slows decay.

Try this

Q1. State how you would recognise a first-order reactant from a concentration-time graph. [1 mark]

  • Cue. The half-life is constant.

Q2. A reaction is first order in X and second order in Y. Write the rate equation and give the units of kk. [2 marks]

  • Cue. rate=k[X][Y]2\text{rate} = k[\text{X}][\text{Y}]^2; overall third order, so kk has units dm6mol−2s−1\text{dm}^6\text{mol}^{-2}\text{s}^{-1}.

Exam-style practice questions

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

OCR 20194 marksThe initial rate of a reaction between P and Q was measured. Doubling [P][\text{P}] at constant [Q][\text{Q}] doubled the rate; doubling [Q][\text{Q}] at constant [P][\text{P}] quadrupled the rate. (a) Deduce the order with respect to P and to Q and write the rate equation. (b) State the overall order.
Show worked answer →

(a) Doubling [P][\text{P}] doubles the rate, so the reaction is first order in P (1). Doubling [Q][\text{Q}] multiplies the rate by four (222^2), so it is second order in Q (1). The rate equation is rate=k[P][Q]2\text{rate} = k[\text{P}][\text{Q}]^2 (1).

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

Markers reward the correct order in each reactant from the proportional reasoning, the full rate equation, and the overall order.

OCR 20213 marksFor the reaction in the previous part, the rate is 4.0×10−3 mol dm−3s−14.0 \times 10^{-3}\ \text{mol dm}^{-3}\text{s}^{-1} when [P]=0.10[\text{P}] = 0.10 and [Q]=0.20 mol dm−3[\text{Q}] = 0.20\ \text{mol dm}^{-3}. Calculate the rate constant kk and give its units.
Show worked answer →

Rearranging rate=k[P][Q]2\text{rate} = k[\text{P}][\text{Q}]^2 gives k=rate[P][Q]2k = \dfrac{\text{rate}}{[\text{P}][\text{Q}]^2} (1).

k=4.0×10−3(0.10)(0.20)2=4.0×10−30.10×0.040=4.0×10−34.0×10−3=1.0 (1)k = \dfrac{4.0 \times 10^{-3}}{(0.10)(0.20)^2} = \dfrac{4.0 \times 10^{-3}}{0.10 \times 0.040} = \dfrac{4.0 \times 10^{-3}}{4.0 \times 10^{-3}} = 1.0\ (1).

Units: mol dm−3s−1(mol dm−3)3=mol−2dm6s−1\dfrac{\text{mol dm}^{-3}\text{s}^{-1}}{(\text{mol dm}^{-3})^3} = \text{mol}^{-2}\text{dm}^6\text{s}^{-1} (1).

Markers reward the rearrangement, the numerical value, and the correct units derived from the rate equation.

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