4.1.2Unit 4. Rates, Equilibria and Further Organic Chemistry
Rate Equations and Orders — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions
5 min read 6 parts Lesson 11 of 56 Unit quiz
Part 1 of 6The Rate Equation
A rate equation shows how the rate depends on the concentration of each species.
Form
- For A + B → products, rate = k[A]ᵐ[B]ⁿ, where m and n are the orders with respect to A and B.
Rate constant
- The constant k links rate to concentrations and is specific to one reaction at one temperature.
Overall order
- This is the sum of the individual orders, m + n.
Experimental only
- The orders can only be found by experiment and cannot be read from the coefficients of the balanced equation.
Part 2 of 6Meaning of Order
Meaning of Order
Zero order
- Changing the concentration has no effect on the rate, so the species does not appear in the rate equation.
First order
- Doubling the concentration doubles the rate, because the rate is directly proportional to the concentration.
Second order
- Doubling the concentration increases the rate by a factor of four.
Catalysts
- A catalyst or an acid that does not appear in the stoichiometric equation can still appear in the rate equation.
| Order | Concentration × 2 | Rate changes by |
|---|---|---|
| 0 | × 2 | × 1 |
| 1 | × 2 | × 2 |
| 2 | × 2 | × 4 |
| 1 | × 3 | × 3 |
Part 3 of 6Units of k
1
Rearrange the rate equation to make k the subject, k = rate ÷ (concentration terms).
2
Substitute the units: rate is mol dm⁻³ s⁻¹ and concentration is mol dm⁻³.
3
Cancel the units algebraically, keeping the powers of each unit.
4
A first-order reaction gives k in s⁻¹ and a second-order reaction gives dm³ mol⁻¹ s⁻¹.
5
A third-order reaction gives dm⁶ mol⁻² s⁻¹, and a zero-order reaction gives mol dm⁻³ s⁻¹.
Part 4 of 6Using the Equation
Calculating rate
- Substitute the concentrations and k into the rate equation to predict the rate of reaction.
Calculating k
- Use known concentrations and a measured rate and rearrange to find k.
Temperature
- k increases with temperature, which is why rate rises as the mixture is heated.
Interpretation
- The order shows how many particles of a species are involved in the rate-determining step.
Part 5 of 6Worked Example
Question
- The rate equation for a reaction is rate = k[A][B]².
- When [A] = 0.20 mol dm⁻³ and [B] = 0.10 mol dm⁻³ the rate is 4.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
- Find k with units, then the rate when [A] = 0.40 and [B] = 0.20 mol dm⁻³.
- k = rate ÷ ([A][B]²) = 4.0 × 10⁻⁴ ÷ (0.20 × 0.10²) = 4.0 × 10⁻⁴ ÷ 2.0 × 10⁻³ = 0.20.
- Unitsmol dm⁻³ s⁻¹ ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹.
- New rate = 0.20 × 0.40 × 0.20² = 0.20 × 0.40 × 0.040 = 3.2 × 10⁻³ mol dm⁻³ s⁻¹.
Answer
- k = 0.20 dm⁶ mol⁻² s⁻¹;
- new rate = 3.2 × 10⁻³ mol dm⁻³ s⁻¹
Part 6 of 6Exam Checklist & Tips
4.1.2 – Key Points for the Exam
- Rate equation An expression relating rate to the concentrations of reactants, each raised to its order.
- Order of reaction The power to which the concentration of a species is raised in the rate equation.
- Overall order The sum of the orders with respect to all species in the rate equation.
- Rate constant The constant k in the rate equation, which depends on temperature.
- Zero order A species whose concentration does not affect the rate.
Common Mistakes
- Taking the orders from the coefficients in the balanced equation instead of from experiment.
- Giving the units of k without working them out from the order of the reaction.
- Forgetting that k changes with temperature, so it is not constant across different temperatures.
- Including a zero-order reactant in the rate equation.
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