Rate law, rate constant and order of a reaction
The rate law links rate to concentrations: rate = k[A]ˣ[B]ʸ. The order x + y is found only by experiment and may be zero or a fraction.
What changes the rate, and the rate law
The rate of a reaction depends on the concentration of reactants (pressure, for gases), the temperature and the presence of a catalyst. At a fixed temperature, the rate depends on the concentrations of the reactants. For aA + bB → products, we write Rate = k[A]ˣ[B]ʸ. This is the rate law (or rate equation). k is the rate constant. The powers x and y may or may not equal a and b. They must be found by experiment, so you cannot write a rate law just by looking at the balanced equation.
Rate laws found by experiment
| Reaction | Rate law | Order |
|---|---|---|
| 2NO(g) + O₂(g) → 2NO₂(g) | k[NO]²[O₂] | 2 + 1 = 3 |
| CHCl₃ + Cl₂ → CCl₄ + HCl | k[CHCl₃][Cl₂]^½ | 1 + ½ = 1.5 |
| CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (H⁺) | k[CH₃COOC₂H₅]¹[H₂O]⁰ (water in large excess) | 1 |
| 2NH₃(g) → N₂(g) + 3H₂(g) (on hot platinum, 1130 K) | k[NH₃]⁰ = k | 0 |
Order of a reaction
The order is the sum of the powers of the concentration terms in the rate law. In rate = k[A]ˣ[B]ʸ, the order with respect to A is x, with respect to B is y, and the overall order is x + y. If the order is zero, the rate does not depend on concentration at all. Order can be a fraction, as in the chloroform example above. An elementary reaction happens in one step. A complex reaction happens in a series of elementary steps, and its order comes from the slowest step.
Units of the rate constant
| Order | Rate law | Unit of k |
|---|---|---|
| Zero | Rate = k | mol L⁻¹ s⁻¹ |
| First | Rate = k[A] | s⁻¹ |
| Second | Rate = k[A]² or k[A][B] | mol⁻¹ L s⁻¹ |
| Third | Rate = k[A]²[B] | mol⁻² L² s⁻¹ |
| n | Rate = k[A]ⁿ | (mol L⁻¹)¹⁻ⁿ s⁻¹ |
Worked example: rate law from initial rates
For A + B → products, these initial rates were measured at 298 K. Exp 1: [A] = 0.10, [B] = 0.10 mol L⁻¹, rate = 2.0 × 10⁻³ mol L⁻¹ min⁻¹. Exp 2: [A] = 0.20, [B] = 0.10, rate = 4.0 × 10⁻³. Exp 3: [A] = 0.10, [B] = 0.20, rate = 8.0 × 10⁻³. Find the rate law, the order and k.
- Let rate = k[A]ˣ[B]ʸ.
- Exp 2 ÷ Exp 1: [B] is the same and [A] doubles. 4.0 × 10⁻³ ÷ 2.0 × 10⁻³ = 2 = 2ˣ, so x = 1.
- Exp 3 ÷ Exp 1: [A] is the same and [B] doubles. 8.0 × 10⁻³ ÷ 2.0 × 10⁻³ = 4 = 2ʸ, so y = 2.
- Rate law: rate = k[A][B]². Overall order = 1 + 2 = 3.
- From Exp 1: k = rate ÷ ([A][B]²) = 2.0 × 10⁻³ ÷ (0.10 × 0.010) = 2.0 × 10⁻³ ÷ 1.0 × 10⁻³.
- k = 2.0 mol⁻² L² min⁻¹.
Rate = k[A][B]²; order 3; k = 2.0 mol⁻² L² min⁻¹
Activity: the starch clock reaction (iodide ion with hydrogen peroxide)
You need potassium iodide solution, dilute sulphuric acid, freshly made starch solution, dilute sodium thiosulphate solution, dilute hydrogen peroxide, beakers, measuring cylinders and a stopwatch.
- 1In a beaker, mix 20 mL KI solution, 10 mL dilute H₂SO₄, 2 mL starch and a small fixed volume (say 5 mL) of Na₂S₂O₃ solution.
- 2Add 10 mL H₂O₂, swirl and start the stopwatch at once.
- 3Slow step: H₂O₂ + 2I⁻ + 2H⁺ → I₂ + 2H₂O. Fast step: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, so iodine is removed as soon as it forms.
- 4★When all the thiosulphate is used up, free iodine meets starch and a blue-black colour appears suddenly. Stop the watch and record the time t.
- 5Repeat with 15, 10 and 5 mL KI, adding water each time so the total volume stays the same.
- 6Observation: the lower the iodide concentration, the longer the time. Since rate ∝ 1/t, a graph of 1/t against volume of KI is nearly a straight line: the rate rises with [I⁻].
- 7Other version (potassium iodate with sodium sulphite): IO₃⁻ + 3SO₃²⁻ → I⁻ + 3SO₄²⁻; then IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O. Sulphite turns the iodine back to iodide (I₂ + SO₃²⁻ + H₂O → 2I⁻ + SO₄²⁻ + 2H⁺) until it runs out; then starch turns blue-black. Time it at different iodate concentrations in the same way.
★ key step
Doctors choose how often to give a medicine from how fast the body breaks it down, which follows a rate law. That is why some tablets are taken every 6 hours and others once a day.
2 marks: “For a reaction, rate = k[A]²[B]^½. What is its overall order? Give the unit of k.” Order = 2 + ½ = 2.5. Unit of k = (mol L⁻¹)^(1 − 2.5) s⁻¹ = mol^(−3/2) L^(3/2) s⁻¹.
3 marks: “For 2A + B → C: doubling [A] at constant [B] makes the rate four times; doubling [B] at constant [A] leaves the rate unchanged. Write the rate law and the order. What happens to the rate if both are doubled?” 2ˣ = 4 so x = 2; 2ʸ = 1 so y = 0. Rate = k[A]²[B]⁰ = k[A]²; second order. Doubling both: rate becomes 2² × 2⁰ = 4 times.
2 marks: “Give two differences between the rate of a reaction and its rate constant.” Rate changes as concentrations change; k does not (it is fixed at a given temperature). Unit of rate is always mol L⁻¹ s⁻¹; unit of k depends on the order.
Order is what the experiment says, not what the equation says. Unit of k: (mol L⁻¹) to the power (1 − n), per second.