Topic 11 · page 7 of 21

Orders from initial-rate data

Compare two experiments in which only one concentration changes. How much the initial rate changes tells you the order for that reagent. Then use any experiment to find k.

14 min read11.5(iii)

Read first: Initial rates and continuous monitoring, The rate constant and its units

The comparison 11.5(iii)

An initial-rate experiment gives a table of initial concentrations and initial rates. To find the order for one reagent, pick two experiments in which only that reagent's concentration changes. Everything else is the same, so any change in rate is caused by that one concentration.

MethodOrder from a table of initial rates

  1. Choose two experiments in which only one concentration changes.
  2. Work out the factor by which that concentration changes, and the factor by which the rate changes.
  3. Match them: rate factor 1 gives order 0; rate factor equal to the concentration factor gives order 1; rate factor equal to its square gives order 2.
  4. Repeat for each reagent, then write the rate equation.
  5. Substitute one whole experiment into the rate equation to find k, and work out its units.
If [A] is multiplied by and the rate is multiplied by the order with respect to A is
2 1 0
2 2 1
2 4 2
3 3 1
3 9 2

A full worked example 11.5(iii)

Worked exampleTwo reagents

The initial rate of a reaction between A and B was measured in four experiments (illustrative data).

Experiment [A] / mol dm⁻³ [B] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 3.0 × 10⁻⁴
2 0.20 0.10 6.0 × 10⁻⁴
3 0.10 0.30 2.7 × 10⁻³
4 0.20 0.30 5.4 × 10⁻³

Find the order with respect to A and to B, write the rate equation and calculate k.

  1. Experiments 1 and 2: [A] doubles, [B] is constant, and the rate doubles (6.0 × 10⁻⁴ ÷ 3.0 × 10⁻⁴ = 2). So the reaction is first order with respect to A.
  2. Experiments 1 and 3: [B] triples, [A] is constant, and the rate becomes 2.7 × 10⁻³ ÷ 3.0 × 10⁻⁴ = 9 times faster, which is 3². So the reaction is second order with respect to B.
  3. Rate equation: rate = k[A][B]², third order overall.
  4. Use experiment 1: k = 3.0 × 10⁻⁴ ÷ (0.10 × 0.10²) = 3.0 × 10⁻⁴ ÷ 1.0 × 10⁻³ = 0.30.
  5. Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)³ = dm⁶ mol⁻² s⁻¹. So k = 0.30 dm⁶ mol⁻² s⁻¹.
  6. Check with experiment 4: 0.30 × 0.20 × 0.30² = 5.4 × 10⁻³, which matches the table.

The check in the last step is worth doing in the exam. If the answer disagrees with an experiment you did not use, an order is wrong.

Zero order and three reagents 11.5(iii)

Worked exampleThree reagents, one of them zero order

(Illustrative data.)

Experiment [P] / mol dm⁻³ [Q] / mol dm⁻³ [R] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.10 0.10 0.10 1.6 × 10⁻³
2 0.20 0.10 0.10 6.4 × 10⁻³
3 0.10 0.20 0.10 3.2 × 10⁻³
4 0.10 0.10 0.20 1.6 × 10⁻³
  1. 1 and 2: [P] ×2, rate × 4, so second order with respect to P.
  2. 1 and 3: [Q] ×2, rate × 2, so first order with respect to Q.
  3. 1 and 4: [R] ×2, rate unchanged, so zero order with respect to R. R does not appear in the rate equation.
  4. rate = k[P]²[Q]. Using experiment 1: k = 1.6 × 10⁻³ ÷ (0.10² × 0.10) = 1.6 dm⁶ mol⁻² s⁻¹.
  5. To predict a new rate, for [P] = 0.30, [Q] = 0.050 and [R] = 0.50 mol dm⁻³: rate = 1.6 × 0.30² × 0.050 = 7.2 × 10⁻³ mol dm⁻³ s⁻¹. The concentration of R makes no difference.

Zero order does not mean R is unnecessary. The reaction may still use it up. It means that the rate-determining step does not involve R, so changing [R] does not change the rate. Page 10 takes this further.

When two concentrations change 11.5(iii)

A table does not always give you a clean pair. If two concentrations change at once, find one order first from a clean pair, divide its effect out of the rate factor, and what is left belongs to the other reagent.

Worked exampleTwo concentrations change

In experiment 1, [A] = 0.10 and [B] = 0.10 mol dm⁻³ and the rate is 3.0 × 10⁻⁴ mol dm⁻³ s⁻¹. In experiment 2, [A] = 0.20 and [B] = 0.10 and the rate is 6.0 × 10⁻⁴. In experiment 3, [A] = 0.40 and [B] = 0.20 and the rate is 4.8 × 10⁻³. Find the order with respect to B.

  1. Experiments 1 and 2 are a clean pair. [A] ×2 gives rate ×2, so A is first order.
  2. From experiment 2 to 3, the rate rises by 4.8 × 10⁻³ ÷ 6.0 × 10⁻⁴ = 8.
  3. [A] doubles, which accounts for a factor of 2. The remaining factor is 8 ÷ 2 = 4.
  4. [B] doubled and caused a factor of 4, so B is second order.

Check yourself

  1. Use these data (illustrative) to find the orders with respect to X and Y, write the rate equation and calculate k with its units.
Experiment [X] / mol dm⁻³ [Y] / mol dm⁻³ Initial rate / mol dm⁻³ s⁻¹
1 0.020 0.020 2.0 × 10⁻⁶
2 0.040 0.020 4.0 × 10⁻⁶
3 0.020 0.040 4.0 × 10⁻⁶
  1. Use your rate equation from question 1 to calculate the initial rate when [X] = 0.060 and [Y] = 0.010 mol dm⁻³.
  2. In a fourth experiment [X] = 0.040 and [Y] = 0.040 mol dm⁻³. Predict the initial rate without calculating k.
  3. For a reaction of P and Q, experiment 1 has [P] = 0.10, [Q] = 0.10 and rate 4.0 × 10⁻³ mol dm⁻³ s⁻¹. Experiment 2 has [P] = 0.10, [Q] = 0.30 and rate 4.0 × 10⁻³. Experiment 3 has [P] = 0.30, [Q] = 0.10 and rate 3.6 × 10⁻². Deduce the rate equation and calculate k, with units.
  4. In a clock reaction the time to the end point is 48 s when [Z] = 0.10 mol dm⁻³ and 12 s when [Z] = 0.20 mol dm⁻³. Deduce the order with respect to Z.
  5. A student compares experiments 2 and 3 in question 1 and says that, since the rates are equal, neither X nor Y affects the rate. Explain why this is wrong.

Answers cover these until you have tried the questions

  1. 1 and 2: [X] ×2, rate ×2, so first order in X. 1 and 3: [Y] ×2, rate ×2, so first order in Y. Rate equation: rate = k[X][Y]. k = 2.0 × 10⁻⁶ ÷ (0.020 × 0.020) = 2.0 × 10⁻⁶ ÷ 4.0 × 10⁻⁴ = 5.0 × 10⁻³. Units: dm³ mol⁻¹ s⁻¹. So k = 5.0 × 10⁻³ dm³ mol⁻¹ s⁻¹.
  2. rate = 5.0 × 10⁻³ × 0.060 × 0.010 = 5.0 × 10⁻³ × 6.0 × 10⁻⁴ = 3.0 × 10⁻⁶ mol dm⁻³ s⁻¹.
  3. Both concentrations double relative to experiment 1, and each is first order, so the rate is × 2 × 2 = × 4: 2.0 × 10⁻⁶ × 4 = 8.0 × 10⁻⁶ mol dm⁻³ s⁻¹.
  4. 1 and 2: [Q] ×3, rate unchanged, so zero order in Q. 1 and 3: [P] ×3, rate × (3.6 × 10⁻² ÷ 4.0 × 10⁻³ = 9), so second order in P. rate = k[P]². k = 4.0 × 10⁻³ ÷ 0.10² = 4.0 × 10⁻³ ÷ 1.0 × 10⁻² = 0.40. Units: (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = dm³ mol⁻¹ s⁻¹. So k = 0.40 dm³ mol⁻¹ s⁻¹.
  5. Rate ∝ 1 ÷ t, so the rate rises by 48 ÷ 12 = 4 when [Z] doubles. 4 = 2², so the reaction is second order with respect to Z.
  6. Between experiments 2 and 3, [X] halves and [Y] doubles at the same time. Because each is first order, the effects cancel (× ½ × 2 = × 1). The rates are equal, but both concentrations still affect the rate. To find an order, only one concentration may change.