Topic 11 · page 8 of 21

Orders from concentration–time graphs

A straight line means zero order. For a curve, measure successive half-lives. If they are constant the reaction is first order, and if each is double the last it is second order.

13 min read11.5(i)11.2

Read first: Half-life, Initial rates and continuous monitoring

Three shapes 11.5(i)

A concentration–time graph from one experiment shows how the concentration of a reactant falls. Its shape depends on the order with respect to that reactant. The three graphs below are for three different reactions, each starting at 0.080 mol dm⁻³, with the times at which the concentration halves marked.

Three graphs of concentration of A against time, each starting at 0.080 mol per dm cubed. Zero order: a straight line falling to zero at 160 seconds, with successive half-lives of 80, 40 and 20 seconds. First order: a curve that flattens, with each half-life 40 seconds. Second order: a curve that falls steeply and then has a long tail, with successive half-lives of 20, 40 and 80 seconds.
Zero, first and second order. The half-lives get shorter, stay constant, or get longer.
Order with respect to A Shape Successive half-lives Why
0 straight line downwards get shorter the same amount reacts each second, so less time is needed to remove half of what is left
1 curve that flattens constant rate is proportional to concentration
2 steeper curve, then a long tail each is double the last rate is proportional to the concentration squared, so it falls off faster than the concentration

Zero order 11.5(i)

If the graph is a straight line, the gradient is constant, so the rate does not change as the concentration falls. The rate does not depend on [A]: the reaction is zero order. You do not need half-lives. The rate equals the size of the gradient, and it equals k.

Worked exampleZero order from a table

[A] is 0.400 mol dm⁻³ at 0 s, 0.300 at 25 s, 0.200 at 50 s and 0.100 at 75 s. State the order and calculate the rate.

  1. [A] falls by 0.100 mol dm⁻³ in every 25 s. The graph is a straight line, so the rate is constant.
  2. The reaction is zero order with respect to A.
  3. Rate = 0.100 ÷ 25 = 4.0 × 10⁻³ mol dm⁻³ s⁻¹, and this is also k, in mol dm⁻³ s⁻¹.

A zero-order line only lasts while some reactant is left. It stops when the concentration reaches zero.

First or second order 11.5(i)11.2

Both give a curve that flattens, so the shape alone does not tell them apart. Measure half-lives.

MethodOrder from a curve

  1. If the graph is a straight line, the order is 0.
  2. If it is a curve, measure at least two successive half-lives, showing the construction lines on the graph.
  3. If they are equal, the order is 1.
  4. If each is about double the one before, the order is 2.
  5. Half-lives that get shorter go with a straight-line graph: the order is 0.

Worked exampleSecond order from half-lives

[A] is 0.080 mol dm⁻³ at 0 s, 0.040 at 20 s, 0.020 at 60 s and 0.010 at 140 s. Find the order.

  1. First half-life: 0.080 → 0.040 takes 20 − 0 = 20 s.
  2. Second half-life: 0.040 → 0.020 takes 60 − 20 = 40 s.
  3. Third half-life: 0.020 → 0.010 takes 140 − 60 = 80 s.
  4. The half-lives double each time, so the reaction is second order with respect to A.

Check yourself

  1. Describe how the half-lives change with time for (a) a zero-order reaction, (b) a first-order reaction and (c) a second-order reaction.
  2. [A] is 0.400 mol dm⁻³ at 0 s, 0.300 at 25 s, 0.200 at 50 s and 0.100 at 75 s. State the order and calculate the rate.
  3. [A] is 0.64 mol dm⁻³ at 0 s, 0.32 at 10 s, 0.16 at 30 s and 0.080 at 70 s. Deduce the order with respect to A.
  4. A first-order reaction starts at 0.200 mol dm⁻³ and has a half-life of 15 s. How long does it take to reach 0.0250 mol dm⁻³?
  5. Explain why you cannot decide between first and second order from the shape of a curve alone, and say what you would measure.
  6. A zero-order reaction has a rate of 2.5 × 10⁻³ mol dm⁻³ s⁻¹. How long does it take for [A] to fall from 0.060 to 0.010 mol dm⁻³?

Answers cover these until you have tried the questions

  1. (a) Shorter each time. (b) Constant. (c) Longer: each is double the previous one.
  2. [A] falls by the same 0.100 in each 25 s, so the graph is a straight line: zero order. Rate = 0.100 ÷ 25 = 4.0 × 10⁻³ mol dm⁻³ s⁻¹.
  3. 0.64 → 0.32 takes 10 s; 0.32 → 0.16 takes 30 − 10 = 20 s; 0.16 → 0.080 takes 70 − 30 = 40 s. The half-lives double each time: second order.
  4. 0.200 → 0.100 → 0.0500 → 0.0250 is three half-lives. Time = 3 × 15 = 45 s.
  5. Both give a curve that flattens, with a tail, so the eye cannot separate them. Measure successive half-lives: constant means first order, and each double the last means second order.
  6. Δ[A] = 0.060 − 0.010 = 0.050 mol dm⁻³. Time = 0.050 ÷ 2.5 × 10⁻³ = 20 s.