Topic 11 · page 3 of 21

Half-life

The half-life is the time for a concentration to halve. Read two or more successive half-lives from a concentration-time graph, and if they are all equal the reaction is first order.

11 min read11.1(vi)11.2

Read first: Rate, rate equation and order

What half-life means 11.1(vi)

Half-life is written t½. It is a time, so its unit is seconds, minutes or hours. It describes how fast a reactant is used up without needing a tangent or a gradient.

Reading half-lives from a graph 11.2

On a concentration–time graph, pick a concentration, then find the time for it to fall to half. Then pick the new concentration and do it again. Each half-life starts where the last one ended.

A graph of concentration of A in mol per dm cubed against time in seconds. The curve starts at 0.080 and falls, getting less steep. Marked points show the concentration halving from 0.080 at 0 seconds to 0.040 at 40 seconds, 0.020 at 80 seconds, 0.010 at 120 seconds and 0.005 at 160 seconds. A row of double arrows under the time axis marks the four intervals between them, each labelled 40 seconds.
Illustrative data. Four successive half-lives, each 40 s.

MethodMeasuring successive half-lives

  1. Draw a smooth curve through the points.
  2. Choose a starting concentration, for example 0.080 mol dm⁻³, and read its time.
  3. Find half of it, 0.040, and read its time from the curve. The difference is the first half-life.
  4. Start again at 0.040 and find the time for it to reach 0.020. That is the second half-life.
  5. Repeat once more. Show the construction lines on the graph.

Worked exampleHalf-lives from the graph

Read the times from the graph above and find each half-life.

  1. [A] = 0.080 at 0 s and 0.040 at 40 s. First half-life = 40 − 0 = 40 s.
  2. [A] = 0.040 at 40 s and 0.020 at 80 s. Second half-life = 80 − 40 = 40 s.
  3. [A] = 0.020 at 80 s and 0.010 at 120 s. Third half-life = 120 − 80 = 40 s.

Constant half-life means first order 11.2

Here every half-life is the same, 40 s, however much reactant is left. That is the signature of a first-order reaction.

Why does this happen? In a first-order reaction the rate is proportional to the concentration. When the concentration halves, the rate halves, so there is half as much to react and it reacts at half the speed. The time to lose half of what is there stays the same.

A constant half-life lets you predict how much remains.

Worked exampleTime for the reactant to fall to one eighth

A first-order reaction has a half-life of 40 s. How long does it take for [A] to fall from 0.080 mol dm⁻³ to 0.010 mol dm⁻³?

  1. 0.080 → 0.040 → 0.020 → 0.010 is three halvings.
  2. Three half-lives = 3 × 40 = 120 s.

After n half-lives, the fraction of the reactant left is (12)n(\tfrac{1}{2})^n. After 4 half-lives it is 116\tfrac{1}{16}, which is 6.25%.

Half-lives that are not constant also tell you something. They shorten for a zero-order reaction and lengthen for a second-order one (page 8).

Check yourself

  1. Define half-life.
  2. The concentration of A in a reaction is 0.160 mol dm⁻³ at 0 s, 0.080 at 30 s, 0.040 at 60 s, 0.020 at 90 s and 0.010 at 120 s. Find the half-lives and state the order with respect to A.
  3. A first-order reaction has a half-life of 25 s. How long does it take for the concentration to fall from 0.048 mol dm⁻³ to 0.0060 mol dm⁻³?
  4. What percentage of the reactant is left after four half-lives of a first-order reaction?
  5. A student reads from a graph that [A] reaches 0.040 mol dm⁻³ at 40 s, 0.020 at 80 s and 0.010 at 120 s from the start. She writes "the half-lives are 40, 80 and 120 s, so they increase and the reaction is second order". Explain her mistake and give the correct conclusion.
  6. Explain why a first-order reaction has a half-life that does not depend on the starting concentration.

Answers cover these until you have tried the questions

  1. The time taken for the concentration of a reactant to fall to half of its initial value.
  2. 0.160 → 0.080 takes 30 s; 0.080 → 0.040 takes 60 − 30 = 30 s; 0.040 → 0.020 takes 90 − 60 = 30 s; 0.020 → 0.010 takes 120 − 90 = 30 s. The half-life is a constant 30 s, so the reaction is first order with respect to A.
  3. 0.048 → 0.024 → 0.012 → 0.0060 is three halvings. Time = 3 × 25 = 75 s.
  4. (½)⁴ = 1/16 = 0.0625, so 6.25% is left.
  5. She measured each time from t = 0, so she found the time for one, two and three half-lives. Measured correctly, each half-life starts where the last ended: 40 − 0 = 40 s, 80 − 40 = 40 s and 120 − 80 = 40 s. They are constant, so the reaction is first order.
  6. The rate of a first-order reaction is proportional to the concentration. When the concentration is halved, the rate is halved, so the reaction still takes the same time to use up half of what is present. The half-life therefore does not depend on the starting concentration.