Topic 11 · page 9 of 21

Orders from rate–concentration graphs

Plot the rate against the concentration. A horizontal line is zero order, a straight line through the origin is first order, and an upward curve is second order. Tangents on a concentration–time graph supply the rates.

13 min read11.5(ii)11.4(ii)

Read first: Orders from concentration–time graphs

Three shapes 11.5(ii)

The rate equation is a relationship between the rate and a concentration, so plotting rate on the y axis against concentration on the x axis shows the order directly.

Three graphs of rate against concentration of A. Zero order: a horizontal line, so the rate does not change when the concentration changes. First order: a straight line through the origin, so doubling the concentration doubles the rate. Second order: a curve rising more and more steeply through the origin, so doubling the concentration gives four times the rate.
Rate against concentration. Zero order, first order and second order. These are shapes only, with no numbers on the rate axis.
Order with respect to A Shape of the rate–concentration graph Double [A]
0 horizontal line rate unchanged
1 straight line through the origin rate × 2
2 upward curve through the origin rate × 4

For a first-order reaction, rate = k[A], so the gradient of the line is k. Its units are s⁻¹, the rate unit divided by the concentration unit.

Getting the graph from one experiment 11.5(ii)11.4(ii)

A continuous-monitoring experiment gives a concentration–time graph, not a rate–concentration one. You build the second graph from the first with tangents, as in Topic 9 and on page 6.

MethodRate–concentration points from a concentration–time graph

  1. Draw a smooth curve through the concentration–time points.
  2. Choose several times, spread along the curve, and read the concentration at each.
  3. Draw a tangent at each time. Find its gradient with a large triangle. The rate is minus the gradient.
  4. Plot rate against concentration for each point, and look at the shape.

Worked exampleTangents give a straight line

Tangents were drawn to a curve at four points (illustrative readings, from a reaction with a half-life of 40 s).

[A] / mol dm⁻³ rate / mol dm⁻³ s⁻¹ rate ÷ [A] / s⁻¹
0.080 1.39 × 10⁻³ 0.0174
0.040 6.93 × 10⁻⁴ 0.0173
0.020 3.47 × 10⁻⁴ 0.0174
0.010 1.73 × 10⁻⁴ 0.0173
  1. Each time [A] halves, the rate halves. The rate is proportional to [A].
  2. The last column is the same within rounding, about 0.0173 s⁻¹, so the points lie on a straight line through the origin.
  3. The reaction is first order with respect to A and k = 0.0173 s⁻¹ (the gradient).
A graph of rate in mol per dm cubed per second, times ten to the minus three, against concentration of A in mol per dm cubed. Four points lie on a straight line through the origin. A dashed gradient triangle on the line gives a gradient of 0.0173 per second.
The four points from the table. They lie on a straight line through the origin, so the reaction is first order and the gradient is k.

Confirming second order 11.5(ii)

A curve through the origin is second order, but a curve alone is hard to judge. To confirm, plot rate against [A]². For rate = k[A]², this is a straight line through the origin with gradient k. Or check numerically that rate ÷ [A]² is constant.

Worked exampleSecond order from rates

Rates were found from tangents at four points (illustrative readings).

[A] / mol dm⁻³ rate / mol dm⁻³ s⁻¹ rate ÷ [A] / s⁻¹ rate ÷ [A]² / dm³ mol⁻¹ s⁻¹
0.080 4.0 × 10⁻³ 0.050 0.625
0.040 1.0 × 10⁻³ 0.025 0.625
0.020 2.5 × 10⁻⁴ 0.0125 0.625
0.010 6.25 × 10⁻⁵ 0.00625 0.625
  1. Each time [A] halves, the rate falls to a quarter.
  2. Rate ÷ [A] is not constant, so the reaction is not first order.
  3. Rate ÷ [A]² is constant, so it is second order, and k = 0.625 dm³ mol⁻¹ s⁻¹.

Check yourself

  1. Describe the shape of the rate–concentration graph for a reaction that is (a) zero order, (b) first order, (c) second order in A.
  2. A rate–concentration graph for A is a straight line through the origin. When [A] = 0.050 mol dm⁻³ the rate is 1.5 × 10⁻³ mol dm⁻³ s⁻¹. State the order and calculate k, with units.
  3. A rate–concentration graph is a horizontal line at 5.0 × 10⁻⁴ mol dm⁻³ s⁻¹. State the order and give k, with units.
  4. Rates are 1.0 × 10⁻⁴, 4.0 × 10⁻⁴ and 9.0 × 10⁻⁴ mol dm⁻³ s⁻¹ at [A] = 0.10, 0.20 and 0.30 mol dm⁻³. Deduce the order and calculate k, with units.
  5. Tangents to a concentration–time graph have gradients of −3.2 × 10⁻³, −1.6 × 10⁻³ and −8.0 × 10⁻⁴ mol dm⁻³ s⁻¹ where [A] is 0.080, 0.040 and 0.020 mol dm⁻³. Deduce the order and calculate k.
  6. A student plots rate against [A] and gets a curve rising through the origin. State the order and describe a second graph that would confirm it.

Answers cover these until you have tried the questions

  1. (a) A horizontal line. (b) A straight line through the origin. (c) An upward curve through the origin, rising more steeply as [A] increases.
  2. A straight line through the origin means first order. k = 1.5 × 10⁻³ ÷ 0.050 = 0.030 s⁻¹ (the rate unit divided by the concentration unit).
  3. The rate does not depend on [A], so the reaction is zero order. rate = k, so k = 5.0 × 10⁻⁴ mol dm⁻³ s⁻¹.
  4. The rate rises by 4 when [A] doubles (0.10 → 0.20) and by 9 when [A] triples (0.10 → 0.30): second order. k = rate ÷ [A]² = 1.0 × 10⁻⁴ ÷ 0.10² = 1.0 × 10⁻² dm³ mol⁻¹ s⁻¹. Check: 4.0 × 10⁻⁴ ÷ 0.20² = 1.0 × 10⁻².
  5. The rates (minus the gradients) are 3.2 × 10⁻³, 1.6 × 10⁻³ and 8.0 × 10⁻⁴, so the rate halves each time [A] halves: first order. k = 3.2 × 10⁻³ ÷ 0.080 = 0.040 s⁻¹. Check: 8.0 × 10⁻⁴ ÷ 0.020 = 0.040.
  6. Second order. Plot rate against [A]²: if the reaction is second order this is a straight line through the origin, with gradient k.