Topic 9 · page 2 of 15

Rate from the gradient of a tangent

On a curve the rate changes from moment to moment. Draw a tangent at the point you want, find its gradient, and you have the rate at that point, including the initial rate at t = 0.

13 min read9.39.3(ii)

Read first: Measuring rate with time

Why a curve needs a tangent 9.39.3(ii)

Concentration does not fall at a steady speed. It falls fastest at the start, when there is most reactant, and more slowly as the reactant is used up. Here are illustrative figures for a reactant A:

Time / s [A] / mol dm⁻³
0 0.100
20 0.0607
40 0.0368
60 0.0223
80 0.0135
100 0.0082

The graph is a curve, and the gradient of a curve is different at every point. The gradient at one point is the gradient of the tangent, the straight line that just touches the curve there and does not cross it.

Two graphs of the concentration of reactant A against time, falling from 0.100 mol per dm cubed. Both curves fall steeply at first and flatten. Left graph: a tangent at time zero touches the curve at 0.100 and reaches the time axis at 40 seconds, so its gradient is minus 2.5 times ten to the minus three mol per dm cubed per second. Right graph: a tangent at 40 seconds is less steep, with gradient minus 9.2 times ten to the minus four. Each tangent has a dashed gradient triangle below it.
Left: the tangent at t = 0 gives the initial rate. Right: a tangent at t = 40 s gives a lower rate.

MethodFinding a rate from a graph

  1. Draw a smooth curve through the points. Do not join them dot to dot.
  2. Choose the time you want. Place a ruler on the curve at that point so that the curve is balanced evenly on both sides of the ruler. Draw the tangent and extend it well past the point.
  3. Draw a large right-angled triangle on the tangent, at least half the width of the graph, using points on the tangent (not on the curve).
  4. Gradient = Δy ÷ Δx. Read both from the axes, with units.
  5. For a reactant the gradient is negative. The rate is minus the gradient.

Worked examples with the reactant graph 9.3(ii)

Worked exampleInitial rate

The tangent at t = 0 passes through (0, 0.100) and reaches the time axis at (40, 0). Use the triangle between (10, 0.075) and (36, 0.010) on the tangent. Find the initial rate.

  1. Δ[A] = 0.010 − 0.075 = −0.065 mol dm⁻³
  2. Δt = 36 − 10 = 26 s
  3. gradient = −0.065 ÷ 26 = −2.5 × 10⁻³ mol dm⁻³ s⁻¹
  4. rate = −gradient = 2.5 × 10⁻³ mol dm⁻³ s⁻¹

Worked exampleRate at 40 s

The tangent at t = 40 s passes through (0, 0.0736) and (80, 0). Use the triangle between (20, 0.0552) and (70, 0.0092). Find the rate at 40 s.

  1. Δ[A] = 0.0092 − 0.0552 = −0.0460 mol dm⁻³
  2. Δt = 70 − 20 = 50 s
  3. gradient = −0.0460 ÷ 50 = −9.2 × 10⁻⁴ mol dm⁻³ s⁻¹
  4. rate = 9.2 × 10⁻⁴ mol dm⁻³ s⁻¹

The rate at 40 s is about 0.37 of the initial rate. It is lower because the reactant has been used up, so collisions between particles are less frequent.

Graphs for a product 9.3(ii)

A product graph rises and flattens, so the gradient is positive and the rate is the gradient itself. Here is the volume of gas collected in an experiment (illustrative data):

Graph of volume of gas collected against time. The curve rises steeply from the origin and levels off near 40 cubic centimetres by 120 seconds. A tangent drawn at time zero passes through the origin and through 40 seconds and 40 cubic centimetres. A dashed gradient triangle on the tangent has sides 20 seconds and 20 cubic centimetres, so the initial rate is 1.0 cubic centimetre per second.
The tangent at t = 0 passes through the origin and (40 s, 40 cm³).

Worked exampleInitial rate from a gas graph

Use the triangle between (20 s, 20 cm³) and (40 s, 40 cm³) on the tangent.

  1. ΔV = 40 − 20 = 20 cm³
  2. Δt = 40 − 20 = 20 s
  3. rate = 20 ÷ 20 = 1.0 cm³ s⁻¹

The unit of a gradient is always the unit of y divided by the unit of x. With volume on y and time on x it is cm³ s⁻¹. With concentration on y it is mol dm⁻³ s⁻¹.

Check yourself

  1. Explain why the rate of a reaction is greatest at the start.
  2. A tangent to a concentration–time curve passes through (0, 0.091) and (60, 0.000). Calculate the rate at the point of contact, in mol dm⁻³ s⁻¹.
  3. On a graph of gas volume against time, a tangent at t = 0 passes through (0, 0) and (50 s, 25 cm³). Calculate the initial rate.
  4. A graph of mass lost against time has a tangent that passes through (20 s, 0.20 g) and (100 s, 0.40 g). Calculate the rate at the point of contact.
  5. A student joins two points on a curve with a ruler and uses the gradient as the rate at t = 0. State what she has actually calculated and why it is wrong.
  6. The gradient of a tangent to a falling reactant graph is −1.5 × 10⁻³ mol dm⁻³ s⁻¹. State the rate.

Answers cover these until you have tried the questions

  1. At the start the reactant concentration is highest, so the particles are closest together and collide most often. More collisions per second means more successful collisions per second. The rate falls as the reactant is used up.
  2. gradient = (0.000 − 0.091) ÷ (60 − 0) = −1.52 × 10⁻³. Rate = 1.5 × 10⁻³ mol dm⁻³ s⁻¹ (2 significant figures).
  3. rate = 25 ÷ 50 = 0.50 cm³ s⁻¹.
  4. Δm = 0.40 − 0.20 = 0.20 g and Δt = 100 − 20 = 80 s. Rate = 0.20 ÷ 80 = 2.5 × 10⁻³ g s⁻¹.
  5. She has calculated the mean rate between the two points (a chord). It is wrong because the rate changes along the curve, so the rate at one instant is the gradient of the tangent at that point.
  6. A rate is positive, so the rate is 1.5 × 10⁻³ mol dm⁻³ s⁻¹.