Topic 11 · page 6 of 21

Initial rates and continuous monitoring

Two ways to collect data for finding orders. The initial-rate method uses many separate experiments and one rate from each. Continuous monitoring uses one experiment and follows the whole curve.

13 min read11.4(i)11.4(ii)

Read first: Choosing a technique to follow a reaction

Page 5 chose a property to measure. This page says how to use it. There are two designs, and each leads to a different way of finding the order.

The initial-rate method 11.4(i)

In this method you carry out separate experiments. In each one, the starting concentration of one reagent is different and everything else is the same. For every experiment you find the initial rate, the rate at t = 0. Then you see how the initial rate depends on the concentration you changed.

MethodInitial-rate method

  1. Choose the reagent to investigate. Prepare several mixtures in which only its initial concentration changes.
  2. Keep the temperature, the total volume and the concentrations of all other reagents the same.
  3. For each mixture, follow a property against time and plot a concentration–time graph.
  4. Draw a tangent at t = 0. The initial rate is minus its gradient.
  5. Compare the initial rates with the initial concentrations. Repeat for each other reagent in turn.
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. A straight tangent is drawn at time zero. It meets the time axis at about 58 seconds. A dashed gradient triangle on the tangent runs from 0 s and 0.080 to 40 s and 0.0246.
The tangent at t = 0 gives the initial rate. Illustrative data.

Worked exampleInitial rate from a tangent

The tangent at t = 0 passes through (0 s, 0.0800 mol dm⁻³) and (40 s, 0.0246 mol dm⁻³). Calculate the initial rate.

  1. Δ[A] = 0.0246 − 0.0800 = −0.0554 mol dm⁻³
  2. Δt = 40 − 0 = 40 s
  3. gradient = −0.0554 ÷ 40 = −1.385 × 10⁻³ mol dm⁻³ s⁻¹
  4. initial rate = 1.4 × 10⁻³ mol dm⁻³ s⁻¹

The initial rate is used because at t = 0 the concentrations are the ones you made up, known exactly. Later, the reactants have changed in ways you do not know, and products may be speeding up, slowing down or reversing the reaction.

Clock reactions 11.4(i)

Drawing a tangent for every experiment is slow. A clock reaction is an acceptable approximation of the initial-rate method, and the specification says so.

In a clock reaction, a fixed, small amount of a product is formed, and a sudden change tells you when it has been made. The time t to the change is measured.

Two conditions make this valid. The amount formed is small, so the concentrations of the reactants hardly change before the end point. And the amount is the same in every experiment. Page 20 gives the iodine clock in full.

Worked exampleOrder from clock times

Experiments with [X] = 0.020, 0.040 and 0.080 mol dm⁻³ give times of 120, 60 and 30 s. Find the order with respect to X.

  1. Rate ∝ 1 ÷ t, so the relative rates are 1 ÷ 120, 1 ÷ 60 and 1 ÷ 30. These are 8.3 × 10⁻³, 1.7 × 10⁻² and 3.3 × 10⁻² s⁻¹.
  2. Each time [X] doubles, the time halves and the rate doubles.
  3. The reaction is first order with respect to X.

Continuous monitoring 11.4(ii)

In this method you carry out one experiment and measure the property at regular intervals throughout. You plot a concentration–time graph, or a volume–time graph if you collect a gas. The shape of the curve and its half-lives give the order (page 8), and tangents at several points give rates to plot against concentration (page 9).

A volume–time graph for a product rises and levels off at the final volume, VfinalV_\text{final}. The amount of reactant left is proportional to Vfinal−VV_\text{final} - V. Use that quantity when you look for half-lives.

Worked exampleHalf-lives from gas volumes

A reaction gives a gas, and the final volume is 40 cm³. The volume collected is 0, 20, 30 and 35 cm³ at 0, 30, 60 and 90 s. What is the order with respect to the reactant?

  1. Reactant left ∝ 40 − V: 40, 20, 10 and 5 cm³ at 0, 30, 60 and 90 s.
  2. The reactant halves from 40 to 20 in 30 s, from 20 to 10 in 30 s and from 10 to 5 in 30 s.
  3. The half-life is constant, so the reaction is first order with respect to the reactant.

One run gives a whole curve, but it only tells you the order for the substance being followed. To find the other orders you must repeat the experiment with that other substance's concentration changed, or keep the others in large excess so that their concentrations barely change.

Comparing the methods 11.4(i)11.4(ii)

Initial-rate method Continuous monitoring
Experiments several, one concentration changed each time one
Data from each one initial rate (or one time, for a clock) a whole concentration–time or volume–time curve
Order found from how the initial rate changes with concentration half-lives, or rate against concentration
Orders found for every reagent, one by one for the monitored substance
Suits clock reactions, techniques with sampling colorimetry, gas volume, mass change

Check yourself

  1. Describe how the initial-rate method could be used to find the order with respect to reagent A in a reaction between A and B.
  2. Explain why initial rates, and not rates later in the reaction, are used to compare experiments.
  3. In a clock reaction the time to the end point increases from 40 s to 80 s when [Y] is halved. State the order with respect to Y and explain your answer.
  4. A tangent drawn at t = 0 on a concentration–time graph passes through (0 s, 0.120 mol dm⁻³) and (80 s, 0.000 mol dm⁻³). Calculate the initial rate.
  5. A gas is given off and its final volume is 60 cm³. Volumes at 0, 50, 100 and 150 s are 0, 30, 45 and 52.5 cm³. Deduce the order with respect to the reactant.
  6. Give one advantage of continuous monitoring over the initial-rate method, and one limitation.

Answers cover these until you have tried the questions

  1. Carry out several experiments with different initial concentrations of A, and the same concentration of B, the same temperature and the same total volume. For each, plot a concentration–time graph and draw a tangent at t = 0 to find the initial rate (or time a clock reaction). Compare how the initial rate changes as [A] changes.
  2. At t = 0 the concentrations are known exactly, and no product has built up to affect the rate. Later in the reaction the concentrations have changed by unknown amounts, so rates cannot be compared fairly.
  3. The rate ∝ 1 ÷ t, so doubling the time halves the rate. Halving [Y] halves the rate, so the reaction is first order with respect to Y.
  4. gradient = (0.000 − 0.120) ÷ (80 − 0) = −1.5 × 10⁻³. Initial rate = 1.5 × 10⁻³ mol dm⁻³ s⁻¹.
  5. Reactant left ∝ 60 − V = 60, 30, 15 and 7.5 cm³ at 0, 50, 100 and 150 s. Each time it halves in 50 s: the half-life is constant, so the reaction is first order.
  6. Advantage (any one): only one experiment is needed; it gives the whole curve; there is no need to draw many tangents at t = 0. Limitation: it gives the order only for the substance being followed (unless the others are in large excess).