Measuring Reaction Rates – Rate Laws and Measurements (Senior 6 Chemistry) | Ganzaa

Rate Laws and Measurements

Lessons

Senior 6 · Chemistry

Rate Laws and Measurements

5 lessons

Lesson 2 of 5

Measuring Reaction Rates

25 min

You cannot count molecules, so follow a property

A rate is a change of concentration with time, but nobody can count the molecules in a flask as they react. Instead you choose a property that you can measure and that changes in step with the concentration: the mass of the flask when a gas escapes, the volume of gas collected, the depth of colour, the electrical conductivity, the pressure of a gas mixture, or the result of a titration. The choice depends on what the reaction does.

In Senior 3 you collected gas in a syringe and timed a cross disappearing. At A-level the aim is different: to turn each reading into a concentration, so that you can find orders and rate constants in the lessons that follow.

Definition

Continuous method: a property of the whole reaction mixture is recorded throughout the reaction without disturbing it (mass, gas volume, colour, conductivity, pressure). Example: reading a colorimeter every 15 s.

Sampling (discontinuous) method: small portions (aliquots) are withdrawn at intervals, the reaction in each is stopped, and each is analysed, usually by titration.

Quenching: stopping, or almost stopping, the reaction in a sample so that its composition stays fixed until it is analysed: by sudden dilution in ice-cold water, or by removing a catalyst or a reactant (e.g. adding sodium hydrogencarbonate to remove an acid catalyst).

Absorbance, $A$ : a measure of how much light of a chosen wavelength a solution absorbs. By the Beer–Lambert law, $A = \varepsilon c l$ , so at fixed path length $l$ the absorbance is proportional to the concentration $c$ of the coloured species.

Choosing a method

Property followed	Use it when…	Example
Loss of mass	a dense gas escapes (not $\mathrm{H_2}$ , whose mass is too small to weigh accurately)	$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}$
Volume of gas	a gas is produced; use a gas syringe if the gas dissolves in water	$\mathrm{Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)}$
Colour (colorimetry)	one reactant or product is coloured	iodine (brown) in the iodination of propanone; manganate(VII) (purple)
Titration of samples	an acid, a base or iodine is used up or formed	$\mathrm{I_2}$ titrated with thiosulfate; $\mathrm{OH^{-}}$ titrated with acid in ester hydrolysis
Electrical conductivity	the number or kind of ions changes	$\mathrm{(CH_3)_3CBr + H_2O \rightarrow (CH_3)_3COH + H^{+} + Br^{-}}$ : ions form, conductivity rises
Pressure (fixed volume)	the number of moles of gas changes	$\mathrm{2N_2O_5(g) \rightarrow 4NO_2(g) + O_2(g)}$ : pressure rises
Clock (time to a fixed point)	you only need initial rates	iodine clock; thiosulfate with acid; initial rate $\propto \frac{1}{t}$

Loss of mass

Figure 2.1: Following the reaction of marble chips with acid by the loss of mass.

The mass lost at any time equals the mass of $\mathrm{CO_2}$ given off, which converts to moles of acid used. The cotton wool matters: without it, fine acid spray carried out by the bubbles would also be "lost" and the rate would be overestimated.

Worked example 1: from mass loss to a rate of change of concentration

Problem. Excess marble chips are added to $50.0\,\mathrm{cm^3}$ of hydrochloric acid. In the first 60 s the mass falls by 0.44 g. Calculate the average rate of production of $\mathrm{CO_2}$ in $\mathrm{mol\,s^{-1}}$ and the average rate of fall of $[\mathrm{HCl}]$ in $\mathrm{mol\,dm^{-3}\,s^{-1}}$ . ( $M(\mathrm{CO_2}) = 44.0\,\mathrm{g\,mol^{-1}}$ )

Solution. $n(\mathrm{CO_2}) = \frac{0.44}{44.0} = 0.0100\,\mathrm{mol}$ , so the rate of production is $\frac{0.0100}{60} = 1.7 \times 10^{-4}\,\mathrm{mol\,s^{-1}}$ .

From the equation, 2 mol HCl give 1 mol $\mathrm{CO_2}$ : $n(\mathrm{HCl}) = 0.0200\,\mathrm{mol}$ . In $0.0500\,\mathrm{dm^3}$ , $\Delta[\mathrm{HCl}] = \frac{0.0200}{0.0500} = 0.400\,\mathrm{mol\,dm^{-3}}$ .

-\frac{\Delta[\mathrm{HCl}]}{\Delta t} = \frac{0.400}{60} = 6.7 \times 10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

Colorimetry

A colorimeter shines light of one colour through the sample and measures how much gets through (Figure 2.2). The filter is chosen to pass the colour that the sample absorbs, which is the complementary colour of the solution: blue light for a yellow-brown iodine solution, for example. The instrument is first set to zero absorbance with a cuvette of solvent, then calibrated with standard solutions of known concentration. The reaction is followed without disturbing it, and readings can be taken every few seconds.

Figure 2.2: The parts of a colorimeter.

Worked example 2: a calibration curve

Problem. Standard iodine solutions of 0.0010, 0.0020, 0.0030 and $0.0040\,\mathrm{mol\,dm^{-3}}$ give absorbances of 0.21, 0.42, 0.63 and 0.84. In a reaction that uses up iodine, the absorbance is 0.70 at the start and 0.49 after 120 s. Find the average rate of fall of $[\mathrm{I_2}]$ .

Solution. The calibration line passes through the origin with gradient $\frac{0.84}{0.0040} = 210\,\mathrm{dm^3\,mol^{-1}}$ , so $[\mathrm{I_2}] = \frac{A}{210}$ .

[\mathrm{I_2}]_0 = \frac{0.70}{210} = 3.33 \times 10^{-3}, \qquad [\mathrm{I_2}]_{120} = \frac{0.49}{210} = 2.33 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}

-\frac{\Delta[\mathrm{I_2}]}{\Delta t} = \frac{1.00 \times 10^{-3}}{120} = 8.3 \times 10^{-6}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

Worked example 3: sampling, quenching and titration

Problem. Propanone reacts with iodine in acid: $\mathrm{CH_3COCH_3 + I_2 \xrightarrow{H^{+}} CH_3COCH_2I + HI}$ . Every 5 minutes a $10.0\,\mathrm{cm^3}$ sample is run into sodium hydrogencarbonate solution and titrated with $0.0100\,\mathrm{mol\,dm^{-3}}$ sodium thiosulfate, $\mathrm{I_2 + 2S_2O_3^{2-} \rightarrow 2I^{-} + S_4O_6^{2-}}$ . The titres at 5 and 10 minutes are 15.6 and $13.2\,\mathrm{cm^3}$ . (a) Why is hydrogencarbonate used? (b) Find $[\mathrm{I_2}]$ in each sample and the average rate.

Solution. (a) It neutralises the $\mathrm{H^{+}}$ catalyst, so the reaction in the sample practically stops (quenching) while it is titrated.

(b) At 5 min: $n(\mathrm{S_2O_3^{2-}}) = 0.0156 \times 0.0100 = 1.56 \times 10^{-4}\,\mathrm{mol}$ , so $n(\mathrm{I_2}) = 7.8 \times 10^{-5}\,\mathrm{mol}$ and $[\mathrm{I_2}] = \frac{7.8 \times 10^{-5}}{0.0100} = 7.8 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ . At 10 min, in the same way, $[\mathrm{I_2}] = 6.6 \times 10^{-3}\,\mathrm{mol\,dm^{-3}}$ .

\text{rate} = \frac{(7.8 - 6.6) \times 10^{-3}}{300\,\mathrm{s}} = 4.0 \times 10^{-6}\,\mathrm{mol\,dm^{-3}\,s^{-1}}

If further samples show the same fall in every 5 minutes, the $[\mathrm{I_2}]$ –time graph is a straight line: the rate does not depend on $[\mathrm{I_2}]$ . Lesson 3 explains what this means.

Warning

Common mistakes

Following the loss of mass for a reaction that gives hydrogen. Hydrogen is so light that the change is within the balance's error.
Collecting carbon dioxide over water in an inverted burette; some dissolves, so the volume is too small. Use a gas syringe.
Forgetting the mole ratio when converting a titre or a mass of gas to a change in concentration.
Using a clock method for anything except an initial rate. " $\frac{1}{t}$ " measures the rate only while little of the reactant has been used.

Safety. Wear eye protection. Hydrochloric acid is an irritant. Propanone is highly flammable: no flames in the room. Iodine solutions stain and are harmful. Thiosulfate with acid releases sulfur dioxide: work in a well-ventilated room or fume cupboard, and pupils with asthma should not breathe the fumes.

Tip

Every rate method is a chain of three links: reading (g, $\mathrm{cm^3}$ , absorbance, titre) → moles (molar mass, molar volume, calibration, titration) → concentration (divide by the volume of the mixture). Exam questions usually break one link; find it.

Pearson UK and International Schools: Monitoring rates of reaction by recording a change in gas volume and a change in colour. Watch how the gas syringe and colorimeter readings are taken at regular intervals.

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Practice

Distinguish a continuous method from a sampling method, with one example of each.
Suggest a suitable method, with a reason, for following each reaction: (a) $\mathrm{Zn(s) + 2HCl(aq) \rightarrow ZnCl_2(aq) + H_2(g)}$ ; (b) $\mathrm{CH_3COOC_2H_5 + OH^{-} \rightarrow CH_3COO^{-} + C_2H_5OH}$ ; (c) $\mathrm{Br_2(aq) + HCOOH(aq) \rightarrow 2Br^{-}(aq) + 2H^{+}(aq) + CO_2(g)}$ .
Explain why a cotton wool plug is placed in the neck of the flask in Figure 2.1.
In 40 s, 36 cm³ of hydrogen (measured at room temperature and pressure, molar volume $24.0\,\mathrm{dm^3\,mol^{-1}}$ ) is collected from magnesium and $25.0\,\mathrm{cm^3}$ of hydrochloric acid. Calculate the average rate of fall of $[\mathrm{HCl}]$ .
A colorimeter gives $A = 0.36$ for $0.0020\,\mathrm{mol\,dm^{-3}}$ bromine. Assuming the Beer–Lambert law, what is $[\mathrm{Br_2}]$ when $A = 0.27$ ?
A $5.00\,\mathrm{cm^3}$ sample of a reaction mixture is quenched and needs $12.0\,\mathrm{cm^3}$ of $0.0200\,\mathrm{mol\,dm^{-3}}$ thiosulfate to react with its iodine. Find $[\mathrm{I_2}]$ in the mixture.
Evaluate the use of a "time for the cross to disappear" to compare the rates of the thiosulfate–acid reaction at different concentrations.

Show answersHide answers

Continuous: the property is followed without disturbing the mixture, e.g. colorimetry. Sampling: aliquots are removed, quenched and titrated, e.g. iodine with thiosulfate.
(a) Gas volume with a syringe (hydrogen is too light for mass loss). (b) Titration of $\mathrm{OH^{-}}$ in quenched samples, or conductivity (fast $\mathrm{OH^{-}}$ ions are replaced by slower ethanoate ions). (c) Colorimetry, because bromine is coloured and the products are colourless.
It lets $\mathrm{CO_2}$ escape but keeps in acid spray, which would otherwise add to the mass lost.
$n(\mathrm{H_2}) = \frac{36}{24\,000} = 1.5 \times 10^{-3}\,\mathrm{mol}$ ; $n(\mathrm{HCl}) = 3.0 \times 10^{-3}\,\mathrm{mol}$ ; $\Delta[\mathrm{HCl}] = \frac{3.0 \times 10^{-3}}{0.0250} = 0.12\,\mathrm{mol\,dm^{-3}}$ ; rate $= \frac{0.12}{40} = 3.0 \times 10^{-3}\,\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$[\mathrm{Br_2}] = 0.0020 \times \frac{0.27}{0.36} = 0.0015\,\mathrm{mol\,dm^{-3}}$ .
$n(\mathrm{S_2O_3^{2-}}) = 0.0120 \times 0.0200 = 2.40 \times 10^{-4}$ ; $n(\mathrm{I_2}) = 1.20 \times 10^{-4}\,\mathrm{mol}$ ; $[\mathrm{I_2}] = \frac{1.20 \times 10^{-4}}{0.00500} = 0.0240\,\mathrm{mol\,dm^{-3}}$ .
It gives an initial rate ( $\propto \frac{1}{t}$ ) because the same small amount of sulfur forms each time, and it needs no special equipment. But judging when the cross disappears is subjective, the depth of liquid must be the same each time, and a light sensor or colorimeter would be more precise.

Sources and credits

Image: AutoColorimeter.jpg by Calmkelp, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Image: Iodine-C Clock experiment.jpg by Daniel J. Lulu, via Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Monitoring rates of reaction by recording a) change in gas volumes b) a change in colour, Pearson UK & International Schools, embedded from YouTube.
The text, tables, apparatus drawings, illustrative data, worked examples and Practice items were written and drawn by Ganzaa.
This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

1You cannot count molecules, so follow a property

2Choosing a method

3Loss of mass

4Colorimetry

You cannot count molecules, so follow a property

Choosing a method

Loss of mass

Colorimetry