Rate of Chemical Reactions

Lessons

Senior 3 · Chemistry

Rate of Chemical Reactions

4 lessons

Lesson 1 of 4

Measuring the Rate of a Reaction

25 min

Fizz, flame and rust

Drop an effervescent vitamin C tablet into a glass of water and the fizzing is over in a minute or two. Strike a match and the head flares in a fraction of a second. A steel hoe left out in the rain shows rust only after days, and takes years to rust through. All three are chemical reactions, yet they run at very different speeds. The speed of a reaction is called its rate. Industry wants products quickly and cheaply; a cook wants food to spoil slowly; a firefighter wants burning to stop. To control a rate, you first have to measure it.

Definition

Rate of reaction: the change in the amount of a reactant or a product per unit time.

$\text{average rate} = \frac{\text{change in amount of reactant or product}}{\text{time taken}}$

Its unit depends on what is measured: $\mathrm{cm^3\,s^{-1}}$ for a volume of gas, $\mathrm{g\,s^{-1}}$ for a mass, and $\mathrm{mol\,dm^{-3}\,s^{-1}}$ for a concentration. Example: if $30\,\mathrm{cm^3}$ of gas is given off in 20 s, the average rate is $\frac{30}{20} = 1.5\,\mathrm{cm^3\,s^{-1}}$ .

Following a reaction as it happens

You cannot watch particles react, so you follow a property that changes steadily as the reaction goes on, and record it against time.

Table 1.1: Four ways to follow the rate of a reaction

Method	What changes	Example reaction
collect the gas in a gas syringe, or in a measuring cylinder over water	volume of gas rises	magnesium and hydrochloric acid give hydrogen
stand the flask on a balance and let the gas escape	mass falls	marble chips (calcium carbonate) and hydrochloric acid give carbon dioxide
time how long a precipitate takes to hide a cross drawn under the flask	solution turns cloudy	sodium thiosulfate and hydrochloric acid give a sulfur precipitate
watch, or measure, a colour change	colour fades or appears	the purple of acidified potassium manganate(VII) fades as it oxidises ethanedioic acid

An experiment: magnesium and hydrochloric acid

0.048 g of magnesium ribbon (about 5 cm) is dropped into $50\,\mathrm{cm^3}$ of hydrochloric acid of concentration $1.0\,\mathrm{mol\,dm^{-3}}$ . The bung is put in at once and the stop-clock started. The hydrogen pushes out the plunger of a gas syringe (Figure 1.1), and the volume is read every 20 seconds.

$\mathrm{Mg(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2(g)}$

A glass gas syringe: the barrel and plunger apart (top two) and assembled (bottom). The scale is printed on the barrel, and the nozzle on the left joins the delivery tube. Photo: Geni, CC BY-SA 3.0, via Wikimedia Commons

Warning

Safety. Hydrogen forms an explosive mixture with air, so keep flames away from the apparatus. Wear eye protection. If acid splashes on your skin, rinse it under plenty of running water for at least 15 minutes and tell your teacher; if it goes in your eye, rinse the eye for at least 15 minutes and get medical help.

Table 1.2: Volume of hydrogen collected (illustrative results)

time (s)	0	20	40	60	80	100	120	140	160	180
volume of hydrogen ( $\mathrm{cm^3}$ )	0	17	29	37	42	45	47	48	48	48

Reading the graph

Steepest at the start. The gradient of the curve is the rate, and it is largest at $t = 0$ , when the acid is most concentrated and the magnesium strip is whole.
Less steep as time goes on. Reactants are used up, so the reaction slows down.
Flat from 140 s. All the magnesium has reacted (it is the limiting reactant, since the acid is in excess), so the rate is zero. The final volume, $48\,\mathrm{cm^3}$ , is set by the amount of magnesium, not by how fast it reacted.

The average rate over an interval is the change in volume divided by the time taken:

$\text{average rate} = \frac{\Delta V}{\Delta t}$

The rate at one instant is the gradient of the tangent to the curve at that time. Draw a straight line that just touches the curve at the chosen time, pick two points on the line far apart, and divide the rise by the run.

Definition

Initial rate: the rate at the very start of a reaction, $t = 0$ . It is the gradient of the tangent drawn to the curve at the origin. Initial rates are used to compare experiments fairly, because at $t = 0$ the starting conditions are known exactly. Example: in Figure 1.2 the initial rate is $1.0\,\mathrm{cm^3\,s^{-1}}$ .

Worked example

Problem. Use Table 1.2 to find the average rate of reaction (a) during the first 20 s; (b) between 60 s and 100 s; (c) over the whole reaction.

Solution.

(a) $\frac{17 - 0}{20 - 0} = 0.85\,\mathrm{cm^3\,s^{-1}}$

(b) $\frac{45 - 37}{100 - 60} = \frac{8}{40} = 0.20\,\mathrm{cm^3\,s^{-1}}$

(c) The reaction ended at 140 s, when the volume reached its final value: $\frac{48}{140} = 0.34\,\mathrm{cm^3\,s^{-1}}$ (to 2 significant figures).

The rate in (b) is less than a quarter of the rate in (a): the reaction slows down as the reactants are used up.

Worked example

Problem. (a) Use the two tangents in Figure 1.2 to find the rate at 0 s and at 40 s. (b) Show that 0.048 g of magnesium should give $48\,\mathrm{cm^3}$ of hydrogen at room temperature and pressure. (Mg = 24; molar volume $24\,\mathrm{dm^3\,mol^{-1}}$ )

Solution. (a) The tangent at 0 s passes through (0, 0) and (40, 40):

$\text{rate} = \frac{40 - 0}{40 - 0} = 1.0\,\mathrm{cm^3\,s^{-1}}$

The tangent at 40 s passes through (10, 14) and (90, 54):

$\text{rate} = \frac{54 - 14}{90 - 10} = \frac{40}{80} = 0.50\,\mathrm{cm^3\,s^{-1}}$

After 40 s the reaction runs at half its starting rate.

(b) $n(\mathrm{Mg}) = \frac{0.048}{24} = 0.0020\,\mathrm{mol}$ . One mole of Mg gives one mole of $\mathrm{H_2}$ , so $n(\mathrm{H_2}) = 0.0020\,\mathrm{mol}$ , and its volume is $0.0020 \times 24\,000 = 48\,\mathrm{cm^3}$ . The acid is in excess: $0.050\,\mathrm{mol}$ of HCl is present and only $0.0040\,\mathrm{mol}$ is needed.

Warning

Common mistakes. (1) The rate is a gradient, not a reading: at 160 s the volume is at its largest, but the rate is zero. (2) For the average over the whole reaction, divide by the time the reaction took (140 s), not by the time the readings stopped (180 s). (3) Read a tangent's gradient from two points far apart on the line, never from the data points, which lie on the curve.

Tip

Steep means fast; flat means finished. On any amount–time graph, the rate is the slope. The curve flattens when the limiting reactant runs out.

Cognito: Rates of Reaction, How to Measure it (5 min). Watch for the units of rate and why every curve is steepest at the start and flat at the end.

✦

Practice

A student adds marble chips to excess hydrochloric acid in a flask on a balance, with a cotton-wool plug in the neck, and records the loss in mass.

time (min)	0	1	2	3	4	5	6
loss in mass (g)	0	0.62	0.98	1.18	1.28	1.32	1.32

Define rate of reaction and give two possible units for it.
Write a balanced equation, with state symbols, for the reaction of calcium carbonate with hydrochloric acid. Why does the flask lose mass?
Why is a cotton-wool plug put in the neck of the flask?
Calculate the average rate of loss of mass (a) in the first minute; (b) between 2 and 4 minutes.
At what time did the reaction stop? Give the reason it stopped.
Calculate the amount, in moles, of carbon dioxide given off, and the mass of calcium carbonate that reacted. (C = 12, O = 16, Ca = 40)
In Figure 1.2, a tangent drawn at 80 s passes through (40, 34) and (120, 50). Find the rate at 80 s, and compare it with the initial rate.
Explain why reading the volume of gas every 20 s gives a more useful picture of the rate than reading only the final volume.

Show answersHide answers

The change in the amount of a reactant or product per unit time; for example $\mathrm{cm^3\,s^{-1}}$ , $\mathrm{g\,min^{-1}}$ or $\mathrm{mol\,dm^{-3}\,s^{-1}}$ .
$\mathrm{CaCO_3(s) + 2HCl(aq) \rightarrow CaCl_2(aq) + H_2O(l) + CO_2(g)}$ . Carbon dioxide escapes into the air, so the mass of the flask and its contents falls.
It lets the gas out but stops droplets of acid spraying out, which would add to the loss in mass.
(a) $\frac{0.62}{1} = 0.62\,\mathrm{g\,min^{-1}}$ ; (b) $\frac{1.28 - 0.98}{4 - 2} = 0.15\,\mathrm{g\,min^{-1}}$ .
At 5 minutes, when the loss in mass stops changing. The marble (the limiting reactant) was used up, since the acid was in excess.
$n(\mathrm{CO_2}) = \frac{1.32}{44} = 0.030\,\mathrm{mol}$ ; $n(\mathrm{CaCO_3}) = 0.030\,\mathrm{mol}$ , mass $= 0.030 \times 100 = 3.0\,\mathrm{g}$ .
$\frac{50 - 34}{120 - 40} = \frac{16}{80} = 0.20\,\mathrm{cm^3\,s^{-1}}$ , one fifth of the initial rate of $1.0\,\mathrm{cm^3\,s^{-1}}$ .
The final volume shows only how much product formed. The readings over time show how fast it formed and how the rate changed, so you can draw the curve and find rates from it.

Sources and credits

Photo: Brausetablette.JPG, Stefanie Wulf (Silberchen), licensed under CC BY-SA 3.0, via Wikimedia Commons.
Photo: Gas syringe.jpg, Geni, licensed under CC BY-SA 3.0, via Wikimedia Commons.
Video: GCSE Chemistry, Rates of Reaction: How to Measure it, Cognito, embedded from YouTube.
Written by Ganzaa. The apparatus diagram, graph, tables, worked examples and Practice items were drawn and written by Ganzaa; the results in Table 1.2 and the Practice table are illustrative.

1Fizz, flame and rust

2Following a reaction as it happens

3An experiment: magnesium and hydrochloric acid

4Reading the graph

Fizz, flame and rust

Following a reaction as it happens

An experiment: magnesium and hydrochloric acid

Reading the graph