Rate experiments — measuring reaction speed — Chemistry

Before this

This leads to

The Maxwell–Boltzmann distribution
Rate laws · coming soon

From "faster" to a number

The last lecture began with a lump of marble fizzing slowly in acid, and the same mass of powdered marble frothing away in moments. "Slowly" and "in moments" are honest words, but they are not measurements. How much faster is the powder: twice as fast, or fifty times? Does warming the acid by 10 °C really double the rate? Words cannot answer these questions. Numbers can.

The last lecture ended with a promise: "The next lecture shows how chemists measure a rate in the laboratory, so that these effects become numbers." Here is how.

The difficulty is that you cannot watch particles react. So every method in this lecture uses the same trick. Find something you can measure that changes as the reaction goes: a volume of gas, a mass, a colour, a concentration. Measure it again and again. The way it changes over time tells you the rate.

What a rate is

The rate of reaction is how quickly a reactant is used up or a product is made: the change in an amount divided by the time it took.

Rate of reaction

$rate = \frac{change in amount}{time taken}$ , for any reactant or product.

For a concentration, the unit is mol/dm³/s. For a gas, it may be cm³/s. For a mass, g/s.

A rate is always given as a positive number, whether a reactant falls or a product rises.

For concentrations, chemists write the rate as $\frac{Δ c}{Δ t}$ , the change in concentration divided by the change in time. The unit mol/dm³/s means "moles per cubic decimetre, every second". Pre-University books write it mol dm⁻³ s⁻¹. The two mean the same thing.

Four ways to follow a reaction

1. Collect the gas. If a product is a gas, lead it into a gas syringe and read the volume every 30 seconds. Marble chips (calcium carbonate) in hydrochloric acid give carbon dioxide:

CaCO₃(s) + 2HCl(aq) \to CaCl₂(aq) + H₂O(l) + CO₂(g)

Magnesium in acid gives hydrogen, and it can be collected in the same way. An upturned measuring cylinder full of water also works, although carbon dioxide dissolves a little in the water, so a syringe is more accurate for it.

2. Weigh the flask. If a gas escapes into the air, the flask gets lighter. Stand it on a balance and read the mass every 30 seconds. A loose plug of cotton wool lets the gas out but keeps in the fine spray of acid. This method needs a heavy gas, as Worked Example 2 shows.

Two ways to follow marble chips reacting with acid. Top: the carbon dioxide is collected in a gas syringe, and the volume is read at regular times. Bottom: the carbon dioxide escapes through a cotton wool plug, and the falling mass is read on a balance that reads to 0.01 g.

3. Measure a colour. If a reactant or product is coloured, a colorimeter can follow it. It shines light of one colour through the solution and measures how much is absorbed. For dilute solutions, the absorbance is proportional to the concentration (some books call this the Beer–Lambert law). So the absorbance gives the concentration directly, many times a second, without disturbing the mixture. Iodine is brown, so the reaction between iodine and propanone can be followed as the colour fades. Bromine is orange, so a reaction that makes bromine can be followed as the colour grows.

4. Take samples and stop them. When nothing convenient changes, take a small sample of the mixture at known times. Stop the reaction in each sample at once, which is called quenching: cool it suddenly, dilute it with cold water, or add a chemical that removes one reactant. Then find the concentration in each sample by titration. This takes more work, but it works for almost any reaction.

Each method suits some reactions and not others. Ask two questions. What changes that I can measure? And can I measure it without stopping the reaction?

Reading a rate from a graph

Here are the readings from a gas syringe. A student added 0.25 g of marble chips to an excess of dilute hydrochloric acid at room temperature (invented, but realistic, data).

Time in s	0	30	60	90	120	150	180	240
Volume of CO₂ in cm³	0	22	36	44	50	54	56	58

In the first 30 seconds, 22 cm³ appeared. In the next 30 seconds, only 14 cm³. Then 8, then 6. The reaction is slowing down. Why? Because the marble is being used up. As the chips shrink, fewer particles of calcium carbonate are exposed to the acid, so there are fewer collisions each second. Fewer collisions each second means a lower rate, as the collision theory lecture showed.

So the rate is different at every moment. How can one number describe it? There are two kinds of answer.

An average rate covers a stretch of time. Between 90 s and 120 s, 6 cm³ was collected, so the average rate over that interval is $6 \div 30 = 0.20 cm^{3} / s$ . On the graph, this is the gradient of a straight line, a chord, joining the two points.

The rate at one instant is the gradient of the curve at that single point. A curve has no single gradient over a stretch, so you draw a tangent: a straight line that touches the curve at that point and has the same steepness there. Its gradient is the rate at that instant. The rate at the very start, $t = 0$ , is called the initial rate.

The gas syringe readings, with a smooth curve through them. The tangent at 0 s (dashed) has a gradient of 0.90 cm³/s: the initial rate. The chord from 0 s to 30 s (solid grey, extended as a dotted line) has a gradient of only about 0.73 cm³/s: the average over the first 30 s. The curve levels off near 60 cm³ when the marble is used up.

That is it. That is the whole idea. An average rate is the gradient of a chord across an interval. The rate at an instant is the gradient of a tangent at a point. Because the rate keeps changing, the two are only equal for a reaction going at a steady rate, and real reactions almost never do. This is the hard part of the lecture, so try it on the graph below.

Explore — tangents on a concentration–time graph

The graph shows the concentration of an invented reactant R as it is used up. Move the time. Watch the tangent touch the curve, and read its gradient. Then switch on the chord to compare the average rate over the next 20 seconds.

Time: 20 s

An invented reactant R is used up, and its concentration falls from 1.00 mol/dm³. The table gives the rate at several times, from the gradients of tangents.

Time in s	Concentration in mol/dm³	Rate from the tangent in mol/dm³/s
0	1.000	0.0400
10	0.670	0.0268
20	0.449	0.0180
40	0.202	0.0081
60	0.091	0.0036

The chord from 20 s to 40 s gives an average rate of about 0.0124 mol/dm³/s, between the rates at its two ends. All values are rounded.

What should you have seen? The tangent is steepest at the start and gets flatter and flatter. The rate falls from 0.040 mol/dm³/s at the start to under 0.004 at 60 s. Every chord lies between the tangents at its two ends, so an average rate always hides how fast the reaction was at the start of the interval.

To find the gradient of a tangent by hand, draw it long, so that two points on it are far apart. Then read those two points and divide the change in $y$ by the change in $x$ . Long tangents make small reading errors matter less.

Now test the difference between a chord and a tangent on real readings.

Spot the error — one student's rates

A student analysed the gas syringe readings above. One line contains a mistake. Tap the line you think is wrong, then read why.

A student's working on the gas syringe readings, with the mistake marked.

Line 1. "The initial rate is the gradient of the tangent at $t = 0$ ." Correct.
Line 2. "In the first 30 s, 22 cm³ of CO₂ was collected." Correct.
Line 3. "So the initial rate is $22 \div 30 \approx 0.73 cm^{3} / s$ ." The mistake. This is the average rate over the first 30 s. The reaction slowed during those 30 s, so the average is lower than the rate at the start. The tangent at $t = 0$ gives 0.90 cm³/s.
Line 4. "From 90 s to 120 s, $50 - 44 = 6 cm^{3}$ was collected." Correct.
Line 5. "So the average rate from 90 s to 120 s is $6 \div 30 = 0.20 cm^{3} / s$ ." Correct, and correctly called an average.
Line 6. "The rate falls because the marble is used up, so there are fewer collisions each second." Correct.

What should you have learned? "Initial" means one instant, and one instant needs a tangent. Dividing a reading by a time always gives an average.

Working with real measurements

Worked Example 1 — Checking the end of the curve

In the gas syringe experiment, 0.25 g of calcium carbonate reacted with excess acid. Why does the curve level off near 60 cm³? Take $M_{r}$ of CaCO₃ as 100 and the molar gas volume as 24 dm³/mol at r.t.p.

Moles of CaCO₃: $n = \frac{0.25}{100} = 0.0025 mol$ .

The equation gives 1 CO₂ for each CaCO₃, so $n (CO_{2}) = 0.0025 \times \frac{1}{1} = 0.0025 mol$ .

Volume: $0.0025 \times 24000 = 60 cm^{3}$ .

So the reaction stops when all the marble has reacted and 60 cm³ has been made. The readings approach 60 cm³, as they should. ✓ This is a useful check on any rate experiment: the final amount must match the stoichiometry.

Stop and think

The student repeats the gas syringe experiment with 0.50 g of the same marble chips instead of 0.25 g. The acid is still in excess. Predict what happens to the initial rate and to the final volume of gas.

Show the answer

Both roughly double, but for different reasons. Twice the mass of chips of the same size has about twice the surface area. So there are about twice as many collisions each second at the start, and the initial rate roughly doubles. The final volume doubles for a different reason, the stoichiometry. Twice as much calcium carbonate makes twice as much carbon dioxide, 120 cm³ instead of 60 cm³. A rate tells you how fast. The final amount tells you how much. Keep the two questions apart.

Worked Example 2 — Why weigh carbon dioxide but not hydrogen?

The same experiment is done on a balance that reads to 0.01 g. Find the total loss in mass. Then compare a reaction that makes the same volume, 60 cm³, of hydrogen. Take $M_{r}$ as 44 for CO₂ and 2 for H₂.

CO₂: $0.0025 mol \times 44 = 0.11 g$ . The balance shows the mass falling by 0.11 g in steps of 0.01 g: eleven steps. That is enough to follow the reaction, though not finely.

H₂: $0.0025 mol \times 2 = 0.005 g$ . That is half of the smallest step the balance can show. The whole reaction would barely change the reading.

So mass loss works for carbon dioxide, a heavy gas, but not for hydrogen, the lightest gas. For hydrogen, collect the gas instead. Check: $0.11 \div 0.005 = 22$ , the ratio of the molar masses, $44 \div 2$ . ✓

Worked Example 3 — Concentration and the disappearing cross

Sodium thiosulfate reacts with hydrochloric acid to make solid sulfur, which turns the solution cloudy:

Na₂S₂O₃(aq) + 2HCl(aq) \to 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)

A flask stands on a paper marked with a cross. The time is measured from mixing until the cross can no longer be seen. Here are invented results for four concentrations of thiosulfate, with the acid and temperature kept the same.

Thiosulfate in mol/dm³	Time in s	1/time in 1/s
0.20	25	0.040
0.15	33	0.030
0.10	50	0.020
0.05	100	0.010

The time is not the rate. A faster reaction gives a shorter time. But each run stops when the same amount of sulfur has formed, enough to hide the cross. So the average rate up to that point is that fixed amount divided by the time, and $\frac{1}{time}$ is proportional to it.

Now look at the last column. Doubling the concentration from 0.10 to 0.20 mol/dm³ doubles $\frac{1}{time}$ , from 0.020 to 0.040. In these results, the rate is proportional to the concentration of thiosulfate. The last lecture could only say "more collisions each second". Here the effect has become a number.

Three classic mistakes

Trap 1 — Calling the time "the rate"

In a clock-type experiment, a longer time means a slower reaction. The time is not the rate. Use $\frac{1}{time}$ to compare rates, and only when every run ends at the same fixed amount of change.

Trap 2 — A tangent that cuts the curve

A tangent must touch the curve at your chosen point and have the same steepness there. A line that crosses the curve, or joins two readings, is a chord, and it gives an average rate. Draw the tangent long, and read two points far apart on it.

Trap 3 — A negative rate, or a missing unit

For a reactant, the gradient of a concentration–time graph is negative, because the concentration falls. The rate is the size of that gradient, a positive number. Always give the unit: mol/dm³/s for concentration, cm³/s for a gas volume, g/s for a mass.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Marble chips react with acid in a flask on a balance (invented data). The readings are: 0 s, 152.40 g; 30 s, 152.32 g; 60 s, 152.27 g; 90 s, 152.24 g; 120 s, 152.22 g. (a) Find the total mass lost at each time. (b) Find the average rate, in g/s, over the first 30 s and from 60 s to 120 s. (c) Explain why the two rates differ. (d) How many moles of CO₂ escaped in the 120 s?

Hint 1
Mass lost = 152.40 minus each reading. The mass lost is the mass of CO₂ that escaped.

Hint 2
Average rate = mass lost in the interval ÷ time of the interval. For (d), use $n = \frac{m}{M _{r}}$ with $M_{r} = 44$ .

Full answer
(a) 0.00, 0.08, 0.13, 0.16, 0.18 g. (b) First 30 s: $0.08 \div 30 \approx 0.0027 g/s$ . From 60 s to 120 s: $(0.18 - 0.13) \div 60 \approx 0.00083 g/s$ . (c) The reactants are being used up, so there are fewer collisions each second and the rate falls. (d) $n = \frac{0.18}{44} \approx 0.0041 mol$ . Check: the masses lost always increase, and by less each time, as a slowing reaction should. ✓
Choose a method to follow the rate of each reaction, and give a reason. (a) Zinc with dilute sulfuric acid, making hydrogen. (b) A reaction in which an orange bromine solution loses its colour. (c) A reaction between two colourless solutions that makes only colourless, dissolved products, one of them an acid. (d) Sodium thiosulfate with hydrochloric acid.

Hint 1
For each, ask: is a gas made, does a colour change, or does nothing visible change?

Hint 2
Remember Worked Example 2: hydrogen is too light to weigh.

Full answer
(a) Collect the hydrogen in a gas syringe. Mass loss would not work: the hydrogen is far too light. (b) Use a colorimeter, because the colour fades as bromine is used up, and absorbance follows its concentration. (c) Take samples at known times, quench them, and titrate the acid. Nothing can be seen or collected. (d) Time the disappearing cross, because solid sulfur makes the solution cloudy; or, better, follow the cloudiness with a light sensor.
A student draws a tangent to a concentration–time curve for a reactant at $t = 15 s$ . Two points on the tangent are (5 s, 0.80 mol/dm³) and (35 s, 0.20 mol/dm³). (a) Find the gradient. (b) Give the rate at 15 s, with its unit. (c) Give the same rate in mol/dm³/min.

Hint 1
Gradient = change in concentration ÷ change in time, taking the second point minus the first.

Hint 2
There are 60 s in a minute, so a rate per minute is 60 times the rate per second.

Full answer
(a) $\frac{0.20 - 0.80}{35 - 5} = \frac{- 0.60}{30} = - 0.020 mol/dm^{3} /s$ . (b) The rate is the size of the gradient: 0.020 mol/dm³/s. The minus sign only shows that a reactant is being used up. (c) $0.020 \times 60 = 1.2 mol/dm^{3} /min$ . Check: the tangent falls by 0.60 over 30 s, which is 0.020 each second. ✓
Use the disappearing-cross results in Worked Example 3. (a) Predict the time for a thiosulfate concentration of 0.12 mol/dm³. (b) A student repeats the experiment with a thicker cross drawn in black ink, and every time is longer. Does this change the conclusion about concentration? (c) Suggest one way to make the timing more reliable.

Hint 1
In the results, $\frac{1}{time}$ is 0.20 times the concentration. Use that to find $\frac{1}{time}$ for 0.12 mol/dm³.

Hint 2
For (b), ask what stays the same between runs: does each run still end at the same amount of sulfur?

Full answer
(a) $\frac{1}{time} = 0.20 \times 0.12 = 0.024$ , so the time is about $1 \div 0.024 \approx 42 s$ . (b) No. A darker cross needs more sulfur to hide it, so every run ends at a larger amount of sulfur, and every time is longer. But all runs still end at the same amount as each other, so $\frac{1}{time}$ is still proportional to the rate, and doubling the concentration still doubles it. The proportion is unchanged, even though the numbers change. (c) Judging when a cross disappears depends on the person's eyes. A light sensor and data logger below the flask can record when the light falls to a fixed level, and it can also follow the cloudiness continuously.

The bigger idea

Every method in this lecture follows a property that can be measured from outside while the reaction goes on inside. That idea is older than most of chemistry's equations. In 1850, Ludwig Wilhelmy made one of the first measurements of a reaction's rate. He followed sugar breaking down in acid by measuring how the solution turned polarised light. In the 1860s at Oxford, Augustus Harcourt and the mathematician William Esson timed reactions and found that the rate depended on how much of each reactant was present. Their reaction between hydrogen peroxide and iodide is one form of the iodine clock still used in schools.

Modern methods follow the same idea at much higher speed. In a stopped-flow instrument, two solutions are forced together and mixed, and a light beam begins following the mixture after about a thousandth of a second. That is fast enough to follow many reactions of enzymes and proteins. The next lecture explains why temperature changes rates so strongly, by looking at how the energies of particles are shared out.

Key concepts in this lesson

The rate of reaction is the change in the amount of a reactant or product divided by the time taken. It is a positive number, with a unit such as mol/dm³/s, cm³/s or g/s.
Rates are measured by following a property that changes during the reaction: the volume of gas, the mass of the flask, the colour, or the concentration in quenched samples.
The rate usually falls during a reaction, because the reactants are used up and there are fewer collisions each second.
An average rate is the gradient of a chord across an interval, and the rate at an instant is the gradient of a tangent at a point.
The initial rate is the gradient of the tangent at the very start, which is larger than the average over the first interval.
In a clock-type experiment, the time is not the rate, but 1/time is proportional to the rate when every run ends at the same amount of change.