Calculating & measuring rate
By the end of this topic you'll work out a mean rate of reaction, choose how to follow a reaction, and read rate off a graph.
Part 1What "rate" means
The rate of reaction is how quickly reactants are used up, or how quickly products are made. A fast reaction (like an explosion) finishes in a fraction of a second; a slow one (like rusting) can take years.
To put a number on it, you measure how much a reactant or product changes over a known time. That gives you the mean (average) rate:
Equation
- mean rate = quantity of reactant used (or product formed) ÷ time taken recall
- The quantity can be a mass (g) or a gas volume (cm³). So units are usually g/s or cm³/s (and mol/s if you're given moles).
Worked example — mean rate from a gas volume
A reaction produces 48 cm³ of gas in 60 s. Calculate the mean rate of reaction.
⚠ Watch out — rate is per second, not a total
Rate is a quantity divided by time — never just the amount made. "48 cm³" is not a rate; "0.8 cm³/s" is. Always include the time, and always give a unit like g/s or cm³/s.
Part 2Following a reaction
To measure rate you have to follow the reaction over time. There are three common methods, and which one you pick depends on what changes during the reaction.
Measuring gas volume — collect the gas given off in a gas syringe (or over water) and record the volume every few seconds. Measuring mass loss — stand the flask on a balance; if a gas escapes, the mass falls, so record the mass every few seconds. Measuring a colour change / cloudiness — time how long a cross takes to disappear as a precipitate forms (a turbidity method).
Marble chips react with acid, giving off carbon dioxide. Which method would not work to follow this reaction?
- ACollecting the gas in a gas syringe and recording its volume
- BStanding the flask on a balance and recording the mass loss
- CTiming how long a cross under the flask takes to disappear
- DRecording the volume of gas given off every 10 seconds
Show answer
Part 3Reading rate off a graph
If you plot the amount of product (or reactant) against time, the steepness of the line tells you the rate. A steep line means fast; a shallow line means slow. The line is steepest at the start (most reactant present) and levels off when the reaction finishes.
Higher tier — finding rate from the gradient of a graph
On Higher tier you may be asked for the rate at a particular moment, not just the mean. For a straight part of the graph the rate is simply the gradient (change in y ÷ change in x).
For a curve, draw a tangent — a straight line that just touches the curve at the point you want — then find the gradient of that tangent. A steeper tangent means a faster instantaneous rate. The tangent at the very start gives the fastest rate of all.
Write the equation for mean rate of reaction.
mean rate = quantity of reactant used (or product formed) ÷ time taken. Units such as g/s or cm³/s.A reaction loses 1.2 g of mass in 40 s. Calculate the mean rate.
mean rate = 1.2 ÷ 40 = 0.03 g/s.Name three ways of following the progress of a reaction.
Measuring the volume of gas given off, measuring the loss in mass, and timing a colour change / cloudiness (disappearing cross).On a graph of gas volume against time, where is the reaction fastest?
Where the line is steepest — at the start, when the most reactant is present. It levels off as the reactants are used up.
Factors & collision theory
The four things that speed reactions up — and the one idea, collisions, that explains every one of them.
Part 1Collision theory
For particles to react, they must collide — and they must collide with enough energy. The minimum energy needed for a collision to lead to a reaction is the activation energy.
So anything that makes collisions happen more often, or with more energy, speeds the reaction up. That single idea explains all four factors below.
The four factors that change rate
- Concentration (of solutions)
- More concentrated → particles closer together → more frequent collisions.
- Pressure (of gases)
- Higher pressure squeezes gas particles closer → more frequent collisions.
- Surface area
- Breaking a solid into smaller pieces exposes more particles → more frequent collisions.
- Temperature
- Hotter particles move faster, so collisions are more frequent and more energetic.
- Catalyst
- Provides an easier route with a lower activation energy (see Topic 4).
⚠ Watch out — temperature does two things
Most factors work by making collisions more frequent. Temperature is special: heating makes particles collide more often and gives a greater fraction of them enough energy to react (more collisions reach the activation energy). Mention both for full marks. Also: pressure only matters for gases, and surface area only for solids.
A student grinds a solid reactant into a fine powder. Why does this speed up the reaction?
- AThe particles gain more energy
- BA larger surface area means more frequent collisions
- CThe activation energy is lowered
- DThe concentration of the solid increases
Show answer
Part 2Explaining a result
Exam questions love to ask why a change sped a reaction up. A good answer always links the change to collisions. Use the chain: the change → particles closer / faster → more frequent (or more energetic) collisions → faster rate.
Worked example — explaining the effect of temperature
Explain, in terms of particles, why a reaction is faster at 50 °C than at 20 °C.
What two things must be true for a collision to cause a reaction?
The particles must collide, and they must collide with at least the activation energy (the minimum energy needed to react).List the four factors (other than a catalyst) that affect rate.
Concentration of solutions, pressure of gases, surface area of solids, and temperature.Why does increasing the concentration of a solution speed up a reaction?
There are more reactant particles in the same volume, so collisions are more frequent → faster rate.In what way is temperature different from the other factors?
It makes collisions both more frequent and more energetic — a greater proportion of collisions reach the activation energy. The others mainly change how often particles collide.
Required practical — concentration & rate
The investigation you'll be examined on — how concentration changes the rate, measured two different ways.
Part 1The disappearing-cross method
The classic version uses sodium thiosulfate and hydrochloric acid. They react to make a pale yellow precipitate of sulfur, which slowly turns the mixture cloudy. You time how long it takes for the cloudiness to hide a cross drawn under the flask.
Effect of concentration on rate (thiosulfate + acid)
Aim: investigate how changing the concentration of a reactant affects the rate of reaction.
- Draw a cross on paper and stand a conical flask on top of it.
- Measure a fixed volume of sodium thiosulfate solution into the flask. Record its concentration.
- Add the hydrochloric acid and start a stopwatch at once.
- Look down through the flask and stop the watch the moment the cross disappears. Record the time.
- Repeat with different concentrations of thiosulfate (diluting with water), keeping everything else the same.
Control / improve: keep the temperature, volumes and the same observer constant so concentration is the only variable. A shorter time means a faster rate, so rate ∝ 1 ÷ time. Judging "when the cross disappears" is subjective — using a light sensor and data logger makes it more reliable.
⚠ Watch out — short time means fast rate
The disappearing-cross method measures a time, not a rate directly. A shorter time means a faster rate, so the relationship is rate ∝ 1 ÷ time. Don't say "the higher concentration took longer" — it takes less time. And it's a fair test only if temperature and volumes are kept constant.
Part 2The gas-volume alternative
You can investigate the same idea with marble chips (calcium carbonate) and hydrochloric acid, which give off carbon dioxide. Here you measure the volume of gas (or the mass lost) at set times and compare how steeply the graph rises for different acid concentrations.
Worked example — comparing two concentrations
With 1.0 mol/dm³ acid, 60 cm³ of gas is collected in 30 s. With 2.0 mol/dm³ acid, the same 60 cm³ comes off in 15 s. Compare the rates.
In the thiosulfate practical, which is the dependent variable you measure?
- AThe concentration of the thiosulfate
- BThe temperature of the mixture
- CThe time taken for the cross to disappear
- DThe volume of acid added
Show answer
In the thiosulfate + acid practical, what do you actually time?
How long it takes for the mixture to turn cloudy enough to hide a cross drawn under the flask.How is the measured time related to the rate?
A shorter time means a faster rate — rate is proportional to 1 ÷ time.Name two variables you must keep the same for a fair test.
Any two of: temperature, total volume, the volume of acid, the same observer/judgement of "disappeared".Suggest one way to make the disappearing-cross method more reliable.
Use a light sensor and data logger instead of judging by eye, removing the subjective decision about when the cross has gone.
Catalysts — a lower energy route
How a catalyst speeds a reaction without being used up — and the energy diagram examiners want you to label.
Part 1What a catalyst does
A catalyst speeds up a reaction without being used up in it. Because it isn't consumed, the same small amount works over and over, and there's the same mass of catalyst at the end as at the start.
It works by providing a different reaction pathway with a lower activation energy. With a lower energy barrier, a greater proportion of collisions now have enough energy to react — so the rate goes up.
⚠ Watch out — what a catalyst does NOT do
A catalyst is not used up and does not appear in the chemical equation. It does not change the products, and for a reversible reaction it does not change the position of equilibrium — it just gets there faster. Don't say it "gives the particles more energy"; it lowers the activation energy they need.
Which statement about catalysts is correct?
- AA catalyst increases the energy of the colliding particles
- BA catalyst is used up, so you must keep adding more
- CA catalyst provides a route with a lower activation energy
- DA catalyst increases the amount of product formed
Show answer
Part 2Enzymes — biological catalysts
Living things use enzymes as catalysts. They speed up reactions in the body (and in industry, like fermentation) at low temperatures, which saves energy. Like all catalysts, they aren't used up and they lower the activation energy.
Worked example — proving a catalyst isn't used up
A reaction is run with 0.5 g of a catalyst. Explain what you'd expect to find about the catalyst at the end, and why.
Define a catalyst.
A substance that speeds up a reaction without being used up in it.How does a catalyst speed up a reaction?
It provides a different pathway with a lower activation energy, so a greater proportion of collisions are successful.Does a catalyst appear in the balanced chemical equation? Explain.
No — it isn't used up, so it doesn't appear as a reactant or product (it's sometimes written above the arrow).What name is given to the biological catalysts found in living organisms?
Enzymes.
Reversible reactions & equilibrium
Reactions that can go both ways, why energy changes are reversed, and how a sealed system settles into a balance.
Part 1Reactions that go both ways
In a reversible reaction, the products can react together to remake the reactants. We show this with a special double arrow (⇌) instead of a single one.
A useful example is hydrated copper sulfate. Heat the blue crystals and they turn white, driving off water; add water back and they turn blue again — the same change, run in reverse.
⚠ Watch out — the energy change is reversed too
If the forward reaction is exothermic (gives out energy), the backward reaction is endothermic (takes in energy) by exactly the same amount — and vice versa. So if heating drives a change one way, cooling lets it go back. Don't say both directions release energy.
Part 2Dynamic equilibrium
If a reversible reaction happens in a closed system (nothing gets in or out), it eventually reaches equilibrium. At equilibrium the forward and backward reactions are still happening, but at the same rate, so the amounts of reactants and products stay constant.
This is called dynamic equilibrium: "dynamic" because the reactions never stop, "equilibrium" because nothing appears to change. It doesn't mean equal amounts of reactant and product — just unchanging amounts.
Higher tier — Le Chatelier's principle
If you change the conditions of a system at equilibrium, the position of equilibrium shifts to oppose the change. This is Le Chatelier's principle, and it lets you predict which way a reaction will move.
Concentration: add more of a reactant and the equilibrium shifts to make more product (using the added reactant up). Remove a product and it shifts to make more product.
Temperature: raise the temperature and equilibrium shifts in the endothermic direction (the one that takes energy in); lower it and it shifts the exothermic way.
Pressure (gases only): raise the pressure and equilibrium shifts to the side with fewer gas molecules; lower it and it shifts to the side with more.
Worked example — applying Le Chatelier (Higher)
For N₂ + 3H₂ ⇌ 2NH₃, predict the effect of increasing the pressure on the amount of ammonia.
A reversible reaction in a sealed flask reaches equilibrium. Which statement is true?
- ABoth reactions have stopped
- BThere are equal amounts of reactants and products
- CThe forward and backward reactions happen at the same rate
- DOnly the forward reaction is still happening
Show answer
What symbol shows that a reaction is reversible?
A double arrow, ⇌, instead of a single arrow.If the forward reaction is exothermic, what can you say about the backward reaction?
It is endothermic, taking in the same amount of energy that the forward reaction gave out.Describe what "dynamic equilibrium" means.
In a closed system, the forward and backward reactions occur at the same rate, so the amounts of reactants and products stay constant (the reactions don't stop).Why must the system be closed to reach equilibrium?
So that no reactant or product can escape — both directions stay possible and can balance.[HT] Using Le Chatelier's principle, what happens if you increase the concentration of a reactant?
The equilibrium shifts towards the products to oppose the change (using up the added reactant), making more product.