OCR GCSE Chemistry: Rates, Calculations and Equilibrium (C5)
OCR GCSE Chemistry: Rates, Calculations and Equilibrium (C5)
Topic C5 (Monitoring and controlling reactions) on the OCR Gateway Science A specification (J248) is the most quantitative topic on the course. It is where chemistry stops being purely descriptive and starts demanding numbers: how concentrated a solution is, how much product a reaction should make and how much it actually does, how fast a reaction goes and how to speed it up, and how to nudge a reaction that runs both ways towards the products you want. Because so much of C5 is calculation, it is a topic where careful, methodical working is rewarded — and where a student who can rearrange an equation and keep their units straight has a real advantage.
This guide works through C5 at GCSE depth. For each idea you will find a clear explanation, a worked example where it helps, the highest-yield exam points, and the misconceptions that lose marks. C5 is examined on both the Foundation and Higher tiers; a great deal of the calculation work — concentration in mol/dm3, titration calculations, atom economy, gas volumes and Le Chatelier's principle — is Higher-tier only, and those points are flagged with [H].
If you want structured practice alongside this guide, work through the LearningBro OCR GCSE Chemistry: Monitoring and Controlling Reactions course, which covers every idea below with exam-style questions that match the OCR format.
How C5 Is Examined on OCR J248
C5 sits on Paper 2 of J248 (which covers C4–C6), a paper lasting 1 hour 45 minutes and worth 90 marks, counting for 50% of the qualification. Expect short recall on collision theory and catalysts, a heavy dose of calculation (concentration, yield, atom economy, titration and gas-volume questions), graph-reading on rates, and extended six-mark responses on how to control a reaction or choose conditions for the Haber process. The maths matters: at least 20% of the marks across J248 reward mathematical skills, and a large share of those live in C5. The periodic table is provided in the exam, so you will have the relative atomic masses you need — but you must know how to use them.
As always, the command word sets the task: "calculate" wants a number with working and units; "describe" wants what happens; "explain" wants why, in terms of particles, collisions or energy; and "suggest" invites you to apply your chemistry to an unfamiliar situation.
Concentration of Solutions
The concentration of a solution tells you how much solute is dissolved in a given volume of solution. The more solute you dissolve in the same volume — or the same amount of solute in a smaller volume — the more concentrated the solution. There are two units you must be able to work with.
The first is grams per cubic decimetre, g/dm3, which simply measures the mass of solute dissolved in each cubic decimetre of solution:
concentration (g/dm3)=volume of solution (dm3)mass of solute (g)
The second, used at Higher tier [H], is moles per cubic decimetre, mol/dm3, which counts the amount in moles of solute per cubic decimetre:
concentration (mol/dm3)=volume of solution (dm3)amount of solute (mol)
The single most important practical point — and the one that costs the most marks — is the unit of volume. Concentrations are quoted per cubic decimetre, but volumes in the lab are usually measured in cubic centimetres (cm3). One cubic decimetre is one litre, which is 1000 cubic centimetres, so:
volume in dm3=1000volume in cm3
Always convert cm3 to dm3 (divide by 1000) before you put a volume into a concentration calculation. To swap between the two concentration units, convert mass and moles using the relative formula mass: moles=Mrmass, so a concentration in g/dm3 divided by Mr gives mol/dm3.
Common misconception: g/dm3 and mol/dm3 are not interchangeable numbers. A solution of 40 g/dm3 sodium hydroxide (Mr=40) is 1 mol/dm3, but a solution of 40 g/dm3 of a substance with Mr=80 is only 0.5 mol/dm3. Always convert through the relative formula mass.
Titrations: The Required Practical
A titration is a precise technique for finding the volume of one solution that exactly reacts with a known volume of another — most often used to find the concentration of an acid or alkali. It is one of the eight required practicals assessed in the written papers, so you must know the method, the apparatus and the sources of error inside out.
The standard acid–alkali titration runs like this:
- Use a pipette (with a pipette filler) to measure a fixed, accurate volume of one solution — typically 25.0 cm3 of alkali — into a clean conical flask.
- Add a few drops of a suitable indicator. A single indicator that gives a sharp colour change is used — for example phenolphthalein (pink in alkali, colourless in acid) or methyl orange (yellow in alkali, red in acid). Universal indicator is not used, because its gradual range of colours does not give a sharp end point.
- Fill a burette with the other solution (the acid) and record the starting reading at eye level, reading from the bottom of the meniscus.
- Add the acid from the burette, swirling the flask constantly, until the indicator just changes colour permanently — the end point. Record the final burette reading; the difference is the titre.
- Repeat until you obtain concordant titres — results that agree within 0.10 cm3 of each other — and take the mean of the concordant titres for your calculation.
The first run is usually a quick rough (trial) titration to find the approximate end point; it is not included in the mean. A few details examiners reward repeatedly: rinse the burette with the acid (and the pipette with the alkali) before use so residual water does not dilute the solution; read both pieces of glassware at eye level to avoid parallax error; and discard any anomalous titres before averaging.
Common misconception: you do not average all your titres. You identify the concordant ones (within 0.10 cm3), ignore the rough run and any anomalies, and average only those. Including the rough titre drags the mean off and loses accuracy marks.
Titration Calculations [H]
Once you have a mean titre, a Higher-tier calculation lets you find an unknown concentration. The logic is always the same three steps, and laying them out clearly is how you secure the method marks even if the final arithmetic slips.
- Find the moles of the solution you know completely (the one with both a known concentration and a known volume), using moles=concentration×volume in dm3.
- Use the balanced equation to find the moles of the other reactant, via the mole ratio.
- Find the unknown concentration by dividing those moles by the volume (in dm3) of the unknown solution.
Worked example: finding an unknown concentration
In a titration, 25.0 cm3 of sodium hydroxide solution is exactly neutralised by 20.0 cm3 of hydrochloric acid of concentration 0.100 mol/dm3. The equation is NaOH+HCl→NaCl+H2O Calculate the concentration of the sodium hydroxide in mol/dm3.
Step 1 — moles of the acid (known completely). Convert the volume: 20.0 cm3=100020.0=0.0200 dm3. Then
moles of HCl=0.100×0.0200=0.00200 mol
Step 2 — mole ratio. The equation shows NaOH and HCl react in a 1 : 1 ratio, so moles of NaOH=0.00200 mol.
Step 3 — concentration of the alkali. Convert its volume: 25.0 cm3=0.0250 dm3. Then
concentration of NaOH=0.02500.00200=0.0800 mol/dm3
So the sodium hydroxide is 0.0800 mol/dm3. A strong answer shows every step, carries the units throughout, and only rounds at the very end. If you needed g/dm3 instead, multiply by Mr of NaOH (23+16+1=40): 0.0800×40=3.2 g/dm3.
Percentage Yield
In the real world a reaction rarely gives you as much product as the equation predicts. The theoretical yield is the maximum mass of product calculable from the balanced equation; the actual yield is the mass you really obtain. Percentage yield compares the two:
% yield=theoretical yieldactual yield×100
The yield is almost always less than 100%, for several reasons you should be able to explain:
- The reaction is reversible and does not go to completion, so some reactants remain.
- Some product is lost during transfer, filtration or purification.
- Unwanted side reactions produce different products, using up some of the reactants.
- The reactants may not be pure, or the reaction may be incomplete.
Worked example: percentage yield
A reaction has a theoretical yield of 8.0 g of product, but a student obtains 6.0 g. Calculate the percentage yield.
% yield=8.06.0×100=75%
A higher percentage yield means less waste and a more efficient, often cheaper, process — which matters both economically and environmentally.
Atom Economy [H]
Atom economy is a Higher-tier idea that measures how much of the mass of the reactants ends up in the desired product, rather than in by-products. It is a measure of how efficient and sustainable a reaction is in terms of raw materials:
% atom economy=total Mr of all reactantsMr of desired product×100
A reaction with high atom economy wastes little of its starting materials, which conserves finite raw materials and reduces waste to dispose of — both important for sustainable development. Even a reaction with a high percentage yield can have a poor atom economy if it produces large amounts of unwanted by-product; the two measures answer different questions, so do not confuse them.
Worked example: atom economy
Hydrogen can be made by reacting carbon with steam: C+2H2O→CO2+2H2 Calculate the atom economy for making hydrogen. (Ar: H = 1, C = 12, O = 16.)
The desired product is hydrogen, 2H2, with mass 2×(2×1)=4. The total mass of reactants is carbon (12) plus two waters (2×18=36), giving 12+36=48. So
% atom economy=484×100=8.3%
That low figure tells you most of the reactant mass ends up as the CO2 by-product rather than as hydrogen — a genuinely poor use of raw materials, and exactly the kind of point an "explain" question wants you to draw out.
Gas Volumes [H]
At Higher tier, you may need to work with the volumes of gases. A useful fact is that equal amounts (in moles) of any gas occupy equal volumes under the same conditions. At room temperature and pressure (rtp), one mole of any gas occupies the molar gas volume of 24 dm3 (equivalently 24,000 cm3). This lets you convert between moles of a gas and its volume:
volume of gas at rtp (dm3)=moles of gas×24
Worked example: volume of a gas
Calculate the volume, at rtp, of 0.50 mol of carbon dioxide.
0.50×24=12 dm3
To go the other way — from a volume to moles — divide by 24. These calculations link neatly to balanced equations: find the moles of a reactant, use the mole ratio to find the moles of gas produced, then multiply by 24 for the volume.
Rates of Reaction and Collision Theory
The rate of a reaction is how quickly reactants are turned into products. Some reactions, such as an explosion, are over in a fraction of a second; others, such as the rusting of iron, take years. The key to controlling and explaining rates is collision theory.
Collision theory states that a reaction can only happen when reacting particles collide with enough energy. Two conditions must be met for a successful collision: the particles must actually collide, and they must collide with at least a minimum amount of energy called the activation energy — the minimum energy needed for the particles to react. Anything that increases the frequency of collisions, or the proportion of collisions that have the activation energy, increases the rate. Four factors are examined repeatedly, each explained in the same terms:
| Factor | Effect on rate | Explanation in terms of collisions |
|---|---|---|
| Temperature | Higher temperature → faster | Particles move faster, so they collide more frequently, and a greater proportion of collisions have the activation energy |
| Concentration (of solutions) | Higher concentration → faster | There are more particles in a given volume, so collisions happen more frequently |
| Pressure (of gases) | Higher pressure → faster | The gas particles are squeezed closer together, so they collide more frequently |
| Surface area (of a solid) | Larger surface area → faster | More particles are exposed at the surface, so collisions happen more frequently (powders react faster than lumps) |
| Catalyst | Faster | Provides an alternative pathway with a lower activation energy, so more collisions are successful |
The unifying idea is that you should always explain a change in rate using collision theory — more frequent collisions, or a higher proportion of collisions with the activation energy. Simply saying a reaction "goes faster" earns nothing; saying "the particles collide more frequently because there are more of them in the same volume" earns the mark.
Common misconception: raising the temperature does not speed a reaction up only by making particles collide more often. The bigger effect is that a greater proportion of collisions now have at least the activation energy. A full-mark answer mentions both.
Measuring and Calculating Rates
To measure a rate you track how a quantity changes over time. The three standard methods each suit different reactions:
- Measuring the volume of gas produced using a gas syringe or an upturned measuring cylinder over water — ideal when a gas is given off, for example marble chips with hydrochloric acid producing carbon dioxide.
- Measuring the loss in mass of the reaction mixture on a balance as a gas escapes — again suited to reactions that produce a gas.
- Measuring the time for a solution to become cloudy — the "disappearing cross" method. Sodium thiosulfate reacts with hydrochloric acid to form a pale-yellow precipitate of sulfur; you place the flask over a paper cross and time how long it takes for the cross to disappear from view. A shorter time means a faster rate.
From a graph of product formed (or reactant used) against time, the rate at any moment is the gradient of the graph at that point. Two features appear on almost every rate graph:
- The line is steepest at the start, because the concentration of reactants is highest then, so collisions are most frequent and the rate is greatest.
- The line gradually levels off and becomes horizontal as reactants are used up and the rate falls; it is flat once the reaction has finished (a reactant has run out).
You can find the mean (overall) rate by dividing the total quantity of product formed (or reactant used) by the total time taken. To find the rate at a particular moment on a curved graph, draw a tangent to the curve at that point and calculate its gradient.
Catalysts
A catalyst is a substance that speeds up a reaction without being used up in the process — so it is not part of the overall equation and can be reused. Catalysts work by providing an alternative reaction pathway with a lower activation energy, which means a greater proportion of collisions have enough energy to react, so the reaction goes faster.
The advantage is industrial as much as chemical: a catalyst lets a process run faster, or at a lower temperature, which saves energy and money and can reduce the environmental impact of a process. Different reactions need different catalysts — there is no universal catalyst.
Enzymes are biological catalysts: large protein molecules that speed up the reactions in living things. They are used in industry too, for example in making foods and in biological washing powders. Like all catalysts, enzymes are not used up, but unlike most they work best within a narrow range of temperature and pH and are denatured outside it.
Common misconception: a catalyst does not lower the activation energy of the original reaction. It provides a different pathway that has a lower activation energy. And because it is not used up, you never include it in the balanced equation — though it is often written above the arrow.
Reversible Reactions and Dynamic Equilibrium
In many reactions the products can react together to re-form the reactants. Such a reaction is reversible, shown with the special double arrow ⇌. A classic example is the heating of hydrated copper(II) sulfate, which is reversible:
hydrated copper sulfate⇌anhydrous copper sulfate+water
Heating drives off the water (blue → white); adding water reverses it (white → blue), releasing heat. A reversible reaction is also energetically balanced: if the forward reaction is exothermic (gives out heat), the backward reaction is endothermic (takes in heat) by exactly the same amount.
When a reversible reaction takes place in a closed system (nothing can enter or leave), it eventually reaches dynamic equilibrium. At equilibrium, the forward and backward reactions are still happening, but at exactly the same rate, so the amounts of reactants and products stay constant. The word "dynamic" is important: equilibrium does not mean the reaction has stopped — it means the two opposing reactions are perfectly balanced.
Common misconception: at equilibrium the amounts of reactants and products are constant but not necessarily equal. Equilibrium means the rates of the forward and backward reactions are equal, not the concentrations.
Le Chatelier's Principle [H]
At Higher tier, you can predict how the position of an equilibrium shifts when conditions change, using Le Chatelier's principle: if a change is made to a system at equilibrium, the position of equilibrium shifts to oppose (counteract) that change. The three changes you must be able to apply are temperature, pressure and concentration.
| Change made | How the equilibrium responds |
|---|---|
| Increase temperature | Equilibrium shifts in the endothermic direction (to take in the added heat) |
| Decrease temperature | Equilibrium shifts in the exothermic direction (to give out heat) |
| Increase pressure (gases) | Equilibrium shifts towards the side with fewer molecules of gas (to reduce the pressure) |
| Decrease pressure (gases) | Equilibrium shifts towards the side with more molecules of gas |
| Increase concentration of a reactant | Equilibrium shifts towards the products (to use up the added reactant) |
| Remove a product | Equilibrium shifts towards the products (to replace it) |
Shifting the equilibrium towards the products increases the yield of those products, which is exactly what an industrial chemist wants. The skill the exam tests is reasoning through each change in turn: decide which direction "opposes" the change, then say what happens to the yield.
The Haber Process
The Haber process is the headline industrial application of equilibrium, and a reliable six-mark target. It makes ammonia (NH3) — the raw material for fertilisers — from nitrogen and hydrogen in a reversible reaction:
N2+3H2⇌2NH3
The nitrogen comes from the air (≈78% N2), and the hydrogen is obtained mainly from natural gas. The forward reaction is exothermic. The conditions used are a careful compromise, because what maximises the yield does not maximise the speed or the cost:
- Temperature: about 450 °C. A lower temperature would shift the exothermic equilibrium towards more ammonia (higher yield), but it would make the reaction too slow. A higher temperature speeds the reaction up but lowers the yield. So 450 °C is a compromise — a reasonable yield at a reasonable rate.
- Pressure: about 200 atmospheres. There are 4 molecules of gas on the left (N2+3H2) and 2 on the right (2NH3), so high pressure shifts the equilibrium towards the side with fewer molecules — the ammonia — increasing the yield. Pressure is kept high to favour the product, but not higher than necessary, because very high pressures are expensive and dangerous to build and run.
- Catalyst: iron. An iron catalyst speeds up the reaction (it reaches equilibrium faster) without affecting the position of equilibrium or the yield — it simply gets you there sooner, saving time and energy.
Finally, the ammonia is cooled and liquefied so it can be removed, and the unreacted nitrogen and hydrogen are recycled back through the reactor. Removing the product and recycling the reactants pushes more raw material towards product overall, improving the efficiency of the process.
Common misconception: the catalyst does not increase the yield of ammonia. A catalyst speeds up both the forward and backward reactions equally, so the system reaches equilibrium faster but at the same position — the same yield, sooner.
Common Mistakes Across C5
The same errors recur every year. Knowing them in advance protects easy marks.
- Forgetting to convert cm3 to dm3. Divide by 1000 before any concentration or titration calculation.
- Averaging all the titres. Use only the concordant ones (within 0.10 cm3); ignore the rough run and anomalies.
- Treating g/dm3 as the same number as mol/dm3. Convert through the relative formula mass.
- Explaining rate changes without collision theory. Always refer to collision frequency and/or the proportion of collisions reaching the activation energy.
- Saying a catalyst lowers the activation energy of the reaction. It provides an alternative pathway with a lower activation energy, and it is not used up.
- Thinking equilibrium means the reaction has stopped, or that amounts are equal. It is dynamic — equal rates, constant (not necessarily equal) amounts.
- Claiming the Haber catalyst raises the yield. It only speeds the reaction up; the yield is unchanged.
Exam Technique for C5 on OCR J248
C5 is on Paper 2, and it rewards students who can combine recall with confident calculation.
- Show every step in calculations. Concentration, yield, atom economy, gas-volume and titration questions carry method marks; a wrong final number with correct working still scores well.
- Watch the units, and round only at the end. Convert volumes to dm3, quote answers in the unit asked for, and keep extra figures until the last step.
- Explain with particles. For any rate question, reach straight for collision frequency and activation energy. For equilibrium, reach for Le Chatelier.
- Read rate graphs carefully. Steepest at the start, levelling off as reactants run out; the gradient is the rate, and a flat line means the reaction has finished.
- Plan six-mark answers. Choosing Haber conditions or explaining how to control a reaction is levels-marked for joined-up reasoning — jot the key trade-offs first.
Prepare with LearningBro
The LearningBro OCR GCSE Chemistry: Monitoring and Controlling Reactions course covers all of C5 — concentration of solutions, titrations and titration calculations, percentage yield and atom economy, gas volumes, rates of reaction and collision theory, catalysts, and reversible reactions, Le Chatelier's principle and the Haber process — with worked examples and exam-style questions that mirror the real OCR papers, plus immediate feedback.
To rehearse whole-paper strategy and the command words, work through the OCR GCSE Chemistry Exam Prep course. For the chemistry that feeds into C5, our Chemical Reactions guide covers the reaction types and energy changes behind the rates and yields here. And for the wider picture of the whole subject, start with our OCR GCSE Chemistry complete revision guide.
C5 is a topic where understanding how to handle numbers matters as much as knowing the chemistry. Master the unit conversions, the three-step titration method, collision theory and Le Chatelier's principle, and C5 becomes one of the most dependable places to score on Paper 2. Good luck with your revision.