How crowded is a solution?

A solution forms when a solute dissolves in a solvent. Its concentration measures the mass — or amount — of solute in a given volume. The more crowded the particles, the more concentrated.

Volume conversions: ÷1000 to go from cm³ → dm³; ×1000 to go from dm³ → cm³. So 250 cm³ = 0.250 dm³.

Your turn — concentration in g/dm³

From g/dm³ to mol/dm³

Higher tier also uses concentration in mol/dm³ (sometimes written mol dm⁻³ — they mean the same thing):

To swap between the two units, use moles = mass ÷ M_r. To turn g/dm³ into mol/dm³, simply divide by the M_r.

Watch out: g/dm³ and mol/dm³ are not the same number. g/dm³ counts grams; mol/dm³ counts moles. Always divide (or multiply) by the M_r to cross between them — never just relabel the units.

Your turn — convert to mol/dm³

Finding an unknown concentration

A titration finds the exact volume of one solution (run in from a burette) needed to react with a measured volume of another. From that volume and a known concentration, we can calculate the unknown concentration.

• Read the burette to the nearest 0.05 cm³.
• Ignore the rough titre. Only average concordant results — those within 0.10 cm³ of each other.
• The balanced equation gives the mole ratio linking the two solutions.

Your turn — titration calculation

How much did we actually make?

The theoretical yield is the maximum mass the equation predicts. The actual yield is what you really collect. They rarely match because of:

• incomplete reactions (some reactant left over);
• practical losses (product stuck on apparatus, lost on filtering);
• unwanted side reactions making something else.

Your turn — percentage yield

How "green" is the reaction?

Atom economy measures how many of the atoms in the reactants end up in the useful product. A high atom economy means little waste — better for the environment and for cost.

A reaction that makes only one product has a 100% atom economy.

Don't confuse them: % yield asks "how much of the possible product did we collect?"; atom economy asks "how much of the reactant mass becomes the wanted product?" A reaction can have a high yield but a low atom economy, or the other way round.

Your turn — atom economy

One mole of any gas = 24 dm³

Avogadro's law: equal volumes of different gases (at the same temperature and pressure) contain equal numbers of molecules. At room temperature and pressure (r.t.p.) one mole of any gas occupies the molar volume:

Rearranged: moles = volume ÷ 24. (Use 24 000 cm³ if the volume is in cm³.)

Your turn — gas volume

How fast is the reaction?

The rate of reaction is how quickly reactants are used up or products are made. You can follow it by measuring a volume of gas made, a mass lost, or how long a solution takes to turn cloudy.

Higher tier: to find the rate at a single instant, draw a tangent to the curve at that time and calculate its gradient (change in y ÷ change in x). A faster reaction is shown by 1/t being larger — 1/t is proportional to rate.

Your turn — mean rate

Why reactions speed up

Collision theory: particles only react when they collide with enough energy and the correct orientation. The minimum energy a collision needs to be successful is the activation energy (E_a).

Anything that makes collisions more frequent or gives particles more energy increases the rate.

Reading collision theory

Five ways to change the rate

• Concentration (of a solution) — more particles per dm³ → more frequent collisions → faster.
• Pressure (of a gas) — squeezes particles closer → more frequent collisions → faster.
• Surface area — breaking a solid into smaller pieces raises the surface-area-to-volume ratio, exposing more particles to collide → faster.
• Temperature — particles move faster and collide with more energy → much faster.
• Catalyst — provides a different pathway with a lower activation energy → faster, without being used up.

Does this change increase the rate?

Tap whether each change makes the reaction faster or slower.

Speeding things up — for free

A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy. Crucially, a catalyst is not used up — it is left chemically unchanged at the end, so a tiny amount works again and again.

• Different reactions need different catalysts.
• Catalysts lower the cost of industrial processes (less energy, lower temperatures needed).
• Enzymes are biological catalysts — proteins that speed up reactions in living things (and in industry, e.g. brewing). Each works best at a particular temperature and pH.

Common error: a catalyst does not get used up, and it does not make more product — it only changes how fast equilibrium/completion is reached. It lowers E_a; it does not change the energy of the reactants or products.

What does a catalyst do?

Reactions 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:

In a closed system (nothing escapes), the forward and reverse reactions eventually happen at the same rate. This is dynamic equilibrium.

Watch out: equilibrium does not mean the reaction has stopped, and it does not mean equal amounts of reactants and products. It means the forward and reverse rates are equal, so the concentrations no longer change.

What is dynamic equilibrium?

The equilibrium fights back

Le Chatelier's principle: if you change the conditions of a system at equilibrium, the position of equilibrium shifts to oppose (counteract) that change.

• Concentration — add more reactant → shifts right (makes more product). Remove product → shifts right.
• Temperature — increase temperature → shifts in the endothermic direction (to take in the extra heat). Decrease temperature → shifts exothermic.
• Pressure (gases) — increase pressure → shifts to the side with fewer gas molecules (to lower the pressure).

Key idea: the equilibrium always moves in the direction that opposes the change you made. Industry uses this to push reactions (like the Haber process) toward more product. A catalyst speeds up reaching equilibrium but does not shift its position.

Which way does it shift?

For the exothermic reaction N₂ + 3H₂ ⇌ 2NH₃, tap a change on the left, then its correct effect on the right.

The C5 toolkit

You've covered all three parts of OCR Gateway C5 — monitoring (C5.1), controlling (C5.2) and equilibria (C5.3). Press Finish to see your score.