IGCSE Chemistry Topic 6: Chemical Reactions

6.1 Physical & Chemical Changes 6.2 Rate of Reaction 6.3 Reversible Reactions & Equilibrium 6.4 Redox

Hey Tara! Welcome to Topic 6 - Chemical Reactions. This is the topic where chemistry stops being a list of facts and starts being a story with a plot. You will learn how to tell a genuine chemical change from a physical one, how to speed reactions up (and why a flour mill can explode but a bag of flour cannot), how some reactions refuse to finish and instead settle into a dynamic equilibrium that industry has to negotiate with, and finally how electrons move from one substance to another in redox reactions. Four subtopics, four ideas, and honestly they show up in more exam questions than almost anything else in the syllabus. Take them one at a time - and use the practice questions at the end of each section to check you really have it. Let us begin!

6.1 Physical and Chemical Changes ▼

The Big Idea

Everything that happens to matter is either a physical change or a chemical change. The difference sounds obvious until you meet a tricky example - and Cambridge examiners love tricky examples. So let us get the definitions exactly right.

Physical change: no new substance

A physical change alters the form or state of a substance, but the substance itself is chemically the same afterwards.

Chemical change: new substance formed

A chemical change (a chemical reaction) rearranges atoms into different substances with different chemical properties.

Physical Changes in Detail

In a physical change the particles themselves are unchanged. Only the arrangement, spacing or energy of those particles changes. The chemical bonds inside each molecule stay intact; only the weaker forces between molecules are affected.

The classic physical changes are the changes of state:

Melting - solid to liquid (ice → water)
Freezing - liquid to solid (water → ice)
Boiling / evaporation - liquid to gas (water → steam)
Condensation - gas to liquid (steam → water)
Sublimation - solid straight to gas (solid iodine → purple iodine vapour)

Other physical changes include dissolving (sugar in tea - you can get the sugar back by evaporating the water), crushing or grinding (chalk into powder), magnetising iron, and mixing two substances that do not react (sand and iron filings).

The Four Fingerprints of a Physical Change

Feature	What happens	Example
No new substance	Chemical formula stays identical before and after	H₂O(s) → H₂O(l): still water
Easily reversible	Usually reversed by simply cooling, heating or evaporating	Freeze the water back into ice
Small energy change	Only intermolecular forces are broken, not covalent bonds	Melting ice needs 6 kJ/mol; burning hydrogen releases 286 kJ/mol
Mass is conserved	Mass stays the same (as it does in chemical changes too)	10 g of ice melts to give 10 g of water

Chemical Changes in Detail

In a chemical change, chemical bonds are broken and new bonds are formed. The atoms are all still there (mass is conserved) but they are now joined up differently, so the products are genuinely different substances with different melting points, colours, densities and reactivities.

Mg + 2HCl → MgCl₂ + H₂

Magnesium ribbon dissolves in hydrochloric acid, fizzing as hydrogen gas is produced. Two brand-new substances appear - this is unmistakably a chemical change.

Signs (Not Proof) of a Chemical Change

These observations suggest a chemical reaction. Be careful - none of them is 100% proof on its own, and the exam loves to exploit that.

Observation	Example	Watch out!
Gas produced (effervescence)	Marble chips + hydrochloric acid fizz with CO₂	Boiling water also produces bubbles - but that is physical
Colour change	Colourless bromine water turns from orange to colourless with an alkene	Mixing paints changes colour but is only physical
Precipitate forms	AgNO₃(aq) + NaCl(aq) gives a white AgCl precipitate	A solid appearing on cooling a hot saturated solution is crystallisation - physical
Temperature change	Neutralisation warms up; dissolving NH₄NO₃ cools down	Dissolving is physical yet still shows a temperature change
Light or sound emitted	Burning magnesium gives a brilliant white light	A light bulb glows without any chemical change in the glass
Hard to reverse	You cannot un-fry an egg or un-burn a match	Some chemical reactions ARE reversible (see 6.3)

Memory Trick

"NEW substance = chemical." Everything else is a clue, not a verdict. Ask yourself one question: if I write down the formula before and the formula after, are they different? Different formula = chemical change. Same formula = physical change.

The Classic Confusing Cases

These six examples come up again and again. Learn them and you will never lose a mark here.

Process	Type	Reasoning
Dissolving salt in water	Physical	NaCl is still NaCl - the ions are separated by water but no new substance forms. Evaporate the water and you recover the salt.
Boiling a kettle	Physical	H₂O(l) → H₂O(g). Bubbles do NOT prove a chemical change here - the gas is still water.
Rusting iron	Chemical	Fe + O₂ + H₂O → hydrated iron(III) oxide. A new orange-brown solid with entirely different properties.
Melting candle wax	Physical	Solid wax → liquid wax. Same molecules. (But the wax burning in the flame IS chemical.)
Cooking an egg	Chemical	Protein molecules are permanently denatured and cross-linked. You cannot reverse it by cooling.
Heating hydrated copper(II) sulfate	Chemical	CuSO₄·5H₂O → CuSO₄ + 5H₂O. Blue → white, and the formula changes. It is chemical even though it is reversible.

Exam Tip

"Is it reversible?" is not a reliable test. Heating hydrated copper(II) sulfate is a reversible chemical change. Dissolving sugar is a reversible physical change. Melting is a reversible physical change. Always go back to the real test: has a new substance with a different chemical formula been made?

Energy and Chemical Change

Chemical changes usually involve much larger energy changes than physical changes, because covalent or ionic bonds must be broken and made. Compare:

Change	Type	Approximate energy
Melting ice	Physical	about 6 kJ per mole
Boiling water	Physical	about 41 kJ per mole
Burning methane	Chemical	about 890 kJ per mole released
Decomposing calcium carbonate	Chemical	about 178 kJ per mole absorbed

This is why a gas cooker can heat a whole pan of dal, but a melting ice cube barely cools your drink by a few degrees.

Chemical Change and the Conservation of Mass

In both physical and chemical changes, the total mass never changes. Atoms are neither created nor destroyed - they are only rearranged. If a reaction in an open beaker appears to lose mass, a gas has escaped. If it appears to gain mass, a gas from the air (usually oxygen) has been absorbed.

2Mg + O₂ → 2MgO

Burn 4.8 g of magnesium in air and you collect 8.0 g of white magnesium oxide. The extra 3.2 g is the oxygen that joined in from the air - mass was conserved all along.

Worked Examples

Worked Example 1 Ananya leaves a glass of water outside in Jaipur in May. By evening the glass is empty. Her brother says "a chemical reaction destroyed the water." Is he right? Explain fully.

Step 1: Identify what happened

The liquid water evaporated in the heat: H₂O(l) → H₂O(g). The water molecules have gained enough energy to escape from the liquid surface into the air.

Step 2: Apply the test

Compare formulae. Before: H₂O. After: H₂O. The formula is unchanged, so no new substance has been made.

Step 3: Check reversibility and energy

Cool the vapour and it condenses straight back to liquid water. Only weak intermolecular forces were overcome, not covalent O–H bonds. Both facts support a physical change.

Her brother is wrong. This is a physical change (evaporation). No new substance was formed - the same H₂O molecules are now spread out in the air as water vapour. Nothing was destroyed; mass is conserved.

Worked Example 2 A student heats 5.00 g of blue hydrated copper(II) sulfate crystals. A white powder is left and droplets of liquid condense on the cool part of the tube. The white powder has a mass of 3.20 g. State the type of change with reasons, and explain what happened to the missing mass.

Step 1: Write the equation

CuSO₄·5H₂O(s) → CuSO₄(s) + 5H₂O(l). Blue hydrated copper(II) sulfate becomes white anhydrous copper(II) sulfate plus water.

Step 2: Apply the new-substance test

The formula changes from CuSO₄·5H₂O to CuSO₄. Water has been chemically released from within the crystal lattice, and the colour changes blue → white. A new substance HAS been formed, so this is a chemical change.

Step 3: Account for the mass

5.00 g − 3.20 g = 1.80 g. That 1.80 g is the water driven off as vapour, which condensed on the cooler part of the tube. Mass is still conserved - the water simply left the solid.

Step 4: Address the reversibility trap

Adding water back turns the white solid blue again. That does NOT make it physical - reversibility is not the test. It is a reversible chemical change.

It is a chemical change because a new substance (anhydrous copper(II) sulfate) with a different formula and colour is formed. The missing 1.80 g is water of crystallisation driven off as vapour. The change is reversible but still chemical.

Worked Example 3 A strip of magnesium ribbon is burned in a crucible with a lid. Its mass rises from 2.4 g to 4.0 g. Explain, in terms of type of change and conservation of mass, why the mass increased.

Step 1: Identify the change

Shiny grey magnesium burns with a brilliant white flame to give a white powder. Equation: 2Mg + O₂ → 2MgO. A new substance (magnesium oxide) has been formed, so this is a chemical change.

Step 2: Explain the mass increase

Oxygen from the air has combined with the magnesium and is now part of the solid product. The mass of oxygen added = 4.0 − 2.4 = 1.6 g.

Step 3: Confirm conservation of mass

Total mass of reactants (2.4 g Mg + 1.6 g O₂) = 4.0 g = mass of product. Nothing was created; the oxygen simply came from the surrounding air, which is outside the crucible and so was not weighed at the start.

A chemical change occurred. The mass increased by 1.6 g because oxygen from the air chemically combined with the magnesium to form magnesium oxide. Mass is conserved overall - the "extra" mass was always present as oxygen gas in the air.

Exam Tips for 6.1

1. Always justify with "new substance": Do not just write "chemical". Write "chemical, because a new substance with a different chemical formula is formed."

2. Bubbles are not proof: Boiling water bubbles furiously and is entirely physical. Only say "gas produced indicates a chemical change" when the gas is a different substance from the reactants.

3. Dissolving is physical: This is the single most common slip. Salt dissolving, sugar dissolving, and copper(II) sulfate dissolving are all physical - the solute can be recovered by evaporation.

4. Reversibility proves nothing: Hydrated/anhydrous copper(II) sulfate is a reversible chemical change. Melting is a reversible physical change. Do not use reversibility as your reason.

5. Mass is always conserved: In both types of change. If mass appears to change in an open container, explain it as gas escaping or gas being absorbed from the air.

6. Learn the standard colour changes: Blue → white (hydrated → anhydrous CuSO₄), pink → blue (hydrated → anhydrous CoCl₂), grey → white (Mg → MgO), grey → orange-brown (Fe → rust). These are quick marks.

🌎 Apply It: Real-World Chemistry

Telling physical from chemical change is a decision that food factories, museums, farmers and forensic scientists all make every single day.

At the Tata Salt works on the Gujarat coast, seawater is pumped into shallow pans and left to evaporate in the sun. Weeks later, workers rake up crystals of common salt. A visiting student claims the sun has "manufactured salt out of seawater."

Is solar salt production a physical or chemical process? Justify your answer with the correct test.

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The Chemistry

The sodium chloride was already dissolved in the seawater as Na⁺(aq) and Cl⁻(aq) ions. Solar heat evaporates only the water: H₂O(l) → H₂O(g). The ions left behind pack back into a crystal lattice: NaCl(aq) → NaCl(s).

Applying the Test

Formula before: NaCl. Formula after: NaCl. No new substance is created, so this is a physical process - specifically crystallisation following evaporation. Dissolve the crystals again and you are straight back to where you started.

Chemistry Connection

The sun did not "make" salt. It separated a mixture. Separation techniques (evaporation, crystallisation, filtration, distillation) are always physical because they never change chemical formulae - that is exactly why they work for purifying substances.

Conservators at the British Museum in London are treating a bronze Roman helmet that has developed a patchy green crust. They must decide whether the green material is a coating that can be dissolved off, or a product formed from the metal itself.

The green layer is copper carbonate hydroxide. Is corrosion a physical or chemical change, and what does this mean for the conservators?

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The Chemistry

Copper in the bronze has reacted with oxygen, water and carbon dioxide from the air over centuries. Copper metal (Cu) has been converted into a copper compound - a completely different substance with a different formula, colour, hardness and density.

Why It Matters

Because it is a chemical change, the green material is not sitting on top of the metal - it IS the metal, transformed. Removing it removes original material permanently. Conservators therefore stabilise the surface rather than stripping it, and control humidity to stop further reaction.

Chemistry Connection

Corrosion (including rusting) is chemical, which is why it cannot simply be "wiped off" and why prevention - barrier coatings, sacrificial protection, dry storage - is the only real strategy. A physical change could be undone; this cannot.

A quality control chemist at a Nestle factory in Vevey, Switzerland, tests chocolate that has developed a white dusty film after a hot delivery journey. Customers complain that the chocolate has "gone mouldy". Laboratory analysis shows the white film is pure cocoa fat.

Has the chocolate undergone a chemical change? What is really happening, and is the chocolate safe?

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The Chemistry

This is "fat bloom". The cocoa butter melted during the hot journey, migrated to the surface, then re-solidified in a different crystal form. Melting and freezing are changes of state.

Applying the Test

The fat molecules are chemically identical before and after - only their physical arrangement and crystal structure changed. No new substance was formed, so this is a physical change. The chocolate is perfectly safe to eat; it just looks unappetising.

Chemistry Connection

A change in appearance is one of the weakest clues to a chemical change. Here the appearance changed dramatically while the chemistry did not change at all. Manufacturers fix it by "tempering" - controlled cooling that forces the fat into the right crystal form.

A forensic scientist in Manchester examines a burned document from a suspected arson case. Only grey ash and a faint smell remain. The defence lawyer argues that the paper "could be restored" if the right solvent were used.

Explain, in terms of bonds and substances, why the paper cannot be restored.

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The Chemistry

Paper is mostly cellulose, a carbohydrate. Combustion breaks the covalent bonds in the cellulose and forms new bonds in carbon dioxide, water vapour and carbon-rich ash: cellulose + O₂ → CO₂ + H₂O + carbon residue.

Why It Cannot Be Undone

Completely new substances have been made, most of which have escaped into the air as gases. There is no solvent that can gather CO₂ and H₂O from the atmosphere and reassemble the exact original cellulose chains. Combustion is a strongly exothermic chemical change with an enormous energy barrier to reversal.

Chemistry Connection

This is the difference between "reversible in principle" and "reversible in practice". Very exothermic chemical changes such as combustion are effectively one-way. Compare with melting wax, where the same molecules simply re-solidify on cooling.

A tea estate manager in Assam explains that fresh green tea leaves are deliberately bruised and left in warm humid air for several hours. The leaves darken from green to coppery brown and develop a completely different aroma before being dried and packed as black tea.

Green tea and black tea come from the same plant. Is the "fermentation" step a physical or a chemical change? Give evidence.

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The Chemistry

Bruising releases enzymes that catalyse the oxidation of colourless catechins in the leaf into coloured theaflavins and thearubigins. New aroma molecules are produced at the same time. Oxygen from the air is a reactant.

The Evidence

Three strong indicators: (1) a permanent colour change green → brown, (2) completely new smell and taste molecules, (3) it cannot be reversed - you cannot turn black tea back into green tea. Different substances with different formulae now exist, so it is a chemical change.

Chemistry Connection

Note that the subsequent drying step IS physical - it only removes water. One production line can therefore contain a chemical change (oxidation) followed by a physical change (drying). Examiners love asking you to separate the two stages.

Practice Questions: 6.1

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

Which statement is the best definition of a chemical change?

A A change that cannot be reversed

B A change in which a gas is produced

C A change in which one or more new substances are formed

D A change in which the mass increases

The only reliable definition is the formation of new substances. Irreversibility, gas production and mass changes are clues, not definitions - and each has physical-change counter-examples.

Question 2

Which of the following is a physical change?

A Rusting of an iron gate

B Dissolving sugar in hot tea

C Burning natural gas

D Digesting starch in the mouth

Dissolving is physical - the sugar molecules are unchanged and could be recovered by evaporating the water. Rusting, burning and digestion all form new substances.

Question 3

Blue hydrated copper(II) sulfate is heated and turns white. This change is best described as:

A Physical, because it can be reversed by adding water

B Chemical, because a new substance with a different formula is formed

C Physical, because only water is removed

D Neither physical nor chemical

CuSO₄·5H₂O becomes CuSO₄ - a different chemical formula, so a chemical change. Being reversible does not make a change physical.

Question 4

A student boils water in a beaker and sees vigorous bubbling. Which conclusion is correct?

A A chemical change occurs because a gas is produced

B A chemical change occurs because hydrogen and oxygen are released

C A physical change occurs because the gas is still water

D A physical change occurs because the mass stays the same

The bubbles are water vapour, H₂O(g) - the same substance in a different state. Option D is wrong as a reason because mass is conserved in chemical changes too.

Question 5

Which observation is the strongest evidence that a chemical reaction has taken place?

A The mixture becomes warmer

B The solid disappears into the liquid

C An insoluble precipitate forms when two clear solutions are mixed

D The liquid becomes more viscous

A precipitate formed from two solutions is a genuinely new insoluble substance. Warming can accompany dissolving (physical), and a solid disappearing is usually dissolving.

Question 6

Iodine crystals are gently warmed and a purple vapour fills the tube. On cooling, dark crystals reform. This is:

A Sublimation, a physical change

B Decomposition, a chemical change

C Oxidation, a chemical change

D Neutralisation, a chemical change

Solid iodine turns directly to gas (sublimation) and back again. It is I₂ throughout - no new substance, so physical.

Question 7

2.4 g of magnesium is burned completely in air to give 4.0 g of magnesium oxide. The mass increase is because:

A Mass is created during chemical reactions

B Oxygen from the air has combined with the magnesium

C The magnesium absorbed heat energy which has mass

D Nitrogen from the air was trapped in the powder

1.6 g of oxygen from the air has become part of the product. Mass is conserved overall: 2.4 + 1.6 = 4.0 g.

Question 8

Which pair contains ONE physical change and ONE chemical change, in that order?

A Rusting iron; burning coal

B Melting candle wax; burning candle wax

C Condensing steam; freezing water

D Cooking an egg; digesting an egg

A burning candle shows both at once: the wax melts (physical) and the vaporised wax then burns in the flame (chemical). This is a classic exam example.

Question 9

A student says "all changes that are easy to reverse are physical." This statement is:

A Correct, reversibility is the definition of a physical change

B Incorrect, because some chemical changes such as hydration of copper(II) sulfate are easily reversed

C Incorrect, because no physical change can be reversed

D Correct, but only for changes of state

Reversible chemical changes exist (anhydrous/hydrated salts, the Haber process, thermal decomposition of ammonium chloride). Reversibility is never the deciding test.

Question 10

Which change involves breaking covalent bonds within molecules?

A Boiling water

B Melting ice

C Cracking a long-chain alkane into ethene and an alkane

D Condensing steam

Changes of state only overcome the weak forces BETWEEN molecules. Cracking breaks strong C–C covalent bonds inside molecules, which is why it needs about 600 °C and a catalyst.

Question 11

Anhydrous cobalt(II) chloride paper is blue. When a drop of water is added it turns pink. This colour change is used as a test for water and is:

A A physical change, since the paper only gets wet

B A chemical change, since a hydrated compound with a new formula is formed

C A physical change, since it can be reversed by heating

D A nuclear change

CoCl₂ (blue) becomes CoCl₂·6H₂O (pink). The water is chemically incorporated into the lattice, giving a new compound - a reversible chemical change.

Question 12

Which process would allow you to recover the original substance unchanged, proving the change was physical?

A Adding water to the ash of burnt paper

B Evaporating the water from a salt solution

C Cooling the carbon dioxide from a burning candle

D Filtering the gas from a fizzing antacid tablet

Evaporating a salt solution returns the identical salt, confirming that dissolving was physical. In the other options new substances have already been formed.

Question 13

Typical energy changes are about 6 kJ/mol for melting ice but about 890 kJ/mol for burning methane. This is mainly because:

A Methane molecules are heavier than water molecules

B Melting only overcomes weak forces between molecules, whereas burning breaks and makes strong covalent bonds

C Melting is endothermic and burning is exothermic

D Methane is a gas and ice is a solid

Physical changes only involve intermolecular forces; chemical changes involve full covalent or ionic bonds, which are roughly 10-100 times stronger.

Question 14

A sealed flask containing calcium carbonate and hydrochloric acid is weighed before and after reaction. The mass:

A Decreases because carbon dioxide is produced

B Increases because a new compound is formed

C Stays exactly the same because no matter can enter or leave

D Cannot be predicted

In a sealed system mass is always conserved. The CO₂ is still inside the flask, so the balance reading does not change. In an OPEN flask the mass would fall as CO₂ escapes.

Question 15

Which of these everyday events is a chemical change?

A Chopping vegetables for a curry

B Ice cream melting in the sun

C Milk turning sour overnight

D Grinding coffee beans

Bacteria convert lactose into lactic acid - new substances with a new taste and smell, and it cannot be reversed. The others only change size, shape or state.

Question 16

Which statement about mass in chemical reactions is correct?

A Mass is always lost because energy is released

B Mass is conserved only in physical changes

C The total mass of reactants equals the total mass of products because atoms are only rearranged

D Mass increases whenever a solid product is formed

Atoms are neither created nor destroyed. Apparent mass changes in open containers are always explained by gases entering or leaving.

Question 17

A silver spoon slowly develops a black coating of silver sulfide when left near boiled eggs. This tarnishing is:

A A chemical change, because silver has reacted to form a new compound

B A physical change, because the coating can be polished off

C A physical change, because the spoon is still a spoon

D Not a change at all, only dirt

Silver metal has reacted with sulfur compounds to form silver sulfide - a new substance with a new formula. Polishing removes the compound; it does not turn it back into silver.

Question 18

In which change is the chemical formula of the main substance UNCHANGED?

A CaCO₃ → CaO + CO₂

B I₂(s) → I₂(g)

C 2H₂O → 2H₂ + O₂

D Zn + CuSO₄ → ZnSO₄ + Cu

Only the state symbol changes in B, so it is physical. A, C and D all show different formulae on each side - chemical changes.

Question 19

Photosynthesis in a mango tree converts carbon dioxide and water into glucose and oxygen. The best reason for calling this a chemical change is:

A It requires sunlight

B It takes place slowly

C Glucose and oxygen are new substances with completely different properties from the reactants

D The leaves are green

Always justify with the formation of new substances. Speed, colour and energy source are not part of the definition.

Question 20

A student mixes sand and iron filings, then separates them with a magnet. Which statement is correct?

A Mixing was chemical and separating was physical

B Both mixing and separating were physical changes

C Both were chemical changes because a mixture is a new substance

D Mixing was physical and separating was chemical

Making or separating a mixture is always physical - the components keep their own properties and formulae, which is exactly why the magnet works.

6.2 Rate of Reaction ▼

What Is Rate of Reaction?

The rate of reaction is a measure of how fast reactants are used up, or how fast products are formed, per unit time.

rate = change in quantity ÷ time taken

The "quantity" can be volume of gas (cm³), mass (g), or concentration (mol/dm³). Typical units: cm³/s, g/s, cm³/min, mol/dm³/s.

Reaction rates vary enormously. An explosion in a firework is over in milliseconds. Rusting of an iron railing at Marine Drive in Mumbai takes years. Both are chemical reactions - they just have wildly different rates.

Collision Theory - the Explanation Behind Everything

Every explanation in this section comes back to one model. Learn it properly and you can answer any rate question.

For a reaction to happen, particles must COLLIDE

with enough energy (at least the activation energy, Eₐ), AND in the correct orientation. A collision that satisfies both conditions is called a successful (effective) collision.

Therefore the rate of a reaction depends on the frequency of successful collisions - how many successful collisions happen each second. Anything that raises that frequency raises the rate.

Memory Trick

Every rate answer follows the same three-step sentence: (1) what physically changes, (2) so the frequency of collisions (or the proportion with E ≥ Eₐ) changes, (3) so the frequency of successful collisions changes, so the rate changes. Write it every time and you will bank the marks.

Factor 1: Concentration of a Solution

Increasing the concentration increases the rate of reaction.

Explanation: a more concentrated solution contains more particles in the same volume (the particles are more crowded). This means the particles collide more frequently, so there are more successful collisions per second and the rate increases.

Exam Tip

Never write "there are more particles" on its own - a bigger beaker also has more particles but the rate does not change. You must say more particles per unit volume (or "the particles are closer together"). That phrase is worth the mark.

Effect of concentration on the volume of gas produced. All three curves level off at the same height because the limiting amount of reactant is unchanged.

Factor 2: Pressure of a Gas

Increasing the pressure of gaseous reactants increases the rate of reaction.

Explanation: squeezing a gas into a smaller volume pushes the gas particles closer together, so there are more particles per unit volume. They therefore collide more frequently, giving more successful collisions per second.

Memory Trick

Pressure is simply "concentration for gases". The explanation is word-for-word the same. Note that pressure only affects reactions involving gases - squeezing a beaker of solution does essentially nothing.

Factor 3: Surface Area of a Solid

Increasing the surface area (using smaller pieces or a powder) increases the rate of reaction.

Explanation: a reaction between a solid and a liquid or gas can only happen at the surface of the solid. Breaking a lump into smaller pieces exposes more surface area for the same mass, so more particles of the solid are available to be hit. This gives a greater frequency of collisions and therefore a faster rate.

Cutting a solid into smaller pieces creates new exposed faces. The mass is identical, but the surface area available for collisions increases sharply.

CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Powder reacts much faster, but because the mass of marble is the same, the total gas produced is identical.

Factor 4: Temperature

Increasing the temperature increases the rate of reaction - usually dramatically. As a rough rule of thumb, a 10 °C rise roughly doubles the rate of many reactions.

Explanation (two parts - you need BOTH for full marks):

The particles gain kinetic energy and move faster, so they collide more frequently.
More importantly, a greater proportion of particles now have energy equal to or greater than the activation energy, so a much larger fraction of collisions are successful.

Exam Tip

The second point is the one students forget, and it is the one that carries most of the marks. The huge effect of temperature is not mainly because particles move faster - it is because far more particles now exceed Eₐ. If a question is worth 3 marks, you almost certainly need both points plus "so more successful collisions per second".

Effect of temperature on the rate of Mg + HCl. The initial rate is the gradient of the tangent at t = 0. All curves plateau at the same volume - temperature changes how FAST, not HOW MUCH.

Factor 5: Catalysts

A catalyst is a substance that increases the rate of a chemical reaction and is not chemically changed at the end of the reaction.

Explanation: a catalyst provides an alternative reaction pathway with a lower activation energy. Because Eₐ is lower, a greater proportion of collisions have enough energy to be successful at the same temperature, so the rate increases.

A catalyst DOES	A catalyst does NOT
Speed up the reaction	Change the amount (yield) of product
Lower the activation energy	Change the enthalpy change ΔH
Provide an alternative pathway	Get used up (it can be recovered and reused)
Work in small amounts	Make an impossible reaction happen

Energy profile with and without a catalyst. The catalyst lowers the activation energy only. The reactant and product energy levels - and therefore ΔH - are completely unaffected.

Supplement

Enzymes - Nature's Catalysts

Enzymes are biological catalysts. They are protein molecules made by living cells that speed up the reactions of metabolism, allowing them to proceed rapidly at body temperature (about 37 °C) rather than requiring the high temperatures a laboratory would need.

Key features of enzymes:

They are highly specific - each enzyme catalyses only one reaction or one type of reaction
They work best at an optimum temperature (about 37 °C in humans) and an optimum pH
Above the optimum temperature the enzyme is denatured - its shape is permanently changed and it stops working, so the rate falls sharply
Like all catalysts they are not used up and are needed only in tiny amounts

Enzyme activity against temperature. Unlike an ordinary catalyst, an enzyme's rate peaks then collapses, because heat denatures the protein and destroys the active site.

Real uses of enzymes you should know: yeast (zymase) converting glucose to ethanol and carbon dioxide in fermentation and in bread-making; proteases and lipases in biological washing powders removing protein and fat stains at low temperatures; enzymes in the digestive system such as amylase breaking starch down to maltose.

Photochemical Reactions

A photochemical reaction is a reaction in which light provides the energy needed for the reaction to take place. Light, rather than heat, supplies the activation energy.

Photosynthesis

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

Carbon dioxide + water → glucose + oxygen. Light energy is absorbed by chlorophyll in the leaf. The reaction is endothermic: the rate increases as light intensity increases (up to a limit).

Photosynthesis is the single most important photochemical reaction on Earth - it makes the glucose that feeds almost every food chain and the oxygen in the air you are breathing right now.

Photography with Silver Salts

Traditional black-and-white photographic film is coated with silver halides such as silver bromide, AgBr, or silver chloride. When light hits the film, the silver halide decomposes to produce tiny grains of metallic silver, which appear black.

2AgBr → 2Ag + Br₂

Pale yellow silver bromide decomposes in light to grey-black silver metal. Where the light was bright, more silver forms and the film goes dark - this is why the film is a negative.

Silver chloride behaves the same way (2AgCl → 2Ag + Cl₂), turning from white to grey on standing in sunlight. This is also why silver halide precipitates in the laboratory darken if left on the bench.

Memory Trick

"Photo" means light. In photosynthesis, light energy is stored in glucose. In photography, light energy breaks apart silver bromide. Both are photochemical - light supplies the energy either way.

Measuring the Rate of a Reaction

To measure a rate you must follow something that changes measurably with time. There are three standard methods in the syllabus.

Method 1: Measuring the Volume of Gas Produced

If the reaction produces a gas, collect it in a gas syringe (or an inverted measuring cylinder over water) and record the volume every 10 or 15 seconds.

Best for: Mg + HCl (hydrogen), CaCO₃ + HCl (carbon dioxide), decomposition of H₂O₂ with MnO₂ (oxygen)
Advantage: a gas syringe is accurate and gives readings directly in cm³
Limitation: if the gas is soluble in water (like CO₂) you should not collect it over water; use a syringe instead

Method 2: Measuring the Loss of Mass

Place the reaction flask on a balance with a loose cotton-wool plug and record the mass every 10 or 15 seconds. As the gas escapes, the mass falls.

Best for: reactions producing a dense gas such as CO₂
Advantage: very simple; no gas leaks to worry about
Limitation: useless for hydrogen, which is so light that the mass change is too small to measure reliably

A mass-loss graph slopes DOWNWARDS. It still becomes horizontal at the end, and the steepest part is still at the start.

Method 3: Formation of a Precipitate (the "disappearing cross")

Some reactions produce an insoluble solid that makes the solution cloudy. The classic example is sodium thiosulfate with hydrochloric acid:

Na₂S₂O₃ + 2HCl → 2NaCl + SO₂ + S + H₂O

The pale yellow sulfur precipitate makes the mixture cloudy. Place the flask over a paper cross and time how long the cross takes to disappear.

A shorter time = a faster reaction. The rate is proportional to 1 ÷ time.
The judgement of when the cross "disappears" is subjective - the same person should judge every run, using the same cross and the same total volume, to make it a fair test.
A light sensor / colorimeter gives a more objective measurement.

Method	What you measure	Graph shape	Ideal reaction
Gas syringe	Volume of gas / cm³	Rises, then plateaus	Mg + HCl
Balance	Mass of flask / g	Falls, then levels off	CaCO₃ + HCl
Disappearing cross	Time for cross to vanish / s	Rate ∝ 1/time	Na₂S₂O₃ + HCl
Colorimeter	Light absorbed / transmitted	Depends on colour change	Bromine + methanoic acid

Interpreting Rate Graphs - the Skill Examiners Test Most

A typical rate graph plots volume of gas (y-axis) against time (x-axis). Reading it correctly is worth a huge number of marks, so read this section twice.

Feature of graph	What it tells you
Gradient (steepness)	The rate at that moment. Steeper = faster.
Steepest at the very start	Concentration of reactants is highest at t = 0, so collisions are most frequent then.
Gradient decreasing	Reactants are being used up, so concentration falls and the rate slows down.
Curve becomes horizontal (plateau)	The reaction has stopped - one reactant is completely used up. The rate is now zero.
Height of the plateau	The total amount of product made - decided by the amount of the limiting reactant, NOT by how fast the reaction went.
Two curves with the same plateau	Same amount of limiting reactant; only a rate factor was changed.
Two curves with different plateaus	A different amount (moles) of limiting reactant was used.

Exam Tip - The Number One Rate Graph Trap

Steepness and height are two completely different things. Steepness answers "how fast?"; height answers "how much?". Doubling the concentration of the acid in excess makes the curve steeper but does NOT raise the plateau. Doubling the mass of magnesium raises the plateau. Read the question carefully to see which one has changed.

Contrast this with the earlier graphs. Here the plateaus are at different heights because a different amount of the limiting reactant (magnesium) was used, not because the rate factor changed.

Supplement

Calculating a Rate from a Graph

To find the average rate over an interval, divide the change in the quantity by the time taken:

average rate = Δvolume ÷ Δtime

Example: 48 cm³ of gas collected in 30 s gives an average rate of 48 ÷ 30 = 1.6 cm³/s.

To find the instantaneous rate (the rate at one particular moment), draw a tangent to the curve at that point and calculate the gradient of the tangent. The initial rate is the gradient of the tangent at t = 0, and it is always the fastest rate of the whole reaction.

Worked Examples

Worked Example 1 In an experiment, 0.12 g of magnesium is added to excess dilute hydrochloric acid. 96 cm³ of hydrogen is collected in 40 s, and the reaction is complete at 60 s. Calculate (a) the average rate over the first 40 s and (b) the average rate over the whole reaction. Explain why the two values differ.

Step 1: Average rate for the first 40 s

rate = volume ÷ time = 96 cm³ ÷ 40 s = 2.4 cm³/s.

Step 2: Find the total volume

The magnesium is the limiting reactant, so once it has all reacted no more gas forms. If the graph plateaus at 96 cm³, the total volume is 96 cm³ and the reaction simply becomes very slow between 40 s and 60 s.

Step 3: Average rate over the whole reaction

rate = 96 cm³ ÷ 60 s = 1.6 cm³/s.

Step 4: Explain the difference

The reaction is fastest at the start when the concentration of acid and the surface area of magnesium are greatest. As the reactants are used up, collisions become less frequent and the rate falls. Including the slow final 20 s therefore drags the average down.

(a) 2.4 cm³/s. (b) 1.6 cm³/s. The whole-reaction average is lower because the rate decreases with time as the reactants are consumed - the graph is steepest at the beginning and flattens towards the end.

Worked Example 2 A student repeats the reaction of 1.0 g of marble chips with 50 cm³ of 1.0 mol/dm³ hydrochloric acid, but this time uses 1.0 g of powdered marble with the same acid. Sketch and describe how the two curves of volume of CO₂ against time compare, and explain the shape using collision theory.

Step 1: Identify what changed

Only the surface area changed. The mass of marble (1.0 g) and the amount of acid are identical in both runs.

Step 2: Predict the gradient

The powder has a much larger surface area, so more calcium carbonate particles are exposed to the acid. The frequency of successful collisions is higher, so the initial rate is greater and the powder curve is steeper.

Step 3: Predict the plateau

The same mass, and therefore the same number of moles, of CaCO₃ is used in both runs. Assuming acid is in excess, the same number of moles of CO₂ is produced, so both curves level off at the same height.

Step 4: Describe the shape

Both curves start steep and gradually flatten as the marble and acid are used up, finally becoming horizontal when the reaction stops. The powder simply reaches the horizontal section sooner.

The powdered marble curve is steeper at the start and plateaus earlier, but both curves finish at exactly the same volume of CO₂. Larger surface area means more exposed particles and therefore a greater frequency of successful collisions, so the rate is higher - but the total amount of product is fixed by the 1.0 g of marble.

Worked Example 3 In the sodium thiosulfate experiment, a cross disappears in 60 s at 20 °C and in 15 s at 40 °C. (a) By what factor has the rate increased? (b) Explain the increase fully in terms of collision theory. (c) Give two variables that must be kept constant.

Step 1: Use rate proportional to 1/time

At 20 °C: rate ∝ 1/60 = 0.0167 s⁻¹. At 40 °C: rate ∝ 1/15 = 0.0667 s⁻¹.

Step 2: Find the factor

0.0667 ÷ 0.0167 = 4. Equivalently, 60 ÷ 15 = 4. The rate has increased four times - consistent with the rule of thumb that rate roughly doubles for each 10 °C rise.

Step 3: Explain using collision theory

At the higher temperature the particles have more kinetic energy, so (i) they move faster and collide more frequently, and (ii) crucially, a much greater proportion of particles now have energy greater than or equal to the activation energy. Both effects raise the frequency of successful collisions per second, so the sulfur precipitate forms sooner.

Step 4: Control variables

Any two of: concentration and volume of sodium thiosulfate; concentration and volume of hydrochloric acid; the same paper cross and the same depth of solution; the same observer judging when the cross disappears; the same conical flask.

(a) The rate increased by a factor of 4. (b) Higher temperature gives particles more kinetic energy, so collisions are more frequent AND a far greater proportion of collisions have energy ≥ Eₐ, so many more collisions are successful each second. (c) Keep the concentration and volume of both solutions constant, and use the same cross, same flask and same observer.

Exam Tips for 6.2

1. Always use the phrase "frequency of successful collisions": Just saying "more collisions" is rarely enough. The mark scheme wants collisions that are both frequent AND energetic enough.

2. Concentration = particles per unit volume: Say "more particles in the same volume" or "particles are closer together", never just "more particles".

3. Temperature needs TWO points: (i) particles move faster so collide more often, and (ii) a greater proportion of particles have energy ≥ Eₐ. Point (ii) is the bigger effect and the more valuable mark.

4. Catalysts change rate, not yield: A catalyst lowers Eₐ and speeds things up. It never changes ΔH, never changes the position of equilibrium and never changes the final amount of product.

5. Gradient = rate, plateau height = amount: The single biggest source of lost marks in this topic. Learn to say "steeper gradient so faster rate" and "same final volume so same amount of product".

6. A flat line means STOPPED, not slow: When a rate curve becomes horizontal the rate is zero because a reactant has been completely used up.

7. Pressure only matters for gases: Do not offer "increase the pressure" as a way to speed up a reaction between a solid and a solution.

8. Rate is proportional to 1/time: In the disappearing-cross experiment, a shorter time means a faster rate. Plot 1/t on the y-axis to get a straight line through the origin against concentration.

🌎 Apply It: Real-World Chemistry

Rate of reaction is not an abstract idea - it decides whether a factory is profitable, whether food is safe, and in one famous case whether a whole building stays standing.

A flour mill near Ludhiana, Punjab, has strict rules: no naked flames, no smoking, and dust extraction fans that run continuously. The manager explains that a 25 kg sack of flour is completely safe next to a flame, but the same flour suspended as dust in the air is one of the most dangerous things in the building.

Explain, using ideas about rate of reaction, why flour dust is explosive but a sack of flour is not.

▼

The Chemistry

Flour is mostly starch, a carbohydrate that burns in oxygen. In a sack, only the flour on the outer surface can touch oxygen, so the surface area exposed to air is tiny compared with the total mass. Combustion is slow and stays local.

Why Dust Is Different

Suspended dust consists of billions of microscopic particles, each completely surrounded by oxygen. The surface area is enormous - thousands of times greater for the same mass. The frequency of collisions between starch and oxygen molecules is therefore vastly higher, so the combustion rate is enormous. Energy is released faster than it can escape, the air expands violently, and the result is a dust explosion.

Chemistry Connection

This is surface area taken to an extreme. The same principle explains why kindling catches fire but a log does not, why powdered marble fizzes faster than chips, and why extraction fans (which remove suspended dust and hence reduce surface area in contact with air) are a legal safety requirement in mills, sugar refineries and coal mines worldwide.

A dairy technologist at Amul in Anand, Gujarat, compares two ways of preserving milk: chilling it to 4 °C in a cold chain, or heating it briefly to 72 °C then chilling. Milk kept at 30 °C in a village spoils in a few hours; the same milk at 4 °C lasts several days.

Use rate of reaction ideas to explain why refrigeration preserves milk, and why heating first makes it last even longer.

▼

The Chemistry of Chilling

Spoilage is a set of chemical reactions catalysed by bacterial enzymes, converting lactose into lactic acid. Lowering the temperature reduces the kinetic energy of the particles, so collisions are less frequent and, crucially, a far smaller proportion of collisions have energy ≥ Eₐ. The rate of spoilage therefore drops sharply - roughly halving for every 10 °C fall.

The Chemistry of Pasteurisation

Heating to 72 °C for 15 seconds does something different: it denatures the bacterial enzymes and kills most of the bacteria. Once the enzyme catalyst is destroyed, the alternative low-activation-energy pathway is gone, so even at room temperature the spoilage reactions are extremely slow.

Chemistry Connection

Two different rate strategies in one carton: refrigeration reduces the energy of the particles, while pasteurisation removes the catalyst. Notice the neat link to the enzyme graph - above the optimum temperature the enzyme is denatured and activity collapses, and here that collapse is exactly what we want.

An automotive engineer at Jaguar Land Rover in Coventry is testing catalytic converters. She notes that for the first 60 seconds after a cold start, the converter removes almost none of the carbon monoxide, but once it reaches about 300 °C it removes over 95%. The catalyst is a thin coating of platinum and rhodium spread over a honeycomb ceramic.

Why is the honeycomb shape used, and why does the converter need to warm up before it works?

▼

The Honeycomb

Platinum is extraordinarily expensive, so the aim is maximum catalytic surface area from minimum metal. A honeycomb of thousands of thin channels gives a surface area equivalent to a couple of football pitches from a few grams of metal. More exposed catalyst surface means more sites where exhaust gas molecules can adsorb and react, so a much greater frequency of successful collisions.

The Warm-Up Period

Even with the catalyst providing a lower-activation-energy pathway, the exhaust particles still need energy ≥ Eₐ. At ambient temperature almost no particles have that energy. As the exhaust heats the converter, the proportion of particles exceeding Eₐ rises steeply and conversion jumps from near zero to over 95%. Engineers call the crossover the "light-off temperature".

Chemistry Connection

This scenario combines two factors at once - surface area and temperature - both acting on the same reaction. It also shows why most urban vehicle emissions occur in the first minute of a journey, and why short trips are disproportionately polluting.

A conservation photographer in Reykjavik still shoots black-and-white film. He stores unexposed film in a light-proof black canister inside a refrigerator, and develops it in a darkroom lit only by a dim red lamp. He says both precautions are about the same piece of chemistry.

Explain the chemistry of the film, and why the red safelight does not ruin it.

▼

The Chemistry

The film is coated with silver bromide. Light supplies the energy for a photochemical decomposition: 2AgBr → 2Ag + Br₂. The metallic silver grains appear black, so bright areas of the scene become dark areas on the negative.

Light-Proofing and Cooling

Any stray light starts the reaction, fogging the film. The canister removes the energy source entirely. Refrigeration slows the slow background decomposition reactions that would gradually degrade the emulsion over months, because at low temperature far fewer particles have energy ≥ Eₐ.

Why Red Light Is Safe

Red light carries the least energy of the visible spectrum. Standard black-and-white emulsions are not sensitive to it, so red photons do not supply enough energy to trigger the decomposition of the silver halide. Blue and ultraviolet light, being far more energetic, would fog the paper instantly.

Chemistry Connection

Photochemical reactions need light of sufficient energy, not just any light - the same activation energy idea, but with photons instead of heat. This is also why silver halide precipitates in a school laboratory darken if you leave them on a sunny bench.

A commercial greenhouse near Almeria in southern Spain grows tomatoes under glass. The operators burn propane inside the greenhouse to raise the carbon dioxide concentration to about three times the atmospheric level, and install supplementary lamps in winter. Yields rise by around 30%.

Explain, in terms of rate of reaction, why raising CO₂ concentration and light intensity increases tomato yield - and why there is a limit to the benefit.

▼

The Chemistry

Photosynthesis (6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂) is a photochemical reaction catalysed by enzymes. Raising the CO₂ concentration means more CO₂ molecules per unit volume, so a greater frequency of collisions with the enzyme active sites and a faster rate of glucose production.

The Role of Light

Light supplies the energy for the reaction. More intense light means more photons arriving per second, so more reaction events per second. In a dark Spanish winter, light is the factor holding the rate back, which is why lamps help most then.

Why There Is a Limit

Eventually another factor becomes limiting - the enzymes are working at maximum capacity, or the temperature or water supply restricts the rate. Beyond that point, adding more CO₂ or more light produces no further increase, and excessive temperature would denature the enzymes and reduce the rate.

Chemistry Connection

This is the concentration factor and the photochemical idea working together in a commercial setting, plus the important insight that rate graphs level off when something else becomes limiting. It is exactly the same shape of curve you see when a reaction plateaus because a reactant runs out.

Practice Questions: 6.2

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

According to collision theory, a reaction occurs only when particles collide:

A At any speed

B With energy equal to or greater than the activation energy and in the correct orientation

C In the presence of a catalyst

D Only at high pressure

Both conditions must be met for a successful collision: sufficient energy (≥ Eₐ) and correct orientation. Most collisions fail one or both tests.

Question 2

Which explanation for the effect of increasing concentration would gain full marks?

A There are more particles, so more reaction

B There are more particles per unit volume, so collisions are more frequent and there are more successful collisions per second

C The particles move faster

D The activation energy is lowered

Concentration does not change particle speed or Eₐ. It changes how crowded the particles are, hence collision frequency. "More particles" alone is too vague.

Question 3

On a graph of volume of gas against time, a horizontal (flat) section means that:

A The reaction is going at a steady slow rate

B The reaction has stopped because a reactant has been used up

C The gas is escaping from the apparatus

D The temperature has fallen

Zero gradient means zero rate. No more gas is being produced because the limiting reactant is exhausted.

Question 4

Excess dilute hydrochloric acid reacts with 0.10 g of magnesium. The experiment is repeated with 0.10 g of magnesium and MORE CONCENTRATED acid, still in excess. Compared with the first curve, the second curve is:

A Steeper and levels off at a higher volume

B Steeper but levels off at the same volume

C Less steep and levels off at the same volume

D Identical to the first curve

Magnesium is the limiting reactant and its mass is unchanged, so the amount of hydrogen is unchanged. Higher acid concentration only makes the reaction faster.

Question 5

Why does increasing temperature have such a large effect on rate?

A Because the activation energy is lowered

B Because there are more particles in the mixture

C Because a greater proportion of particles have energy ≥ Eₐ, as well as colliding more frequently

D Because the particles become larger

Only a catalyst lowers Eₐ. Heating raises the energy of the particles, so many more collisions clear the energy barrier - this dominates over the modest increase in collision frequency.

Question 6

Which change would NOT increase the rate of the reaction between magnesium ribbon and dilute sulfuric acid?

A Cutting the ribbon into smaller pieces

B Warming the acid from 20 °C to 40 °C

C Increasing the pressure above the solution

D Using more concentrated sulfuric acid

Pressure only affects reactions between gases. Liquids and solids are almost incompressible, so pressure does not change their particle spacing.

Question 7

A catalyst increases the rate of a reaction because it:

A Increases the energy of the particles

B Provides an alternative pathway of lower activation energy

C Increases the concentration of the reactants

D Makes the reaction more exothermic

The particles keep the same energy distribution; the barrier they must clear is simply lower, so a larger proportion of collisions succeed.

Question 8

Which statement about catalysts is FALSE?

A A catalyst is not chemically changed at the end of the reaction

B A catalyst can be used in small amounts

C A catalyst increases the total amount of product formed

D A catalyst lowers the activation energy

A catalyst changes only how quickly the product is made, never how much. The final plateau height on a rate graph is unchanged.

Question 9

In the reaction Na₂S₂O₃ + 2HCl, a cross under the flask takes 40 s to disappear at one concentration and 20 s at another. The second run has:

A Half the rate

B Twice the rate

C The same rate

D Four times the rate

Rate is proportional to 1/time. 1/20 is twice 1/40, so halving the time doubles the rate.

Question 10

Which method is LEAST suitable for following the rate of Mg + HCl → MgCl₂ + H₂?

A Collecting the gas in a gas syringe

B Measuring the loss in mass on a balance

C Collecting the gas over water in an inverted measuring cylinder

D Timing how long the magnesium takes to disappear

Hydrogen has such a low density that the mass loss is far too small for a school balance to detect. Mass loss works well for dense gases like CO₂.

Question 11

A rate graph is always steepest at the very start of the reaction because:

A The apparatus is coldest then

B The concentration of reactants (and surface area of solid) is greatest then

C The catalyst has not yet been used up

D The gas syringe is empty

At t = 0 nothing has been consumed yet, so collision frequency is at its maximum. From then on the reactants are progressively used up and the gradient falls.

Question 12

Two experiments use the same mass of marble but one uses powder and one uses chips, with excess acid. Which statement is correct?

A The powder produces more carbon dioxide in total

B Both produce the same total volume, but the powder produces it faster

C The chips produce more carbon dioxide in total

D Neither reaction reaches completion

Total volume depends only on the moles of limiting reactant, which is the same in both. Surface area affects rate, not yield.

Question 13

Enzymes differ from ordinary catalysts because they:

A Are used up during the reaction

B Are denatured above an optimum temperature, so the rate falls sharply

C Increase the activation energy

D Work only at very high pressures

Enzymes are proteins. Above the optimum temperature the active site loses its shape permanently, so activity collapses instead of continuing to rise.

Question 14

Which is a photochemical reaction?

A Neutralisation of HCl by NaOH

B Thermal decomposition of calcium carbonate

C Decomposition of silver bromide on photographic film

D Rusting of iron

2AgBr → 2Ag + Br₂ is driven by light energy. Thermal decomposition is driven by heat, so it is not photochemical.

Question 15

In photosynthesis, increasing the light intensity increases the rate because:

A More light energy is supplied per second to drive the endothermic reaction

B The activation energy is reduced

C The concentration of chlorophyll increases

D The reaction becomes exothermic

Light is the energy source for this photochemical reaction. Beyond a certain intensity another factor (CO₂, temperature or enzymes) becomes limiting and the rate levels off.

Question 16

A student collects 60 cm³ of gas in 25 s. The average rate of reaction is:

A 0.42 cm³/s

B 2.4 cm³/s

C 1500 cm³/s

D 35 cm³/s

rate = volume ÷ time = 60 ÷ 25 = 2.4 cm³/s. Option A is the mistake of dividing time by volume.

Question 17

Two rate curves for the same reaction level off at DIFFERENT heights. This must mean that:

A One experiment used a catalyst

B One experiment was carried out at a higher temperature

C Different amounts (moles) of the limiting reactant were used

D One experiment used a powder instead of lumps

Catalysts, temperature and surface area change the gradient only. A different final height can only come from a different amount of limiting reactant.

Question 18

Why is a fine spray of petrol used in a car engine rather than a stream of liquid?

A To lower the activation energy of combustion

B To increase the surface area in contact with oxygen so combustion is much faster

C To make the reaction more exothermic

D To reduce the pressure in the cylinder

Tiny droplets have an enormous combined surface area, so oxygen molecules collide with fuel molecules far more frequently and combustion completes within milliseconds.

Question 19

In a mass-loss experiment with calcium carbonate and acid, the graph of mass against time:

A Rises steeply then levels off

B Falls steeply at first, then becomes horizontal

C Is a straight line with a constant negative gradient

D Rises and then falls

The flask loses mass as CO₂ escapes. It is steepest at the start (fastest rate) and horizontal at the end when the reaction has stopped.

Question 20

Manganese(IV) oxide is added to hydrogen peroxide solution and oxygen is rapidly produced. At the end, the manganese(IV) oxide is filtered off, dried and weighed. Its mass is unchanged. This shows that manganese(IV) oxide:

A Is a reactant

B Is a catalyst

C Is a product

D Took part in the reaction and was regenerated as a different compound

Being recovered unchanged in both mass and chemical identity while dramatically speeding up 2H₂O₂ → 2H₂O + O₂ is the definition of a catalyst.

6.3 Reversible Reactions and Equilibrium ▼

What Is a Reversible Reaction?

Most reactions you have met go to completion: reactants turn into products and that is the end of the story. A reversible reaction is different - the products can react together to re-form the original reactants.

A + B ⇌ C + D

The special double arrow ⇌ means the reaction is reversible. Left to right is the forward reaction. Right to left is the backward (reverse) reaction.

Exam Tip

Draw the symbol properly: two half arrows, one pointing right on top and one pointing left underneath (⇌). A pair of full double-headed arrows or an equals sign will not be accepted. If a question uses ⇌, that is a huge hint that equilibrium ideas are being tested.

Key Example 1: Hydrated and Anhydrous Copper(II) Sulfate

CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O

Forward (heat): blue hydrated crystals → white anhydrous powder + water. Endothermic. Backward (add water): white powder + water → blue crystals. Exothermic - the test tube gets noticeably warm.

	Hydrated copper(II) sulfate	Anhydrous copper(II) sulfate
Formula	CuSO₄·5H₂O	CuSO₄
Colour	Blue	White
Made by	Adding water to the white powder	Heating the blue crystals
Energy change	Exothermic when formed (adding water)	Endothermic when formed (heating)

This colour change is used as a chemical test for the presence of water: water turns anhydrous copper(II) sulfate from white to blue.

Key Example 2: Hydrated and Anhydrous Cobalt(II) Chloride

CoCl₂·6H₂O ⇌ CoCl₂ + 6H₂O

Forward (heat): pink hydrated crystals → blue anhydrous solid + water. Backward (add water): blue solid + water → pink crystals.

Cobalt(II) chloride paper is dried until it is blue, then used as a test for water: water turns blue cobalt(II) chloride paper pink.

Memory Trick

Two tests for water, two colour changes. Copper: white → blue. Cobalt: blue → pink. Remember "Copper goes blue, Cobalt goes pink". Careful - blue appears on both lists but means opposite things, so always name the compound.

Note that these tests show water is present. To prove it is pure water you must also show it boils at exactly 100 °C and freezes at 0 °C.

Dynamic Equilibrium

If a reversible reaction takes place in a closed system (nothing can enter or leave), something remarkable happens. At the start, only the forward reaction can occur. As products build up, the backward reaction speeds up. Eventually the two rates become equal, and the reaction reaches equilibrium.

At equilibrium: rate of forward reaction = rate of backward reaction

The concentrations of reactants and products remain constant - but they are NOT necessarily equal. Both reactions continue - this is why it is called a dynamic equilibrium.

Reaching dynamic equilibrium. The forward rate falls as reactants are used up; the backward rate rises as products build up. When they meet, concentrations stop changing - but both reactions keep going.

Three conditions are needed for equilibrium to be established:

The reaction must be reversible
The system must be closed - no reactants or products can escape
The temperature must be constant

Exam Tip - The "It Has Stopped" Misconception

At equilibrium the reaction has not stopped. Reactant particles are still turning into products and product particles are still turning back, at exactly the same rate, so nothing appears to change. The correct phrase is: "the concentrations remain constant because the rate of the forward reaction equals the rate of the backward reaction." Never write "the reaction has finished."

Also: equilibrium does NOT mean 50:50. The position of equilibrium may lie far to the left or far to the right.

Changing the Position of Equilibrium

The position of equilibrium tells you whether the mixture contains mostly reactants (position lies to the left) or mostly products (position lies to the right). We can shift it by changing the conditions. The guiding idea is that a system at equilibrium responds so as to oppose the change you imposed.

Supplement

Effect of Changing Concentration

Change	Equilibrium shifts	Reason
Increase concentration of a reactant	To the right (more products)	The system removes some of the added reactant
Increase concentration of a product	To the left (more reactants)	The system removes some of the added product
Remove a product as it forms	To the right	The system tries to replace what was taken away - this is why ammonia is condensed out in the Haber process

Effect of Changing Temperature

This is the one that requires real thought. Look at whether the forward reaction is exothermic or endothermic.

Change	Equilibrium shifts	Reason
Increase temperature	In the endothermic direction	The system absorbs the extra heat you supplied, opposing the rise
Decrease temperature	In the exothermic direction	The system releases heat to oppose the fall

Memory Trick

"Heat it, it beats it." Raise the temperature and the equilibrium shifts the way that soaks heat up - the endothermic direction. Cool it and the equilibrium shifts the exothermic way to make heat. Write "endo" next to the endothermic arrow on the equation before you answer, and you will never get the direction wrong.

Effect of Changing Pressure (gases only)

First count the moles of gas on each side of the equation. Then:

Change	Equilibrium shifts	Reason
Increase pressure	Towards the side with fewer moles of gas	Fewer gas molecules exert less pressure, opposing the increase
Decrease pressure	Towards the side with more moles of gas	More gas molecules raise the pressure, opposing the decrease
Equal moles of gas on both sides	No shift at all	Neither side relieves the pressure, so position is unaffected

Effect of a Catalyst

A catalyst speeds up the forward and backward reactions equally. It therefore has no effect whatsoever on the position of equilibrium or on the yield. It only makes equilibrium be reached sooner. This is a guaranteed exam question - learn it word for word.

Industrial Application 1: The Haber Process

The Haber process manufactures ammonia, the starting point for nitrogen fertilisers that feed roughly half the world's population. Nitrogen comes from the air; hydrogen comes from natural gas.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

The forward reaction is exothermic. 4 moles of gas on the left → 2 moles of gas on the right.

Condition	Value used	Why this value is chosen
Temperature	about 450 °C	A low temperature would give a higher yield (forward reaction is exothermic) but the rate would be far too slow. 450 °C is a compromise: an acceptable yield in an acceptable time.
Pressure	about 200 atm	High pressure shifts the equilibrium right, towards the side with fewer moles of gas (4 → 2), increasing the yield, and also increases the rate. Pressures much higher than this are avoided because of the cost and danger of the equipment.
Catalyst	Iron	Increases the rate so equilibrium is reached sooner. It does NOT increase the yield.
Recycling	Unreacted N₂ and H₂ are recycled	The ammonia is cooled and condensed out as a liquid, which also shifts the equilibrium to the right and improves the overall conversion.

Exam Tip - The Compromise Answer

The classic 3-mark question is: "Explain why 450 °C is used even though a lower temperature would give a greater yield." Your answer must contain all three ideas: (1) the forward reaction is exothermic, so a lower temperature would shift the equilibrium right and give a higher yield; (2) BUT at a lower temperature the rate would be too slow and the process uneconomic; (3) so 450 °C is a compromise between yield and rate.

Industrial Application 2: The Contact Process

The Contact process manufactures sulfuric acid, arguably the most important industrial chemical in the world. The key equilibrium step is the oxidation of sulfur dioxide.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

The forward reaction is exothermic. 3 moles of gas on the left → 2 moles of gas on the right.

Condition	Value used	Why this value is chosen
Temperature	about 450 °C	Again a compromise. Lower temperature favours a higher yield (exothermic forward reaction) but is too slow.
Pressure	about 2 atm (only slightly above atmospheric)	Although higher pressure would shift the equilibrium right (3 → 2 moles), the yield at 2 atm is already about 96%. Building high-pressure plant would cost far more than the tiny extra yield is worth.
Catalyst	Vanadium(V) oxide, V₂O₅	Increases the rate so equilibrium is reached sooner. It does NOT change the yield.

Memory Trick

Both processes: 450 °C, exothermic forward reaction, fewer moles of gas on the right. The single difference to remember is the pressure and the catalyst: Haber = 200 atm + iron (High pressure, Ha-ber); Contact = 2 atm + V₂O₅ (low pressure, "Contact is Calm").

The remaining steps of the Contact process, for context: sulfur is burned to make SO₂ (S + O₂ → SO₂); the SO₃ is absorbed into concentrated sulfuric acid to make oleum; the oleum is then diluted carefully with water to give sulfuric acid. SO₃ is not added directly to water because the reaction is dangerously violent and produces an uncontrollable acid mist.

Worked Examples

Worked Example 1 For the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) (forward reaction exothermic), predict and explain the effect on the yield of ammonia of (a) increasing the temperature, (b) increasing the pressure, (c) adding an iron catalyst.

Step 1: Label the equation

Forward = exothermic, so backward = endothermic. Moles of gas: left = 1 + 3 = 4; right = 2.

Step 2 (a): Increasing temperature

The system opposes the rise by absorbing heat, so it shifts in the endothermic direction - which here is backward, to the left. The yield of ammonia decreases. (The rate increases, but that is a separate matter from yield.)

Step 3 (b): Increasing pressure

The system opposes the increase by moving to the side with fewer moles of gas. Right has 2 moles, left has 4, so the equilibrium shifts to the right. The yield of ammonia increases.

Step 4 (c): Adding a catalyst

A catalyst speeds up the forward and backward reactions by the same amount, so the position of equilibrium is unchanged. The yield is exactly the same; equilibrium is simply reached in a shorter time.

(a) Yield decreases - the equilibrium shifts left (endothermic direction) to oppose the temperature rise. (b) Yield increases - the equilibrium shifts right, towards the 2 moles of gas, to oppose the pressure rise. (c) Yield unchanged - a catalyst changes only the time taken to reach equilibrium, never its position.

Worked Example 2 Blue hydrated copper(II) sulfate crystals are heated in a test tube until the solid is white. The tube is then cooled and a few drops of water are added. Describe and explain the observations, and state what happens to the temperature.

Step 1: On heating

CuSO₄·5H₂O → CuSO₄ + 5H₂O. The blue crystals turn white as the water of crystallisation is driven off. Colourless droplets of water condense on the cooler upper part of the tube. This forward step is endothermic - it needs continuous heating.

Step 2: On adding water

CuSO₄ + 5H₂O → CuSO₄·5H₂O. The white powder turns blue again as the hydrated compound re-forms.

Step 3: Temperature

The backward reaction is exothermic, so heat is released and the test tube feels noticeably warm (it can hiss and steam if a lot of anhydrous solid is used). Whenever a reversible reaction is exothermic one way, it must be endothermic the other way, with the same magnitude of energy change.

Step 4: Note the use

This colour change is the standard chemical test for water: water turns white anhydrous copper(II) sulfate blue.

Heating: blue → white with water condensing at the top of the tube (endothermic). Adding water: white → blue and the tube becomes warm (exothermic). The reaction CuSO₄·5H₂O ⇌ CuSO₄ + 5H₂O is reversible, and the two directions have opposite energy changes.

Worked Example 3 In the Contact process, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the forward reaction is exothermic. Explain why a temperature of 450 °C and a pressure of only about 2 atm are used, when equilibrium theory would suggest a low temperature and a high pressure.

Step 1: What theory predicts for temperature

The forward reaction is exothermic, so lowering the temperature would shift the equilibrium to the right and give a higher yield of SO₃.

Step 2: Why a low temperature is not used

At low temperature very few particles have energy ≥ Eₐ, so the rate would be extremely slow and equilibrium would take far too long to reach. The plant would be uneconomic. 450 °C is therefore a compromise between an acceptable yield and an acceptable rate, and the V₂O₅ catalyst helps the rate further without affecting the yield.

Step 3: What theory predicts for pressure

There are 3 moles of gas on the left and 2 on the right, so increasing the pressure would shift the equilibrium right and raise the yield.

Step 4: Why only 2 atm is used

At just 2 atm the yield is already about 96%. Raising the pressure could add only a percent or two, but would require enormously thick, expensive and potentially dangerous pressure vessels and huge compressor running costs. The extra yield does not justify the extra cost - this is an economic decision, not a chemical one.

450 °C is a compromise: a lower temperature would raise the yield of the exothermic forward reaction but make the rate uneconomically slow. Only 2 atm is used because the yield is already about 96% at low pressure, so the extra cost and hazard of high-pressure equipment would not be repaid by the very small increase in yield.

Exam Tips for 6.3

1. Equilibrium is dynamic, not dead: Say "the rate of the forward reaction equals the rate of the backward reaction, so the concentrations remain constant." Never "the reaction stops."

2. Equal rates, not equal amounts: Equilibrium says nothing about whether there is more reactant or more product.

3. Annotate before answering: Write "exo →" and "endo ←" on the equation, and count the moles of gas on each side. Then the temperature and pressure answers become mechanical.

4. Temperature: shift towards endothermic when heated. Cooling shifts towards exothermic. Get this the right way round and half of 6.3 is done.

5. Pressure needs GASES and unequal moles: If the moles of gas are equal on both sides, pressure has no effect on the position of equilibrium. Solids and liquids are not counted.

6. Catalysts never change yield: Not in rates, not in equilibrium, not ever. They change only the time taken.

7. Learn the four industrial conditions cold: Haber 450 °C / 200 atm / iron. Contact 450 °C / 2 atm / V₂O₅. These are straight recall marks.

8. "Compromise" is the magic word: Whenever a question asks why a moderate temperature is used, the answer is a compromise between yield and rate. Say both halves.

🌎 Apply It: Real-World Chemistry

Equilibrium is the reason fertiliser is affordable, why divers get the bends, and why a weather ornament can predict rain.

An engineer at an IFFCO fertiliser plant in Kalol, Gujarat, is asked by a visiting minister why the reactor runs at 450 °C. The minister has read that the reaction is exothermic and argues that running the plant cooler would produce more ammonia and save fuel.

Is the minister right in theory? Explain why the plant does not do this in practice.

▼

The Theory

The minister is correct about the equilibrium. N₂ + 3H₂ ⇌ 2NH₃ is exothermic forwards, so lowering the temperature shifts the equilibrium to the right (the exothermic direction) and gives a higher equilibrium yield of ammonia.

The Practice

At low temperature very few molecules have energy ≥ Eₐ, so the rate is desperately slow. A reactor might take weeks to approach that high yield. A plant that produces a lot of ammonia very slowly makes less ammonia per day than one producing a moderate yield quickly, and the capital tied up in the plant earns nothing meanwhile.

The Engineering Answer

450 °C is the compromise. An iron catalyst raises the rate further at no cost in yield, 200 atm pushes the equilibrium right (4 moles → 2 moles of gas), and the ammonia is condensed out and removed, which pulls the equilibrium further right while unreacted gases are recycled.

Chemistry Connection

Industrial chemistry is always yield versus rate versus cost. Equilibrium tells you what is possible; kinetics tells you how long you would wait; accountancy decides. This is why "compromise" earns the mark and "low temperature gives more ammonia" alone does not.

A Cornish seaside shop sells "weather houses" - small ornaments containing a strip of paper treated with cobalt(II) chloride. The label claims the ornament turns blue for fine weather and pink for rain. Tourists are sceptical and think it is a trick.

Explain the real chemistry, and say whether the ornament genuinely responds to weather.

▼

The Chemistry

CoCl₂·6H₂O (pink) ⇌ CoCl₂ (blue) + 6H₂O. In dry air, water leaves the paper and the blue anhydrous form dominates. In humid air, water is absorbed and the pink hydrated form dominates.

Does It Work?

Yes - though it measures humidity, not weather. Humid air often precedes rain in coastal Cornwall, so the correlation is real but indirect. The ornament is genuinely a reversible chemical reaction responding to the concentration of water vapour, shifting the position of equilibrium back and forth.

Chemistry Connection

This is the concentration effect on equilibrium in your living room. Add water (a product of the forward reaction) and the equilibrium shifts left towards pink; remove water and it shifts right towards blue. The same compound in the laboratory is your standard test for water: blue cobalt(II) chloride paper turns pink.

A bottling plant for sparkling water in Bavaria dissolves carbon dioxide under a pressure of about 4 atm and caps the bottles immediately. A quality technician notes that an unopened bottle stays fizzy for years, but an opened one goes flat within a day even if the cap is replaced.

Use equilibrium ideas to explain why sealing matters so much, and why the drink fizzes when opened.

▼

The Chemistry

CO₂(g) ⇌ CO₂(aq) in the sealed bottle. Because the bottle is a closed system, a genuine dynamic equilibrium is established: CO₂ molecules leave the solution and dissolve back at exactly equal rates, so the fizziness stays constant.

Opening the Bottle

Opening it releases the pressure. Lowering the pressure shifts the equilibrium towards the side with more moles of gas - the gaseous CO₂ - so dissolved CO₂ comes out of solution and you hear the hiss and see bubbles.

Why It Goes Flat

Once opened, gas escapes into the room. The system is no longer closed, so equilibrium can never be re-established. The forward loss of CO₂ continues unopposed until almost none is left dissolved. Replacing the cap helps a little by allowing a small pressure to build up again, but much CO₂ has already gone.

Chemistry Connection

This is a perfect demonstration of why a closed system is one of the three requirements for equilibrium. It also shows the pressure rule in action: reduce the pressure and the equilibrium moves towards the greater number of moles of gas. Divers experience the same chemistry in their bloodstream, which is why they must surface slowly.

A chemical engineer at a sulfuric acid plant on Teesside in north-east England is challenged by an intern: "If high pressure improves the yield of SO₃, why does this plant run at barely 2 atm when the Haber plant next door runs at 200 atm?"

Give the engineer's full answer, comparing the two processes.

▼

The Chemistry Is Similar

Both have an exothermic forward reaction and fewer moles of gas on the right: 2SO₂ + O₂ ⇌ 2SO₃ goes 3 → 2, and N₂ + 3H₂ ⇌ 2NH₃ goes 4 → 2. In both cases high pressure would shift the equilibrium right.

The Decisive Difference

The starting yield is completely different. The Contact process already achieves about 96% conversion at just 2 atm, so high pressure could gain at most a few percent - not worth the enormous cost of high-pressure vessels, compressors and safety systems. The Haber process, by contrast, gives a poor yield at low pressure, so 200 atm buys a genuinely large improvement that repays the investment.

Chemistry Connection

Equilibrium principles tell you the direction of the benefit, but not whether the benefit is worth having. Cambridge questions on the Contact process very often ask exactly this, and the expected answer is "the yield is already very high at low pressure, so high pressure is not economically justified."

A physiologist in Nepal studies trekkers ascending to Everest Base Camp at 5,364 m, where atmospheric pressure is about half that at sea level. Haemoglobin binds oxygen reversibly: Hb + 4O₂ ⇌ Hb(O₂)₄. Trekkers who ascend too fast suffer altitude sickness, but those who acclimatise for several days cope well.

Explain, using equilibrium ideas, why altitude causes problems and how the body compensates.

▼

The Chemistry

Oxygen binding to haemoglobin is a reversible reaction that reaches equilibrium in the lungs. At sea level the high concentration of oxygen pushes the equilibrium to the right, so most haemoglobin is loaded with oxygen.

Why Altitude Hurts

At Base Camp there is roughly half as much oxygen per unit volume. Decreasing the concentration of a reactant shifts the equilibrium to the left, so less oxyhaemoglobin forms and less oxygen reaches the tissues - causing headache, breathlessness and nausea.

Acclimatisation

Over days the body produces more red blood cells and more haemoglobin. Increasing the concentration of the other reactant (Hb) pushes the equilibrium back to the right, restoring oxygen delivery. Breathing rate also rises, raising oxygen concentration in the lungs.

Chemistry Connection

The concentration rule is not just an industrial trick - your bloodstream runs on it every second. Note also that this reaction MUST be reversible: haemoglobin has to release the oxygen again in the tissues, which is precisely why carbon monoxide is so lethal. CO binds essentially irreversibly and destroys the equilibrium altogether.

Practice Questions: 6.3

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

The symbol ⇌ in a chemical equation shows that the reaction is:

A Exothermic

B Reversible

C Catalysed

D Complete

The two half-arrows show the reaction can proceed in both directions. It says nothing about energy changes or catalysts.

Question 2

At dynamic equilibrium in a closed system:

A Both reactions have stopped

B The concentration of reactants equals the concentration of products

C The rate of the forward reaction equals the rate of the backward reaction

D Only the forward reaction is occurring

Equal RATES, not equal amounts. Both reactions continue - that is what "dynamic" means.

Question 3

Anhydrous copper(II) sulfate is:

A Blue, and turns white when water is added

B White, and turns blue when water is added

C Pink, and turns blue when heated

D Green, and turns colourless when heated

"Anhydrous" means without water. CuSO₄ is white; adding water gives blue CuSO₄·5H₂O. This is the standard test for water.

Question 4

Hydrated cobalt(II) chloride is heated. The colour change observed is:

A Blue to pink

B Pink to blue

C White to blue

D Blue to white

Heating drives water off: CoCl₂·6H₂O (pink) → CoCl₂ (blue). Adding water reverses it, which is why blue cobalt chloride paper turning pink is a test for water.

Question 5

Which condition is NOT required for a dynamic equilibrium to be established?

A The reaction must be reversible

B The system must be closed

C A catalyst must be present

D The temperature must be constant

A catalyst only shortens the time taken to reach equilibrium. Equilibrium is established perfectly well without one.

Question 6

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), forward reaction exothermic, increasing the temperature will:

A Increase the yield of ammonia

B Decrease the yield of ammonia

C Have no effect on the yield

D Stop the reaction

Raising the temperature shifts the equilibrium in the endothermic direction, which here is backwards. Yield falls (although the rate rises).

Question 7

Increasing the pressure on the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) shifts it:

A To the right, because there are fewer moles of gas on the right

B To the left, because there are fewer moles of gas on the right

C To the left, because the forward reaction is exothermic

D Nowhere, because a catalyst is present

3 moles of gas on the left, 2 on the right. The system opposes the pressure increase by moving to the side with fewer gas molecules.

Question 8

For the equilibrium H₂(g) + I₂(g) ⇌ 2HI(g), increasing the pressure will:

A Shift it to the right

B Shift it to the left

C Have no effect on the position of equilibrium

D Convert all the HI back to elements

2 moles of gas on the left and 2 on the right. Neither side relieves the pressure, so the position is unchanged (the rate does increase, though).

Question 9

Adding a catalyst to a system at equilibrium:

A Shifts the equilibrium to the right

B Increases the yield of product

C Makes equilibrium be reached more quickly, with no change in position

D Shifts the equilibrium towards the endothermic side

A catalyst speeds up the forward and backward reactions equally, so the balance point is unchanged - only the time to get there is shorter.

Question 10

The conditions used in the Haber process are:

A 450 °C, 2 atm, vanadium(V) oxide catalyst

B 450 °C, 200 atm, iron catalyst

C 1000 °C, 200 atm, nickel catalyst

D 25 °C, 1 atm, platinum catalyst

Haber: about 450 °C, about 200 atm, iron catalyst. Option A gives the Contact process conditions - a very common mix-up.

Question 11

Why is a temperature of 450 °C used in the Haber process rather than a lower temperature?

A A lower temperature would reduce the yield

B A lower temperature would give a higher yield but the rate would be far too slow, so 450 °C is a compromise

C The iron catalyst only works above 400 °C

D Nitrogen does not become a gas below 450 °C

The compromise answer needs both halves: better yield at low temperature, unacceptable rate at low temperature.

Question 12

Why does the Contact process use a pressure of only about 2 atm?

A Because high pressure would shift the equilibrium to the left

B Because the yield is already about 96% at low pressure, so high-pressure equipment is not economically justified

C Because sulfur trioxide decomposes above 2 atm

D Because there are more moles of gas on the right

Higher pressure WOULD help slightly, but the gain is tiny and the cost enormous. It is an economic decision.

Question 13

If the forward reaction of a reversible reaction releases 92 kJ/mol, then the backward reaction:

A Also releases 92 kJ/mol

B Absorbs 92 kJ/mol

C Involves no energy change

D Absorbs 46 kJ/mol

The two directions have equal magnitude but opposite sign. Exothermic one way must be endothermic the other way by the same amount.

Question 14

In the Haber process, ammonia is cooled and removed as a liquid while unreacted gases are recycled. Removing the ammonia:

A Shifts the equilibrium to the right, increasing the overall conversion

B Shifts the equilibrium to the left

C Has no effect because ammonia is a product

D Lowers the activation energy

Removing a product makes the system try to replace it, driving the forward reaction. This is why continuous removal is used in industry.

Question 15

A sealed flask of brown NO₂ and colourless N₂O₄ is at equilibrium. The colour stops changing. This means:

A All the NO₂ has been converted to N₂O₄

B The molecules have stopped moving

C The concentrations are constant because forward and backward rates are equal

D The reaction has permanently finished

Constant colour means constant concentrations - which is exactly what equal forward and backward rates produce. Molecules continue to interconvert.

Question 16

Which is the correct test to show that a colourless liquid contains water?

A It turns blue litmus paper red

B It turns white anhydrous copper(II) sulfate blue

C It turns limewater milky

D It relights a glowing splint

White → blue with anhydrous CuSO₄ (or blue → pink with cobalt(II) chloride paper) shows water is present. To show it is PURE water, also check the boiling point is 100 °C.

Question 17

Adding more N₂ to the equilibrium mixture N₂ + 3H₂ ⇌ 2NH₃ at constant temperature and pressure will:

A Shift the equilibrium to the right and produce more ammonia

B Shift the equilibrium to the left

C Have no effect

D Lower the temperature of the mixture

Increasing the concentration of a reactant makes the system oppose the change by using some of it up, so the forward reaction is favoured.

Question 18

The catalyst used in the Contact process is:

A Iron

B Nickel

C Vanadium(V) oxide

D Manganese(IV) oxide

V₂O₅ for the Contact process; iron for the Haber process; nickel for hydrogenation of alkenes; MnO₂ for decomposing hydrogen peroxide.

Question 19

A reversible reaction is carried out in an OPEN beaker and one product is a gas. The system will:

A Reach equilibrium more quickly

B Never reach equilibrium, because the escaping gas keeps driving the forward reaction

C Reach equilibrium with a position further to the left

D Behave exactly as in a closed system

A closed system is essential. If a product escapes, the backward reaction can never build up to match the forward rate, so the reaction goes essentially to completion.

Question 20

For the equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (forward exothermic), which combination would theoretically give the HIGHEST yield of SO₃?

A High temperature, high pressure

B High temperature, low pressure

C Low temperature, high pressure

D Low temperature, low pressure

Exothermic forward reaction favours low temperature; fewer moles of gas on the right favours high pressure. Industry does not use these ideal conditions because the rate would be too slow and the plant too expensive.

6.4 Redox ▼

Three Definitions, One Idea

Redox is short for reduction and oxidation. The two always happen together - if something is oxidised, something else must be reduced at the same time. You need three levels of definition, and you must be able to use whichever the question makes possible.

In terms of...	Oxidation is...	Reduction is...
Oxygen	Gain of oxygen	Loss of oxygen
Electrons	Loss of electrons	Gain of electrons
Oxidation number	Increase in oxidation number	Decrease in oxidation number

Memory Trick

OIL RIG - Oxidation Is Loss (of electrons), Reduction Is Gain (of electrons).

And for oxidation numbers: oxidation number goes UP = OXidised (both have an "up-beat"). Reduction reduces the number - the word tells you.

Level 1: Redox in Terms of Oxygen

This is the oldest definition and still the easiest to spot. Look at the classic extraction of iron in the blast furnace:

Fe₂O₃ + 3CO → 2Fe + 3CO₂

Fe₂O₃ has lost oxygen → iron(III) oxide has been reduced. CO has gained oxygen → carbon monoxide has been oxidised.

Notice how one gain of oxygen is exactly matched by one loss. That is the "redox always comes in pairs" rule.

Level 2: Redox in Terms of Electrons

The electron definition is more powerful because it works even when there is no oxygen in the equation at all. Consider a strip of zinc placed in blue copper(II) sulfate solution - the zinc becomes coated with pink-brown copper and the blue colour fades.

Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

Ionically: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

Splitting this into half-equations makes the electron transfer visible:

Zn → Zn²⁺ + 2e⁻

Oxidation - zinc loses 2 electrons. Its oxidation number rises 0 → +2.

Cu²⁺ + 2e⁻ → Cu

Reduction - copper(II) ions gain 2 electrons. Oxidation number falls +2 → 0.

Exam Tip

In a half-equation, electrons on the right = oxidation (they have been lost). Electrons on the left = reduction (they have been gained). Check that both the atoms and the total charge balance on each side.

Oxidising Agents and Reducing Agents

This is where most marks are lost in the whole topic, so read it slowly.

An OXIDISING AGENT oxidises something else - and is itself REDUCED

It takes electrons away from the other substance, so it gains electrons itself. Examples: oxygen, chlorine, acidified potassium manganate(VII), hydrogen peroxide, concentrated nitric acid.

A REDUCING AGENT reduces something else - and is itself OXIDISED

It gives electrons to the other substance, so it loses electrons itself. Examples: carbon, carbon monoxide, hydrogen, reactive metals such as zinc and magnesium, potassium iodide.

Memory Trick

An agent does the job to someone else, and gets the opposite done to itself. A travel agent books your holiday, not their own. So an oxidising agent is reduced, and a reducing agent is oxidised. Always the opposite.

Trick to lock it in: in Fe₂O₃ + 3CO → 2Fe + 3CO₂, the CO is oxidised - so CO is the reducing agent. If you ever write "CO is the oxidising agent because it is oxidised", stop and re-read this box.

Substance	What happens to it	Its role	Electrons	Oxidation number
Zn in Zn + Cu²⁺	Oxidised	Reducing agent	Loses 2e⁻	0 → +2 (up)
Cu²⁺ in Zn + Cu²⁺	Reduced	Oxidising agent	Gains 2e⁻	+2 → 0 (down)
CO in Fe₂O₃ + CO	Oxidised	Reducing agent	Loses electrons	+2 → +4 (up)
Fe₂O₃ in Fe₂O₃ + CO	Reduced	Oxidising agent	Gains electrons	+3 → 0 (down)

Supplement

Level 3: Oxidation Numbers

The oxidation number (or oxidation state) is the charge an atom would have if all the bonds in the compound were completely ionic. It is written with the sign first: +2, −1, +7.

The Rules - Learn These Exactly

Rule	Oxidation number	Example
Any uncombined element	0	Fe, Cu, O₂, Cl₂, H₂, S₈ all = 0
A simple monatomic ion	= the charge on the ion	Na⁺ = +1, Cl⁻ = −1, Mg²⁺ = +2, O²⁻ = −2
Oxygen in a compound	−2	Except in peroxides such as H₂O₂, where it is −1
Hydrogen in a compound	+1	Except in metal hydrides such as NaH, where it is −1
Group I metals	+1	Na, K, Li in any compound
Group II metals	+2	Mg, Ca in any compound
Fluorine	−1	Always, in every compound
Sum in a neutral compound	0	In H₂O: 2(+1) + (−2) = 0 ✓
Sum in a polyatomic ion	= the charge on the ion	In SO₄²⁻: the total must be −2

Worked Method: Finding an Unknown Oxidation Number

Find the oxidation number of manganese in the manganate(VII) ion, MnO₄⁻:

Let the oxidation number of Mn be x.
Each oxygen is −2, and there are four of them: 4 × (−2) = −8.
The total must equal the charge on the ion, which is −1.
So x + (−8) = −1, giving x = +7.

Try the same method on the dichromate(VI) ion, Cr₂O₇²⁻: 2x + 7(−2) = −2, so 2x = +12 and x = +6.

Roman Numerals in Names

The Roman numeral in a chemical name gives the oxidation number of the element immediately in front of it.

Name	Formula	What the numeral means
iron(II) chloride	FeCl₂	Iron has oxidation number +2
iron(III) chloride	FeCl₃	Iron has oxidation number +3
copper(II) oxide	CuO	Copper has oxidation number +2
lead(IV) oxide	PbO₂	Lead has oxidation number +4
manganate(VII)	MnO₄⁻	Manganese has oxidation number +7
manganese(IV) oxide	MnO₂	Manganese has oxidation number +4

Exam Tip

If Fe²⁺ changes into Fe³⁺, the oxidation number has gone from +2 to +3 - an increase, so it has been oxidised (it lost an electron). Students often see the ion becoming "more positive" and wrongly call it reduction because "it gained a plus". Think electrons: it has lost a negative particle, so oxidation. And remember +3 is bigger than +2 - going from −2 to 0 is also an increase.

Identifying Redox Reactions from an Equation

Use this four-step routine every time:

Assign oxidation numbers to every atom on both sides
Look for any element whose number has changed
If nothing changed, it is not a redox reaction
If something changed: increase = oxidised (that substance is the reducing agent); decrease = reduced (that substance is the oxidising agent)

Reaction	Redox?	Reasoning
Mg + 2HCl → MgCl₂ + H₂	Yes	Mg: 0 → +2 (oxidised). H: +1 → 0 (reduced).
HCl + NaOH → NaCl + H₂O	No	Every oxidation number is unchanged. Neutralisation is never redox.
CaCO₃ → CaO + CO₂	No	Ca stays +2, C stays +4, O stays −2. Thermal decomposition here is not redox.
2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻	Yes	Fe: +2 → +3 (oxidised). Cl: 0 → −1 (reduced). Chlorine is the oxidising agent.
AgNO₃ + NaCl → AgCl + NaNO₃	No	Precipitation - the ions simply swap partners, no electron transfer.
2H₂O₂ → 2H₂O + O₂	Yes	O goes −1 → −2 (reduced) AND −1 → 0 (oxidised). The same element does both - this is called disproportionation.

Chemical Tests for Oxidising and Reducing Agents

Two colour-change tests are named in the syllabus. Learn the colours, the direction of the change, and what conclusion you may draw.

Test 1: Acidified Potassium Manganate(VII) - Tests for a REDUCING Agent

purple → colourless

Acidified potassium manganate(VII), KMnO₄, is a powerful oxidising agent and is deep purple. If it meets a reducing agent it is itself reduced: MnO₄⁻ (Mn = +7, purple) → Mn²⁺ (Mn = +2, colourless / very pale pink). A purple-to-colourless change therefore shows that a reducing agent is present.

The solution must be acidified (usually with dilute sulfuric acid) because the reduction to Mn²⁺ requires H⁺ ions. Without acid you get a brown MnO₂ precipitate instead.

Test 2: Potassium Iodide Solution - Tests for an OXIDISING Agent

colourless → brown

Potassium iodide solution, KI, contains I⁻ ions and is colourless. It is a reducing agent. If it meets an oxidising agent it is oxidised: 2I⁻ → I₂ + 2e⁻. Iodine is brown in solution. A colourless-to-brown change therefore shows that an oxidising agent is present.

Memory Trick

The reagent tells you the opposite of what it is. Manganate(VII) is an oxidising agent, so it detects a REDUCING agent. Iodide is a reducing agent, so it detects an OXIDISING agent.

Colours: "Purple disappears, brown appears." Purple → colourless with manganate(VII). Colourless → brown with iodide.

Reagent	Colour change	What is detected	What happens to the reagent
Acidified potassium manganate(VII)	Purple → colourless	A reducing agent	The reagent is reduced (Mn +7 → +2)
Potassium iodide solution	Colourless → brown	An oxidising agent	The reagent is oxidised (I −1 → 0)

Common Redox Reactions You Should Recognise

Reaction type	Example	Oxidised / reduced
Combustion	CH₄ + 2O₂ → CO₂ + 2H₂O	Carbon oxidised (−4 → +4); oxygen reduced (0 → −2)
Metal extraction	Fe₂O₃ + 3CO → 2Fe + 3CO₂	Iron reduced; carbon oxidised
Displacement	Zn + CuSO₄ → ZnSO₄ + Cu	Zinc oxidised; copper(II) reduced
Metal + acid	Mg + 2HCl → MgCl₂ + H₂	Magnesium oxidised; hydrogen reduced
Rusting	Iron + oxygen + water → hydrated iron(III) oxide	Iron oxidised (0 → +3); oxygen reduced
Electrolysis	At the cathode: Cu²⁺ + 2e⁻ → Cu	Reduction at the cathode; oxidation at the anode
Respiration	C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O	Glucose oxidised; oxygen reduced

Memory Trick for Electrolysis

"Red Cat, An Ox" - Reduction at the Cathode, Anode is where Oxidation happens. Positive ions travel to the cathode and gain electrons; negative ions travel to the anode and lose them.

Worked Examples

Worked Example 1 Deduce the oxidation number of the underlined element in each of: (a) MnO₄⁻, (b) Cr₂O₇²⁻, (c) SO₄²⁻, (d) NH₃, (e) H₂O₂.

(a) MnO₄⁻

Let Mn = x. Oxygen is −2 each: 4 × (−2) = −8. Total must equal the ion charge, −1. So x − 8 = −1, giving Mn = +7. This is why it is called manganate(VII).

(b) Cr₂O₇²⁻

2x + 7(−2) = −2, so 2x − 14 = −2, so 2x = +12 and Cr = +6. Careful with the 2 in Cr₂ - divide at the end.

x + 4(−2) = −2, so x − 8 = −2 and S = +6.

(d) NH₃

Hydrogen is +1 each and the compound is neutral: x + 3(+1) = 0, so N = −3. Nitrogen can be negative because it is more electronegative than hydrogen.

(e) H₂O₂

This is a peroxide - the exception. Hydrogen is +1, so 2(+1) + 2(oxygen) = 0 gives oxygen = −1 (and hydrogen = +1 as usual). Watch for peroxides in exam questions.

(a) +7 (b) +6 (c) +6 (d) −3 (e) H = +1 and O = −1

Worked Example 2 For the reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), state which species is oxidised and which is reduced, identify the oxidising agent and the reducing agent, write the two half-equations, and describe the colour change you would see.

Step 1: Assign oxidation numbers

Fe³⁺ = +3 → Fe²⁺ = +2 (a decrease). I⁻ = −1 → I₂ = 0 (an increase).

Step 2: Apply the rules

Decrease in oxidation number = reduction, so Fe³⁺ is reduced. Increase in oxidation number = oxidation, so I⁻ is oxidised.

Step 3: Name the agents (the opposite!)

Fe³⁺ is reduced, so Fe³⁺ is the oxidising agent. I⁻ is oxidised, so the iodide ion is the reducing agent.

Step 4: Half-equations

Reduction (electrons on the left): Fe³⁺ + e⁻ → Fe²⁺
Oxidation (electrons on the right): 2I⁻ → I₂ + 2e⁻
Multiply the first by 2 so that the electrons cancel when the half-equations are added.

Step 5: Observation

The colourless iodide solution turns brown as iodine is formed, and the yellow-brown iron(III) solution becomes pale green iron(II). The brown colour is the standard indication that an oxidising agent was present.

I⁻ is oxidised (−1 → 0) and is the reducing agent. Fe³⁺ is reduced (+3 → +2) and is the oxidising agent. Half-equations: 2Fe³⁺ + 2e⁻ → 2Fe²⁺ and 2I⁻ → I₂ + 2e⁻. Observation: the solution turns brown as iodine forms.

Worked Example 3 A student adds a few drops of acidified potassium manganate(VII) to two unknown solutions. With solution X the purple colour disappears; with solution Y the purple colour remains. She then adds potassium iodide to solution Y and it turns brown. Deduce what X and Y are, explaining each observation in terms of electron transfer.

Step 1: Interpret the manganate(VII) result for X

Purple → colourless means the MnO₄⁻ ion has been reduced from Mn(+7) to Mn²⁺(+2). Something must have given it those electrons, so X is a reducing agent.

Step 2: Interpret the manganate(VII) result for Y

The purple colour remains, so nothing reduced the manganate(VII). Y is therefore not a reducing agent. This is a negative result - it does not by itself prove Y is an oxidising agent.

Step 3: Interpret the iodide result for Y

Colourless → brown means iodide ions have been oxidised to iodine: 2I⁻ → I₂ + 2e⁻. Something must have accepted those electrons, so Y is an oxidising agent.

Step 4: Summarise the electron transfer

With X: X loses electrons (oxidised); MnO₄⁻ gains them (reduced). With Y: I⁻ loses electrons (oxidised); Y gains them (reduced). In both cases the reagent does the opposite of what it detects.

X is a reducing agent - it donated electrons to the manganate(VII), reducing Mn from +7 to +2 and discharging the purple colour. Y is an oxidising agent - it accepted electrons from iodide ions, oxidising them from −1 to 0 and forming brown iodine.

Exam Tips for 6.4

1. OIL RIG, every single time: Oxidation Is Loss of electrons, Reduction Is Gain. Write it in the margin at the start of the paper.

2. Agents are always the opposite: The oxidising agent is the substance that is reduced. The reducing agent is the substance that is oxidised. This is the most common error in the whole topic.

3. State the numbers, not just the words: "Iron is oxidised from +2 to +3" scores where "iron is oxidised" alone may not. Quote both values.

4. Sign first: Write +2, not 2+, for oxidation numbers. Ionic charges are written the other way round (Fe²⁺), and examiners do notice.

5. If no oxidation number changes, it is not redox: Neutralisation and precipitation reactions are never redox. Do not be fooled by a dramatic colour change.

6. Learn the two tests as a pair: Acidified manganate(VII) purple → colourless detects a REDUCING agent. Potassium iodide colourless → brown detects an OXIDISING agent.

7. Remember the exceptions: Oxygen is −2 except in peroxides (−1). Hydrogen is +1 except in metal hydrides (−1). Uncombined elements are always 0, including O₂ and Cl₂.

8. Half-equations must balance twice: Balance the atoms AND the total charge. Electrons on the right = oxidation; electrons on the left = reduction.

🌎 Apply It: Real-World Chemistry

Redox runs the world: it extracts our metals, powers our phones, purifies our water, keeps our food fresh and even catches drink-drivers.

A metallurgist at the Tata Steel plant in Jamshedpur explains the blast furnace to visiting students. Iron ore (haematite, Fe₂O₃), coke and limestone are fed in at the top, hot air is blasted in at the bottom, and molten iron runs out below. She says the entire furnace exists to perform one redox reaction.

Identify the redox reaction, state what is oxidised and reduced, and name the reducing agent.

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The Chemistry

Coke burns to CO₂, which then reacts with more hot coke to form carbon monoxide. The key step is Fe₂O₃ + 3CO → 2Fe + 3CO₂.

Assigning the Roles

Iron goes from +3 in Fe₂O₃ to 0 in Fe - a decrease, so iron is reduced. Carbon goes from +2 in CO to +4 in CO₂ - an increase, so carbon monoxide is oxidised. Therefore CO is the reducing agent and Fe₂O₃ is the oxidising agent.

Why It Matters

Iron in nature is almost always found as an oxide, because iron is reactive enough to have been oxidised over geological time. Extracting the metal means reversing that oxidation - which is exactly why metal extraction always costs energy and why recycling steel saves so much of it.

Chemistry Connection

Notice the trap this scenario sets. CO is oxidised, so students often call it the oxidising agent. It is the reducing agent. The agent always does the opposite to itself.

A traffic police officer in Birmingham uses an electronic breathalyser at a roadside stop. She explains that the earliest models, still occasionally seen in training museums, used a tube of orange crystals of potassium dichromate(VI) that turned green when a driver over the limit blew into them.

Explain the redox chemistry of the original breathalyser, including oxidation number changes.

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The Chemistry

Ethanol in the breath is oxidised to ethanoic acid by acidified potassium dichromate(VI). The dichromate(VI) ion, Cr₂O₇²⁻, is itself reduced to the chromium(III) ion, Cr³⁺.

Oxidation Numbers

In Cr₂O₇²⁻: 2x + 7(−2) = −2 gives Cr = +6 (orange). In Cr³⁺ the oxidation number is +3 (green). A decrease from +6 to +3 means chromium has been reduced and has gained electrons. Ethanol therefore lost electrons and was oxidised.

The Roles

Dichromate(VI) is the oxidising agent; ethanol is the reducing agent. The more alcohol in the breath, the further the orange-to-green change travels along the tube, which is how the reading was estimated.

Chemistry Connection

This is the same logic as the acidified manganate(VII) test in the syllabus: a coloured oxidising agent changes colour when it is reduced, revealing the presence of a reducing agent. Modern breathalysers use a fuel cell, but that too is a redox device - it oxidises the ethanol at an electrode and measures the current.

An engineer maintaining an offshore wind turbine in the North Sea inspects the steel monopile. Bolted to the submerged section are large blocks of zinc, badly corroded and due for replacement. The steel underneath is untouched. She calls them "sacrificial anodes".

Explain, using half-equations and oxidation numbers, why the zinc corrodes and the steel does not.

▼

The Chemistry

Rusting is the oxidation of iron: Fe → Fe²⁺ + 2e⁻ (0 → +2, then on to +3 in hydrated iron(III) oxide). Zinc is more reactive than iron, meaning it loses electrons more readily: Zn → Zn²⁺ + 2e⁻.

Why the Steel Is Protected

Because zinc is oxidised in preference, it supplies a continuous flow of electrons to the steel. Any Fe²⁺ that starts to form is immediately reduced back to iron by those electrons. The iron is effectively forced to stay in oxidation state 0.

The Roles

The zinc is oxidised, so zinc is the reducing agent - it is protecting the steel by reducing it. Dissolved oxygen in the seawater is the oxidising agent, being reduced from 0 to −2.

Chemistry Connection

"Sacrificial" is a perfect description: the zinc is deliberately allowed to be oxidised so that the iron is not. The same chemistry protects ships' hulls, underground pipelines and the steel reinforcement in bridges - and it is why galvanised (zinc-coated) iron keeps working even when the coating is scratched.

A water treatment chemist in Singapore doses reservoir water with chlorine, then tests the outflow with starch-iodide paper, which turns blue-black if excess chlorine remains. Downstream, sulfur dioxide is added to remove any surplus chlorine before the water enters the supply network.

Explain the redox chemistry of both the test and the de-chlorination step.

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The Test

Chlorine is a strong oxidising agent. It oxidises iodide ions in the paper: Cl₂ + 2I⁻ → 2Cl⁻ + I₂. Iodine goes from −1 to 0 (oxidised) while chlorine goes from 0 to −1 (reduced). The iodine formed gives the intense blue-black colour with starch.

De-chlorination

Sulfur dioxide is a reducing agent. It donates electrons to the chlorine, reducing it to harmless chloride ions while the sulfur is oxidised from +4 in SO₂ to +6 in sulfate. The chlorine is removed by being reduced.

The Roles

In the test, chlorine is the oxidising agent and iodide the reducing agent. In the de-chlorination, sulfur dioxide is the reducing agent and chlorine the oxidising agent. Chlorine plays the same role in both - it is always the electron acceptor here.

Chemistry Connection

This is exactly the syllabus test for an oxidising agent - potassium iodide turning from colourless to brown - deployed on an industrial scale. It also shows why chlorine kills bacteria: it oxidises the molecules inside their cells, exactly as it oxidises iodide ions.

A packaging technologist at a snack factory in Chennai seals small sachets labelled "Do Not Eat" into packets of banana chips. The sachets contain iron powder, salt and activated carbon. Without them the chips turn rancid within weeks; with them they stay fresh for six months.

Explain how an iron sachet keeps food fresh, in redox terms, and state which substance is the reducing agent.

▼

The Chemistry

The sachet is an oxygen scavenger. The iron powder is deliberately allowed to rust: 4Fe + 3O₂ → 2Fe₂O₃. Iron is oxidised from 0 to +3 and oxygen is reduced from 0 to −2. The salt and moisture speed the reaction up, and the huge surface area of the powder makes it fast enough to be useful.

Why This Preserves the Chips

Rancidity is itself an oxidation - oxygen attacking the fats in the chips. By reacting with essentially all the oxygen inside the sealed packet, the iron removes the oxidising agent that would otherwise attack the food. No oxygen, no oxidation of the fats.

The Roles

The iron is oxidised, so iron is the reducing agent; oxygen is the oxidising agent. The iron is sacrificed to protect the food, in the same spirit as a sacrificial anode protects a steel pile.

Chemistry Connection

Rusting is usually the villain of chemistry lessons. Here it is deliberately encouraged because it is useful. The same reaction also powers disposable hand warmers - and in every case iron powder is chosen precisely because its enormous surface area gives a fast enough rate.

Practice Questions: 6.4

20 multiple choice questions. Click an option to check your answer.

Your Score 0 / 20

Question 1

In terms of electrons, oxidation is:

A Loss of electrons

B Gain of electrons

C Loss of protons

D Gain of oxygen only

OIL RIG: Oxidation Is Loss, Reduction Is Gain (of electrons). The oxygen definition is valid but narrower.

Question 2

In the reaction Fe₂O₃ + 3CO → 2Fe + 3CO₂, the reducing agent is:

A Fe₂O₃

B CO

C Fe

D CO₂

CO gains oxygen, so CO is oxidised - and the substance that is oxidised is the reducing agent. Fe₂O₃ is reduced, so it is the oxidising agent.

Question 3

The oxidation number of manganese in MnO₄⁻ is:

A +2

B +4

C +7

D −1

x + 4(−2) = −1, so x = +7. That is why the ion is named manganate(VII).

Question 4

Which change represents oxidation?

A Cu²⁺ → Cu

B Fe²⁺ → Fe³⁺

C Cl₂ → 2Cl⁻

D Mn⁷⁺ → Mn²⁺

+2 → +3 is an increase in oxidation number, caused by losing an electron. The other three are all decreases, so they are reductions.

Question 5

An oxidising agent is a substance that:

A Is oxidised and loses electrons

B Is reduced and gains electrons

C Always contains oxygen

D Has no change in oxidation number

It oxidises something else by taking its electrons, so it gains electrons and is reduced. Chlorine is an oxidising agent containing no oxygen at all.

Question 6

Acidified potassium manganate(VII) turns from purple to colourless. This shows the presence of:

A An oxidising agent

B A reducing agent

C An acid

D Water

The manganate(VII) is itself reduced from Mn(+7) to Mn²⁺(+2), so something must have donated electrons to it - a reducing agent.

Question 7

Colourless potassium iodide solution turns brown when a substance is added. The substance is:

A An oxidising agent

B A reducing agent

C A catalyst

D An alkali

Iodide ions have been oxidised to brown iodine (2I⁻ → I₂ + 2e⁻), so the added substance must have accepted those electrons.

Question 8

Which reaction is NOT a redox reaction?

A Mg + 2HCl → MgCl₂ + H₂

B HCl + NaOH → NaCl + H₂O

C Zn + CuSO₄ → ZnSO₄ + Cu

D 2Mg + O₂ → 2MgO

In neutralisation no oxidation number changes at all - Na stays +1, Cl stays −1, H stays +1, O stays −2. Neutralisation is never redox.

Question 9

The oxidation number of an atom of an uncombined element, such as the chlorine in Cl₂, is:

A −1

B +1

C 0

D +2

Any element in its uncombined form has oxidation number 0, including diatomic molecules like Cl₂, O₂ and H₂.

Question 10

In the half-equation Zn → Zn²⁺ + 2e⁻, the zinc is:

A Oxidised, and acts as a reducing agent

B Reduced, and acts as an oxidising agent

C Oxidised, and acts as an oxidising agent

D Unchanged

Electrons on the right means they have been lost: oxidation, 0 → +2. Whatever is oxidised is the reducing agent.

Question 11

What is the oxidation number of sulfur in H₂SO₄?

A +2

B +4

C +6

D −2

2(+1) + x + 4(−2) = 0, so 2 + x − 8 = 0 and x = +6.

Question 12

In the name iron(III) oxide, the Roman numeral III tells you that:

A There are three iron atoms in the formula

B Iron has an oxidation number of +3

C There are three oxygen atoms

D The compound has a charge of 3−

The Roman numeral always gives the oxidation number of the element just before it. Fe₂O₃ happens to contain two iron atoms, not three.

Question 13

In 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻, which species is reduced?

A Fe²⁺

B Cl₂

C Fe³⁺

D Cl⁻

Chlorine goes from 0 to −1, a decrease, so it is reduced and is the oxidising agent. Iron goes from +2 to +3 and is oxidised.

Question 14

The oxidation number of oxygen in hydrogen peroxide, H₂O₂, is:

A −2

B −1

C 0

D +2

Peroxides are the exception to the "oxygen is −2" rule. With H at +1: 2(+1) + 2x = 0, so x = −1.

Question 15

Which statement about redox reactions is correct?

A Oxidation can occur without reduction

B Oxidation and reduction always occur together

C Redox reactions must involve oxygen

D The oxidising agent is always oxidised

Electrons lost by one species must be gained by another - they cannot vanish. Many redox reactions, such as Zn + Cu²⁺, contain no oxygen.

Question 16

During electrolysis, reduction takes place at the:

A Cathode, where positive ions gain electrons

B Anode, where positive ions gain electrons

C Cathode, where negative ions lose electrons

D Anode, where positive ions lose electrons

Red Cat, An Ox: Reduction at the Cathode, Oxidation at the Anode. Cations move to the cathode and gain electrons.

Question 17

Magnesium burns in carbon dioxide: 2Mg + CO₂ → 2MgO + C. In this reaction:

A Magnesium is reduced and CO₂ is oxidised

B Magnesium is oxidised and CO₂ is the oxidising agent

C Neither species changes oxidation number

D Carbon is oxidised from +4 to 0

Mg goes 0 → +2 (oxidised, so it is the reducing agent). Carbon goes +4 → 0, a DECREASE, so it is reduced and CO₂ is the oxidising agent.

Question 18

Which half-equation represents a reduction?

A 2Br⁻ → Br₂ + 2e⁻

B Fe → Fe²⁺ + 2e⁻

C Cl₂ + 2e⁻ → 2Cl⁻

D Mg → Mg²⁺ + 2e⁻

Electrons on the LEFT means electrons are being gained: reduction. Chlorine goes from 0 to −1.

Question 19

Why must the potassium manganate(VII) solution be acidified before use as a test reagent?

A To make the purple colour brighter

B Because H⁺ ions are needed for the reduction to colourless Mn²⁺

C To act as a catalyst

D To stop the solution decomposing in light

In acidic conditions MnO₄⁻ is reduced all the way to Mn²⁺, which is colourless. Without acid you get a brown MnO₂ precipitate instead, which obscures the result.

Question 20

A piece of iron is protected from rusting by attaching a block of zinc. In this arrangement:

A The zinc is reduced and acts as an oxidising agent

B The zinc is oxidised in preference to the iron, so zinc acts as a reducing agent

C The iron is oxidised faster than normal

D No redox reaction takes place

Zinc is more reactive so it loses electrons more readily (Zn → Zn²⁺ + 2e⁻). Those electrons keep the iron in oxidation state 0, preventing rust. This is sacrificial protection.

Made with care for Tara at Bangalore International School

Cambridge IGCSE Chemistry (0620) | 2026-2028 Syllabus | Extended Tier

Topic 6: Chemical Reactions - Complete Study Guide