ScienceCBSE Class 10 · Chapter 1

Chemical Reactions and Equations Class 10: Complete Guide

Updated 14 October 2026 · Fact-checked

Chemical Reactions and Equations is the first chemistry chapter of Class 10 Science. It teaches you to write and balance chemical equations and to classify reactions as combination, decomposition, displacement, double displacement, precipitation, exothermic, endothermic and redox. The board asks you to balance and classify equations, explain lab observations, and identify the substances oxidised and reduced.

What Chemical Reactions and Equations is about

This chapter is about what happens when substances change into new substances. You learn how to spot that a reaction has happened, how to write it as a short chemical equation, and why that equation must be balanced. You then sort reactions into types: combination, decomposition, displacement, double displacement and precipitation. You also learn whether they give out or take in energy, and whether they involve oxidation and reduction.

It is the base for the rest of chemistry. Reactions of acids with metals and bases come back in Acids, Bases and Salts. The idea that a more reactive metal displaces a less reactive one is used again in Metals and Non-metals. Respiration as an energy-releasing reaction links to Life Processes.

Board questions take a few clear shapes. You may be asked to balance an equation and name its type, or to say what happens when something is heated or mixed and write the equation. You may get an MCQ on what is oxidised and what is reduced, or an assertion-reason on exothermic and endothermic reactions. Case-based and lab-based questions are built on the NCERT activities and the listed practicals. Newer questions also ask you to find the error in an equation or judge which student in a dialogue is right.

Weightage: This chapter is part of Unit I, Chemical Substances: Nature and Behaviour, which carries 25 of the 80 theory marks together with Acids, Bases and Salts, Metals and Non-metals and Carbon and its Compounds.Source: CBSE Science syllabus 2026-27, Unit I Chemical Substances: Nature and Behaviour

In real board papers: asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.8 marks per paper. See the questions

Key concepts

Chemical reactions and how to recognise them

A chemical reaction happens whenever a chemical change takes place. The substances you start with change their nature and identity, and new substances form. Milk turning sour, food being cooked or digested, and respiration are all everyday examples.

You cannot see atoms rearranging, so you judge from what you observe. Any one of these tells you a reaction has taken place:

a change in state
a change in colour
evolution (giving off) of a gas
a change in temperature

For example, a magnesium ribbon burns with a dazzling white flame and turns into a white powder, magnesium oxide. Both state and colour change. When zinc granules are added to dilute sulphuric acid, gas bubbles form around the zinc. When quick lime is added to water, the beaker becomes hot.

In the exam, always link the observation to the reaction. Do not just list the four signs.

Worked example

Three activities are done in a lab. (i) A cleaned magnesium ribbon is burnt and the product is collected in a watch-glass. (ii) Zinc granules are added to dilute sulphuric acid in a conical flask. (iii) Water is slowly added to quick lime in a beaker. For each, state one observation that shows a chemical reaction has occurred and name one product formed. (3 marks)

Try it first, then show the solutionSolution

(i) The ribbon burns with a dazzling white flame and changes into a white powder (change in state and colour). Product: magnesium oxide. (1 mark)
(ii) Bubbles of gas form around the zinc granules (evolution of a gas). Product: hydrogen gas (zinc sulphate also forms). (1 mark)
(iii) The beaker becomes hot (change in temperature). Product: calcium hydroxide (slaked lime). (1 mark)

Answer: (i) White flame and white powder; magnesium oxide. (ii) Gas bubbles; hydrogen. (iii) Beaker becomes hot; calcium hydroxide.

Writing a chemical equation

NCERT section 1.1.1 Writing a Chemical Equation

Describing a reaction in a full sentence is long. A word-equation is shorter, for example Magnesium + Oxygen → Magnesium oxide. Reactants go on the left-hand side and products on the right. Plus signs separate the substances on each side. The arrow points towards the products and shows the direction of the reaction.

A chemical equation is shorter still because it uses formulae, as in $M g + O_{2} \to M g O$ . If the atoms are not equal on both sides, it is only a skeletal equation.

To make an equation more informative, you add:

state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous, meaning dissolved in water
conditions above or below the arrow, such as heat, sunlight, a catalyst or pressure

For example, $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$ uses (g) for water because the water is steam. State symbols are usually added only when they are needed, but board questions often ask for them.

Worked example

Limestone (calcium carbonate, a solid) is heated strongly. It breaks down into solid calcium oxide and carbon dioxide gas. (a) Write the word-equation. (b) Write the balanced chemical equation with state symbols and the condition. (c) What does the arrow in the equation tell you? (3 marks)

Try it first, then show the solutionSolution

(a) Calcium carbonate → Calcium oxide + Carbon dioxide (1 mark)
(b) $C a C O_{3} (s) heat C a O (s) + C O_{2} (g)$ . Atoms: Ca 1 = 1, C 1 = 1, O 3 = 1 + 2. (1 mark)
(c) The arrow points from the reactant to the products, so it shows the direction in which the reaction goes. (1 mark)

Answer: (a) Calcium carbonate → Calcium oxide + Carbon dioxide. (b) $C a C O_{3} (s) heat C a O (s) + C O_{2} (g)$ . (c) It shows the direction of the reaction, from reactants to products.

Balanced chemical equations

NCERT section 1.1.2 Balanced Chemical Equations

The law of conservation of mass says mass is neither created nor destroyed in a chemical reaction. Atoms are only rearranged, so the number of atoms of each element must be the same before and after. A balanced equation shows this. An unbalanced one wrongly suggests that mass has appeared or vanished.

You balance by the hit-and-trial method:

Draw a box around each formula in your mind. Never change anything inside it.
Count the atoms of each element on both sides.
Start with the compound that has the most atoms, and with the element that has the most atoms in it.
Put whole-number coefficients in front of formulae. Use the smallest numbers that work.
Leave single elements such as $F e$ or $H_{2}$ till last, because changing them disturbs nothing else.
Recount every element to check.

To balance oxygen you write $4 H_{2} O$ , never $H_{2} O_{4}$ or $(H_{2} O)_{4}$ . Changing a subscript makes a different substance.

Worked example

Balance the following equations. Show your atom count. (a) $H_{2} + N_{2} \to N H_{3}$ (b) $H_{2} S + O_{2} \to H_{2} O + S O_{2}$ (3 marks)

Try it first, then show the solutionSolution

(a) N: 2 on the left, 1 on the right, so write $2 N H_{3}$ . Now H: 2 on the left, 6 on the right, so write $3 H_{2}$ . (½ mark)
Balanced: $3 H_{2} + N_{2} \to 2 N H_{3}$ . Check: H 6 = 6, N 2 = 2. (½ mark)
(b) H and S are already 2 and 1 on each side. O: 2 on the left, 1 + 2 = 3 on the right. An odd count, so double the products first: $2 H_{2} O + 2 S O_{2}$ gives O = 2 + 4 = 6. (½ mark)
Now balance H and S on the left: $2 H_{2} S$ gives H 4 and S 2, matching the right. (½ mark)
Balance O on the left: 6 atoms are needed, so write $3 O_{2}$ . (½ mark)
Balanced: $2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$ . Check: H 4 = 4, S 2 = 2, O 6 = 6. (½ mark)

Answer: (a) $3 H_{2} + N_{2} \to 2 N H_{3}$ (b) $2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$

Combination reaction

NCERT section 1.2.1 Combination Reaction

In a combination reaction, two or more substances combine to form a single product. The reactants can be elements or compounds. The general pattern is A + B → AB.

Examples from the chapter:

Quick lime and water: $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q) + heat$ . The product is slaked lime.
Burning of coal: $C (s) + O_{2} (g) \to C O_{2} (g)$
Formation of water: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$
Burning of magnesium: $2 M g + O_{2} \to 2 M g O$

The test is simple. Count the products. Exactly one product means combination, however many reactants there are.

Many combination reactions also give out a lot of heat. The quick lime reaction makes the beaker hot, so it is also exothermic. A single reaction can belong to more than one type, and board questions often want both names. The burning of magnesium is combination, exothermic and also oxidation, because magnesium gains oxygen.

Worked example

A cleaned magnesium ribbon is burnt in air and turns into a white powder. (a) Write the balanced chemical equation. (b) Name the type of reaction based on the number of products and justify. (c) Is it exothermic or endothermic? Is magnesium oxidised or reduced? Give a reason for each. (3 marks)

Try it first, then show the solutionSolution

(a) $2 M g (s) + O_{2} (g) \to 2 M g O (s)$ . Check: Mg 2 = 2, O 2 = 2. (1 mark)
(b) Combination reaction, because two reactants, magnesium and oxygen, form a single product, magnesium oxide. (1 mark)
(c) Exothermic, because heat (and light) is given out as the ribbon burns. (½ mark)
Magnesium is oxidised, because it gains oxygen to become magnesium oxide. (½ mark)

Answer: (a) $2 M g + O_{2} \to 2 M g O$ (b) Combination: one product from two reactants. (c) Exothermic; magnesium is oxidised because it gains oxygen.

Decomposition reaction

NCERT section 1.2.2 Decomposition Reaction

In a decomposition reaction, a single reactant breaks down into two or more simpler products: AB → A + B. It is the opposite of combination, which builds one product from many.

Breaking bonds needs energy, so decomposition needs energy in one of three forms:

Heat (thermal decomposition): $2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$ . The green crystals first lose water and change colour, and you smell burning sulphur. Others: $C a C O_{3} heat C a O + C O_{2}$ , and $2 P b (N O_{3})_{2} heat 2 P b O + 4 N O_{2} + O_{2}$ , which gives brown fumes of nitrogen dioxide.
Electricity: water with a few drops of dilute sulphuric acid breaks into hydrogen and oxygen. The volume of one gas collected is double the other, because each water molecule has two hydrogen atoms for one oxygen atom.
Light: white silver chloride turns grey in sunlight as it splits into silver and chlorine. Silver bromide behaves the same way. These reactions are used in black and white photography.

Worked example

(a) About 2 g of ferrous sulphate crystals are heated in a dry boiling tube. State two observations, write the balanced equation and name the type of reaction. Is it exothermic or endothermic? (b) Write the balanced equation for the decomposition of water by electricity. In this experiment, which gas is collected in double the volume, and why? (5 marks)

Try it first, then show the solutionSolution

(a) Observations: the green colour of the crystals changes, and a smell of burning sulphur is noticed. (1 mark)
$2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$ . Check: Fe 2 = 2, S 2 = 1 + 1, O 8 = 3 + 2 + 3. (1 mark)
Thermal decomposition. It is endothermic because heat must be supplied to break the ferrous sulphate down. (1 mark)
(b) $2 H_{2} O (l) electricity 2 H_{2} (g) + O_{2} (g)$ (1 mark)
Hydrogen is collected in double the volume. Each water molecule has two hydrogen atoms for every oxygen atom, so two molecules of hydrogen form for every one molecule of oxygen. (1 mark)

Answer: (a) Green colour changes and smell of burning sulphur; $2 F e S O_{4} heat F e_{2} O_{3} + S O_{2} + S O_{3}$ ; thermal decomposition; endothermic. (b) $2 H_{2} O electricity 2 H_{2} + O_{2}$ ; hydrogen is double because water has H and O atoms in the ratio 2 : 1.

Exothermic and endothermic reactions

NCERT section 1.2.1 Combination Reaction

Reactions also differ in what they do with energy.

Exothermic reactions give out heat along with the products, so the surroundings get warmer. Examples:

quick lime with water
burning of natural gas: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$
respiration: $C_{6} H_{12} O_{6} (a q) + 6 O_{2} (a q) \to 6 C O_{2} (a q) + 6 H_{2} O (l) + energy$ . Carbohydrates in rice, potatoes and bread are broken down to glucose. Glucose combines with oxygen in our cells and releases the energy we need to stay alive.
decomposition of vegetable matter into compost

Endothermic reactions absorb energy. Most decomposition reactions are endothermic because heat, light or electricity must be supplied to break the reactant apart.

The energy label is separate from the type label. The quick lime reaction is both combination and exothermic. Heating ferrous sulphate is both decomposition and endothermic. Compost formation is a decomposition that is exothermic. So 'decomposition' only usually means endothermic, not always.

Worked example

Assertion (A): The burning of natural gas is a combination reaction.
Reason (R): Heat is released when natural gas burns.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true. (1 mark)

Try it first, then show the solutionSolution

Burning of natural gas: $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ . There are two products, so it is not a combination reaction. A is false. (½ mark)
Heat is given out, so the reaction is exothermic. R is true. Option (d). (½ mark)

Answer: (d) A is false but R is true.

Displacement reaction

NCERT section 1.2.3 Displacement Reaction

In a displacement reaction, a more reactive element pushes a less reactive element out of its compound: A + BC → AC + B.

In the NCERT activity, cleaned iron nails are kept in copper sulphate solution for about 20 minutes:
$F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$
The nails become brownish because copper is deposited on them. The blue colour of the solution fades because copper sulphate is used up. Zinc and lead displace copper in the same way:

$Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)$
$P b (s) + C u C l_{2} (a q) \to P b C l_{2} (a q) + C u (s)$

The reaction only goes one way. A less reactive metal cannot displace a more reactive one, so copper placed in iron sulphate solution shows no change.

In the practical, you put Zn, Fe, Cu and Al into solutions of each other's sulphates. You note which metals displace which, and from that you arrange them in decreasing order of reactivity.

Worked example

In a practical, strips of Al, Zn, Fe and Cu are placed in solutions of $A l_{2} (S O_{4})_{3}$ , $Z n S O_{4}$ , $F e S O_{4}$ and $C u S O_{4}$ . A student records these results: Al displaces Zn, Fe and Cu from their solutions. Zn displaces Fe and Cu. Fe displaces only Cu. Cu displaces none. (a) Arrange the four metals in decreasing order of reactivity. (b) Write the balanced equation for iron in copper sulphate solution, with state symbols. (c) State two changes seen in that test tube. (3 marks)

Try it first, then show the solutionSolution

(a) The metal that displaces the most others is the most reactive, so the order is Al > Zn > Fe > Cu. (1 mark)
(b) $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$ (1 mark)
(c) The iron becomes coated brownish with copper (½ mark), and the blue colour of the copper sulphate solution fades (½ mark).

Answer: (a) Al > Zn > Fe > Cu (b) $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$ (c) A brownish coating forms on the iron, and the blue colour of the solution fades.

Double displacement and precipitation reactions

NCERT section 1.2.4 Double Displacement Reaction

In a double displacement reaction, two compounds swap ions with each other: AB + CD → AD + CB. Nothing is pushed out by an element. The partners are simply exchanged.

In the NCERT activity, sodium sulphate solution is mixed with barium chloride solution:
$N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$
The $B a^{2 +}$ ions and $S O_{4}^{2 -}$ ions join to make barium sulphate. It does not dissolve in water, so it appears as a white solid. Sodium chloride stays dissolved.

An insoluble solid formed in a solution is called a precipitate. Any reaction that forms one is a precipitation reaction. So this single reaction has two names: double displacement (by how it happens) and precipitation (by what you see).

Not every double displacement gives a precipitate. Sodium hydroxide solution and hydrochloric acid exchange ions to give sodium chloride solution and water: $N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$ .

Worked example

Barium chloride solution reacts with aluminium sulphate solution to give aluminium chloride solution and a precipitate of barium sulphate. (a) Write the balanced equation with state symbols. (b) Name two types this reaction belongs to and justify each. (3 marks)

Try it first, then show the solutionSolution

Skeletal: $B a C l_{2} + A l_{2} (S O_{4})_{3} \to B a S O_{4} + A l C l_{3}$ . Start with $A l_{2} (S O_{4})_{3}$ : it has 3 sulphate groups, so write $3 B a S O_{4}$ , which needs $3 B a C l_{2}$ . (½ mark)
That gives 6 Cl on the left, so write $2 A l C l_{3}$ , which also balances Al at 2. (½ mark)
$3 B a C l_{2} (a q) + A l_{2} (S O_{4})_{3} (a q) \to 3 B a S O_{4} (s) + 2 A l C l_{3} (a q)$ . Check: Ba 3 = 3, Cl 6 = 6, Al 2 = 2, S 3 = 3, O 12 = 12. (1 mark)
(b) Double displacement: the reactants exchange ions, with $B a^{2 +}$ joining $S O_{4}^{2 -}$ and $A l^{3 +}$ joining $C l^{-}$ . (½ mark)
Precipitation: insoluble barium sulphate forms as a solid. (½ mark)

Answer: (a) $3 B a C l_{2} (a q) + A l_{2} (S O_{4})_{3} (a q) \to 3 B a S O_{4} (s) + 2 A l C l_{3} (a q)$ (b) Double displacement (exchange of ions) and precipitation (insoluble $B a S O_{4}$ forms).

Oxidation and reduction (redox reactions)

NCERT section 1.2.5 Oxidation and Reduction

For this chapter, oxidation and reduction are defined by oxygen and hydrogen:

Oxidation: gain of oxygen or loss of hydrogen.
Reduction: loss of oxygen or gain of hydrogen.

When copper powder is heated in a china dish, its surface becomes coated with black copper(II) oxide: $2 C u + O_{2} heat 2 C u O$ . Copper is oxidised. If hydrogen is passed over the heated black material, it turns brown again as copper forms: $C u O + H_{2} heat C u + H_{2} O$ . Here CuO loses oxygen and is reduced. Hydrogen gains oxygen and is oxidised.

One substance is oxidised while another is reduced in the same reaction, so these are called redox reactions. More examples:

$Z n O + C \to Z n + C O$ : carbon is oxidised, ZnO is reduced.
$M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$ : HCl loses hydrogen and is oxidised to $C l_{2}$ ; $M n O_{2}$ loses oxygen and is reduced.

Always name the reactant that is oxidised or reduced, not the product.

Worked example

(a) In $2 P b O (s) + C (s) \to 2 P b (s) + C O_{2} (g)$ , name the substance oxidised and the substance reduced. (b) In $M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$ , identify the substance oxidised and the substance reduced, giving a reason in terms of oxygen or hydrogen for each. (3 marks)

Try it first, then show the solutionSolution

(a) Carbon is oxidised because it gains oxygen to form $C O_{2}$ . Lead oxide (PbO) is reduced because it loses oxygen to form Pb. (1 mark)
(b) HCl is oxidised: it loses hydrogen and becomes $C l_{2}$ . (1 mark)
$M n O_{2}$ is reduced: it loses oxygen and becomes $M n C l_{2}$ . (1 mark)

Answer: (a) Oxidised: C; reduced: PbO. (b) Oxidised: HCl (loses hydrogen); reduced: $M n O_{2}$ (loses oxygen).

Science in real life: Chemical Reactions and Equations

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Combination reaction

Suppose a mason working at your home puts lumps of quick lime in a metal bucket and slowly pours water over them. The water hisses, and the bucket becomes too hot to hold. He warns you not to touch it.

(a) Name the product formed and write the balanced equation with state symbols. (b) Classify the reaction in two ways and justify each. (c) Explain the hot bucket.

Try it first, then show the solutionSolution

Quick lime is calcium oxide. With water it forms calcium hydroxide (slaked lime).
$C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q) + heat$
Combination reaction: two reactants form a single product.
Exothermic reaction: heat is released along with the product.
The large amount of heat given out passes into the bucket, so it becomes hot.

Answer: Calcium hydroxide; $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q) + heat$ ; combination and exothermic. The heat released makes the bucket hot.

Memory hook: Hot lime bucket: two things joined into one and gave out heat.

Balanced chemical equations

Imagine your kitchen stove runs on natural gas, which is mainly methane. While the dal cooks, you wonder how much oxygen the flame uses.

(a) Write the balanced equation for the burning of methane. (b) How many molecules of oxygen are needed to burn 4 molecules of methane completely, and how many molecules of water are formed? (c) Is this reaction exothermic or endothermic?

Try it first, then show the solutionSolution

Skeletal: $C H_{4} + O_{2} \to C O_{2} + H_{2} O$ . C is already balanced at 1.
H: 4 on the left, so write $2 H_{2} O$ on the right.
O on the right: 2 (in $C O_{2}$ ) + 2 (in $2 H_{2} O$ ) = 4, so write $2 O_{2}$ on the left.
Balanced: $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ .
1 methane needs 2 oxygen molecules, so 4 methane need 4 × 2 = 8 oxygen molecules.
1 methane gives 2 water molecules, so 4 methane give 8 water molecules.
Heat is given out to cook the food, so the reaction is exothermic.

Answer: (a) $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O$ (b) 8 molecules of oxygen; 8 molecules of water (c) Exothermic

Memory hook: One gas flame: 1 methane eats 2 oxygen and gives 2 water.

Exothermic and endothermic reactions

Suppose your school garden has a compost pit filled with vegetable peels and leaves. When you push your hand into the heap after a week, it feels warm inside. Your friend says decomposition reactions always absorb heat, so the warmth must come from the sun.

Is your friend right? Classify compost formation by energy change and compare it with the heating of calcium carbonate.

Try it first, then show the solutionSolution

The breaking down of vegetable matter into compost is a decomposition that gives out heat, so it is exothermic.
That is why the inside of the heap feels warm, even away from sunlight.
Heating calcium carbonate, $C a C O_{3} heat C a O + C O_{2}$ , needs heat to be supplied, so it is endothermic.
So most decomposition reactions are endothermic, but not all. Compost formation is an exception.
Your friend is wrong to say 'always'.

Answer: No. Compost formation is an exothermic decomposition, while heating calcium carbonate is an endothermic decomposition.

Memory hook: A warm compost pit is proof that 'decomposition' does not always mean 'absorbs heat'.

Decomposition reaction

Imagine you visit the school chemistry lab and see silver chloride stored in a dark-coloured bottle. A little of it left in an open dish on the window sill has turned grey.

(a) Why did the silver chloride turn grey? Write the balanced equation. (b) Name the type of reaction and the form of energy involved. (c) Why is it kept in a dark bottle?

Try it first, then show the solutionSolution

White silver chloride breaks down in sunlight into silver, which looks grey, and chlorine gas.
$2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$
It is a decomposition reaction: one reactant gives two products.
Light energy is absorbed to break it down, so it is endothermic.
A dark bottle keeps out light, so the silver chloride does not decompose.

Answer: It decomposes in sunlight: $2 A g C l sunlight 2 A g + C l_{2}$ ; decomposition caused by light; the dark bottle blocks light to prevent this.

Memory hook: Dark bottle, no light, no break-up.

Displacement reaction

Suppose, for a science exhibition, your friend plans to keep blue copper sulphate solution in an iron container overnight. Another friend suggests storing iron sulphate solution in a copper vessel instead.

Which plan is safe for the container? Explain with equations, and say what would be seen in the unsafe case.

Try it first, then show the solutionSolution

Iron is more reactive than copper, so iron displaces copper from copper sulphate.
$F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$
So the iron container would react: a brownish coating of copper would form on it, and the blue colour of the solution would fade.
Copper is less reactive than iron, so copper cannot displace iron from iron sulphate.
No reaction takes place, so the copper vessel is safe for iron sulphate solution.

Answer: Storing iron sulphate in a copper vessel is safe. Copper sulphate in an iron container reacts: $F e + C u S O_{4} \to F e S O_{4} + C u$ , with a brownish coating on the iron and the blue colour fading.

Memory hook: The stronger metal steals the partner. Iron steals from copper, never the other way.

Oxidation and reduction (redox reactions)

Imagine your science teacher heats a pinch of brown copper powder in a china dish during a demonstration. The surface turns black. She then says that passing hydrogen over the hot black powder would bring the brown colour back.

Write both equations. For the first reaction, name the substance oxidised. For the second reaction, name the substance oxidised and the substance reduced. Give the colour of the product each time.

Try it first, then show the solutionSolution

Heating in air: $2 C u + O_{2} heat 2 C u O$ . The product is black copper(II) oxide.
Copper gains oxygen, so copper is oxidised.
With hydrogen: $C u O + H_{2} heat C u + H_{2} O$ . Brown copper forms again.
CuO loses oxygen, so it is reduced. Hydrogen gains oxygen, so it is oxidised.
Oxidation and reduction occur together in the second reaction, so it is a redox reaction.

Answer: $2 C u + O_{2} \to 2 C u O$ (black): Cu oxidised. $C u O + H_{2} \to C u + H_{2} O$ (brown copper): CuO reduced, $H_{2}$ oxidised.

Memory hook: Brown to black means oxygen added; black to brown means oxygen taken away.

Topics in this chapter

Chemical Reactions and Chemical EquationsComing soon
Balancing Chemical Equations
Combination and Decomposition Reactions
Displacement and Double Displacement ReactionsComing soon
Oxidation and Reduction (Redox Reactions)

Chemical Reactions and Equations: formulas and key facts

Law of conservation of mass: Mass is neither created nor destroyed in a chemical reaction, so the number of atoms of each element is the same on both sides.; Use this to answer 'Why must equations be balanced?'
Signs that a reaction has occurred: Change in state, change in colour, evolution of a gas, change in temperature.; Link the right sign to the activity named in an observation question.
State symbols: $(s)$ solid, $(l)$ liquid, $(g)$ gas, $(a q)$ dissolved in water; Add these whenever a question says 'with state symbols'. Steam is $H_{2} O (g)$ .
Burning of magnesium (combination, exothermic, oxidation): $2 M g + O_{2} \to 2 M g O$; Activity 1.1. The skeletal form $M g + O_{2} \to M g O$ is unbalanced.
Iron and steam (hit-and-trial model): $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$; NCERT's worked balancing example. Practise the method on it.
Quick lime and water (combination, exothermic): $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q) + heat$; Practical 2. The product is slaked lime.
Burning of natural gas and respiration (exothermic): $C H_{4} + 2 O_{2} \to C O_{2} + 2 H_{2} O; C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O + energy$; Standard examples of exothermic reactions. Neither is a combination reaction.
Heating ferrous sulphate (thermal decomposition): $2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$; Practical 2. The green crystals change colour and smell of burning sulphur.
Heating limestone and lead nitrate (thermal decomposition): $C a C O_{3} heat C a O + C O_{2}; 2 P b (N O_{3})_{2} heat 2 P b O + 4 N O_{2} + O_{2}$; Lead nitrate gives brown fumes of $N O_{2}$ .
Decomposition by electricity and by light: $2 H_{2} O electricity 2 H_{2} + O_{2}; 2 A g C l sunlight 2 A g + C l_{2}$; Hydrogen is double the volume of oxygen. White AgCl turns grey. AgBr behaves the same way.
Displacement: $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$; Also written for Zn and Pb. A more reactive metal displaces a less reactive one. Practical order: Al > Zn > Fe > Cu.
Double displacement and precipitation: $N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$; Practical 2. White insoluble $B a S O_{4}$ is the precipitate.
Copper oxidised and copper oxide reduced: $2 C u + O_{2} heat 2 C u O; C u O + H_{2} heat C u + H_{2} O$; Brown copper turns black, then back to brown when hydrogen is passed.
Oxidation and reduction (this chapter's definitions): Oxidation: gain of O or loss of H. Reduction: loss of O or gain of H.; Apply to the reactants. Redox means both happen together.
Endothermic and exothermic: Exothermic: heat is given out. Endothermic: energy is absorbed.; Most decomposition reactions are endothermic. Compost formation is an exothermic exception.

Chemical Reactions and Equations: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Electrolysis of water (Activity 1.7)

Draw a labelled diagram of the set-up used to decompose water by electricity, or questions that show the set-up in words and ask which gas is in which tube and why one volume is double.

Draw

A plastic mug with two holes drilled in its base, each fitted with a rubber stopper
A carbon electrode pushed up through each rubber stopper into the mug
Water in the mug, covering both electrodes, with a note that a few drops of dilute sulphuric acid are added
Two test tubes, first filled with water, turned upside down over the two electrodes
Wires from the two electrodes to a 6 volt battery, with a switch in the circuit
Bubbles rising from both electrodes; show one test tube with about twice as much gas as the other

Label

Plastic mug
Rubber stoppers
Carbon electrodes
Water with a few drops of dilute sulphuric acid
Test tubes (inverted)
Hydrogen gas (larger volume)
Oxygen gas (smaller volume)
6 V battery
Switch

Watch out: The common slip is drawing equal gas volumes in both tubes. The hydrogen tube must show about double the volume of the oxygen tube, because each water molecule has two hydrogen atoms for one oxygen atom ( $2 H_{2} O \to 2 H_{2} + O_{2}$ ). Also remember the test tubes are filled with water and inverted, not empty.

Heating ferrous sulphate crystals in a boiling tube (Activity 1.5)

Draw and label the set-up for the thermal decomposition of ferrous sulphate, or show the correct way to heat the tube and smell the gas.

Draw

A dry boiling tube containing a small amount (about 2 g) of ferrous sulphate crystals at the bottom
A test tube holder or pair of tongs gripping the tube near its mouth
The tube held at a slant over the flame of a burner or spirit lamp
The mouth of the tube pointing away from you and others
A hand wafting the gas gently towards the nose instead of smelling directly from the mouth of the tube

Label

Boiling tube (dry)
Ferrous sulphate crystals
Test tube holder / tongs
Burner or spirit lamp flame
Gases given off ( $S O_{2}$ and $S O_{3}$ )
Wafting the odour

Watch out: Do not draw the tube pointing at a person or the nose held over the mouth of the tube. The tube must be dry. In the answer, note that the green crystals change colour and that the smell is like burning sulphur.

Iron nails in copper sulphate solution (Activity 1.9)

Draw and label the set-up comparing iron nails dipped in copper sulphate solution with a nail kept aside, and state the changes.

Draw

Two test tubes marked A and B, each with about 10 mL of blue copper sulphate solution
In tube B, two cleaned iron nails tied with a thread and dipped into the solution
In tube A, solution only, kept for comparison
One cleaned iron nail kept outside on the side for comparison
After about 20 minutes: show the solution in B paler blue than in A, and the nails in B with a brownish coating

Label

Test tube A (copper sulphate solution, for comparison)
Test tube B
Thread
Iron nails (cleaned)
Copper sulphate solution (blue)
Brownish coating of copper on nails
Iron nail kept aside

Watch out: Do not forget the comparison tube A and the spare nail, which are what make the colour change visible. The brownish coating is copper deposited on iron, not rust.

Heating copper powder in a china dish, then passing hydrogen (Activity 1.11)

Draw and label the set-up for heating copper powder, or describe the colour changes when copper is heated and when hydrogen is later passed over the hot product.

Draw

A china dish containing a small amount (about 1 g) of brown copper powder
A flame below the china dish heating it
After heating: the surface of the powder shown with a black coating
A separate labelled arrow or note showing hydrogen gas being passed over the heated black material, which turns brown again

Label

China dish
Copper powder (brown)
Flame / heat
Black coating of copper(II) oxide
Hydrogen gas passed over hot CuO
Copper formed again (brown)

Watch out: Label the black substance as copper(II) oxide, $C u O$ , not soot. Write both equations beside the diagram: $2 C u + O_{2} heat 2 C u O$ and $C u O + H_{2} heat C u + H_{2} O$ . Do not forget the heat condition on both arrows.

Heating lead nitrate in a boiling tube (Activity 1.6)

Draw and label the set-up for the thermal decomposition of lead nitrate, or name the brown fumes seen.

Draw

A boiling tube containing about 2 g of lead nitrate powder at the bottom
A pair of tongs holding the boiling tube
The tube held over a flame
Brown fumes rising out of the mouth of the tube

Label

Boiling tube
Lead nitrate powder
Pair of tongs
Flame
Brown fumes of nitrogen dioxide ( $N O_{2}$ )

Watch out: Shade the fumes brown and label them nitrogen dioxide, $N O_{2}$ , not oxygen. Oxygen is also a product ( $2 P b (N O_{3})_{2} heat 2 P b O + 4 N O_{2} + O_{2}$ ), but the brown fumes are of nitrogen dioxide. Show the tube held with tongs, not by hand.

Mixing sodium sulphate and barium chloride solutions (Activity 1.10)

Draw and label the set-up for the double displacement (precipitation) reaction, or state what is seen on mixing.

Draw

One test tube with about 3 mL of sodium sulphate solution
A second test tube with about 3 mL of barium chloride solution
An arrow showing one solution being poured into the other
The mixed tube with a white solid (precipitate) formed in the liquid

Label

Sodium sulphate solution
Barium chloride solution
White precipitate of barium sulphate ( $B a S O_{4}$ )
Sodium chloride (stays in solution)

Watch out: Draw the precipitate as a white solid in the liquid, not as gas bubbles. Sodium chloride stays dissolved, so do not draw it as a solid. Write the equation beside it: $N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$ .

Quick lime and water in a beaker (Activity 1.4)

Draw and label the set-up showing the formation of slaked lime, or explain what you feel when you touch the beaker.

Draw

A beaker containing a small amount of calcium oxide (quick lime)
Water being added slowly to the beaker
A hand touching the outside of the beaker to feel the change in temperature

Label

Beaker
Calcium oxide (quick lime)
Water (added slowly)
Slaked lime, calcium hydroxide
Hand touching the beaker (feels hot)

Watch out: Write that water is added slowly. Label the product slaked lime, $C a (O H)_{2}$ , and note that the beaker becomes hot: $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q) + heat$ , a combination reaction that is exothermic.

How the board asks Chemical Reactions and Equations

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Balance and classify	A word statement or a skeletal equation is given. You balance it and name the type of reaction, sometimes with state symbols.	Write the correct formulae first. Balance by hit-and-trial, starting with the biggest compound, then check every element. Classify by counting reactants and products and asking: is it combine, break, push out or swap?
'What happens when...' with equations	For example, what happens when a salt is heated, a metal is dipped in a solution, or a substance is left in sunlight. Usually 1 mark for each part.	Give the visible observation (colour, gas, smell, heat), then the balanced equation with the condition over the arrow. Name the type if asked.
Activity and practical-based	Set on quick lime and water, heating ferrous sulphate, iron nails in copper sulphate, sodium sulphate with barium chloride, or the Zn, Fe, Cu and Al reactivity practical.	Answer in order: what was done, what was seen, what it shows (the type of reaction), the equation. Mention a precaution where asked, such as using a dry boiling tube or cleaning the nails with sandpaper.
Oxidised and reduced identification	An MCQ or 2-mark question: which substance is oxidised or reduced, or which statements about a redox equation are incorrect.	For each reactant, check whether it gained or lost oxygen or hydrogen. Name reactants only. A product such as $C O_{2}$ is never 'getting oxidised'.
Assertion-reason	Usually on exothermic and endothermic reactions, or on why a reaction belongs to a type.	Judge A and R separately as true or false first. Only then ask whether R actually explains A. A true fact that is off-topic gives option (b).
Case-based or source-based	A lab story, for example electrolysis of water or heating copper, followed by 3-4 sub-parts with an internal choice.	Pull each fact from the passage, then apply the chapter idea, such as gas volume ratio, type of reaction or energy change. Keep each sub-part short and complete.
Error-finding and dialogue evaluation	An incorrect equation to correct, or a conversation between students about a reaction type, where you say who is right and why.	Check formulae first (for example $A g C l$ , not $A g C l_{2}$ ), then the balance. For dialogues, test each claim against the definition and give one reason per student.

Tricks to remember Chemical Reactions and Equations

Short memory aids. Say each one aloud once, then use it on a question the same day.

Four reaction patterns
Combination: A + B → AB. Decomposition: AB → A + B. Displacement: A + BC → AC + B. Double displacement: AB + CD → AD + CB. Join, break, push out, swap.
Use it when: Classify any equation in seconds by matching its shape.
O for Oxidation
Oxidation starts with O and means Oxygen in (or hydrogen out). Reduction is the reverse: oxygen out or hydrogen in.
Use it when: Redox MCQs: check each reactant against this rule.
Heat, Light, Electricity
Decomposition always needs a supply of energy: H-L-E. Heat for ferrous sulphate, calcium carbonate and lead nitrate; light for silver chloride and silver bromide; electricity for water.
Use it when: 'Write one decomposition for each form of energy' questions.
Reactivity line for the practical
'A Zebra Fears Cows' gives Al > Zn > Fe > Cu. A metal displaces only those written after it.
Use it when: Predict which metal-salt pairs react in the reactivity practical.
Water splits 2 : 1
$H_{2} O$ has 2 H for 1 O, so electrolysis gives twice as much hydrogen as oxygen.
Use it when: Gas volume questions in electrolysis case studies.
Balancing order
Big compound first, lone elements last, recount at the end. Never touch the small numbers inside a formula.
Use it when: Every balancing question.

Chemical Reactions and Equations: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Physical or chemical change? Judge by the substance, not the heat · 1 mark · MCQHarder board level

Two changes are seen in a lab. In the first, brown copper powder heated in a china dish gets a black coating. In the second, water in a beaker is heated until it turns into steam. Which statement is correct?
(a) Both are chemical changes, because heat is supplied in both.
(b) Only the first is a chemical change, because a new substance, copper(II) oxide, forms, while steam is still water, $H_{2} O$ .
(c) Only the second is a chemical change, because there is a change in state.
(d) Both are physical changes, because no gas is given off.

Try it first, then show the solutionSolution

Supplying heat does not decide the type of change; the test is whether the identity of the substance changes and a new substance forms
First change: $2 C u + O_{2} heat 2 C u O$ ; black copper(II) oxide is a new substance, so it is a chemical change
Second change: liquid water and steam are both $H_{2} O$ (steam is written $H_{2} O (g)$ ); only the state changes, the substance stays the same, so it is a physical change
So (a) and (d) are wrong, and (c) is wrong because a change in state alone does not prove a new substance formed

Answer: (b) Only the first is a chemical change, because a new substance, copper(II) oxide, forms, while steam is still water, $H_{2} O$ .

Example 2: Decomposition and colour change: does R explain A? · 1 mark · Assertion-reasonBoard level

Assertion (A): Heating ferrous sulphate crystals is a decomposition reaction.
Reason (R): The green colour of ferrous sulphate crystals changes on heating.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

A: $2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$ ; a single reactant breaks into simpler products, so A is true
R: the green crystals lose water and change colour on heating, so R is true
A colour change only shows that a reaction has happened; it does not show that one substance breaks into several, which is what makes it decomposition. So R does not explain A

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 3: Balance, add states and classify three reactions · 3 marks · Short answerHarder board level

Write balanced chemical equations with state symbols and name the type of each reaction:
(a) Aluminium metal reacts with dilute sulphuric acid to give aluminium sulphate solution and hydrogen gas.
(b) Solid silver bromide is kept in sunlight and gives silver metal and bromine.
(c) Carbon monoxide gas combines with hydrogen gas at 340 atm to give liquid methanol, $C H_{3} O H$ . (3 marks)

Try it first, then show the solutionSolution

(a) Skeletal: $A l + H_{2} S O_{4} \to A l_{2} (S O_{4})_{3} + H_{2}$ . Start with $A l_{2} (S O_{4})_{3}$ , the formula with most atoms: put 3 before $H_{2} S O_{4}$ and 2 before $A l$ , then 3 before $H_{2}$
Check (a): Al 2 = 2, H 6 = 6, S 3 = 3, O 12 = 12. $2 A l (s) + 3 H_{2} S O_{4} (a q) \to A l_{2} (S O_{4})_{3} (a q) + 3 H_{2} (g)$ : displacement reaction (Al displaces hydrogen from the acid)
(b) $2 A g B r (s) sunlight 2 A g (s) + B r_{2} (l)$ ; Ag 2 = 2, Br 2 = 2: decomposition reaction caused by light
(c) $C O (g) + 2 H_{2} (g) 340 atm C H_{3} O H (l)$ ; C 1 = 1, O 1 = 1, H 4 = 4: combination reaction (two reactants, one product)

Answer: (a) $2 A l (s) + 3 H_{2} S O_{4} (a q) \to A l_{2} (S O_{4})_{3} (a q) + 3 H_{2} (g)$ , displacement. (b) $2 A g B r (s) sunlight 2 A g (s) + B r_{2} (l)$ , decomposition. (c) $C O (g) + 2 H_{2} (g) 340 atm C H_{3} O H (l)$ , combination.

Example 4: What happens when: five reactions with equations · 5 marks · Long answerBoard level

State what happens in each case and write the balanced chemical equation:
(i) Dilute sulphuric acid is added to zinc granules.
(ii) Copper powder is heated in a china dish.
(iii) Hydrogen gas is passed over the hot product of (ii).
(iv) Silver chloride is left in sunlight.
(v) Dilute hydrochloric acid is added to iron filings. (5 marks)

Try it first, then show the solutionSolution

(i) Bubbles of hydrogen gas form around the zinc and the temperature of the flask changes: $Z n (s) + H_{2} S O_{4} (a q) \to Z n S O_{4} (a q) + H_{2} (g)$
(ii) The brown copper powder gets a black coating of copper(II) oxide as copper gains oxygen: $2 C u + O_{2} heat 2 C u O$
(iii) The black coating turns brown again as copper oxide loses oxygen and copper forms: $C u O + H_{2} heat C u + H_{2} O$
(iv) The white silver chloride turns grey as it decomposes into silver and chlorine: $2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$
(v) Hydrogen gas and iron chloride are produced: $F e (s) + 2 H C l (a q) \to F e C l_{2} (a q) + H_{2} (g)$

Answer: (i) Hydrogen bubbles, displacement. (ii) Black CuO forms, oxidation/combination. (iii) Black turns brown, CuO reduced and $H_{2}$ oxidised (redox). (iv) White to grey, decomposition by light. (v) Hydrogen and iron chloride form, displacement. Equations as in the steps.

Example 5: Activity: heating lead nitrate · 3 marks · Short answerHarder board level

In Activity 1.6, about 2 g of lead nitrate powder is heated in a boiling tube held with a pair of tongs.
(a) What is observed? Name the substance responsible.
(b) Write the balanced chemical equation.
(c) Name the type of reaction and say whether it is exothermic or endothermic, with a reason.
(d) Why is the tube held with tongs? (3 marks)

Try it first, then show the solutionSolution

(a) Brown fumes are given off; these are nitrogen dioxide, $N O_{2}$
(b) $2 P b (N O_{3})_{2} (s) heat 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$ ; check Pb 2 = 2, N 4 = 4, O 12 = 12
(c) A single reactant breaks into simpler products using heat: thermal decomposition; heat is absorbed, so it is endothermic
(d) The tube becomes very hot; tongs keep the hand away from the heat and the fumes

Answer: Brown fumes of $N O_{2}$ appear. $2 P b (N O_{3})_{2} (s) heat 2 P b O (s) + 4 N O_{2} (g) + O_{2} (g)$ . Thermal decomposition, endothermic. Tongs protect the hand from the hot tube.

Example 6: Displacement, double displacement and redox compared · 5 marks · Long answerHarder board level

(a) Distinguish between a displacement reaction and a double displacement reaction.
(b) Classify each reaction below, giving a reason, and say which one is also a precipitation reaction and which one is also a redox reaction:
(i) $P b (s) + C u C l_{2} (a q) \to P b C l_{2} (a q) + C u (s)$
(ii) $N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$
(iii) $N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$
(iv) $Z n O + C \to Z n + C O$ (5 marks)

Try it first, then show the solutionSolution

(a) Displacement: a more reactive element pushes out another element from its compound, e.g. $F e + C u S O_{4} \to F e S O_{4} + C u$
(a) Double displacement: two compounds exchange ions to form two new compounds, e.g. (iii)
(b)(i) Displacement: lead is more reactive than copper and removes it from copper chloride
(b)(ii) Double displacement: $N a^{+}$ and $H^{+}$ swap partners; no precipitate forms, so it is not a precipitation reaction
(b)(iii) Double displacement and precipitation: $B a^{2 +}$ and $S O_{4}^{2 -}$ combine to give insoluble white $B a S O_{4}$
(b)(iv) Redox: carbon gains oxygen (oxidised) and zinc oxide loses oxygen (reduced)

Answer: (i) Displacement. (ii) Double displacement. (iii) Double displacement and precipitation. (iv) Redox: C oxidised, ZnO reduced. Displacement swaps one element; double displacement exchanges ions between two compounds.

Example 7: Find the errors in a student's work · 3 marks · Short answerHOTS / competency

A student wrote the following in a test. Find the error in each and write the correct version.
(i) $N a + H_{2} O \to N a O H + H_{2}$ (marked by the student as balanced)
(ii) To balance $H_{2} + O_{2} \to H_{2} O$ , the student wrote $H_{2} + O_{2} \to H_{2} O_{2}$ .
(iii) $C a C O_{3} \to C a O + C O_{2}$ , labelled as a combination reaction. (3 marks)

Try it first, then show the solutionSolution

(i) Count H: 2 on the left, 1 + 2 = 3 on the right, so it is not balanced. Correct: $2 N a + 2 H_{2} O \to 2 N a O H + H_{2}$ (Na 2 = 2, H 4 = 4, O 2 = 2)
(ii) The student changed the formula of water to $H_{2} O_{2}$ , a different substance. Only coefficients may change. Correct: $2 H_{2} + O_{2} \to 2 H_{2} O$
(iii) One reactant gives two products, so it is decomposition, not combination. The heat condition is also missing. Correct: $C a C O_{3} (s) heat C a O (s) + C O_{2} (g)$ , thermal decomposition

Answer: (i) Unbalanced H: $2 N a + 2 H_{2} O \to 2 N a O H + H_{2}$ . (ii) Formula changed: $2 H_{2} + O_{2} \to 2 H_{2} O$ . (iii) It is thermal decomposition: $C a C O_{3} heat C a O + C O_{2}$ .

Example 8: Evaluate a lab discussion on energy changes · 3 marks · Short answerHOTS / competency

Three students discuss energy changes.
Riya: Respiration is endothermic, because our body needs energy.
Kabir: In respiration, glucose is oxidised, because it combines with oxygen.
Meera: Heating calcium carbonate to get quick lime is endothermic, because heat must be supplied to break it down.
Who is correct? Justify each decision and write the balanced equation for respiration. (3 marks)

Try it first, then show the solutionSolution

Riya is wrong: in respiration glucose combines with oxygen in our cells and energy is released, so it is exothermic; we need energy, and respiration is how we get it
Kabir is correct: glucose combines with oxygen, which is oxidation; the equation is $C_{6} H_{12} O_{6} (a q) + 6 O_{2} (a q) \to 6 C O_{2} (a q) + 6 H_{2} O (l)$ + energy (C 6 = 6, H 12 = 12, O 18 = 18)
Meera is correct: $C a C O_{3} (s) heat C a O (s) + C O_{2} (g)$ is a thermal decomposition that takes place only when heat is supplied and absorbed, so it is endothermic

Answer: Kabir and Meera are correct. Respiration releases energy (exothermic) and oxidises glucose: $C_{6} H_{12} O_{6} + 6 O_{2} \to 6 C O_{2} + 6 H_{2} O$ + energy. Heating calcium carbonate absorbs heat (endothermic): $C a C O_{3} heat C a O + C O_{2}$ .

Example 9: Balance iron and steam by hit-and-trial, step by step · 3 marks · Short answerBoard level

Balance $F e + H_{2} O \to F e_{3} O_{4} + H_{2}$ by the hit-and-trial method. Show the atom count at each stage, then add state symbols, given that the water is used as steam. (3 marks)

Try it first, then show the solutionSolution

Box each formula and never change what is inside. Count atoms: Fe 1 (LHS) and 3 (RHS); H 2 and 2; O 1 and 4. (½ mark)
Start with the biggest compound, $F e_{3} O_{4}$ , and its element with most atoms, oxygen: 1 on the left, 4 on the right, so write $4 H_{2} O$ . Now O 4 = 4. (½ mark)
Hydrogen is now 8 on the left (4 × 2) and 2 on the right, so write $4 H_{2}$ . Now H 8 = 8. (½ mark)
Iron is 1 on the left and 3 on the right, so write $3 F e$ . Now Fe 3 = 3. (½ mark)
Final check: $3 F e + 4 H_{2} O \to F e_{3} O_{4} + 4 H_{2}$ ; Fe 3 = 3, H 8 = 8, O 4 = 4. (½ mark)
States: $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$ ; (g) with water shows steam. (½ mark)

Answer: $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$

Example 10: Dialogue and error correction: reaction types, redox and formulae · 5 marks · Long answerHOTS / competency

(I) Three students discuss reactions in the lab.
Ankit: Heating copper powder in air, $2 C u + O_{2} \to 2 C u O$ , is a decomposition reaction, because the colour changes.
Sana: In $Z n O + C \to Z n + C O$ , zinc oxide is reduced, because it loses oxygen.
Dev: The electrolysis of water is exothermic, because gas bubbles form at both electrodes.
Who is right? Justify your decision for each student. (3 marks)
(II) Identify the error in each equation and rewrite it correctly.
(i) $N a C l + A g N O_{3} \to A g C l_{2} + N a N O_{3}$
(ii) $F e + H C l \to F e C l_{2} + H_{2}$ (written by a student as balanced) (2 marks)

Try it first, then show the solutionSolution

(I) Ankit is wrong: two reactants give a single product, so it is a combination reaction (and copper is oxidised as it gains oxygen); a colour change only shows that a reaction happened. (1 mark)
(I) Sana is right: ZnO loses oxygen to become Zn, so it is reduced, while carbon gains oxygen and is oxidised. (1 mark)
(I) Dev is wrong: water breaks down only while electricity is supplied, so energy is absorbed and the reaction is endothermic; bubbles show evolution of gas, not release of heat. (1 mark)
(II)(i) Error: the formula of silver chloride is $A g C l$ , not $A g C l_{2}$ . Correct: $N a C l + A g N O_{3} \to A g C l + N a N O_{3}$ (Na 1, Cl 1, Ag 1, N 1, O 3 each side). (1 mark)
(II)(ii) Error: H is 1 on the left and 2 on the right, and Cl is 1 on the left and 2 on the right. Correct: $F e + 2 H C l \to F e C l_{2} + H_{2}$ (Fe 1, H 2, Cl 2 each side). (1 mark)

Answer: (I) Only Sana is right: Ankit's reaction is combination (and oxidation of copper), and electrolysis of water is endothermic. (II)(i) $N a C l + A g N O_{3} \to A g C l + N a N O_{3}$ (ii) $F e + 2 H C l \to F e C l_{2} + H_{2}$

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 6-mark question split into 1 + 1 + 2 + 2 marks.

Which metal pushes out which?

Suppose a student, Tanvi, sets up three test tubes in the school lab, each with about 10 mL of solution and a freshly cleaned metal piece.

Tube P: blue copper sulphate solution with an iron nail.
Tube Q: blue copper sulphate solution with a zinc strip.
Tube R: ferrous sulphate solution with a copper strip.

After 20 minutes she records her observations. In P, the nail has a brownish coating and the blue colour of the solution has faded. In Q, the zinc strip has a brownish coating and the blue colour has faded. In R, she sees no change at all, even after waiting longer. She wants to use these results to compare how reactive the three metals are. Answer A, B and either C or D.

(i) · 1 mark
(A) Why is there no change in tube R?
Answer
Answer: Copper is less reactive than iron, so it cannot displace iron from ferrous sulphate.
A metal displaces another only if it is more reactive. Tube P shows iron displaces copper, so copper is the less reactive one and no reaction happens in R.
(ii) · 1 mark
(B) Write the balanced chemical equation, with state symbols, for the reaction in tube Q.
Answer
Answer: $Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)$
Zinc displaces copper. The brownish coating is copper metal, and the blue fades as copper sulphate is used up. Atoms: Zn 1, Cu 1, S 1, O 4 on both sides.
(iii) · 2 marks
(C) From these results, what can Tanvi conclude about the reactivity of Fe and Cu, and of Zn and Cu? Can she decide whether zinc or iron is more reactive? If not, suggest one extra test.
Answer
Answer: Fe is more reactive than Cu, and Zn is more reactive than Cu. She cannot compare Zn and Fe from these results; she should put a zinc strip in ferrous sulphate solution.
P and R both show Fe above Cu; Q shows Zn above Cu. No tube compares Zn with Fe directly. If zinc gets coated in ferrous sulphate solution, zinc is more reactive than iron (1 mark for the conclusions, 1 mark for seeing the gap and the test).
(iv) · 2 marks
(D) OR (choice with C): Tanvi's friend says the brownish coating on the nail in tube P is rust. Is he right? Write the equation for tube P and explain both changes seen in it.
Answer
Answer: No. The coating is copper. $F e (s) + C u S O_{4} (a q) \to F e S O_{4} (a q) + C u (s)$ ; copper deposits on the nail, and the blue fades as copper sulphate is used up.
Iron is more reactive than copper, so it displaces copper from copper sulphate (displacement reaction) (1 mark for the equation with the reason). The displaced copper deposits on the nail as a brownish coating, and the blue colour fades because copper sulphate is used up while iron sulphate forms (1 mark).

Two tubes, two volumes: electrolysis of water

Suppose Arjun sets up Activity 1.7. He fits two carbon electrodes through rubber stoppers in the base of a plastic mug, fills it with water containing a few drops of dilute sulphuric acid, and connects the electrodes to a 6 volt battery. Two test tubes filled with water are inverted over the electrodes, labelled X and Y.

When he switches on the current, bubbles form at both electrodes and push water out of the tubes. After some minutes, tube X holds 16 mL of gas and tube Y holds 8 mL. Answer A, B and either C or D.

(i) · 1 mark
(A) Name the gas in tube X and the gas in tube Y. Give a reason.
Answer
Answer: X: hydrogen; Y: oxygen. Hydrogen forms in double the volume of oxygen.
Each water molecule has two hydrogen atoms for one oxygen atom, so two molecules of hydrogen form for each molecule of oxygen. The tube with twice the gas, X, holds hydrogen.
(ii) · 1 mark
(B) Write the balanced chemical equation with the condition, and name the type of reaction.
Answer
Answer: $2 H_{2} O (l) electricity 2 H_{2} (g) + O_{2} (g)$ ; decomposition reaction.
A single reactant, water, breaks into two simpler substances using electrical energy. Check: H 4 = 4, O 2 = 2 (½ mark equation, ½ mark type).
(iii) · 2 marks
(C) Later, tube X holds 28 mL of gas. What volume of gas is in tube Y at that time? Arjun's friend says the reaction is exothermic because bubbles are seen. Correct him with a reason.
Answer
Answer: 14 mL in tube Y. The reaction is endothermic, not exothermic.
Hydrogen and oxygen form in the ratio 2 : 1, so oxygen = 28 ÷ 2 = 14 mL (1 mark). Water decomposes only while electrical energy is supplied, so energy is absorbed and the reaction is endothermic; bubbles show evolution of a gas, not release of heat (1 mark).
(iv) · 2 marks
(D) OR (choice with C): Write the equation for the formation of water from hydrogen and oxygen. Using it and the reaction in this activity, explain why decomposition is called the opposite of combination.
Answer
Answer: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ . Combination joins two substances into one; electrolysis splits that one substance back into the same two.
Formation of water: two reactants give a single product, so it is combination (1 mark for the balanced equation and type). Electrolysis: $2 H_{2} O \to 2 H_{2} + O_{2}$ starts from that single product and gives back hydrogen and oxygen, so the two reactions are reverses of each other (1 mark).

Brown to black and back again

Suppose Nisha heats a china dish containing about 1 g of brown copper powder. After some time, the surface of the powder becomes coated with a black substance. Her teacher then passes hydrogen gas over the heated black material. The black coating turns brown again. Answer A, B and either C or D.

(i) · 1 mark
(A) Name the black substance and write the balanced equation for its formation.
Answer
Answer: Copper(II) oxide. $2 C u + O_{2} heat 2 C u O$
Oxygen from the air is added to copper on heating, forming black copper(II) oxide. Check: Cu 2 = 2, O 2 = 2.
(ii) · 1 mark
(B) Classify the reaction in (A) by the number of products and by gain or loss of oxygen.
Answer
Answer: Combination reaction; copper is oxidised.
Two reactants give a single product, so it is combination (½ mark). Copper gains oxygen, so copper is oxidised (½ mark).
(iii) · 2 marks
(C) Write the balanced equation for the reaction when hydrogen is passed. Name the substance oxidised and the substance reduced, with a reason for each.
Answer
Answer: $C u O + H_{2} heat C u + H_{2} O$ . $H_{2}$ is oxidised; CuO is reduced.
Balanced equation with heat over the arrow: Cu 1, O 1, H 2 on each side (1 mark). Hydrogen gains oxygen to form water, so it is oxidised; copper(II) oxide loses oxygen to form copper, so it is reduced (½ + ½ mark).
(iv) · 2 marks
(D) OR (choice with C): Nisha writes: 'In the reaction with hydrogen, only reduction takes place, because copper is obtained.' Evaluate her statement.
Answer
Answer: She is wrong. CuO is reduced, but hydrogen is oxidised at the same time, so it is a redox reaction.
In $C u O + H_{2} \to C u + H_{2} O$ , copper(II) oxide loses oxygen and is reduced (1 mark). The hydrogen gains that oxygen and is oxidised, so oxidation and reduction happen together, which makes it a redox reaction (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

White silver chloride is kept in a china dish in sunlight. Write the balanced chemical equation for the change, state what you would see, and give one use of such reactions. (2 marks) (2 marks)

Step	Marks
Balanced equation with state symbols and condition: $2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$	1
Observation: the white silver chloride turns grey as silver metal forms	0.5
Use: such light-driven decomposition reactions are used in black and white photography	0.5

Classify each reaction as combination, decomposition or displacement and give a reason for each: (i) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ (ii) $2 H_{2} O (l) electricity 2 H_{2} (g) + O_{2} (g)$ (iii) $Z n (s) + C u S O_{4} (a q) \to Z n S O_{4} (a q) + C u (s)$ (3 marks) (3 marks)

Step	Marks
(i) Combination reaction	0.5
Reason: two reactants join to give a single product, water	0.5
(ii) Decomposition reaction (by electricity)	0.5
Reason: a single reactant breaks down into two simpler substances, hydrogen and oxygen	0.5
(iii) Displacement reaction	0.5
Reason: zinc is more reactive than copper and pushes copper out of copper sulphate	0.5

(a) What is a balanced chemical equation? Why must chemical equations be balanced? (b) Balance the following and add state symbols: (i) methane burns in oxygen to give carbon dioxide and water vapour; (ii) iron reacts with steam to give $F e_{3} O_{4}$ and hydrogen gas. (c) In (ii), why is $(g)$ written with water? What does $(a q)$ mean? (5 marks) (5 marks)

Step	Marks
(a) Definition: an equation in which the number of atoms of each element is the same on the reactant side and the product side	1
(a) Reason: by the law of conservation of mass, mass is neither created nor destroyed in a reaction, so atoms cannot appear or disappear	1
(b)(i) $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ (C 1 = 1, H 4 = 4, O 4 = 4)	1
(b)(ii) $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$ (Fe 3 = 3, H 8 = 8, O 4 = 4)	1
(c) $(g)$ shows that water reacts as steam, not as liquid water	0.5
(c) $(a q)$ means the substance is present as a solution in water	0.5

Where students lose marks in Chemical Reactions and Equations

The slips examiners see most often, and what to write instead.

Mistake: Changing subscripts to balance an equation
It seems the quickest way to make the atom counts match.
Wrong working
$H_{2} + O_{2} \to H_{2} O_{2}$
Right working
$2 H_{2} + O_{2} \to 2 H_{2} O$
Do this: Only change the coefficients in front of formulae. A changed subscript is a different substance.
Usually costs: The full mark for the equation, usually 1 mark.
Mistake: Naming a product as the substance oxidised or reduced
Students see Pb or $C O_{2}$ on the right and label those.
Wrong working
In $2 P b O + C \to 2 P b + C O_{2}$ : lead is reduced and carbon dioxide is oxidised.
Right working
PbO is reduced because it loses oxygen. C is oxidised because it gains oxygen.
Do this: Look only at reactants. Ask what each one gains or loses as it becomes a product.
Usually costs: 1 mark in an MCQ, or 1 mark in a short answer.
Mistake: Leaving out the condition over the arrow
The condition seems like extra detail.
Wrong working
$2 A g C l \to 2 A g + C l_{2}$
Right working
$2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$
Do this: For decomposition, always write heat, sunlight or electricity over the arrow.
Usually costs: ½ mark in each part where the question tests the energy form.
Mistake: Getting the electrolysis gas ratio the wrong way round
Students guess that oxygen, being heavier, must be more.
Wrong working
Oxygen is collected in double the volume because it is heavier.
Right working
Hydrogen is collected in double the volume because water has two H atoms for each O atom: $2 H_{2} O \to 2 H_{2} + O_{2}$ .
Do this: Read the ratio straight from the balanced equation.
Usually costs: 1 mark, often in a case-based sub-part.
Mistake: Writing a displacement that cannot happen
Students copy the A + BC pattern without checking reactivity.
Wrong working
$C u + F e S O_{4} \to C u S O_{4} + F e$
Right working
No reaction. Copper is less reactive than iron, so it cannot displace iron.
Do this: Check the order Al > Zn > Fe > Cu before writing any displacement equation.
Usually costs: 1 to 2 marks in practical-based or error-finding questions.
Mistake: Misclassifying $F e_{2} O_{3} + 2 A l \to A l_{2} O_{3} + 2 F e$
Two compounds appear in the equation, so it looks like a swap.
Wrong working
Double displacement, because two compounds are present.
Right working
Displacement: the element Al pushes Fe out of its oxide. Only one compound is on the left. It is also redox.
Do this: Double displacement needs two compounds on the left exchanging partners. If one reactant is a free element, it is displacement.
Usually costs: 1 mark.

NCERT exercise guide: Chemical Reactions and Equations

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

In-text questions (first box): sections 1.1 to 1.1.2

What it tests: Why reactants must be clean before a reaction, turning word-equations into balanced chemical equations, and adding state symbols $(s)$ , $(l)$ , $(g)$ , $(a q)$ .

How to approach it: First write the correct formula of every substance, then balance by changing only the coefficients in front, never the small numbers inside a formula. Count every element on both sides at the end. For state symbols, read the sentence carefully: 'solution in water' means $(a q)$ , 'insoluble' or 'precipitate' means $(s)$ .

Mirrors the questions asking for balanced equations with state symbols

Write balanced chemical equations with state symbols: (i) Solid carbon (coal) burns in oxygen gas to form carbon dioxide gas. (ii) Hydrogen gas combines with oxygen gas to form liquid water.

Show the solutionSolution

(i) Formulae: carbon $C$ , oxygen $O_{2}$ , carbon dioxide $C O_{2}$
Atom count: C 1 = 1, O 2 = 2, so it is already balanced: $C (s) + O_{2} (g) \to C O_{2} (g)$
(ii) Skeletal equation: $H_{2} + O_{2} \to H_{2} O$ ; O is 2 on the left, 1 on the right
Put 2 before $H_{2} O$ : O becomes 2 = 2, but H is now 2 on the left and 4 on the right
Put 2 before $H_{2}$ : H 4 = 4, O 2 = 2
Add states: $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$

Answer: (i) $C (s) + O_{2} (g) \to C O_{2} (g)$ (ii) $2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l)$ . Both are combination reactions.

In-text questions (second box): section 1.2.2

What it tests: Identifying a substance from its use, writing its reaction with water, and reasoning about the volumes of gases collected in the electrolysis of water (Activity 1.7).

How to approach it: For 'identify X' questions, give the name, the formula and the balanced equation, because each earns a mark. For the electrolysis question, use the balanced equation $2 H_{2} O \to 2 H_{2} + O_{2}$ to explain the 2 : 1 ratio of volumes.

Mirrors the 'identify substance X' question

A white solid X is formed when limestone is heated strongly, and X is used in the manufacture of cement. (i) Name X and write its formula. (ii) Write the balanced equation for the reaction of X with water and name the type of reaction.

Show the solutionSolution

Limestone is calcium carbonate; on heating it gives calcium oxide and carbon dioxide
So X is calcium oxide (quick lime), formula $C a O$
Reaction with water: $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q)$ + heat
Two reactants form a single product, calcium hydroxide (slaked lime), so it is a combination reaction; heat is given out, so it is also exothermic

Answer: X is calcium oxide (quick lime), $C a O$ . $C a O (s) + H_{2} O (l) \to C a (O H)_{2} (a q)$ + heat: a combination reaction that is exothermic.

In-text questions (third box): sections 1.2.3 to 1.2.5

What it tests: Explaining a colour change in a displacement reaction, giving a new example of double displacement, and naming the substances oxidised and reduced.

How to approach it: For colour change, name the reaction and say which metal is more reactive. For oxidised and reduced, look at each reactant: gaining oxygen or losing hydrogen means oxidised; losing oxygen or gaining hydrogen means reduced. Always name the substance and give the reason.

Mirrors the question on identifying substances oxidised and reduced

Identify the substance oxidised and the substance reduced in $Z n O + C \to Z n + C O$ . Give reasons.

Show the solutionSolution

Carbon $C$ becomes $C O$ : carbon gains oxygen
So carbon is oxidised
Zinc oxide $Z n O$ becomes $Z n$ : zinc oxide loses oxygen
So zinc oxide is reduced; since one is oxidised and the other reduced, it is a redox reaction

Answer: Carbon is oxidised (gains oxygen to form CO); zinc oxide is reduced (loses oxygen to form Zn).

End-of-chapter exercises

What it tests: MCQs on redox and reaction types, balancing and translating statements into equations, defining exothermic and endothermic reactions, decomposition by heat, light and electricity, displacement versus double displacement, precipitation, and oxidation and reduction with examples. A few items on painting iron, nitrogen-flushed food, corrosion and rancidity belong to section 1.3; in the 2026-27 syllabus, corrosion is studied under Metals and Non-metals.

How to approach it: For MCQs with several statements, test each statement one by one before choosing. For 'balance and identify type', give the balanced equation first, then the type with a one-line reason. For 'difference' questions, write the definitions side by side and give one balanced equation for each.

Mirrors the MCQ on incorrect statements about a redox reaction

Which of the statements about $C u O + H_{2} \to C u + H_{2} O$ are incorrect?
(a) Copper oxide is getting reduced.
(b) Hydrogen is getting oxidised.
(c) Copper is getting oxidised.
(d) Water is getting reduced.
(i) (a) and (b)
(ii) (c) and (d)
(iii) (a), (b) and (c)
(iv) all

Show the solutionSolution

(a) $C u O$ loses oxygen to become $C u$ : reduced, so (a) is correct
(b) $H_{2}$ gains oxygen to become $H_{2} O$ : oxidised, so (b) is correct
(c) Copper is a product formed by reduction, it is not oxidised here, so (c) is incorrect
(d) Water is a product formed by oxidation of hydrogen, so (d) is incorrect

Answer: Option (ii): statements (c) and (d) are incorrect.

Activities 1.4 to 1.11 (with Practicals 2 and 3)

What it tests: Observing and classifying reactions in the lab: quick lime with water, heating ferrous sulphate and lead nitrate, electrolysis of water, silver salts in sunlight, iron nails in copper sulphate, sodium sulphate with barium chloride, and heating copper powder; plus arranging Al, Zn, Fe and Cu by reactivity.

How to approach it: For each activity, learn four things in order: what is done, what is seen (colour, gas, heat, precipitate), the balanced equation, and the type of reaction. Add one precaution where the activity has one, such as using a dry tube or holding it with tongs.

Mirrors Activity 1.10 (double displacement and precipitation)

About 3 mL of sodium sulphate solution is mixed with about 3 mL of barium chloride solution. (a) What is observed? (b) Write the balanced equation with state symbols. (c) Which ions combine to form the insoluble product, and why is this called a double displacement reaction?

Show the solutionSolution

(a) A white, insoluble substance (precipitate) forms
(b) $N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$ ; check Na 2 = 2, S 1 = 1, O 4 = 4, Ba 1 = 1, Cl 2 = 2
(c) $B a^{2 +}$ and $S O_{4}^{2 -}$ ions combine to form barium sulphate, the white precipitate; sodium chloride stays in solution
The reactants exchange ions, so it is double displacement; it gives a precipitate, so it is also a precipitation reaction

Answer: A white precipitate of barium sulphate forms. $N a_{2} S O_{4} (a q) + B a C l_{2} (a q) \to B a S O_{4} (s) + 2 N a C l (a q)$ . It is a double displacement (exchange of ions) and precipitation reaction.

Chemical Reactions and Equations: previous year questions

Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 5.8 marks per paper; 81 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Chemical Reactions and Equations

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026, set 31/1/1, Q17 · 1 mark · MCQ

Which of the following options clearly describes both the reactions?
(i) AgNO₃ + NaCl → NaNO₃ + AgCl
(ii) K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl

(A)(i) is double displacement, (ii) is displacement reaction.
(B)Both, (i) and (ii) are displacement reactions and precipitation reactions.
(C)Both, (i) and (ii) are double displacement reactions and precipitation reactions.
(D)(i) is displacement, (ii) is double displacement reaction.

Try it first, then show our solutionSolution

In both reactions two compounds exchange their ions:
(i) $AgNO_{3} + NaCl \to NaNO_{3} + AgCl$
(ii) $K_{2} SO_{4} + BaCl_{2} \to BaSO_{4} + 2 KCl$
So both are double displacement reactions.
Each reaction forms an insoluble product: white $AgCl$ in (i) and white $BaSO_{4}$ in (ii). So both are also precipitation reactions. Correct option: (C).
1 mark

Answer: (C) Both are double displacement and precipitation reactions.

Where the marks are

Chooses (C) (1 mark)

See it in the full 2026 board paper, set 31/1/1

2026, set 31/1/1, Q26 · 3 marks · Very short answer

Answer the following.

(a)
Name the substance oxidised and reduced in the following reaction: ZnO + C → Zn + CO
1 mark
(b)
Balance the following chemical reaction: Pb(NO₃)₂ + KI → PbI₂ + KNO₃
1 mark
(c)
Give one example each of electrolytic decomposition and decomposition by sunlight.
1 mark

Try it first, then show our solutionSolution

(a) In $ZnO + C \to Zn + CO$ , carbon gains oxygen, so C is oxidised.
½ mark
(a) ZnO loses oxygen, so ZnO is reduced.
½ mark
(b) Balanced equation: $Pb (N O_{3})_{2} + 2KI \to Pb I_{2} + 2KN O_{3}$
1 mark
(c) Electrolytic decomposition: $2 H_{2} O electricity 2 H_{2} + O_{2}$
½ mark
(c) Decomposition by sunlight: $2 AgCl sunlight 2 Ag + C l_{2}$
½ mark

Answer: (a) C oxidised, ZnO reduced; (b) Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃; (c) electrolysis of water; 2AgCl → 2Ag + Cl₂ in sunlight

Where the marks are

Substance oxidised: C (½ mark)
Substance reduced: ZnO (½ mark)
Correctly balanced equation (1 mark)
Example of electrolytic decomposition (½ mark)
Example of photolytic decomposition (½ mark)

See it in the full 2026 board paper, set 31/1/1

2025, set 31/1/1, Q21 · 2 marks · Very short answer

A student performs the following experiment in his school laboratory. List two observations to justify that in this experiment a chemical change has taken place.

Figure in the paper: A conical flask containing dilute sulphuric acid with zinc (Zn) granules at the bottom, fitted with a cork through which a delivery tube/rod passes. See it in the official paper.

Try it first, then show our solutionSolution

Reaction: $Zn (s) + H_{2} S O_{4} (aq) \to ZnS O_{4} (aq) + H_{2} (g)$
Observation 1: Bubbles of a gas (hydrogen) form on the zinc granules, which shows that a gas is evolved.
1 mark
Observation 2: The conical flask becomes warm, which shows a change (rise) in temperature.
1 mark

Answer: Evolution of gas (H₂ bubbles) and rise in temperature

Where the marks are

States that a gas is evolved (1 mark)
States that the temperature changes or rises (1 mark)

See it in the full 2025 board paper, set 31/1/1

Every Chemical Reactions and Equations question, by year

For 2027

Sample paper 2026-27: Q17 (1 mark, solved), Q29(A) (5 marks, solved), Q29(B) (5 marks, solved)

2026

Set 31/1/1: Q17 (1 mark, solved), Q19 (1 mark, solved), Q26 (3 marks, solved)

Set 31/2/1: Q17 (1 mark, solved), Q19 (1 mark, solved), Q27 (3 marks, solved)

Set 31/3/1: Q17 (1 mark, solved), Q19 (1 mark, solved), Q21 (1 mark, solved), Q26 (3 marks, solved)

Set 31/4/1: Q17 (1 mark, solved), Q18 (1 mark, solved), Q26 (3 marks, solved)

Set 31/5/1: Q17 (1 mark, solved), Q24 (1 mark, solved), Q26 (3 marks, solved)

Sample paper 2025-26: Q17 (1 mark, solved), Q19 (1 mark, solved), Q22 (1 mark, solved), Q23 (1 mark, solved), Q27 (3 marks, solved)

2025

Set 31/1/1: Q1 (1 mark, solved), Q17 (1 mark, solved), Q21 (2 marks, solved)

Set 31/2/1: Q1 (1 mark, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 31/3/1: Q1 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved)

Set 31/4/1: Q1 (1 mark, solved), Q21 (2 marks, solved), Q27(A) (3 marks, solved), Q27(B) (3 marks, solved)

Set 31/5/1: Q1 (1 mark, solved), Q5 (1 mark, solved), Q17 (1 mark, solved)

Set 31/6/1: Q1 (1 mark, solved), Q3 (1 mark, solved), Q5 (1 mark, solved), Q27 (3 marks, solved)

Sample paper 2024-25: Q1 (1 mark, solved), Q7 (1 mark, solved), Q21 (2 marks, solved)

2024

Set 31/1/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q7 (1 mark, solved), Q21 (2 marks, solved), Q27 (3 marks, solved)

Set 31/2/1: Q1 (1 mark, solved), Q3 (1 mark, solved), Q6 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q27 (3 marks, solved)

Set 31/3/1: Q1 (1 mark, solved), Q5 (1 mark, solved), Q7 (1 mark, solved), Q21 (2 marks, solved)

Set 31/4/1: Q1 (1 mark, solved), Q27 (3 marks, solved), Q34(a) (5 marks, solved), Q34(b) (5 marks, solved)

Set 31/5/1: Q1 (1 mark, solved), Q2 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q27 (3 marks, solved)

Sample paper 2023-24: Q1 (1 mark, solved), Q17 (1 mark, solved), Q21 (2 marks, solved)

2023

Set 31/1/1: Q2 (1 mark, solved), Q3 (1 mark, solved), Q21(b) (2 marks, solved), Q27 (3 marks, solved)

Set 31/2/1: Q1 (1 mark, solved), Q3 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q27 (3 marks, solved)

Set 31/4/1: Q2 (1 mark, solved), Q3 (1 mark, solved), Q17 (1 mark, solved), Q27 (3 marks, solved)

Set 31/5/1: Q2 (1 mark, solved), Q3 (1 mark, solved), Q17 (1 mark, solved), Q27 (3 marks, solved)

Set 31/6/1: Q3 (1 mark, solved), Q12 (1 mark, solved), Q21 (2 marks, solved), Q30 (3 marks, solved)

Practice questions on Chemical Reactions and Equations

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In the electrolysis of water, 12 mL of oxygen gas is collected in one test tube. What volume of hydrogen gas is collected in the other test tube in the same time?
(a) 6 mL
(b) 12 mL
(c) 24 mL
(d) 36 mL
Hint
Look at the coefficients of $H_{2}$ and $O_{2}$ in the balanced equation.
Answer
(c) 24 mL. From $2 H_{2} O \to 2 H_{2} + O_{2}$ , hydrogen and oxygen form in a 2 : 1 ratio, so hydrogen = 2 × 12 = 24 mL.
Q2 · 2 marks · Very short answerBoard level
Lead nitrate solution is mixed with potassium iodide solution and an insoluble lead iodide is formed. Write the balanced chemical equation with state symbols and name two types this reaction belongs to.
Hint
Which product is insoluble? That one gets $(s)$ .
Answer
$P b (N O_{3})_{2} (a q) + 2 K I (a q) \to P b I_{2} (s) + 2 K N O_{3} (a q)$ (Pb 1, N 2, O 6, K 2, I 2 on each side). Ions are exchanged, so it is double displacement; an insoluble product forms, so it is also a precipitation reaction.
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): In the electrolysis of water, the volume of hydrogen collected is double the volume of oxygen.
Reason (R): A molecule of water contains two oxygen atoms for every hydrogen atom.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Check R against the formula $H_{2} O$ .
Answer
(c) A is true but R is false. $2 H_{2} O \to 2 H_{2} + O_{2}$ gives hydrogen and oxygen in the ratio 2 : 1, so A is true. But $H_{2} O$ has two hydrogen atoms for one oxygen atom, the reverse of what R says, so R is false.
Q4 · 3 marks · Short answerHarder board level
Define exothermic and endothermic reactions. Give one balanced equation for each, and classify the decomposition of vegetable matter into compost.
Hint
Burning fuels and thermal decomposition are your easiest examples.
Answer
Exothermic: heat is released as products form, e.g. $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (g)$ . Endothermic: energy is absorbed, e.g. $C a C O_{3} (s) heat C a O (s) + C O_{2} (g)$ . Decomposition of vegetable matter into compost is exothermic.
Q5 · 3 marks · Short answerHarder board level
Identify the substance oxidised and the substance reduced in each reaction, with a reason in terms of oxygen or hydrogen:
(i) $3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$
(ii) $C u O (s) + H_{2} (g) \to C u (s) + H_{2} O (l)$
(iii) $2 H_{2} S + 3 O_{2} \to 2 H_{2} O + 2 S O_{2}$
Hint
In (iii), follow hydrogen: who loses it and who gains it?
Answer
(i) Fe gains oxygen: oxidised; $H_{2} O$ loses oxygen: reduced. (ii) $H_{2}$ gains oxygen: oxidised; $C u O$ loses oxygen: reduced. (iii) $H_{2} S$ loses hydrogen (and S gains oxygen): oxidised; $O_{2}$ gains hydrogen to form water: reduced.
Q6 · 5 marks · Long answerHarder board level
Decomposition reactions need energy. Write one balanced equation each for decomposition by heat, by light and by electricity. Using the formation and electrolysis of water, explain why decomposition is called the opposite of combination. Are decomposition reactions exothermic or endothermic?
Hint
Pick one activity from the chapter for each form of energy.
Answer
Heat: $2 F e S O_{4} (s) heat F e_{2} O_{3} (s) + S O_{2} (g) + S O_{3} (g)$ . Light: $2 A g C l (s) sunlight 2 A g (s) + C l_{2} (g)$ . Electricity: $2 H_{2} O (l) electricity 2 H_{2} (g) + O_{2} (g)$ .
Combination: $2 H_{2} + O_{2} \to 2 H_{2} O$ joins two substances into one; decomposition splits one substance into two, the reverse.
They absorb energy, so they are endothermic.
Q7 · 3 marks · Short answerHarder board level
In Practical 3, a student records: Al gives a deposit in $Z n S O_{4}$ solution; Zn gives a deposit in $F e S O_{4}$ solution; Fe gives a deposit in $C u S O_{4}$ solution; Cu gives no change in $F e S O_{4}$ solution, and Fe gives no change in $Z n S O_{4}$ solution. (a) Arrange Al, Zn, Fe and Cu in decreasing order of reactivity. (b) Write the balanced equation for zinc in ferrous sulphate solution and name the type of reaction.
Hint
Each 'deposit' means the added metal is above the metal in the solution.
Answer
(a) Al > Zn > Fe > Cu (Al beats Zn, Zn beats Fe, Fe beats Cu). (b) $Z n (s) + F e S O_{4} (a q) \to Z n S O_{4} (a q) + F e (s)$ : displacement reaction, since zinc is more reactive and removes iron from its salt.
Q8 · 3 marks · Short answerHarder board level
Aluminium and zinc each react with dilute hydrochloric acid to give their chlorides ( $A l C l_{3}$ and $Z n C l_{2}$ ) and a gas. Write both balanced equations with state symbols, name the gas and the type of reaction.
Hint
For aluminium, balance Cl first: you need a common multiple of 2 and 3 for hydrogen.
Answer
$Z n (s) + 2 H C l (a q) \to Z n C l_{2} (a q) + H_{2} (g)$ . $2 A l (s) + 6 H C l (a q) \to 2 A l C l_{3} (a q) + 3 H_{2} (g)$ (Al 2, H 6, Cl 6 on each side). The gas is hydrogen. Both are displacement reactions: the metal displaces hydrogen from the acid.
Q9 · 3 marks · Short answerHOTS / competency
A student says: ' $N a O H (a q) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$ is not a double displacement reaction, because no precipitate is formed.' Evaluate this statement with a reason, and state whether the reaction is a precipitation reaction.
Hint
What defines double displacement: a precipitate, or an exchange of ions?
Answer
The student is wrong. Double displacement means an exchange of ions between reactants: here $N a^{+}$ pairs with $C l^{-}$ and $H^{+}$ ends up in water, so ions are exchanged. A precipitate is not required for double displacement. Since all products stay dissolved or are water, it is not a precipitation reaction.
Q10 · 3 marks · Short answerHOTS / competency
Suppose a jeweller wants to recover silver from a waste solution of silver nitrate using scrap copper wire. (a) Write the balanced chemical equation and name the type of reaction. (b) Which metal is more reactive, copper or silver? (c) Could a silver strip be used to recover copper from copper sulphate solution? Justify.
Hint
A metal can only displace a less reactive metal.
Answer
(a) $C u (s) + 2 A g N O_{3} (a q) \to C u (N O_{3})_{2} (a q) + 2 A g (s)$ (Cu 1, Ag 2, N 2, O 6 each side): displacement. (b) Copper is more reactive, since it displaces silver. (c) No: silver is less reactive than copper, so it cannot displace copper from copper sulphate.
Q11 · 5 marks · Long answerHarder board level
Balance each skeletal equation. Show the atom check and name the type of reaction in each case.
(i) $A l + C u C l_{2} \to A l C l_{3} + C u$
(ii) $H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + H_{2} O$
(iii) $N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + H_{2} O$
(iv) $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + H C l$
(v) $Z n + A g N O_{3} \to Z n (N O_{3})_{2} + A g$
Hint
Balance the group or element that appears in uneven numbers first (Cl in (i), nitrate in (ii) and (v)), and leave water or the lone metal till last.
Answer
(i) $2 A l + 3 C u C l_{2} \to 2 A l C l_{3} + 3 C u$ (Al 2, Cu 3, Cl 6 each side): displacement, as Al pushes Cu out.
(ii) $2 H N O_{3} + C a (O H)_{2} \to C a (N O_{3})_{2} + 2 H_{2} O$ (H 4, N 2, O 8, Ca 1 each side): double displacement.
(iii) $2 N a O H + H_{2} S O_{4} \to N a_{2} S O_{4} + 2 H_{2} O$ (Na 2, O 6, H 4, S 1 each side): double displacement.
(iv) $B a C l_{2} + H_{2} S O_{4} \to B a S O_{4} + 2 H C l$ (Ba 1, Cl 2, H 2, S 1, O 4 each side): double displacement, and precipitation because insoluble $B a S O_{4}$ forms.
(v) $Z n + 2 A g N O_{3} \to Z n (N O_{3})_{2} + 2 A g$ (Zn 1, Ag 2, N 2, O 6 each side): displacement, as Zn pushes Ag out.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Name the four observations that show a chemical reaction has taken place.Show answer
Change in state, change in colour, evolution of a gas, change in temperature.
2.What do the state symbols (aq) and (g) mean?Show answer
(aq): dissolved in water. (g): gas, for example steam for water.
3.Write the balanced equation for iron reacting with steam.Show answer
$3 F e (s) + 4 H_{2} O (g) \to F e_{3} O_{4} (s) + 4 H_{2} (g)$
4.What gas forms as brown fumes when lead nitrate is heated?Show answer
Nitrogen dioxide, $N O_{2}$ .
5.What is a precipitate?Show answer
An insoluble solid formed in a solution during a reaction, such as white $B a S O_{4}$ .
6.Why does the blue colour of copper sulphate fade when iron nails are dipped in it?Show answer
Iron displaces copper, so copper sulphate is used up and iron sulphate forms.
7.Define reduction in terms of oxygen and hydrogen.Show answer
Loss of oxygen or gain of hydrogen.
8.Why is respiration called exothermic?Show answer
Glucose combines with oxygen in our cells and energy is released.

Best order to study it

Start with the signs of a chemical reaction and how to write word and chemical equations with state symbols and conditions.
Practise balancing by hit-and-trial until you can do any NCERT equation without help.
Learn combination and decomposition together as opposites. Attach the exothermic or endothermic label and the energy source (heat, light, electricity) to each example.
Move to displacement and double displacement. Link them to the practicals: iron nails in copper sulphate, sodium sulphate with barium chloride, and the Al, Zn, Fe, Cu reactivity order.
Learn oxidation and reduction using the copper heating activity, then apply the rule to ZnO + C and $M n O_{2}$ + HCl.
Finish with mixed practice: classify, correct wrong equations, and answer assertion-reason and case-based questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state why a chemical equation must be balanced, using the law of conservation of mass?
Can you balance $F e + H_{2} O \to F e_{3} O_{4} + H_{2}$ and $H_{2} S + O_{2} \to H_{2} O + S O_{2}$ from memory?
Can you write the quick lime and water equation and name its two types?
Can you write one decomposition each by heat, light and electricity, with the condition over the arrow?
Can you describe what you see when ferrous sulphate crystals are heated?
Can you explain why hydrogen is double the volume of oxygen in the electrolysis of water?
Can you arrange Al, Zn, Fe and Cu in decreasing order of reactivity and say which pairs react?
Can you write the sodium sulphate and barium chloride equation and name the precipitate?
Can you name what is oxidised and what is reduced in $C u O + H_{2} \to C u + H_{2} O$ and $2 P b O + C \to 2 P b + C O_{2}$ ?

Chemical Reactions and Equations: frequently asked questions

Do I need to write state symbols in every equation?

Not always. But when a question says 'with state symbols', or the reaction involves a precipitate, steam or a solution, add them. Writing them in practical-based answers shows full understanding and costs nothing.

Is the decomposition of vegetable matter into compost endothermic like other decomposition reactions?

No. Most decomposition reactions are endothermic because energy must be supplied, but the NCERT text gives compost formation as an exothermic reaction.

Can one reaction belong to more than one type?

Yes. Burning magnesium is combination, exothermic and oxidation at once. Sodium sulphate with barium chloride is both double displacement and precipitation. Give every name the question asks for.

Are corrosion and rancidity part of this chapter's syllabus for 2026-27?

They appear in NCERT section 1.3, but they are not named in this chapter's syllabus line. Corrosion and its prevention are studied under Metals and Non-metals instead.

What is the easiest way to tell displacement from double displacement?

Look at the reactants. If one of them is a free element, such as a metal, it is displacement. If both are compounds exchanging ions, it is double displacement.

Based on: CBSE Science (086) syllabus 2026-27; NCERT Science, Textbook for Class X (Reprint 2026-27)