Chemical Reactions and Equations - Balancing chemical equations, combination, decomposition, displacement, and redox reactions

Welcome to Class 10 Science! Chemistry might sometimes feel like a language written in secret symbols, but once you learn the alphabet and grammar, it becomes one of the most exciting subjects.

Think of chemical reactions like baking a cake. You start with individual ingredients (reactants), mix them up, heat them, and end up with something completely new and delicious (products). You can't turn the baked cake back into raw eggs and flour!

In this guide, we will master the core concepts of Chapter 1 of your NCERT textbook: Balancing Chemical Equations and the Four Major Types of Chemical Reactions. Let’s dive in!

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1. Why Do We Need to Balance Chemical Equations?

Before learning *how* to balance, let's understand *why* we must do it.

The Law of Conservation of Mass

In Class 9, you learned Lavoisier’s Law of Conservation of Mass:

*"Mass can neither be created nor destroyed in a chemical reaction."*

This means that the total mass of the elements present in the products of a chemical reaction must equal the total mass of the elements present in the reactants.

In simple words: The total number of atoms of each element MUST remain the same before and after the reaction.

Skeletal vs. Balanced Equations

Skeletal Equation (Unbalanced): Just shows the formulas of reactants and products without balancing atoms.

$$\text{Mg} + \text{O}_2 \rightarrow \text{MgO}$$

*(Notice: There are 2 Oxygen atoms on the left, but only 1 on the right! Mass is not conserved here.)*

Balanced Equation: Shows equal numbers of atoms on both sides.

$$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

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2. Step-by-Step Guide to Balancing Chemical Equations

We use the Hit-and-Trial Method recommended by NCERT. Let's balance a classic NCERT example step-by-step:

$$\text{Fe} + \text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$$

Step 1: Count the atoms on both sides

Draw a simple table to keep track of the atoms.

Element	Reactants (LHS)	Products (RHS)	Balanced?
Fe (Iron)	1	3	No
H (Hydrogen)	2	2	Yes
O (Oxygen)	1	4	No

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Step 2: Pick the compound with the maximum number of atoms

Look at the formula with the most atoms: $\text{Fe}_3\text{O}_4$. It contains 4 Oxygen atoms.

To balance Oxygen:

LHS has 1 Oxygen atom (in $\text{H}_2\text{O}$).

Multiply $\text{H}_2\text{O}$ by 4.

$$\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + \text{H}_2$$

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Step 3: Balance Hydrogen atoms

Now, LHS has $4 \times 2 = 8$ Hydrogen atoms, while RHS has only 2 Hydrogen atoms (in $\text{H}_2$).

Multiply $\text{H}_2$ on RHS by 4.

$$\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$$

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Step 4: Balance Iron (Fe) atoms

LHS has 1 Iron atom, while RHS has 3 Iron atoms (in $\text{Fe}_3\text{O}_4$).

Multiply $\text{Fe}$ on LHS by 3.

$$3\text{Fe} + 4\text{H}_2\text{O} \rightarrow \text{Fe}_3\text{O}_4 + 4\text{H}_2$$

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Step 5: Final Verification & Adding Physical States

Let's re-count:

Fe: $3 \text{ (LHS)} = 3 \text{ (RHS)}$

H: $8 \text{ (LHS)} = 8 \text{ (RHS)}$

O: $4 \text{ (LHS)} = 4 \text{ (RHS)}$

It's balanced! Finally, add the symbols for physical states:

Solid = $(s)$, Liquid = $(l)$, Gas = $(g)$, Aqueous solution (dissolved in water) = $(aq)$

$$3\text{Fe}(s) + 4\text{H}_2\text{O}(g) \rightarrow \text{Fe}_3\text{O}_4(s) + 4\text{H}_2(g)$$

Pro-Tip: Never change the subscripts (small numbers like the '2' in $\text{H}_2\text{O}$) while balancing! You can only change the coefficients (big numbers in front of the molecules).

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3. Types of Chemical Reactions

Now that we know how to write reactions properly, let's explore the four main types of chemical reactions.

```

Types of Chemical Reactions

│

┌────────────────────────┼────────────────────────┐

▼ ▼ ▼

Combination Decomposition Displacement

Reaction Reaction Reaction

(A + B → AB) (AB → A + B) ┌────┴────┐

▼ ▼

Single Double

│

▼

Redox Reaction

(Oxidation + Reduction)

```

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A. Combination Reaction

Analogy: Two solo artists coming together to form a music duo.

In a combination reaction, two or more reactants combine to form a single product.

$$\text{A} + \text{B} \rightarrow \text{AB}$$

NCERT Classic Example: Whitewashing Walls

When quicklime (Calcium oxide) reacts vigorously with water, it produces slaked lime (Calcium hydroxide) releasing a large amount of heat.

$$\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(aq) + \text{Heat}$$

*(Quicklime)* $\qquad \qquad \qquad \qquad$ *(Slaked lime)*

Did You Know? The slaked lime solution is applied to walls. It reacts slowly with $\text{CO}_2$ in the air to form a thin, shiny layer of Calcium Carbonate ($\text{CaCO}_3$) after 2–3 days!

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B. Decomposition Reaction

Analogy: A single team breaking up into multiple independent players.

In a decomposition reaction, a single reactant breaks down to give two or more simpler products. These reactions require energy (heat, light, or electricity) to break chemical bonds.

$$\text{AB} \xrightarrow{\text{Energy}} \text{A} + \text{B}$$

Depending on the source of energy, there are three types:

1. Thermal Decomposition (Uses Heat)

Example: Heating green ferrous sulphate crystals ($\text{FeSO}_4 \cdot 7\text{H}_2\text{O}$).

$$2\text{FeSO}_4(s) \xrightarrow{\Delta \text{ (Heat)}} \text{Fe}_2\text{O}_3(s) + \text{SO}_2(g) + \text{SO}_3(g)$$

*(Green color changes to reddish-brown, and suffocating sulphur smell is released).*

2. Electrolytic Decomposition (Uses Electricity)

Example: Electrolysis of water.

$$2\text{H}_2\text{O}(l) \xrightarrow{\text{Electric Current}} 2\text{H}_2(g) + \text{O}_2(g)$$

*(Volume of Hydrogen gas collected is double the volume of Oxygen gas!).*

3. Photolytic Decomposition (Uses Light)

Example: Silver chloride turns grey in sunlight.

$$2\text{AgCl}(s) \xrightarrow{\text{Sunlight}} 2\text{Ag}(s) + \text{Cl}_2(g)$$

*(This reaction is used in black-and-white photography).*

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C. Displacement Reaction

1. Single Displacement Reaction

Analogy: A stronger wrestler enters the ring and replaces a weaker wrestler.

A reaction in which a more reactive element displaces a less reactive element from its compound.

$$\text{A} + \text{BC} \rightarrow \text{AC} + \text{B}$$

NCERT Classic Example: Iron nail in Copper Sulphate solution.

When an iron nail is dipped in a blue copper sulphate solution, the blue color fades to pale green, and a brown coating of copper settles on the nail.

$$\text{Fe}(s) + \text{CuSO}_4(aq) \rightarrow \text{FeSO}_4(aq) + \text{Cu}(s)$$

*(Blue)* $\qquad \qquad \qquad \quad$ *(Pale Green)* $\qquad$ *(Brown)*

*Why?* Iron ($\text{Fe}$) is more reactive than Copper ($\text{Cu}$), so it kicks Copper out!

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2. Double Displacement Reaction

Analogy: Two dance couples switching partners at the same time!

A reaction in which there is an exchange of ions between the reactants to form two new compounds. Often, an insoluble solid called a precipitate is formed.

$$\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$$

NCERT Classic Example: Mixing Sodium Sulphate and Barium Chloride.

$$\text{Na}_2\text{SO}_4(aq) + \text{BaCl}_2(aq) \rightarrow \text{BaSO}_4(s)\downarrow + 2\text{NaCl}(aq)$$

*(White Precipitate)*

*What happens?* $\text{Ba}^{2+}$ ions combine with $\text{SO}_4^{2-}$ ions to form a white precipitate of Barium Sulphate ($\text{BaSO}_4$).

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D. Redox Reactions (Oxidation & Reduction)

The word Redox comes from Reduction + Oxidation. These two processes always happen together!

Definitions:

Oxidation:

Gain of Oxygen OR

Loss of Hydrogen

Reduction:

Loss of Oxygen OR

Gain of Hydrogen

Agents:

Oxidizing Agent: The substance that *gives oxygen* or *removes hydrogen* (it gets reduced itself).

Reducing Agent: The substance that *takes oxygen* or *gives hydrogen* (it gets oxidized itself).

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Analyzing a Redox Reaction Step-by-Step:

$$\text{CuO} + \text{H}_2 \xrightarrow{\Delta} \text{Cu} + \text{H}_2\text{O}$$

Let's trace what happens to each reactant:

Copper Oxide ($\text{CuO}$): Loses Oxygen to become $\text{Cu}$.

Process: Reduction

Role: Oxidizing Agent

Hydrogen ($\text{H}_2$): Gains Oxygen to become $\text{H}_2\text{O}$.

Process: Oxidation

Role: Reducing Agent

```

Loses Oxygen (Reduction)

┌──────────────────────────────┐

▼ │

CuO(s) + H₂(g) ──► Cu(s) + H₂O(l)

│ ▲

└──────────────────┘

Gains Oxygen (Oxidation)

```

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4. Quick Recap Checklist

Reaction Type	Pattern	Key Identifying Feature
Combination	$A + B \rightarrow AB$	Single product formed
Decomposition	$AB \rightarrow A + B$	Single reactant breaks down
Single Displacement	$A + BC \rightarrow AC + B$	One element replaces another
Double Displacement	$AB + CD \rightarrow AD + CB$	Exchange of ions / Precipitate formed
Redox	Loss/Gain of O or H	Simultaneous oxidation & reduction

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5. Practice Time! (Questions with Detailed Solutions)

Now, let's test your understanding with 3 typical exam-style questions. Try solving them on your own before checking the solutions!

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Question 1: Equation Balancing

Balance the following chemical equation step-by-step and identify its physical states:

$$\text{HNO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Ca(NO}_3)_2 + \text{H}_2\text{O}$$

Solution:

Count atoms on both sides:

$\text{Ca}$: 1 (LHS), 1 (RHS) $\rightarrow$ Balanced.

$\text{N}$: 1 (LHS), 2 (RHS) $\rightarrow$ Unbalanced.

$\text{H}$: 3 (1 from $\text{HNO}_3$ + 2 from $\text{Ca(OH)}_2$) (LHS), 2 (RHS) $\rightarrow$ Unbalanced.

$\text{O}$: 5 (3 + 2) (LHS), 7 (6 + 1) (RHS) $\rightarrow$ Unbalanced.

Balance Nitrate ($\text{NO}_3$) / Nitrogen ($\text{N}$):

Multiply $\text{HNO}_3$ by 2.

$$2\text{HNO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Ca(NO}_3)_2 + \text{H}_2\text{O}$$

Re-count Hydrogen ($\text{H}$):

LHS now has $2 + 2 = 4$ Hydrogen atoms.

RHS has 2 Hydrogen atoms. Multiply $\text{H}_2\text{O}$ by 2.

$$2\text{HNO}_3 + \text{Ca(OH)}_2 \rightarrow \text{Ca(NO}_3)_2 + 2\text{H}_2\text{O}$$

Verify Oxygen ($\text{O}$):

LHS: $(2 \times 3) + 2 = 8$

RHS: $(2 \times 3) + 2 = 8$ $\rightarrow$ Balanced!

Final Balanced Equation with Physical States:

$$2\text{HNO}_3(aq) + \text{Ca(OH)}_2(aq) \rightarrow \text{Ca(NO}_3)_2(aq) + 2\text{H}_2\text{O}(l)$$

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Question 2: Reaction Identification

A shiny brown-coloured element 'X' on heating in air becomes black in colour.

Name the element 'X' and the black-coloured compound formed.

Write the balanced chemical equation for the reaction.

Classify the type of reaction.

Solution:

Identification:

Element 'X' is Copper ($\text{Cu}$) (which is shiny brown).

The black-coloured compound formed is Copper(II) oxide ($\text{CuO}$).

Balanced Chemical Equation:

$$2\text{Cu}(s) + \text{O}_2(g) \xrightarrow{\Delta} 2\text{CuO}(s)$$

Classification:

This is a Combination Reaction (two reactants combine to form a single product) as well as an Oxidation Reaction (Copper gains Oxygen).

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Question 3: Redox Reaction Analysis

Consider the following reaction:

$$\text{MnO}_2 + 4\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{H}_2\text{O} + \text{Cl}_2$$

Identify:

The substance oxidized.

The substance reduced.

The oxidizing agent.

The reducing agent.

Solution:

Let's analyze the changes:

$\text{MnO}_2$ loses oxygen to become $\text{MnCl}_2$.

Therefore, $\text{MnO}_2$ is reduced.

$\text{HCl}$ loses hydrogen to form $\text{Cl}_2$.

Therefore, $\text{HCl}$ is oxidized.

Answers:

Substance oxidized: $\text{HCl}$

Substance reduced: $\text{MnO}_2$

Oxidizing agent: $\text{MnO}_2$ *(because it causes $\text{HCl}$ to be oxidized)*

Reducing agent: $\text{HCl}$ *(because it causes $\text{MnO}_2$ to be reduced)*

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Keep Practicing!

Chemistry is best learned with a pen and paper. Keep practicing balancing equations from your NCERT textbook exercises, and try explaining these concepts to a classmate. You've got this!