Chapter 3 - Chemical Reactions and Equations
1. Choose the correct option from the bracket and explain the statement giving reason.
(Oxidation, displacement, electrolysis, reduction, zinc, copper, double displacement, decomposition)
a. To prevent rusting, a layer of ______ metal is applied on iron sheets.
Ans: zinc
Reason: Zinc is more reactive than iron. Hence, when a layer of zinc is applied on iron sheets, it will undergo corrosion instead of iron. Thus, zinc prevents rusting (corrosion) of iron.
b. The conversion of ferrous sulphate to ferric sulphate is ______ reaction.
Ans: oxidation
Reason: The net ionic reaction for the conversion of ferrous sulphate to ferric sulphate is: Fe²⁺ → Fe³⁺
Here, the positive charge on Fe increases from +2 to +3. Therefore, it is an oxidation reaction.
c. When electric current is passed through acidulated water ______ of water takes place.
Ans: electrolysis
Reason: Water decomposes into hydrogen gas and oxygen gas when electric current is passed through acidulated water. Since decomposition of water takes place by means of electrical energy, it is called electrolysis.
d. Addition of an aqueous solution of ZnSO₄ to an aqueous solution of BaCl₂ is an example of ______ reaction.
Ans: double displacement
Reason: The reactions in which the ions in the reactants are exchanged to form a precipitate are called double displacement reactions. When an aqueous solution of ZnSO₄ is added to an aqueous solution of BaCl₂, ZnCl₂ and white precipitate of BaSO₄ are formed.
2. Write answers to the following.
a. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
Ans:
(i) The reaction in which oxidation and reduction takes place simultaneously is called a redox reaction.
(ii) Example:
CuO + H₂ → Cu + H₂O
In this reaction, CuO loses oxygen to form Cu. This means that reduction of CuO takes place. H₂ molecule takes up oxygen to form H₂O. This means that oxidation of H₂ takes place. Both the oxidation and reduction reactions always occur simultaneously.
b. How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?
Ans:
(i) The decomposition of hydrogen peroxide (H₂O₂) into water and oxygen takes place slowly at room temperature.
(ii) However, hydrogen peroxide decomposes rapidly in the presence of manganese dioxide (MnO₂) powder, which acts as a catalyst.
(iii) Thus, by adding manganese dioxide (MnO₂) powder, the rate of chemical reaction can be increased.
c. Explain the term reactant and product giving examples.
Ans:
(i) The substances taking part in chemical reactions are called reactants.
(ii) The substances formed as a result of a chemical reaction by formation of new bonds are called products.
(iii) Examples:
a) Formation of carbon dioxide gas by combustion of coal in air is a chemical reaction. In this reaction, coal (carbon) and oxygen (from air) are the reactants while carbon dioxide is the product.
b) Decomposition of calcium carbonate by heating to form calcium oxide and carbon dioxide is a chemical reaction. In this reaction, calcium carbonate is the reactant while calcium oxide and carbon dioxide are the products.
d. Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples.
Ans:
(i) The types of reaction with reference to oxygen and hydrogen are oxidation and reduction reactions.
(ii) The chemical reaction in which a reactant combines with oxygen or loses hydrogen to form the product is called oxidation reaction.
(iii) For example:
a) 2Mg + O₂ → 2MgO
Mg combines with oxygen to form MgO, Here, Mg undergoes oxidation.
b) MgH₂ → Mg + H₂ ↑
MgH₂ loses hydrogen to form Mg. Here, MgH₂ undergoes oxidation.
(iv) The chemical reaction in which a reactant gains hydrogen or loses oxygen to form the product is called a reduction reaction.
(vi) For example:
(a) C + 2H₂ → CH₄
C gains hydrogen to form CH₄. Here, C undergoes reduction.
(b) 2Ag₂O → 4Ag + O₂
Ag₂O loses oxygen to form Ag. Here, Ag₂O undergoes reduction.
e. Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.
Ans:
| Similarity | Difference | |
|---|---|---|
Adding NaOH to water | (i) Heat is given away during this process. So, it is an exothermic process. | (ii) No new substances are formed as the process involves only dissolution. The aqueous solution of NaOH is a strong base. |
Adding CaO to water | (i) Heat is given away during this reaction. So, it is an exothermic reaction. | (ii) New substance (calcium hydroxide) is formed which is less soluble and its aqueous solution is weak base. |
3. Explain the following terms with examples.
a. Endothermic reaction
Ans:
(i) The chemical reactions which are accompanied by absorption of heat are called endothermic reactions.
(ii) For example:
When calcium carbonate (CaCO₃) is strongly heated, it undergoes decomposition to form calcium oxide (CaO) and carbon dioxide (CO₂).
CaCO₃\(_{(s)}\) + Heat → CaO\(_{(s)}\) + CO₂↑
Calcium Calcium Carbon
carbonate oxide dioxide
b. Combination reaction
Ans:
(i) When two or more reactants combine in a reaction to form a single product, it is called a combination reaction.
(ii) For example:
When a magnesium (Mg) strip is burnt in air, a white powder of magnesium oxide (MgO) is formed.
2Mg\(_{(s)}\) + O₂\(_{(g)}\) \(\xrightarrow\triangle\) 2MgO\(_{(s)}\)
Magnesium Oxygen Magnesium
oxide
c. Balanced equation
Ans:
(i) A chemical equation in which the number of atoms of the elements in the reactants is the same as the number of atoms of those elements in the products is called a balanced chemical equation.
(ii) For example:
When a magnesium (Mg) strip is burnt in air, a white powder of magnesium oxide (MgO) is formed.
It can be written as,
2Mg\(_{(s)}\) + O₂\(_{(g)}\) \(\xrightarrow\triangle\) 2MgO\(_{(s)}\)
Magnesium Oxygen Magnesium
oxide
This is a balanced equation as 2 atoms of magnesium and oxygen are present on either side of the equation.
d. Displacement reaction
Ans:
(i) The reaction in which the place of the ion of a less reactive element in a compound is taken by another more reactive element by formation of its own ions is called a displacement reaction.
(ii) For example:
When zinc (Zn) dust is added to blue coloured copper sulphate (CuSO₄) solution, a colourless solution of zinc sulphate (ZnSO₄) is formed along with copper (Cu) metal.
CuSO₄\(_{(aq)}\) + Zn\(_{(s)}\) → ZnSO₄\(_{(aq)}\) + Cu\(_{(s)}\)
Copper Zinc Zinc Copper
sulphate sulphate
4. Give scientific reasons.
a. When the gas formed on heating limestone is passed through freshly prepared lime water, the lime water turns milky.
Ans:
(i) When limestone is heated, calcium oxide and carbon dioxide gas are formed.
CaCO₃\(_{(s)}\) + Heat → CaO\(_{(s)}\) + CO₂↑
Calcium Calcium Carbon
carbonate oxide dioxide
(ii) When carbon dioxide gas is passed through freshly prepared lime water, the solution turns milky due to the formation of calcium carbonate, which is insoluble in water.
Ca(OH)₂\(_{(aq)}\) + CO₂\(_{(g)}\) → CaCO₃\(_{(s)}\) + H₂O\(_{(l)}\)
Calcium Carbon Calcium Water
hydroxide dioxide carbonate
b. It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.
Ans:
(i) The rate of a chemical reaction depends on the size of the reactant particles taking part in the reaction.
(ii) The smaller the size of the reactant particles, the higher the rate of reaction.
(iii) The size of reactant particles is larger in pieces of Shahabad tile as compared to powder of Shahabad tile.
(iv) When HCl is added to pieces of Shahabad tile, the CO₂ effervescence is formed slowly. However, when HCI is added to Shahabad powder, the CO₂ effervescence is formed at a faster rate.
(v) Hence, it takes time for pieces of Shahabad tile to disappear in HCI, but its powder disappears rapidly.
c. While preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is added slowly to water with constant stirring.
Ans:
(i) In the process of dilution of concentrated sulphuric acid with water, a very large amount of heat is liberated. This process is highly exothermic.
(ii) As a result, if water is added to concentrated sulphuric acid, then water gets evaporated instantaneously. The mixture may splash out, causing an accident.
(iii) In order to prevent this, the required amount of water is taken in a glass container, and a small quantity of concentrated sulphuric acid at a time is added with constant stirring. So, only a small amount of heat is liberated at a time.
(iv) Hence, while preparing dilute sulphuric acid from concentrated sulphuric acid in the laboratory, the concentrated sulphuric acid is slowly added to water with constant stirring.
d. It is recommended to use an air tight container for storing oil for a long time.
Ans:
(i) When edible oil is left aside for a long time, it undergoes air oxidation.
(ii) Due to this, the taste and smell of oil changes and it becomes rancid. If food is cooked in this oil, its taste also changes.
(iii) Thus, the oil will become unfit for consumption.
(iv) The process of oxidation reaction of oil can be slowed down by storing it in an airtight container. Hence, it is recommended to use an air tight container for storing oil for a long time.
5. Observe the following picture and write down the chemical reaction with explanation.
Ans:
(i) The given picture represents the electrochemical reaction taking place during the corrosion (rusting) of iron.
(ii) Different regions on the surface of iron become anode and cathode.
(iii) In the anode region, Fe is oxidized to Fe²⁺
Fe\(_{s}\) → Fe²⁺ \(_{aq}\) + 2e⁻
(iv) The electrons released in the anode region flow through the metal surface to the cathode region where they reduce oxygen.
(v) In the cathode region, O₂ is reduced to form water.
O₂\(_{g}\) + 4 H⁺\(_{aq}\) + 4e⁻ → 2H₂O\(_{l}\)
(vi) When Fe²⁺ ions migrate from the anode region, they react with water (or OH⁻ ions) and further get oxidized to Fe³⁺ ions.
(vii) Fe³⁺ ions form an insoluble hydrated oxide (Fe₂O₃ • H₂O) which is deposited as a reddish brown layer on the surface. It is called rust.
2 Fe³⁺\(_{(aq)}\) + 4 H₂O\(_{(l)}\) → Fe₂O₃ • H₂O\(_{(s)}\) + 6 H⁺\(_{(aq)}\)
Rust
6. Identify from the following reactions the reactants that undergo oxidation and reduction.
a. Fe\(_{(s)}\) + S\(_{(s)}\) ⟶ FeS\(_{(s)}\)
Ans:
The oxidation state of Fe is 0 and that of Fe in the product side FeS\(_{(s)}\) is 2. This means Fe\(_{(s)}\) undergoes oxidation.
The oxidation state of S\(_{(s)}\) is 0 and that in FeS\(_{(s)}\) is -2. This means S\(_{(s)}\) undergoes reduction.
b. 2Ag₂O\(_{(s)}\) ⟶ 4 Ag\(_{(s)}\) + O₂\(_{(g)}\)
Ans:
The oxidation state of Ag in Ag₂O\(_{(s)}\) is +1 and that of Ag in the product side Ag\(_{(s)}\) is 0. This means Ag undergoes reduction.
The oxidation state of O in Ag₂O\(_{(s)}\) is -2 and O₂ has oxidation state 0. This means O₂ undergoes oxidation.
c. 2Mg\(_{(s)}\) + O₂\(_{(g)}\) \(\xrightarrow\triangle\) 2MgO\(_{(s)}\)
Ans:
The oxidation state of Mg in Mg\(_{(s)}\) is 0 and that of Mg in the product side, MgO\(_{(s)}\), is +2. This means Mg undergoes oxidation.
The oxidation state of O₂\(_{(g)}\) is 0 and on the product side, its oxidation state is -2. This means O₂ undergoes reduction.
d. NiO\(_{(s)}\) + H₂\(_{(g)}\) ⟶ Ni\(_{(s)}\) + H₂O\(_{(l)}\)
Ans:
The oxidation state of Ni in NiO\(_{(s)}\) is +2 and that of Ni in the product side, Ni\(_{(s)}\), is 0. This means Ni undergoes reduction.
The oxidation state of H₂ is 0 and on the product side, H₂O\(_{(l)}\), its oxidation state is +1. This means H₂ undergoes reduction.
7. Balance the following equation stepwise.
a. H₂S₂O₇\(_{(l)}\)+ H₂O\(_{(l)}\) → H₂SO₄\(_{(l)}\)
Ans:
Step 1: Count the number of each atom in reactant side
H = 4
S = 2
O = 8
Step 2: Count the number of each atom in product side
H = 2
S = 1
O = 4
Step 3: Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value.
If we multiply product side by 2, the number of atoms in product and reactant side gets balanced.
Hence the balanced equation is,
H₂S₂O₇\(_{(l)}\)+ H₂O\(_{(l)}\)→ 2 H₂SO₄\(_{(l)}\)
b. SO₂\(_{(g)}\)+ H₂S\(_{(aq)}\) → S\(_{(s)}\) + H₂O\(_{(l)}\)
Ans:
Step 1: Count the number of each atom in reactant side
H = 2
S = 2
O = 2
Step 2: Count the number of each atom in product side
H = 2
S = 1
O = 2
Step 3: Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value.
If we multiply H₂S by 2 in the reactant side and S by 3 and H₂O by 2 in the product side, then the number of atoms in the product and reactant side gets balanced.
Hence the balanced equation is,
SO₂\(_{(g)}\)+ 2H₂S\(_{(aq)}\) → 3S\(_{(s)}\) + 2H₂O\(_{(l)}\)
c. Ag\(_{(s)}\) + HCl\(_{(aq)}\) → AgCl\(_{(s)}\) + H₂\(_{(g)}\)
Ans:
Step 1: Count the number of each atom in reactant side
H = 1
Ag = 1
Cl = 1
Step 2: Count the number of each atom in product side
H = 2
Ag = 1
Cl = 1
Step 3: Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value
If we multiply Ag by 2 and HCl by 2 in the reactant side and AgCl by 2 in the product side, then the number of atoms in product and reactant side gets balanced.
Hence the balanced equation is,
2Ag\(_{(s)}\) + 2HCl\(_{(aq)}\) → 2AgCl\(_{(s)}\) ↓+ H2\(_{(g)}\) ↑
d. NaOH \(_{(aq)}\) + H₂SO₄ \(_{(aq)}\) ⟶ Na2SO₄ \(_{(aq)}\) + H₂O\(_{(l)}\)
Ans:
Step 1: Count the number of each atom in reactant side
Na = 1
H = 3
O = 5
S = 1
Step 2: Count the number of each atom in product side
Na = 2
H = 2
O = 5
S = 1
Step 3: Then balance the number of each atom in an equation by multiplying reactant and product side with numeral value.
If we multiply NaOH by 2 in the reactant side and H₂O by 2 in the product side, then the number of atoms in product and reactant side gets balanced.
Hence the balanced equation is,
2 NaOH\(_{(aq)}\) + H₂SO4\(_{(aq)}\) → Na₂SO4\(_{(aq)}\) +2 H₂O\(_{(l)}\)
8. Identify the endothermic and exothermic reaction.
a. HCl\(_{(l)}\) + NaOH\(_{(s)}\) ⟶ NaCl\(_{(s)}\) + H₂O\(_{(l)}\) + heat
Ans: Heat is released in the product side, as it mentioned in the above reaction. So, It is an exothermic reaction because heat is evolved in an exothermic reaction.
b. 2KClO₃\(_{(s)}\) \(\xrightarrow\triangle\) 2KCl\(_{(s)}\) + 3O₂ \(_{(g)}\) ↑
Ans: Heat is given in the product side to break the compound into simpler substances, as it mentioned in the above reaction. So, it is an endothermic reaction because heat is supplied in an exothermic reaction.
c. CaO\(_{(s)}\) + H₂O\(_{(l)}\) ⟶ Ca(OH)₂\(_{(s)}\) + heat
Ans: Heat is released in the product side, as it mentioned in the above reaction. So, It is an exothermic reaction because heat is evolved in an exothermic reaction.
d. CaCO₃ \(_{(s)}\) \(\xrightarrow\triangle\) CaO \(_{(s)}\) + CO₂ \(_{(g)}\)↑
Ans: Heat is provided in the product side to break the compound into simpler substances, as it mentioned in the above reaction. So, It is an endothermic reaction because heat is supplied in an exothermic reaction.
9. Match the column in the following table.
| Reactants | Products | Type of chemical reaction |
|---|---|---|
BaCl₂ \(_{(aq)}\) + ZnSO₄ \(_{(aq)}\) | H₂CO₃ \(_{(aq)}\) | Displacement |
2AgCl \(_{(s)}\) | FeSO₄ \(_{(aq)}\) + Cu \(_{(s)}\) | Combination |
CuSO₄ \(_{(aq)}\) + Fe \(_{(s)}\) | BaSO₄ ↓ + ZnCl₂ \(_{(aq)}\) | Decomposition |
H₂O\(_{(l)}\)+ CO₂\(_{(g)}\) | 2Ag\(_{(s)}\) + Cl₂\(_{(g)}\) | Double displacement |
Ans:
| Reactants | Products | Type of chemical reaction |
|---|---|---|
BaCl₂ \(_{(aq)}\) + ZnSO₄ \(_{(aq)}\) | BaSO₄ ↓ + ZnCl₂ \(_{(aq)}\) | Double displacement |
2AgCl \(_{(s)}\) | 2Ag\(_{(s)}\) + Cl₂\(_{(g)}\) | Decomposition |
CuSO₄ \(_{(aq)}\) + Fe \(_{(s)}\) | FeSO₄ \(_{(aq)}\) + Cu \(_{(s)}\) | Displacement |
H₂O\(_{(l)}\)+ CO₂\(_{(g)}\) | H₂CO₃ \(_{(aq)}\) | Combination |