Chemical Reactions and Equations
Q1. Choose the correct option from the bracket and explain the statement giving reason.
A. To prevent rusting, a layer of zinc metal is applied on iron sheets.
B. The conversion of ferrous sulphate to ferric sulphate is oxidation reaction.
C. When electric current is passed through acidulated water decomposition of water takes place.
(Explanation when electric current is passed through acidulated water, the process of passing electric current through acidulated water results in both electrolysis and decomposition of water, leading to the production of hydrogen gas and oxygen gas as separate products.)
D. Addition of an aqueous solution of ZnSO4 to an aqueous solution of BaCl2 form a white precipitate is an example of double displacement reaction.
Q2. Write answers to the following.
A. What is the reaction called when oxidation and reduction take place simultaneously? Explain with one example.
Ans:- The reaction in which oxidation and reduction occur simultaneously is called a redox (reduction-oxidation) reaction. In a redox reaction, there is a transfer of electrons between species involved, leading to changes in oxidation states.
One example of a redox reaction is the combustion of methane (CH4) in the presence of oxygen (O2) to produce carbon dioxide (CO2) and water (H2O). The reaction can be represented as follows:
CH4 + 2O2 -> CO2 + 2H2O
In this reaction, methane is oxidized to carbon dioxide, and oxygen is reduced to water. Methane loses electrons, resulting in an increase in oxidation state (oxidation), while oxygen gains electrons, resulting in a decrease in oxidation state (reduction).
B. How can the rate of the chemical reaction, namely, decomposition of hydrogen peroxide be increased?
Ans:- To increase the rate of the chemical reaction, specifically the decomposition of hydrogen peroxide, the use of catalysts is effective. In this case, manganese dioxide (MnO2) acts as a catalyst. The presence of MnO2 provides an alternative pathway for the reaction with lower activation energy, allowing the hydrogen peroxide to decompose at a faster rate.
C. Explain the term reactant and product giving examples.
Reactants are the substances present at the start of a chemical reaction, while products are the substances formed as a result of the reaction.
Example: in the reaction between hydrogen gas (H2) and oxygen gas (O2) to form water (H2O), the reactants are hydrogen gas (H2) and oxygen gas (O2) and H2O is the product.
H2 + O2 → H2O
D. Explain the types of reaction with reference to oxygen and hydrogen. Illustrate with examples.
Ans:-There are several types of reactions that involve oxygen and hydrogen
1.Combustion Reaction: In a combustion reaction, a substance reacts with oxygen to produce heat and light. Hydrogen and oxygen can undergo a combustion reaction to form water. For example: 2H2 + O2 → 2H2O
2.Oxidation-Reduction (Redox) Reaction: In a redox reaction, there is a transfer of electrons between the reactants. Hydrogen and oxygen can participate in redox reactions. An example is the reaction between hydrogen peroxide and potassium permanganate:
2KMnO4 + 3H2O2 → 2MnO2 + 2KOH + 2H2O + O2
3.Hydrogenation Reaction: In a hydrogenation reaction, hydrogen is added to a compound, typically in the presence of a catalyst. One example is the hydrogenation of unsaturated fats:
C=C + H2 → C-C
E. Explain the similarity and difference in two events, namely adding NaOH to water and adding CaO to water.
Ans:-When adding NaOH (sodium hydroxide) to water and adding CaO (calcium oxide) to water, both events involve the reaction of a base with water. However, there are some similarities and differences between these two reactions.
Similarities:
1.Both reactions are exothermic, meaning they release heat energy during the process.
2.Both reactions involve the formation of hydroxide ions (OH-) in the resulting solution.
3.Both reactions result in an increase in the pH of the solution, making it more basic.
Differences:
1.The chemical formulas of the substances involved are different. NaOH is a strong base, while CaO is a metal oxide.
Sodium hydroxide readily dissolves in water to form sodium ions (Na+) and hydroxide ions (OH-). It is a highly soluble compound.
2.CaO, on the other hand, reacts with water through a process called hydrolysis to form calcium hydroxide [Ca(OH)2]. Calcium hydroxide is sparingly soluble in water and forms a white precipitate.
3.The reaction between NaOH and water is rapid and highly exothermic, releasing a large amount of heat. In contrast, the reaction between CaO and water is slower and less exothermic.
4.The pH of the resulting solutions may vary. Sodium hydroxide typically produces a strongly alkaline solution with a high pH, while calcium hydroxide produces a moderately alkaline solution with a lower pH.
Overall, both reactions involve the interaction of a base with water, leading to the formation of hydroxide ions. However, the specific properties and characteristics of the substances involved result in some differences in the reaction process and the properties of the resulting solutions.
Q3. Explain the following terms with examples.
A. Endothermic reaction
Ans:-An endothermic reaction is a chemical reaction that absorbs heat from its surroundings. In other words, it requires an input of energy in the form of heat to proceed. During an endothermic reaction, the energy of the products is higher than the energy of the reactants.
Photosynthesis: In the process of photosynthesis, plants convert carbon dioxide and water into glucose and oxygen, using energy from sunlight. This reaction is endothermic as it requires energy from the sun to drive the reaction.
6CO2 + 6H2O + energy (sunlight) → C6H12O6 + 6O2
B. Combination reaction
Ans:-A combination reaction is a type of chemical reaction in which two or more substances combine to form a single product. In a combination reaction, the reactants come together and react to produce a new compound.
2Mg + O2 → 2MgO
In this reaction, two atoms of magnesium combine with one molecule of oxygen to produce two molecules of magnesium oxide. The combination of magnesium and oxygen atoms results in the formation of a new compound, magnesium oxide.
C. Balanced equation
Ans:-In a balanced equation, the number of atoms of each element in the reactants is equal to the number of atoms of the same element in the products. This ensures that the law of conservation of mass is satisfied, which states that matter cannot be created or destroyed in a chemical reaction.
The balanced equation for the reaction between sodium hydroxide (NaOH) and sulfuric acid (H2SO4) is:
2NaOH + H2SO4 → Na2SO4 + 2H2O
D. Displacement reaction
Ans:-A displacement reaction is a type of chemical reaction where one element displaces another element from a compound. This occurs when a more reactive element replaces a less reactive element in a compound. The displaced element is released as a free element, while the displacing element forms a new compound with the rest of the original compound.
example, zinc (Zn) reacts with copper sulfate (CuSO4). The zinc atoms displace the copper ions in the copper sulfate compound. The zinc atoms lose electrons and form Zn2+ ions, while the copper ions (Cu2+) are released as free copper atoms. The resulting reaction can be represented as:
Zn + CuSO4 → ZnSO4 + Cu
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:-The gas formed on heating limestone is carbon dioxide (CO2). When this carbon dioxide gas is passed through freshly prepared lime water (calcium hydroxide solution), a milky white precipitate of calcium carbonate (CaCO3) is formed.
Lime water (calcium hydroxide) reacts with carbon dioxide to form calcium carbonate. The equation for this reaction is:
Ca(OH)2 + CO2 → CaCO3 + H2O
B. It takes time for pieces of Shahabad tile to disappear in HCl, but its powder disappears rapidly.
Ans :- The powder form of Shahabad tile has a significantly larger surface area compared to the pieces. When a substance is in powder form, it is finely divided, increasing the total surface area available for the reaction. This increased surface area allows for a larger contact area between the reactants (Shahabad tile and HCl), leading to more efficient and faster reactions.
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:- The addition of concentrated sulphuric acid to water is an exothermic reaction, meaning it releases heat. If the concentrated acid is added to water too quickly or without stirring, localized heating can occur, leading to the formation of hot spots. This can cause the water to boil and potentially lead to the acid splattering. By adding the acid slowly and stirring constantly, the heat is distributed more evenly, preventing localized heating and ensuring a controlled and gradual mixing of the acid with water.
D. It is recommended to use air tight container for storing oil for long time.
Ans:-Storing the oil or ghee in an airtight container helps to minimize its exposure to air, reducing the contact with oxygen and slowing down the oxidation process. By limiting the availability of oxygen, the rate of oxidation is reduced, and the development of rancidity is delayed.
Q5. Observe the following picture a write down the chemical reaction with explanation.
Ans:- (i) Oxygen reduction reaction:
O2 + 4e- + 2H2O ⟶ 4OH-
Oxygen from the air gains electrons and reacts with water to form hydroxide ions.
(ii) Iron oxidation reaction:
Fe ⟶ Fe2+ + 2e-
Iron atoms lose electrons and form iron ions.
(iii) Formation of iron ions:
4Fe2+ + O2 ⟶ 4Fe3+ + 2O2-
The iron(II) ions formed in the previous reaction react with oxygen to form iron(III) ions and oxide ions.
(iv) Formation of iron(II) hydroxide and iron(III) hydroxide:
Fe2+ + 2H2O ⇌ Fe(OH)2 + 2H+
Fe3+ + 3H2O ⇌ Fe(OH)3 + 3H+
The iron(II) ions react with water to form iron(II) hydroxide, and the iron(III) ions react with water to form iron(III) hydroxide.
(v) Conversion to iron(III) oxide:
Fe(OH)2 ⇌ FeO + H2O
Fe(OH)3 ⇌ FeO(OH) + H2O
2FeO(OH) ⇌ Fe2O3 + H2O
The iron hydroxides undergo further reactions to form iron(III) oxide and water.
6. Identify from the following reactions the reactants that undergo oxidation and reduction.
A. Fe + S ⟶ FeS
Ans:- In this reaction, iron (Fe) undergoes oxidation by losing electrons to form Fe2+ ions, while sulfur (S) undergoes reduction by gaining those electrons to form S2- ions. Therefore, Fe undergoes oxidation, and S undergoes reduction.
B. 2Ag2O ⟶ 4Ag + O2
Ans:-In the reaction 2Ag2O ⟶ 4Ag + O2, silver oxide (Ag2O) is indeed decomposed into silver (Ag) and oxygen gas (O2). In this reaction, silver undergoes reduction since it changes from a higher oxidation state (+1 in Ag2O) to a zero oxidation state in Ag. Oxygen, on the other hand, undergoes oxidation as it changes from a zero oxidation state in Ag2O to a zero oxidation state in O2.
C. 2Mg + O2 ⟶ 2MgO
Ans:-In this reaction, magnesium (Mg) undergoes oxidation by losing electrons to form Mg2+ ions, while oxygen (O2) undergoes reduction by gaining those electrons to form oxide ions (O2-). Therefore, Mg undergoes oxidation, and O2 undergoes reduction.
D. NiO + H2 ⟶ Ni + H2O
Ans:-In this reaction, nickel(II) oxide (NiO) is reduced by hydrogen gas (H2) to form nickel (Ni) and water (H2O). Nickel(II) oxide undergoes reduction by gaining electrons to form nickel, and hydrogen gas undergoes oxidation by losing those electrons to form water. Therefore, NiO undergoes reduction, and H2 undergoes oxidation.
Q7. Balance the following equation stepwise.
A) H2S2O7(l) + H2O(l) → H2SO4(l)
Ans:-To balance this equation, let's start by counting the number of atoms on each side of the equation:
On the left side: 2 hydrogen (H) atoms, 2 sulfur (S) atoms, and 7 oxygen (O) atoms.
On the right side: 2 hydrogen (H) atoms, 1 sulfur (S) atom, and 4 oxygen (O) atoms.
To balance the hydrogen atoms, we need to put a coefficient of 2 in front of H2O:
H2S2O7(l) + 2H2O(l) → H2SO4(l)
Now, let's balance the sulfur atoms by putting a coefficient of 2 in front of H2SO4:
H2S2O7(l) + 2H2O(l) → 2H2SO4(l)
Finally, let's balance the oxygen atoms. On the left side, we have 7 oxygen atoms, and on the right side, we have 4 + 8 = 12 oxygen atoms. To balance them, put a coefficient of 6 in front of H2O:
H2S2O7(l) + 6H2O(l) → 2H2SO4(l)
The balanced equation is:
H2S2O7(l) + 6H2O(l) → 2H2SO4(l)
B) SO2(g) + H2S(aq) → S(s) + H2O(l)
Ans:-To balance the equation SO2(g) + H2S(aq) = S(s) + H2O(l), let's follow these steps:
Step 1: Count the number of atoms for each element on both sides of the equation:
On the left side:
1 sulfur (S) atom
2 oxygen (O) atoms
1 hydrogen (H) atom
On the right side:
1 sulfur (S) atom
2 oxygen (O) atoms
2 hydrogen (H) atoms
Step 2: Balance the sulfur atoms by putting a coefficient of 1 in front of H2S:
SO2(g) + 1H2S(aq) = S(s) + H2O(l)
Step 3: Balance the hydrogen atoms by putting a coefficient of 2 in front of H2O:
SO2(g) + 1H2S(aq) = S(s) + 2H2O(l)
Step 4: Balance the oxygen atoms by adjusting the coefficient in front of SO2:
2SO2(g) + 1H2S(aq) = 2S(s) + 2H2O(l)
The balanced equation is:
2SO2(g) + H2S(aq) = 2S(s) + 2H2O(l)
C) Ag(s) + HCl(aq) → AgCl + H2
Ans:-To balance the equation Ag(s) + HCl(aq) → AgCl + H2, let's follow these steps:
Step 1: Count the number of atoms for each element on both sides of the equation:
On the left side:
1 silver (Ag) atom
1 hydrogen (H) atom
1 chlorine (Cl) atom
On the right side:
1 silver (Ag) atom
1 chlorine (Cl) atom
1 hydrogen (H) atom
Step 2: Balance the silver atoms by putting a coefficient of 2 in front of AgCl:
Ag(s) + HCl(aq) → 2AgCl + H2
Step 3: Balance the chlorine atoms by adjusting the coefficient in front of HCl:
Ag(s) + 2HCl(aq) → 2AgCl + H2
The balanced equation is:
Ag(s) + 2HCl(aq) → 2AgCl + H2
D) NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
Ans:- To balance the equation NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l), let's follow these steps:
Step 1: Count the number of atoms for each element on both sides of the equation:
On the left side:
1 sodium (Na) atom
1 oxygen (O) atom
1 hydrogen (H) atom
1 sulfur (S) atom
On the right side:
2 sodium (Na) atoms
1 sulfur (S) atom
4 oxygen (O) atoms
2 hydrogen (H) atoms
Step 2: Balance the sodium atoms by putting a coefficient of 2 in front of NaOH:
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + H2O(l)
Step 3: Balance the hydrogen atoms by putting a coefficient of 2 in front of H2O:
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
Step 4: The equation is now balanced with respect to sodium, hydrogen, and sulfur atoms.
The balanced equation is:
2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
Q8. Identify the endothermic and exothermic reaction.
A. HCl + NaOH → NaCl + H2O + heat
Ans:- Exothermic reaction.
B. 2KClO3(s) + heat → 2KCl(s) + 3O2
Ans:- Endothermic reaction.
C. CaO + H2O → Ca(OH)2 + heat
Ans:- Exothermic reaction.
D. CaCO3(s) + heat → CaO(s) + CO2
Ans:- Endothermic reaction.
Q9.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 |
Solution:-
| Reactants | Products | Type of chemical reaction |
|---|---|---|
| BaCl2 (aq) + ZnSO4 (aq) | BaSO4(s)+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 |
Practice test
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