CBSE Class 10 Science Advanced Chemistry Chemical Reactions and Equations Long Answer Questions Set 01

Long Answer Type

Question 1. Why is respiration considered as an exothermic reaction ? Explain.
Answer:
Energy is needed to support life, and our body gets it from the food we eat.
During digestion, large food molecules are broken down to simpler ones such as glucose. In the cells, glucose combines with the oxygen we breathe and gives energy. This special slow combustion is called respiration.
\( \ce{C6H12O6(aq) + 6O2(g) -> 6CO2(g) + 6H2O(l) + "energy"} \)
Because energy is released during the whole process, respiration is an exothermic reaction.
In simple words: Cells burn glucose with oxygen, and since the process gives out energy it is exothermic.

Teacher's Note:
1. Mention digestion, glucose, oxygen and energy in your answer.
2. The same equation is used for the combustion of glucose outside the body.

Question 2. Write the balanced chemical equations for the following reactions. (a) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogencarbonate. (b) Sodium hydrogencarbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide. (c) Copper sulphate on treatment with potassium iodide precipitates cuprous iodide (\( \ce{Cu2I2} \)), liberates iodine gas and also forms potassium sulphate.
Answer:
(a) \( \ce{Na2CO3 + HCl -> NaCl + NaHCO3} \)
(b) \( \ce{NaHCO3 + HCl -> NaCl + H2O + CO2 ^} \)
(c) \( \ce{2CuSO4 + 4KI -> Cu2I2 v + I2 + 2K2SO4} \)
Check (c): Cu 2 and 2, S 2 and 2, K 4 and 4, I 4 and 2 + 2, O 8 and 8.
In simple words: Write all formulae correctly, then place coefficients on each side until every element is equal.

Teacher's Note:
1. Part (a) is the first step of the stepwise neutralisation of sodium carbonate.
2. Iodine is usually produced as a solid, but the question states 'gas', so it is written as given.

Question 3. Balance the following chemical equations and identify the type of chemical reaction. (a) \( \ce{Mg(s) + Cl2(g) -> MgCl2(s)} \) (b) \( \ce{HgO(s) ->[heat] Hg(l) + O2(g)} \) (c) \( \ce{Na(s) + S(s) ->[fuse] Na2S(s)} \) (d) \( \ce{TiCl4(l) + Mg(s) -> Ti(s) + MgCl2(s)} \) (e) \( \ce{CaO(s) + SiO2(s) -> CaSiO3(s)} \) (f) \( \ce{H2O2(l) ->[U.V.] H2O(l) + O2(g)} \)
Answer:
(a) \( \ce{Mg(s) + Cl2(g) -> MgCl2(s)} \) is already balanced: combination.
(b) \( \ce{2HgO(s) ->[heat] 2Hg(l) + O2(g)} \): decomposition (thermal).
(c) \( \ce{2Na(s) + S(s) ->[fuse] Na2S(s)} \): combination.
(d) \( \ce{TiCl4(l) + 2Mg(s) -> Ti(s) + 2MgCl2(s)} \): displacement.
(e) \( \ce{CaO(s) + SiO2(s) -> CaSiO3(s)} \) is already balanced: combination.
(f) \( \ce{2H2O2(l) ->[U.V.] 2H2O(l) + O2(g)} \): decomposition (photodecomposition).
In simple words: Balance each equation by atom counting and then classify it by what happens to the reactants.

Teacher's Note:
1. Combination gives one product, and decomposition has one reactant.
2. Some equations are already balanced, so check before changing anything.

Question 4. A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by emission of light. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y. (a) Write the chemical formulae of X and Y. (b) Write a balanced chemical equation, when X is dissolved in water.
Answer:
(a) X is magnesium oxide, MgO: \( \ce{2Mg + O2 -> 2MgO} \). Y is magnesium nitride, \( \ce{Mg3N2} \): \( \ce{3Mg + N2 -> Mg3N2} \).
(b) \( \ce{MgO + H2O -> Mg(OH)2} \)
In simple words: Burning magnesium makes white MgO in oxygen and Mg3N2 in nitrogen.

Teacher's Note:
1. Magnesium is so reactive that it even burns in nitrogen.
2. Mg(OH)2 is slightly soluble and the solution turns red litmus blue.

Question 5. You are provided with two containers made up of copper and aluminium. You are also provided with solutions of dilute HCl, \( \ce{ZnCl2} \) and \( \ce{H2O} \). In which of the above containers can these solutions be kept ?
Answer:
The reactivity order is Al > Zn > H > Cu.
Copper container: copper is below hydrogen and zinc, so it does not react with dilute HCl, \( \ce{ZnCl2} \) solution or water. All three can be kept in it.
Aluminium container: aluminium displaces hydrogen from dilute HCl, \( \ce{2Al + 6HCl -> 2AlCl3 + 3H2} \), and displaces zinc from its salt, \( \ce{2Al + 3ZnCl2 -> 2AlCl3 + 3Zn} \). So these cannot be kept in it. Water does not react with aluminium because of its protective oxide layer, so only water can be kept in the aluminium container.
In simple words: Copper can hold all three, whereas aluminium can hold only water.

Teacher's Note:
1. Use the reactivity series to decide, since only a more reactive metal displaces a less reactive one.
2. Aluminium containers hold water safely thanks to the oxide layer.

Question 6. What happens when a piece of (a) zinc metal is added to copper sulphate solution ? (b) Aluminium metal is added to dilute hydrochloric acid ? (c) Silver metal is added to copper sulphate solution ? Also, write the balanced chemical equation if the reaction occurs.
Answer:
(a) Zinc displaces copper. The blue colour fades and reddish brown copper settles: \( \ce{Zn + CuSO4 -> ZnSO4 + Cu} \).
(b) Aluminium dissolves and bubbles of hydrogen are given off: \( \ce{2Al + 6HCl -> 2AlCl3 + 3H2 ^} \).
(c) No reaction takes place because silver is less reactive than copper.
In simple words: A metal reacts only if it is more reactive than the other metal or than hydrogen.

Teacher's Note:
1. Remember: Zn is more reactive than Cu, Al is more reactive than H, but Ag is less reactive than Cu.
2. Always write 'no reaction' when that is the case.

Question 7. Explain why ? (a) Digestion of food is a decomposition reaction. (b) All decomposition reactions are endothermic reactions. (c) A popping sound is produced when a burning candle is brought near mouth of a test tube used in electrolysis of water. [CBSE 10th Board 2015, 3 Marks]
Answer:
(a) During digestion, large complex food molecules are broken into simpler, smaller molecules (for example starch into glucose). One substance splits into simpler ones, so it is a decomposition reaction.
(b) Breaking a compound into simpler substances needs energy, usually heat, light or electricity, to break the bonds. Hence decomposition reactions are generally endothermic, as in \( \ce{CaCO3 -> CaO + CO2} \). (A few decompositions, such as that of hydrogen peroxide, release heat, so the word 'all' is not strictly true, but this is the board's expected explanation.)
(c) Hydrogen gas is formed at the cathode during electrolysis of water: \( \ce{2H2O -> 2H2 + O2} \). Hydrogen burns with a pop sound when a flame is brought near it.
In simple words: Digestion splits big molecules, decomposition takes in energy, and hydrogen burns with a pop.

Teacher's Note:
1. Give reasons with an equation wherever possible.
2. Hydrogen is collected in a test tube at the negative electrode.

Question 8. (a) What happens when a strip of lead metal is placed in a solution of copper chloride ? Write the balanced chemical equation for the reaction along with the colour changes observed during the reaction. (b) What are precipitation reactions ? Give one example of a precipitation reaction. [CBSE 10th Board 2015, 3 Marks]
Answer:
(a) Lead is more reactive than copper, so it displaces copper from the solution: \( \ce{Pb(s) + CuCl2(aq) -> PbCl2(aq) + Cu(s)} \). The blue-green colour of the solution fades and reddish brown copper deposits on the lead.
(b) Reactions in which an insoluble solid (precipitate) is formed on mixing solutions are precipitation reactions. Example: \( \ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)} \) gives a white precipitate of barium sulphate.
In simple words: Lead pushes copper out of the solution, and a precipitation reaction gives an insoluble solid.

Teacher's Note:
1. Always describe the colour changes in such questions.
2. Choose the most familiar precipitation reaction for the example.

Question 9. A solution of a substance 'X' is used for white washing. (a) Name the substance 'X' and write its formula. (b) Write the reaction of the substance 'X' named in (a) above with water. (c) Write the balanced equation for the following chemical reaction : Barium chloride + Aluminium sulphate → Barium sulphate + Aluminium chloride [CBSE 10th Board 2013/2015, 3 Marks]
Answer:
(a) X is calcium oxide (quick lime), CaO.
(b) \( \ce{CaO + H2O -> Ca(OH)2 + "heat"} \). The product is calcium hydroxide, slaked lime, which is the actual whitewashing solution. The reaction is highly exothermic.
(c) \( \ce{3BaCl2 + Al2(SO4)3 -> 3BaSO4 + 2AlCl3} \)
In simple words: Quick lime is mixed with water to give slaked lime for whitewash.

Teacher's Note:
1. The board key accepts CaO as X, and the solution contains its product Ca(OH)2.
2. Balance chlorine and sulphate groups as units in (c).

Question 10. Write balanced chemical equations for the following chemical reactions. (a) Hydrogen + Chlorine → Hydrogen chloride (b) Lead + Copper chloride → Lead chloride + Copper (c) Zinc oxide + Carbon → Zinc + Carbon monoxide [CBSE 10th Board 2016, 3 Marks]
Answer:
(a) \( \ce{H2 + Cl2 -> 2HCl} \)
(b) \( \ce{Pb + CuCl2 -> PbCl2 + Cu} \)
(c) \( \ce{ZnO + C -> Zn + CO} \)
In simple words: Replace each name by its formula and check that the atoms are equal on both sides.

Teacher's Note:
1. Only (a) needs a coefficient.
2. Reactions (b) and (c) are already balanced.

Question 11. The colour of copper sulphate solution changes when an iron nail is dipped in it. State, giving the chemical equation for the reaction involved. Write the name of the reaction involved. [CBSE 10th Board 2013/2017, 1 and 3 Marks]
Answer:
The blue colour of the solution fades to pale green, because iron displaces copper: \( \ce{Fe(s) + CuSO4(aq) -> FeSO4(aq) + Cu(s)} \).
A reddish brown coating of copper forms on the nail.
The reaction is a displacement (single displacement) reaction.
In simple words: Iron pushes copper out, so the blue solution becomes pale green and the nail turns brown.

Teacher's Note:
1. Name the type of reaction, since it is asked.
2. Iron(II) sulphate gives the pale green colour.

Question 12. Write the balanced chemical equations for the following reactions. (i) Aqueous solution of sulphuric acid and sodium hydroxide react to form aqueous sodium sulphate and water. (ii) Phosphorus burns in chlorine gas to form phosphorus pentachloride.
Answer:
(i) \( \ce{H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l)} \)
(ii) \( \ce{2P(s) + 5Cl2(g) -> 2PCl5(s)} \)
In simple words: In (i) two NaOH are needed for one H2SO4, and in (ii) five Cl2 for two P.

Teacher's Note:
1. Balance the sodium and hydrogen in (i) and use even numbers for chlorine in (ii).
2. Phosphorus pentachloride is a pale yellow solid.

Question 13. Write the balanced chemical equations for the following reactions. (i) \( \ce{ZnCO3(s) -> ZnO(s) + CO2(g)} \) (ii) Potassium bromide(aq) + Barium iodide(aq) → Potassium iodide(aq) + Barium bromide(aq)
Answer:
(i) \( \ce{ZnCO3(s) ->[\Delta] ZnO(s) + CO2(g)} \) is already balanced.
(ii) \( \ce{2KBr(aq) + BaI2(aq) -> 2KI(aq) + BaBr2(aq)} \)
In simple words: In (ii) two potassium halides are needed for every one barium halide.

Teacher's Note:
1. Treat each ion as a unit while balancing.
2. Check the potassium atoms first in (ii).

Question 14. Balance the following equations. (i) \( \ce{HgO -> Hg + O2} \) (ii) \( \ce{C4H10(g) + O2(g) -> CO2(g) + H2O(l)} \) (iii) \( \ce{BaCl2(aq) + Al2(SO4)3(aq) -> AlCl3(aq) + BaSO4(aq)} \)
Answer:
(i) \( \ce{2HgO -> 2Hg + O2} \)
(ii) \( \ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(l)} \)
(iii) \( \ce{3BaCl2(aq) + Al2(SO4)3(aq) -> 2AlCl3(aq) + 3BaSO4(s)} \) (barium sulphate is the insoluble precipitate).
In simple words: Balance by hit and trial, counting each element on both sides.

Teacher's Note:
1. In (iii) treat the sulphate group as one unit.
2. BaSO4 is insoluble, so (s) is more correct than (aq).

Question 15. What information do we get from a chemical equation ? Explain with the help of examples.
Answer:
A balanced chemical equation tells us:
1. The names of the reactants and products.
2. The number of atoms or molecules of each substance that react or form, from the coefficients, for example \( \ce{2H2 + O2 -> 2H2O} \).
3. The masses: in \( \ce{CaCO3 -> CaO + CO2} \), 100 g of CaCO₃ gives 56 g of CaO and 44 g of CO₂, so mass is conserved.
4. The volume of a gas (22.4 litres of CO₂ per mole at STP) in this case.
5. The valency of an element, for example in \( \ce{Mg + 2HCl -> MgCl2 + H2} \) one Mg atom displaces two H atoms, so Mg has valency 2.
With state symbols and conditions written on the arrow we can also tell the physical state, the concentration, the conditions and whether heat is released or absorbed.
In simple words: An equation is a short summary of what reacts, what forms and how much of each.

Teacher's Note:
1. Learn the CaCO3 example with 100 = 56 + 44 for the mass relation.
2. A skeletal equation gives less information than a balanced one.

Question 16. Identify the substances that are oxidized and the substances that are reduced in the following reactions. (a) \( \ce{3Mg + N2 -> Mg3N2} \) (b) \( \ce{MnO2 + 4HCl -> MnCl2 + 2H2O + Cl2} \) (c) \( \ce{2FeCl3 + H2S -> 2FeCl2 + S + 2HCl} \)
Answer:
(a) Mg is oxidised (loses electrons, 0 to +2) and N₂ is reduced (0 to -3).
(b) HCl is oxidised (Cl^- becomes Cl₂, by losing hydrogen) and MnO₂ is reduced (Mn from +4 to +2, loses oxygen).
(c) H₂S is oxidised (loses hydrogen, forming S) and FeCl₃ is reduced (Fe from +3 to +2).
In simple words: The substance that loses electrons or hydrogen is oxidised and the one that gains them is reduced.

Teacher's Note:
1. Work out the change in oxidation number for every element.
2. In (b) manganese dioxide acts as the oxidising agent.

Question 17. (i) What is observed when a solution of potassium iodide is added to a solution of lead nitrate taken in a test tube ? (ii) Which type of reaction is this ? (iii) Write a balanced chemical equation to represent the above reaction.
Answer:
(i) A yellow precipitate of lead iodide appears.
(ii) It is a double displacement (precipitation) reaction.
(iii) \( \ce{Pb(NO3)2(aq) + 2KI(aq) -> PbI2(s) + 2KNO3(aq)} \)
In simple words: Lead nitrate and potassium iodide exchange ions to give yellow lead iodide.

Teacher's Note:
1. This question is repeated in the book as well as in the Check Your Progress section.
2. Note the coefficient 2 for KI and KNO3.

Question 18. Classify the following reactions. (i) \( \ce{N2 + O2 -> 2NO - "Heat"} \) (ii) \( \ce{2HgO -> 2Hg + O2} \) (iii) \( \ce{Na2SO4 + BaCl2 -> 2NaCl + BaSO4} \) (iv) \( \ce{CuSO4(aq) + Zn -> ZnSO4(aq) + Cu} \) (v) \( \ce{NH3 + HCl -> NH4Cl} \)
Answer:
(i) Combination reaction, and it is also endothermic (heat is absorbed).
(ii) Decomposition (thermal; it also breaks into elements, so it is an analysis reaction).
(iii) Double displacement (precipitation) reaction.
(iv) Displacement (single displacement) reaction.
(v) Combination (addition) reaction.
In simple words: Look at how many substances react and how many form to name the type.

Teacher's Note:
1. A reaction can belong to more than one class, such as combination and endothermic.
2. Reaction (v) forms white fumes of ammonium chloride.

Question 19. Oxidation reactions have some harmful effects. Comment on the sentence.
Answer:
Oxidation gives us energy in respiration and combustion, but it also has harmful effects:
1. Corrosion: metals are slowly eaten away by air and moisture, for example rusting of iron, the green coating on copper and the black coating on silver. Bridges, railings and vehicles are damaged and a lot of money is spent on replacing them.
2. Rancidity: oxidation of fats and oils in food gives a bad smell and taste and spoils the food.
3. Pollution: combustion of fuels releases poisonous gases such as CO, \( \ce{SO2} \), \( \ce{SO3} \) and oxides of nitrogen into the air.
In simple words: Oxidation is useful in giving energy but harmful when it causes rust, stale food and pollution.

Teacher's Note:
1. Give at least two harmful effects with examples.
2. Prevention includes painting, galvanisation, antioxidants and cold storage.