Chemical Equations – Long Answer Questions (CBSE Class 10 Science, Chemistry)
Medium Level (Application & Explanation)
Q1. Convert the following word equations into balanced chemical equations and explain each step:
a) Hydrogen + Oxygen → Water
b) Iron + Sulfur → Iron sulfide
Answer:
- In a word equation, the substances on the left are the reactants and those on the right are the products. We translate names into symbols and then balance atoms using coefficients (not by changing subscripts).
- a) Hydrogen + Oxygen → Water
• Symbols: H₂ + O₂ → H₂O (skeletal; not balanced)
• Count atoms: H = 2 (LHS), 2 (RHS); O = 2 (LHS), 1 (RHS)
• Balance O by making water coefficient 2: H₂ + O₂ → 2 H₂O
• Now H is 2 (LHS) vs 4 (RHS). Balance H by multiplying H₂ by 2: 2 H₂ + O₂ → 2 H₂O (balanced)
- b) Iron + Sulfur → Iron sulfide
• Symbols: Fe + S → FeS
• Fe = 1 (LHS), 1 (RHS); S = 1 (LHS), 1 (RHS).
• This is already balanced.
- This shows how a skeletal equation becomes a balanced equation by adjusting coefficients.
Q2. Explain the law of conservation of mass using the reaction Zn + H₂SO₄ → ZnSO₄ + H₂. Show how atom counting supports this law.
Answer:
- The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. Therefore, the total number of atoms of each element must be the same on both sides of a balanced equation.
- Consider: Zn + H₂SO₄ → ZnSO₄ + H₂
• Count atoms (LHS → RHS):
– Zn: 1 → 1
– H: 2 → 2
– S: 1 → 1
– O: 4 → 4
• Since the number of atoms of each element is equal, the equation is balanced, upholding the law.
- In simple terms, atoms rearrange to form new substances (products) but their total count remains constant, so the overall mass stays the same. This is why balancing is essential: it ensures the equation follows the law of conservation of mass and correctly represents what happens in a real reaction.
Q3. What is a skeletal chemical equation? Using Na + Cl₂ → NaCl as an example, explain the difference between a skeletal and a balanced equation.
Answer:
- A skeletal chemical equation shows the correct formulas of reactants and products but does not ensure equal numbers of atoms on both sides. It is the first draft of an equation before balancing.
- Example: Na + Cl₂ → NaCl
• Count atoms (LHS → RHS):
– Na: 1 → 1
– Cl: 2 → 1
• Chlorine is not balanced. This is a skeletal equation.
- To balance, adjust coefficients:
• Make NaCl coefficient 2 to balance Cl: Na + Cl₂ → 2 NaCl
• Now Na is 1 (LHS) vs 2 (RHS). Balance Na by making it 2 on the LHS: 2 Na + Cl₂ → 2 NaCl
- The final equation is balanced: equal atoms of each element on both sides.
- Key idea: never change subscripts (which change the substance); only change coefficients to keep the identity of substances intact.
Q4. Describe a systematic method to balance Fe + H₂O → Fe₃O₄ + H₂ and explain why starting with the most complex molecule helps.
Answer:
- A clear, stepwise approach prevents confusion:
• Step 1: Write the skeletal equation: Fe + H₂O → Fe₃O₄ + H₂
• Step 2: Count atoms (LHS → RHS):
– Fe: 1 → 3
– H: 2 → 2
– O: 1 → 4
• Step 3: Start with the most complex molecule (Fe₃O₄) because it contains multiple elements and appears in only one place on each side.
• Step 4: Balance Fe by placing 3 before Fe: 3 Fe + H₂O → Fe₃O₄ + H₂
• Step 5: Balance O by placing 4 before H₂O: 3 Fe + 4 H₂O → Fe₃O₄ + H₂
• Step 6: Now balance H: LHS has 8 H, RHS has 2 H. Put 4 before H₂: 3 Fe + 4 H₂O → Fe₃O₄ + 4 H₂
- This equation is now balanced.
- Balancing the most complex molecule first reduces trial-and-error because it anchors multiple elements at once.
Q5. Why is it important to add physical states and conditions to chemical equations? Illustrate using examples.
Answer:
- Adding physical states—(s), (l), (g), (aq)—and conditions (temperature, pressure, catalyst) makes equations more informative and realistic. It tells us how a reaction actually occurs in practice.
- Example 1: 3 Fe(s) + 4 H₂O(g) → Fe₃O₄(s) + 4 H₂(g)
• States show that steam (g) is needed, not liquid water, making the context of the reaction clear.
- Example 2: CO(g) + 2 H₂(g) → CH₃OH(l) at 340°C and 300 atm
• The specified temperature and pressure are crucial for efficient conversion to methanol; without them the reaction is slow or incomplete.
- Example 3: Aqueous reactions: states like (aq) indicate ions dissolved in water, which affects reaction rates and pathways.
- Thus, states and conditions help predict feasibility, rate, and the observed form of reactants/products.
High Complexity (Analytical & Scenario-Based)
Q6. A friend asks about combustion of hydrocarbons. Explain how to predict the products and balance any complete combustion reaction. Provide one worked example.
Answer:
- In complete combustion, a hydrocarbon (CₓHᵧ) reacts with O₂ to form CO₂ and H₂O. The steps:
• Step 1: Write the skeletal form: CₓHᵧ + O₂ → CO₂ + H₂O
• Step 2: Balance C first by putting x before CO₂.
• Step 3: Balance H next by putting y/2 before H₂O.
• Step 4: Count total O on RHS and adjust the O₂ coefficient accordingly. If you get a fraction, multiply the entire equation by 2 to remove it.
- Example: Combustion of methane: CH₄ + O₂ → CO₂ + H₂O
• C: 1 → put 1 CO₂; H: 4 → put 2 H₂O.
• O on RHS: 2 (in CO₂) + 2 (in 2 H₂O) = 4 → need 2 O₂ on LHS.
• Balanced: CH₄ + 2 O₂ → CO₂ + 2 H₂O
- Key ideas: products are CO₂ and H₂O; balance in the sequence C → H → O.
Q7. In a lab, you observe heat release when magnesium burns. Design a verification plan to check if your recorded equation is balanced and includes complete information.
Answer:
- Start with the observed reaction: Magnesium burns in oxygen to form magnesium oxide.
- Step 1: Write the word equation: Magnesium + Oxygen → Magnesium oxide.
- Step 2: Convert to symbols: Mg + O₂ → MgO (skeletal).
- Step 3: Balance atoms: O is 2 on LHS and 1 on RHS; put 2 before MgO: Mg + O₂ → 2 MgO. Now Mg is 1 vs 2, so put 2 before Mg: 2 Mg + O₂ → 2 MgO.
- Step 4: Add physical states and note observation (bright white flame; heat): 2 Mg(s) + O₂(g) → 2 MgO(s); the reaction is exothermic.
- Step 5: Check the law of conservation of mass by counting atoms again: Mg = 2, O = 2 on both sides.
- Step 6: Record essential details: reactants, products, states, and that energy is released (exothermic).
Q8. You face a hard-to-balance equation with multiple elements. Explain your strategy and apply it to balance Al + O₂ → Al₂O₃, justifying each step.
Answer:
- Strategy for complex equations:
• Start with the compound that has the most different atoms (often a product like oxides).
• Balance elements that appear in fewer compounds first (e.g., metals before O/H).
• Use coefficients only; avoid changing subscripts.
• If odd/even issues arise, use the least common multiple to settle them.
- Application: Al + O₂ → Al₂O₃
• Skeletal counts: Al: 1 → 2; O: 2 → 3 (odd/even mismatch).
• Make Al₂O₃ coefficient 2 to get an even number of O: Al + O₂ → 2 Al₂O₃ (now O = 6 on RHS).
• Balance O: need 3 O₂ on LHS: Al + 3 O₂ → 2 Al₂O₃.
• Balance Al: RHS has 4 Al, so put 4 on LHS: 4 Al + 3 O₂ → 2 Al₂O₃.
- Final check: Al = 4, O = 6 on both sides. This method reduces trial-and-error and respects the law of conservation of mass.
Q9. A student changes subscripts instead of coefficients to balance equations. Analyze why this is wrong with examples, and propose correct fixes.
Answer:
- Subscripts are part of a compound’s chemical formula and define its identity; changing them creates a different substance. Coefficients indicate the number of molecules/moles and should be used for balancing.
- Wrong approach example: To balance H₂ + O₂ → H₂O, changing H₂O to H₂O₂ makes O count “match,” but you now have hydrogen peroxide, not water. This violates the reaction’s meaning.
- Another mistake: Changing NaCl to NaCl₂ to match Cl₂ on the LHS invents a compound that doesn’t represent table salt.
- Correct fixes use coefficients:
• Water formation: 2 H₂ + O₂ → 2 H₂O (balanced using multiples).
• Sodium chloride formation: 2 Na + Cl₂ → 2 NaCl.
- Conclusion: Always adjust coefficients to preserve the identity of reactants and products while satisfying conservation of mass.
Q10. Evaluate why adding conditions (temperature, pressure, catalyst) and states is essential for predicting reaction behavior. Use CO(g) + 2 H₂(g) → CH₃OH(l) and steam-iron reaction as illustrations.
Answer:
- Conditions and states make equations predictive rather than just descriptive. They tell us when a reaction is feasible, how fast it proceeds, and which phase each substance is in, affecting collision frequency and energy.
- Example 1 (Methanol synthesis): CO(g) + 2 H₂(g) → CH₃OH(l) at about 340°C and 300 atm.
• High pressure favors the side with fewer gas molecules; appropriate temperature ensures sufficient energy without excessive decomposition.
• Without these conditions, conversion is low; with them, yield improves.
- Example 2 (Steam with iron): 3 Fe(s) + 4 H₂O(g) → Fe₃O₄(s) + 4 H₂(g).
• The (g) state of water indicates steam, necessary to oxidize iron and release hydrogen gas efficiently.
- Thus, specifying states and conditions helps explain observed outcomes and guides experimental setup for reliable results.
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