9.05 Past Paper Practice
Questions
1.Sulfur dioxide is formed when sulfur burns in air.
Sulfur dioxide reacts with oxygen in the presence of a catalyst to form sulfur trioxide. This is a reversible reaction.
Complete the equation for this reaction by writing the sign for a reversible reaction in the box.
2SO2 + O2 ____ 2SO3
Answer:
The complete equation with the reversible reaction sign is: 2SO2 + O2 ⇌ 2SO3
Explanation:
This reaction happens in the Contact Process, with a catalyst (V2O5), and is essential for producing sulfuric acid.
Use ⇌ right left harpoons ⇌ whenever a reaction is described as reversible.
The reaction 2SO2 + O2 ⇌ 2SO3 is reversible because the products (SO3) can break down into reactants (SO2) under certain conditions.
The reversible sign ⇌ indicates dynamic equilibrium, where the forward and reverse reactions occur at the same rate.
2.
Cobalt(II) chloride can be used to test for the presence of water:
CoCl2 + 6H2O ⇌ CoCl2⋅6H2O
(anhydrous cobalt(II) chloride) ⇌ (hydrated cobalt(II) chloride).
State the meaning of the symbol ⇌.
Answer:
The symbol ⇌ means that the reaction is reversible.
Explanation:
Dynamic equilibrium occurs when the forward and reverse reactions happen at the same rate.
⇌ indicates a reversible reaction where the reactants (CoCl2+6H2O can convert to products (CoCl2⋅6H2O and vice versa.
In this case:
Forward reaction: Anhydrous cobalt(II) chloride absorbs water to form hydrated cobalt(II) chloride.
Reverse reaction: Hydrated cobalt(II) chloride loses water to form anhydrous cobalt(II) chloride.
3. This question is about zinc and compounds of zinc.
(i) State what happens to CO2 in this reaction.
(ii) Name the type of reaction.
Answer:
(i): CO2 loses oxygen.
(ii): Redox.
Explanation:
Redox reactions always involve electron transfer or changes in oxidation states.
(i) What happens to CO2?
CO2 losing oxygen is an example of reduction, as reduction involves the loss of oxygen in a chemical reaction.
(ii) Type of Reaction:
This is a redox reaction because both reduction (loss of oxygen by CO2) and oxidation (gain of oxygen by another substance) occur simultaneously.
4. The equation for the reaction between methanoic acid and ethanol in the presence of a catalyst:
HCOOH + CH3CH2OH ⇌ X + H2O ΔH = −29.5 kJ/mol.
The reaction is reversible and occurs in a closed system.
(i) State what is meant by a closed system.
(ii) State two characteristics of an equilibrium.
(iii) Complete the table to show the effect on the concentration of X at equilibrium for each change of condition.
Answer:
(i) A closed system means that no reactants or products can enter or leave the system.
(ii) Characteristics of equilibrium:
- The forward and reverse reactions occur at the same rate.
- The concentrations of reactants and products remain constant.
(iii) Table Completion (Effect on X):
| Change of Condition | Effect on the Concentration of X |
|---|---|
| Temperature is decreased | Increases (favours exothermic reaction) |
| Concentration of HCOOH is decreased | Decreases (fewer reactants available for the reaction) |
| Concentrations of both HCOOH and CH3CH2OH are decreased | Decreases (both reactants reduced) |
| The catalyst is removed | No change (catalyst does not affect equilibrium concentrations) |
Explanation:
Catalyst Removal: Catalysts only affect the speed at which equilibrium is reached, not the equilibrium concentrations.
Closed System:
A closed system prevents any exchange of matter with the surroundings, allowing equilibrium to be established.
Characteristics of Equilibrium:
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.
At equilibrium, the concentrations of reactants and products remain constant, although reactions still occur.
Effects of Changes on X:
Temperature Decrease: Favors the exothermic reaction (ΔH=−29.5 kJ/mol), increasing X.
Decreased Reactant Concentration: Fewer reactants reduce the forward reaction rate, lowering X.
5. This question is about the extraction of iron and related chemical processes.
(i) Calculate the molar mass and percentage of iron in iron(III) oxide (Fe2O3).
(ii) Name the main ore of iron.
(iii) Describe how iron(III) oxide is reduced in the blast furnace.
(iv) Write the balanced equation for the reduction of iron(III) oxide with carbon monoxide.
(v) State the type of reaction involved in the reduction of iron(III) oxide.
(c) State the term used to describe the decomposition of calcium carbonate in the blast furnace.
Answer:
(i)
- Molar mass of Fe2O3: 160 g/mol
- Percentage of iron: 70.0%
(ii) Hematite
(iii) By reduction of iron(III) oxide using carbon monoxide (CO) gas in the blast furnace.
(iv) Balanced equation: Fe2O3 + 3CO → 2Fe + 3CO2
(v) Reduction
(c) Thermal decomposition
Explanation (Summarised):
- Molar Mass Calculation (Fe2O3):
- Iron (Fe): 2×56=112
- Oxygen (O3): 3×16=48
- Total molar mass = 112+48=160 g/mol
- Percentage of Iron in Fe2O3:
- Percentage
Reaction: CaCO3 → CaO + CO2
Reduction of Fe2O3:
Reduction is the removal of oxygen from Fe2O3, facilitated by carbon monoxide (CO) in the blast furnace.
Produces molten iron and carbon dioxide (CO2).
Balanced Equation:
Fe2O3 + 3CO → 2Fe + 3CO2
This shows the stoichiometry of reactants and products.
Type of Reaction:
Reduction: Iron(III) oxide loses oxygen, which defines a reduction process.
Thermal Decomposition:
Calcium carbonate (CaCO3) decomposes under heat in the blast furnace to form calcium oxide (CaO) and carbon dioxide (CO2).
6. Identify the element with the symbol “Ne”.
Neon:
Ne is the symbol for neon, a noble gas element in Group 18 of the periodic table.
7. (i): Balance the equation for the reaction of iron pyrites (FeS2) with oxygen: _ FeS2 +_ O2→_ Fe2O3 +_ SO2
(ii): Name Fe2O3, including the oxidation number of iron.
Answers:
7(a)(i): Balanced equation: 4FeS2 + 11O2 → 2Fe2O3 + 8SO2
7(a)(ii): Iron(III) oxide.
Explanation:
Fe2O3 is called iron(III) oxide because the oxidation number of iron is +3 in this compound.
7(a)(i) Balancing Equation:
FeS2 reacts with O2 to produce Fe2O3 (iron oxide) and SO2 (sulfur dioxide).
Balancing ensures equal numbers of each atom on both sides:
4FeS2 + 11O2 → 2Fe2O3 + 8SO2.
7(a)(ii) Naming Fe2O3:
8(i). Determine the oxidation state of phosphorus (P) in a compound where P is combined with four atoms of oxygen (O).
Answer:
- The oxidation state of O is −2, so the total for four oxygen atoms is: 4 × (−2) = −8
- Let the oxidation state of PPP be xxx. Using the rule that the sum of oxidation states equals the overall charge (000 for neutral compounds): x+(−8)=0x + (-8) = 0
Oxidation state of phosphorus (P): +5
Explanation:
Final Answer: P=+5P = +5P=+5.
Rules for Oxidation States:
Oxygen (O) always has an oxidation state of −2 in most compounds.
The total oxidation states in a neutral compound must sum to 0.
Calculation Process:
Four oxygen atoms contribute −8.
Phosphorus (P) must balance this, so x+(−8)=0.
Solving gives x=+5x = +5x=+5.
9 (i). The reaction involves copper and oxygen.
- State the change in oxidation number for copper and oxygen.
- Define oxidation and reduction based on changes in oxidation numbers.
Answer:
- Oxidation Number Changes:
- Copper: +2→+1 (reduction).
- Oxygen: −2→0 (oxidation).
- Definition of Oxidation and Reduction:
- A decrease in oxidation number is reduction.
- An increase in oxidation number is oxidation.
Explanation:
Copper is reduced, and oxygen is oxidized. The definitions of oxidation and reduction align with changes in their oxidation states.
Oxidation Numbers:
The oxidation number represents the charge an atom would have if all bonds were ionic.
Here, copper’s oxidation number decreases (+2→+1 ), meaning it gains an electron (reduction).
Oxygen’s oxidation number increases (−2→0), meaning it loses electrons (oxidation).
Oxidation and Reduction Rules:
Reduction: Gain of electrons or decrease in oxidation number.
Oxidation: Loss of electrons or increase in oxidation number.
Final Answer:
10. Hydrated copper(II) sulfate is a coloured compound that contains water molecules in its crystal structure.
(i)
- State the term given to water molecules present in hydrated crystals.
- State the colour of hydrated copper(II) sulfate crystals.
- Write the formula of hydrated copper(II) sulfate.
Answers:
- Term for water molecules in hydrated crystals: Water of crystallization.
- Colour of hydrated copper(II) sulfate crystals: Blue.
- Formula of hydrated copper(II) sulfate: CuSO4⋅5H2O .
Explanation:
This indicates it contains five moles of water of crystallization for each mole of copper(II) sulfate.
Water of Crystallization:
These are water molecules chemically bound within the crystal structure of a compound, giving it specific physical properties.
Hydrated Copper(II) Sulfate Crystals:
Hydrated copper(II) sulfate appears blue due to the presence of water molecules in its structure.
Formula:
The chemical formula for hydrated copper(II) sulfate includes 5 water molecules per copper(II) sulfate unit: CuSO4⋅5H2O .