Published on June 16th 2024 | 15 mins , 2887 words
1. Tri-iron tetraoxide is an oxide of iron which can be produced in the laboratory. (a) Write an equation for the reaction which can be used to produce the oxide. (1 mark)
\(3Fe(s) + 4H_{2O(g)}\) \( \rightarrow\) \(Fe_{3}O_{4(s)} + 4H_{2(g)} \) (b) Write an equation for the reaction between the oxide and hydrochloric acid. (1 mark)
2. State the advantage of using a boiling tube while heating compared to a test tube. (1/2 mark)
It has a bigger surface area which spreads heat energy.
3. When preparing a gas in the laboratory, explain why the following is done: (a) Heating is not required when using a flat-bottomed flask. (1/2 mark)
It could crack.
(b) The delivery tube must always be well above the level of the reacting chemicals. If the tube to supply a liquid or solution reagent is a dropping funnel, the tap must be closed. (1/2 mark)
To prevent the gas from coming out of the flask.
(c) If a thistle funnel is used, it must be dipped below the surface of the reacting chemicals. (1/2 mark)
5. Element X reacts with dilute acids but not with cold water. Element Y does not react with dilute acids. Element Z displaces element W from its oxide. W reacts with cold water. Name the element that is likely to have the largest atomic radius. (1 mark)
Element Z
Since elements with larger atomic radii tend to be more reactive due to weaker attraction between the nucleus and outermost electrons thus loses electrons easily
6. Starting with 2M sodium hydroxide solution, describe how you can prepare crystals of an acid salt formed when sodium hydroxide reacts with 2M sulphuric (VI) acid. (3 marks)
To 2M sodium hydroxide solution, add an equal volume of 2M sulphuric (VI) acid solution while stirring.
Heat the solution to saturation and allow it to cool for crystals to form.
Dry between filter papers.
7. Determine the molecular mass of the gas B which diffuses 1/2 times faster than oxygen. (1 mark)
(c) (i) Name one other reagent that can be used to identify F. (1 mark)
Acidified potassium manganate (VII)
(ii) State what would be observed if the reagent is used on F. (1 mark)
Purple colour of acidified potassium manganate (VII) changes to colourless.
11. Under suitable laboratory conditions, ethene can be converted to a compound with a general formula (−CH₂−CH₂−). Name one other compound of the category (−CH₂−CH₂−) which is: (i) Man-made. (1/2 mark)
Polyethene/polythene
(ii) Not man-made. (1/2 mark)
Rubber
12. Substances G and H are represented by the formulae ROH and RCOOH respectively. They belong to two different homologous series of organic compounds. If both G and H react with potassium metal: State the observation made when each of the samples G and H are reacted with sodium hydrogen carbonate. (1 mark)
G-No effervescence
H- Effervescence
13. Explain why dilute sulphuric acid is a stronger acid than concentrated sulphuric acid. (1 mark)
In dilute sulphuric (VI) acid, the acid dissociates completely forming higher number of hydrogen ions.
14. Explain why particles collide and products are formed during a chemical reaction. (1 mark) The particles may not have the necessary activation energy.
The particles may collide in the wrong orientation
15. Explain what would happen to the position of equilibrium in the following reaction if the pressure is increased. (2 marks) CaCO₃(s) ⇌ CaO(s) + CO₂(g) ΔH = +ve kJ mol⁻¹
The position of the equilibrium shifts to the left.
This is because increase in pressure favours the direction with fewer numbers of gaseous
16. The graph below shows the variation of volume of hydrogen with time when excess magnesium was added to 100 cm³ of 1.0 M sulphuric acid at room temperature.
Rate of Reaction16. Sketch on the same axes of the graph in the figure and label the graph that would be obtained if the following were used at room temperature: (1 mark) (i) 100 cm³ of 0.5 M sulphuric acid. (ii) 200 cm³ of 2.0 M sulphuric acid.
Sketch - The graph for 100 cm³ of 0.5 M sulphuric acid would show a slower reaction rate and lower final volume of hydrogen gas compared to the original graph. - The graph for 200 cm³ of 2.0 M sulphuric acid would show a faster reaction rate and higher final volume of hydrogen gas compared to the original graph.
SECTION B
17. Carbon burns in oxygen according to the following equation: \[ \text{C(s) + O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \quad \Delta H = -393 \, \text{kJ/mol} \]
(a) Calculate the: (i) Amount of heat evolved when 5.6 g carbon is burnt completely in oxygen. (1 mark) - Molar mass of carbon (C) = 12 g/mol - Moles of carbon = \( \frac{5.6 \, \text{g}}{12 \, \text{g/mol}} = 0.467 \, \text{mol} \) - Heat evolved = \( 0.467 \, \text{mol} \times (-393 \, \text{kJ/mol}) = -183.5 \, \text{kJ} \)
(ii) Volume of oxygen at s.t.p that would be required to produce 78.6 kJ of heat. (1 mole of gas occupies 22.4 dm³ at s.t.p) (1 mark) - Heat per mole of CO₂ = 393 kJ - Moles of CO₂ for 78.6 kJ = \( \frac{78.6 \, \text{kJ}}{393 \, \text{kJ/mol}} = 0.2 \, \text{mol} \) - Volume of oxygen required = \( 0.2 \, \text{mol} \times 22.4 \, \text{dm³/mol} = 4.48 \, \text{dm³} \)
(b) Calculate the enthalpy of hydration of bromide ions given the following: - Enthalpy of hydration of magnesium ions = -1,891 kJ/mol - Lattice energy of MgBr₂ = -2,421 kJ/mol - Enthalpy of solution of MgBr₂ = -186 kJ/mol (1 mark)
The arrow for the lattice energy should be pointing in the opposite direction since the reaction is exothermic. Being exothermic means that it is enthalpy of lattice formation and not of lattice dissociation. Here is the new answer though the examiner should take his or her time to look into this question again because authorities point to the fact that lattice dissociation energy should be used always ;
(c) Explain why the molar heat of neutralization of KOH and ethanoic acid of equal volume and molarity would be less than the value obtained in the molar heat of neutralization of KOH and sulphuric (VI) acid. (1 mark)
Some of the heat energy generated is used to dissociate the acid molecules before neutralization occurs.
18. (a) A piece of iron wire of mass 2.225 g was put into a conical flask containing dilute sulphuric (VI) acid. The flask was fitted with a bung carrying a Bunsen valve to allow the hydrogen gas generated to escape but prevent air from entering. The mixture was warmed. When the effervescence stopped, the solution was cooled to room temperature and made up to 250 cm³ in a graduated flask. 25 cm³ of the solution were acidified and titrated against a 0.0185 mol/dm³ solution of potassium dichromate (VI); the volume required was 31 cm³. Calculate the percentage of iron in the iron wire. (1 mark)
+ 2 Cr³⁺ + 7 H₂O - Moles of Fe = \( 6 \times 5.735 \times 10^{-4} = 3.441 \times 10^{-3} \, \text{mol} \) - Total moles of Fe in 250 cm³ solution = \( 3.441 \times 10^{-3} \times 10 = 3.441 \times 10^{-2} \, \text{mol} \) - Mass of Fe = \( 3.441 \times 10^{-2} \, \text{mol} \times 55.85 \, \text{g/mol} = 1.92 \, \text{g} \) - Percentage of Fe in wire = \( \frac{1.92 \, \text{g}}{2.225 \, \text{g}} \times 100\% = 86.29\% \)
(b) A sample containing ammonium sulphate was warmed with 250 cm³ of 0.8 mol/dm³ sodium hydroxide solution. After the evolution of ammonia had ceased, the solution was neutralized by 85 cm³ of hydrochloric acid of concentration 0.5 mol/dm³. What mass of ammonium sulphate did the sample contain? (1 mark)
19. (a) 5.125 g of washing soda crystals are dissolved and made up to 250 cm³ of solution. A 25 cm³ portion requires 35.8 cm³ of 0.05 mol/dm³ sulphuric (VI) acid for neutralization. Calculate the percentage of sodium carbonate in the crystals. (1 mark) - Moles of H₂SO₄ = \( 0.05 \, \text{mol/dm}³ \) \(\times 0.0358 \, \) \(\text{dm}³\) = \1.79 \times 10^{-3} \, \text{mol} \) - Reaction: Na₂CO₃ + H₂SO₄ → Na₂SO₄ + CO₂ + H₂O - Moles of Na₂CO₃ = \( 1.79 \times 10^{-3} \, \) \(\text{mol} \) - Total moles of Na₂CO₃ in 250 cm³ = \( 1.79 \times 10^{-3} \, \) \(\text{mol} \times 10 \) \(= 1.79 \times 10^{-2} \, \) \(\text{mol} \) - Mass of Na₂CO₃ = \( 1.79 \times 10^{-2}\) \(\, \text{mol}\) \( \times 106 \, \) \(\text{g/mol} = 1.897 \, \text{g} \) - Percentage of Na₂CO₃ in crystals = \( \frac{1.897 \, \) \(\text{g}}{5.125 \, \text{g}}\) \(\times 100\% = 37.01\% \)
(b) From 23 g of ethanol, 36 g of ethyl ethanoate are obtained by esterification with ethanoic acid in the presence of concentrated sulphuric (VI) acid. What is the percentage yield of the reaction? (1 mark)
\(4FeS_{2(s)}\) + \(11O_{2(g)} \rightarrow\) \(2Fe_{2}O_{3(s) }\)+ \(8SO_{2(g)} \) (b) Why is step II necessary? (1 mark) (c) Name the: (i) Drying agent in step III. (1 mark)
Concentrated sulphuric (VI) acid
(ii) Catalyst in step IV. (1/2 mark)
(d) Describe the process that takes place in step V in order to produce sulphuric acid. (1 mark) (e) Sulphur(IV) oxide combines with air to form sulphur(VI) oxide. (i) Write an equation for the reaction. (1 mark)
(ii) State the conditions for the maximum yield of sulphur(VI) oxide. (1 mark)
21. In an experiment, dry chlorine gas was reacted with aluminium as shown in Figure 2 below.
(a) Name substance R. (1/2 mark) Aluminium chloride (b) Write an equation for the reaction that took place in the combustion tube. (1 mark) \(2Al(s) + 3Cl_{2(g)} \) \(\rightarrow\) \(2AlCl_{3(s)}\) (c) State the function of the anhydrous calcium chloride in the setup above. (1 mark)
Prevent entry of moisture.
(d) Name another substance that can be used instead of anhydrous calcium chloride. (1/2 mark)
Calcium oxide.
(e) Why is the substance named above in (d) more suitable than anhydrous calcium chloride? (1/2 mark)
It prevents emission of chlorine gas/pollution of chlorine
(f) Name another metal that can be used instead of aluminium. (1/2 mark)
Iron metal.
20. The following Figure 1 shows the steps in the manufacture of sulphuric acid by the contact process:
(a) Write an equation for the reaction that takes place in step I. (1 mark) \( \text{S}(s)\) + \( \text{O}_2(g)\) \( \rightarrow \) \(\text{SO}_2(g) \)
(b) Why is step II necessary? (1 mark)
To remove the impurities which poison/reduces surface area of the catalyst
. (c) Name the: (i) Drying agent in step III. (1 mark)
Concentrated sulphuric (VI) acid
(ii) Catalyst in step IV. (1/2 mark)
Vanadium(V) oxide (V₂O₅) or Platinum
(d) Describe the process that takes place in step V in order to produce sulphuric acid. (1 mark)
Sulphur (VI) oxide gas is dissolved in concentrated sulphuric (VI) acid to form oleum.
The oleum is diluted with water to form concentrated sulphuric (VI) acid
In step V, Sulphur (VI) oxide (SO₃) is absorbed in concentrated sulphuric acid (H₂SO₄) to form oleum (H₂S₂O₇), which is then diluted with water to produce sulphuric (VI) acid (H₂SO₄).
(e) Sulphur (IV) oxide combines with air to form sulphur (VI) oxide. (i) Write an equation for the reaction. (1 mark) \( 2\text{SO}_2(g)\) +\( \text{O}_2(g)\) \(\rightarrow \) \(2\text{SO}_3(g) \)
(ii) State the conditions for the maximum yield of sulphur(VI) oxide. (1 mark)
High temperature \(450-500^{o}C \)
Presence of a catalyst: Vanadium (V) oxide/ Platinum
Pressure 2-3atm
The maximum yield of sulphur(VI) oxide is achieved at a temperature of about 450°C, a pressure of 1-2 atm, and in the presence of a vanadium(V) oxide (V₂O₅) catalyst.
21. In an experiment, dry chlorine gas was reacted with aluminum as shown in Figure 2 below.
(a) Name substance R. (1/2 mark)
Aluminium chloride Aluminum chloride (AlCl₃)
(b) Write an equation for the reaction that took place in the combustion tube. (1 mark) \(2\text{Al}(s)\) + \(3\text{Cl}_2(g)\) \(\rightarrow \) \(2\text{AlCl}_3(s) \)
(c) State the function of the anhydrous calcium chloride in the setup above. (1 mark)
Prevent entry of moisture.
Anhydrous calcium chloride acts as a drying agent to remove moisture from the chlorine gas.
(d) Name another substance that can be used instead of anhydrous calcium chloride. (1/2 mark)
Calcium oxide.
(e) Why is the substance named above in (d) more suitable than anhydrous calcium chloride? (1/2 mark)
It prevents emission of chlorine gas/pollution of chlorine.
(f) Name another metal that can be used instead of aluminum. (1/2 mark)
Iron (Fe)/Iron metal.
(g) What property makes substance R to be collected in the flask? (1/2 mark)
It sublimes
Aluminum chloride (R) is collected in the flask because it sublimes at high temperatures.
(h) Write the equation for the reaction between excess phosphorus and chlorine. (1 mark)
(a) (i) Name two substances that are liquid at room temperature. (1/2 mark)
K and N
(ii) Which of the two is more volatile? (1/2 mark)
S
(b) Which substances would dissolve in water and could be separated from the solution by:
(i) Fractional distillation. (1/2 mark)
K
(ii) By evaporation of the water? (1/2 mark)
J
(c) Which of the substances: (i) Has the structure consisting of ions? (1/2 mark)
J
(ii) Is a metal? (1/2 mark)
L
(iii) Is a liquid which would form a separate layer with water? (1/2 mark)
N
(iv) Would the water be above or below? (1/2 mark) Water would be above.
Above
(d) Which substance is a gas which: (i) Would not be collected efficiently over water? (1/2 mark) M
(ii) Would be collected efficiently over water? (1/2 mark) S
SECTION C
23. You are given solid T (aluminum sulfate). Carry out the tests below, writing observations and inferences accordingly. Take a spatula end full of T and put it in a boiling tube. Add about 10 cm³ of water and shake. Keep the mixture for the tests below.
(a) To about 2 cm³ of solution T, add about 5 drops of Nitric Acid (HNO₃(aq)) followed by 2 drops of Barium nitrate.
Observations; A white precipitate forms.
No effervescence
Inference: Presence of sulphate ions (SO₄²⁻) in the solution.
(b) To about 2 cm³ of solution of T, add Sodium Hydroxide (2M NaOH) dropwise until in excess.
Observations: A white precipitate forms, which dissolves upon adding excess NaOH.
White precipitate soluble in excess
Inference:
\(Al^{3+} , Zn^{2+}\) present
(c) To about 2 cm³ of solution T, add Ammonia solution (2M NH₃(aq)) dropwise until in excess.
Observations:
White precipitate insoluble in excess A white precipitate forms, which remains insoluble in excess ammonia.
Inference:
\(Al^{3+}\) present Presence of aluminum ions (Al³⁺) in the solution.
