WAEC SSCE - Chemistry - 1990 (Essay)
Question 1 Report
(a) Give two reasons why carbon(IV) oxide is used for extinguishing fire
(b) Explain briefly the water softening action of cation exchane resins.
(a) Why carbon(IV) oxide is used to extinguish fire
- It is denser (heavier) than air, so it settles over the burning material and forms a blanket that cuts off the supply of oxygen (air) needed for burning.
- It does not support combustion and is itself non-combustible, so once it covers the flame the fire goes out.
(b) Water-softening action of a cation-exchange resin
A cation-exchange resin carries loosely held sodium ions, \(Na^+\) (or \(H^+\)) on its surface. When hard water containing the \(Ca^{2+}\) and \(Mg^{2+}\) ions that cause hardness passes through the resin, the resin exchanges its \(Na^+\) ions for the \(Ca^{2+}\) and \(Mg^{2+}\) ions, which are held by the resin: \[2NaR + Ca^{2+} \to CaR_2 + 2Na^+\] The water leaving the resin now contains only \(Na^+\) ions, which do not cause hardness, so the water is softened. The exhausted resin is regenerated by running concentrated brine (sodium chloride solution) through it.
Answer Details
(a) Why carbon(IV) oxide is used to extinguish fire
- It is denser (heavier) than air, so it settles over the burning material and forms a blanket that cuts off the supply of oxygen (air) needed for burning.
- It does not support combustion and is itself non-combustible, so once it covers the flame the fire goes out.
(b) Water-softening action of a cation-exchange resin
A cation-exchange resin carries loosely held sodium ions, \(Na^+\) (or \(H^+\)) on its surface. When hard water containing the \(Ca^{2+}\) and \(Mg^{2+}\) ions that cause hardness passes through the resin, the resin exchanges its \(Na^+\) ions for the \(Ca^{2+}\) and \(Mg^{2+}\) ions, which are held by the resin: \[2NaR + Ca^{2+} \to CaR_2 + 2Na^+\] The water leaving the resin now contains only \(Na^+\) ions, which do not cause hardness, so the water is softened. The exhausted resin is regenerated by running concentrated brine (sodium chloride solution) through it.
Question 2 Report
Benzene contains six carbon atoms and six hydrogen atoms.
(a)(i) Draw two stable structures of benzene to show how these atoms are arranged.
(ii) What is the concept behind these structures?
(b) Give (i) two uses of benzene. (ii) one industrial source of benzene.
(a)(i) Kekulé structures of benzene, C6H6
(a)(ii) The two structures illustrate resonance. The actual benzene molecule is a resonance hybrid in which the c0-electrons are delocalized over all six carbon atoms. Thus, all the carbon-carbon bonds are identical and are intermediate between a single and a double bond.
(b)(i) Uses of benzene
- It is used in the manufacture of chemicals such as disinfectants.
- It is used in the manufacture of perfumes.
(b)(ii) Industrial source of benzene
Benzene is obtained from coal tar, produced during the destructive distillation of coal.
Answer Details
(a)(i) Kekulé structures of benzene, C6H6
(a)(ii) The two structures illustrate resonance. The actual benzene molecule is a resonance hybrid in which the c0-electrons are delocalized over all six carbon atoms. Thus, all the carbon-carbon bonds are identical and are intermediate between a single and a double bond.
(b)(i) Uses of benzene
- It is used in the manufacture of chemicals such as disinfectants.
- It is used in the manufacture of perfumes.
(b)(ii) Industrial source of benzene
Benzene is obtained from coal tar, produced during the destructive distillation of coal.
Question 3 Report
(a) Write the electronic configuration of an element with atomic number 15, indicating the distribution of electrons in the energy sub-levels
(b) Give the formula and the colour of the complex formed between ammonia and copper (II) ions
(a) Electronic configuration of the element with atomic number 15 (phosphorus)
Filling the sub-levels in order of increasing energy:
\[1s^2\,2s^2\,2p^6\,3s^2\,3p^3\]That is, \(2, 8, 5\) in the main shells. The three 3p electrons occupy the three p-orbitals singly (Hund's rule): \(3p_x^{1}\,3p_y^{1}\,3p_z^{1}\).
(b) Complex formed between ammonia and copper(II) ions
Formula: \([Cu(NH_3)_4]^{2+}\) (the tetraamminecopper(II) ion).
Colour: deep (royal) blue.
Answer Details
(a) Electronic configuration of the element with atomic number 15 (phosphorus)
Filling the sub-levels in order of increasing energy:
\[1s^2\,2s^2\,2p^6\,3s^2\,3p^3\]That is, \(2, 8, 5\) in the main shells. The three 3p electrons occupy the three p-orbitals singly (Hund's rule): \(3p_x^{1}\,3p_y^{1}\,3p_z^{1}\).
(b) Complex formed between ammonia and copper(II) ions
Formula: \([Cu(NH_3)_4]^{2+}\) (the tetraamminecopper(II) ion).
Colour: deep (royal) blue.
Question 4 Report
(a)(i) Define oxidation in terms of electron transfer.
(ii) Write balanced equations for the half reactions for the following changes in acidic solution: Mn0\(^-_4\) + Fe\(^{2+}\) —> Mn\(^{2+}\) + Fe\(^{3+}\)
(b)(i) Distinguish between an electrolytic celI and an electrochemical cell.
(ii) Sketch a cell for the electrolysis of molten magnesium chloride. Lable the anode and the cathode and indicate the direction of electron flow. Give the electrode reactions.
(iii) Give one reason why a platinum anode is not suitable for the eloctrolysis in (b)(i) above.
(c) Calculate the mass of lead that would be deposited from a solution of lead (II) trioxonitrate by the same quantity of electrically depositing 1.35g of copper. (Cu = 63.5, Pb = 207)
(a)(i) Oxidation in terms of electron transfer
Oxidation is the process in which an atom, ion or molecule loses electron(s).
Example:
\[Fe^{2+} \rightarrow Fe^{3+} + e^-\]
(a)(ii) Balanced half-reactions in acidic solution
Oxidation half-reaction:
\[Fe^{2+} \rightarrow Fe^{3+} + e^-\]
Reduction half-reaction:
\[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\]
(b)(i) Difference between an electrolytic cell and an electrochemical cell
| Electrolytic cell | Electrochemical cell (voltaic/galvanic cell) |
|---|---|
| Uses electrical energy from an external source to cause a non-spontaneous chemical reaction. | Uses a spontaneous chemical reaction to produce electrical energy. |
| Converts electrical energy to chemical energy. | Converts chemical energy to electrical energy. |
| The cathode is negative and the anode is positive. | The anode is negative and the cathode is positive. |
(b)(ii) Electrolysis of molten magnesium chloride
Two inert electrodes, such as graphite electrodes, are dipped into molten magnesium chloride and connected to a direct-current source.
DC source (battery)
(+) (−)
|
Answer Details
(a)(i) Oxidation in terms of electron transfer
Oxidation is the process in which an atom, ion or molecule loses electron(s).
Example:
\[Fe^{2+} \rightarrow Fe^{3+} + e^-\]
(a)(ii) Balanced half-reactions in acidic solution
Oxidation half-reaction:
\[Fe^{2+} \rightarrow Fe^{3+} + e^-\]
Reduction half-reaction:
\[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\]
(b)(i) Difference between an electrolytic cell and an electrochemical cell
| Electrolytic cell | Electrochemical cell (voltaic/galvanic cell) |
|---|---|
| Uses electrical energy from an external source to cause a non-spontaneous chemical reaction. | Uses a spontaneous chemical reaction to produce electrical energy. |
| Converts electrical energy to chemical energy. | Converts chemical energy to electrical energy. |
| The cathode is negative and the anode is positive. | The anode is negative and the cathode is positive. |
(b)(ii) Electrolysis of molten magnesium chloride
Two inert electrodes, such as graphite electrodes, are dipped into molten magnesium chloride and connected to a direct-current source.
DC source (battery)
(+) (−)
|
Question 5 Report
Consider the reaction represented by the equation: 2SO\(_{2(g)}\) + O\(_{2(g)}\) 
2SO\(_{3(g)}\). \(\Delta\)H = 188KJ.
(a) Write an expression for the equilibrium constant
(b) Sketch an energy diagram for the forward reaction, showing the profile for the catalyzed and non-catalyzed systems.
(c) state the reason, the effect of the following on the position of equilibrium of the system:
(i) increase in temperature
(ii) increase in pressure;
(iii) removal of some of the SO\(_3\) produced;
(iv) presence of V\(_2\)O\(_5\)
(d)(i) Write equations to show how the sulphur(VI) oxide is converted to tetraoxosulphate(VI) acid in the contact process
(ii) Give two uses of tetraoxosulphate(VI) acid.
Important correction: the equation gives \(\Delta H=+188\text{ kJ}\), which means that the forward reaction is endothermic. Therefore, the energy diagram and temperature effect below follow the equation as written. In the Contact Process, this reaction is more commonly written with \(\Delta H=-188\text{ kJ}\), meaning it is exothermic. If the intended sign was negative, the temperature effect would be the opposite.
(a) Equilibrium constant expression
For
\[2SO_{2(g)}+O_{2(g)}\rightleftharpoons2SO_{3(g)}\]
the concentration equilibrium constant is:
\[K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}\]
The powers are taken from the balanced equation. The formula must contain \(SO_2\), not \(SO\).
(b) Energy profile for the forward reaction
Because \(\Delta H\) is positive as stated, the products are at a higher energy level than the reactants. A catalyst, \(V_2O_5\), provides an alternative pathway with a lower activation energy. It does not change \(\Delta H\), the reactant energy, or the product energy.
(c) Effects on the position of equilibrium
- Increase in temperature: equilibrium shifts to the right, producing more \(SO_3\). The forward reaction is endothermic as written, so it absorbs heat; increasing temperature favours the heat-absorbing direction.
If the intended value was \(\Delta H=-188\text{ kJ}\), the forward reaction would be exothermic and increasing temperature would shift equilibrium to the left, decreasing the yield of \(SO_3\). - Increase in pressure: equilibrium shifts to the right. There are three moles of gaseous reactants, \(2SO_2+O_2\), but only two moles of gaseous product, \(2SO_3\). Higher pressure favours the side with fewer moles of gas.
- Removal of some \(SO_3\): equilibrium shifts to the right. Removing a product lowers its concentration, so the forward reaction occurs to replace some of the removed \(SO_3\).
- Presence of \(V_2O_5\): there is no change in the position of equilibrium. \(V_2O_5\) is a catalyst: it increases the rates of both forward and backward reactions and allows equilibrium to be reached faster, but it does not change the equilibrium composition.
(d)(i) Conversion of sulphur(VI) oxide to tetraoxosulphate(VI) acid in the Contact Process
Sulphur(VI) oxide, \(SO_3\), is absorbed in concentrated tetraoxosulphate(VI) acid to form oleum. Oleum is then diluted with water:
\[SO_{3(g)}+H_2SO_{4(l)}\rightarrow H_2S_2O_{7(l)}\]
\[H_2S_2O_{7(l)}+H_2O_{(l)}\rightarrow2H_2SO_{4(aq)}\]
\(H_2S_2O_7\) is oleum. Direct addition of \(SO_3\) to water is avoided because it forms a fine mist of acid.
(d)(ii) Two uses of tetraoxosulphate(VI) acid, \(H_2SO_4\)
- Manufacture of fertilisers, for example superphosphate and ammonium sulfate.
- Manufacture of detergents, paints, and pigments.
Examination reminder: a catalyst changes the activation energy and reaction rate, but it does not change \(\Delta H\), \(K_c\), or the position of equilibrium.
Answer Details
Important correction: the equation gives \(\Delta H=+188\text{ kJ}\), which means that the forward reaction is endothermic. Therefore, the energy diagram and temperature effect below follow the equation as written. In the Contact Process, this reaction is more commonly written with \(\Delta H=-188\text{ kJ}\), meaning it is exothermic. If the intended sign was negative, the temperature effect would be the opposite.
(a) Equilibrium constant expression
For
\[2SO_{2(g)}+O_{2(g)}\rightleftharpoons2SO_{3(g)}\]
the concentration equilibrium constant is:
\[K_c=\frac{[SO_3]^2}{[SO_2]^2[O_2]}\]
The powers are taken from the balanced equation. The formula must contain \(SO_2\), not \(SO\).
(b) Energy profile for the forward reaction
Because \(\Delta H\) is positive as stated, the products are at a higher energy level than the reactants. A catalyst, \(V_2O_5\), provides an alternative pathway with a lower activation energy. It does not change \(\Delta H\), the reactant energy, or the product energy.
(c) Effects on the position of equilibrium
- Increase in temperature: equilibrium shifts to the right, producing more \(SO_3\). The forward reaction is endothermic as written, so it absorbs heat; increasing temperature favours the heat-absorbing direction.
If the intended value was \(\Delta H=-188\text{ kJ}\), the forward reaction would be exothermic and increasing temperature would shift equilibrium to the left, decreasing the yield of \(SO_3\). - Increase in pressure: equilibrium shifts to the right. There are three moles of gaseous reactants, \(2SO_2+O_2\), but only two moles of gaseous product, \(2SO_3\). Higher pressure favours the side with fewer moles of gas.
- Removal of some \(SO_3\): equilibrium shifts to the right. Removing a product lowers its concentration, so the forward reaction occurs to replace some of the removed \(SO_3\).
- Presence of \(V_2O_5\): there is no change in the position of equilibrium. \(V_2O_5\) is a catalyst: it increases the rates of both forward and backward reactions and allows equilibrium to be reached faster, but it does not change the equilibrium composition.
(d)(i) Conversion of sulphur(VI) oxide to tetraoxosulphate(VI) acid in the Contact Process
Sulphur(VI) oxide, \(SO_3\), is absorbed in concentrated tetraoxosulphate(VI) acid to form oleum. Oleum is then diluted with water:
\[SO_{3(g)}+H_2SO_{4(l)}\rightarrow H_2S_2O_{7(l)}\]
\[H_2S_2O_{7(l)}+H_2O_{(l)}\rightarrow2H_2SO_{4(aq)}\]
\(H_2S_2O_7\) is oleum. Direct addition of \(SO_3\) to water is avoided because it forms a fine mist of acid.
(d)(ii) Two uses of tetraoxosulphate(VI) acid, \(H_2SO_4\)
- Manufacture of fertilisers, for example superphosphate and ammonium sulfate.
- Manufacture of detergents, paints, and pigments.
Examination reminder: a catalyst changes the activation energy and reaction rate, but it does not change \(\Delta H\), \(K_c\), or the position of equilibrium.
Question 6 Report
(a)(i) List two physical properties used as criteria for purity of substances
(ii) describe how you would prepare a pure, dry sample of sodium chloride crystals by a neutralization reaction, using bench reagents.
(iii) Give two other general methods for preparing soluble salts.
(b) Explain the following observations:
(i) a sheet of iron placed in dilute copper (II) tetraoxosulphate (VI) solution reddish brown;
(ii) the white gelatinous precipitate formed when a few drops of sodium hydroxide solution are added to a solution of aluminium salt dissolves in excess alkali;
(iii) the pale green prepared iron(II) chloride solution changes to brown on bubbling chlorine gas through it.
(iv) Write a balanced equation for the reaction of dilute hydrochloric acid with marble. List two industrial process in which limestone is used as a raw material.
(a)(i) Two physical properties used as criteria for purity
- Melting point (a pure solid melts sharply at a fixed temperature).
- Boiling point (a pure liquid boils at a fixed temperature). Impurities lower the melting point and raise the boiling point, and both then occur over a range.
(a)(ii) Preparation of pure, dry sodium chloride by neutralization
- Using a pipette, place a known volume of dilute hydrochloric acid in a conical flask and add a few drops of an indicator (e.g. methyl orange).
- Titrate with sodium hydroxide solution from a burette until the indicator just changes colour (neutral point). Record the volume of alkali used.
- Repeat the experiment using the same volumes of acid and alkali but without the indicator, so that the salt is not contaminated by the dye.
- Evaporate the resulting neutral solution over a water bath to the point of crystallization (until a saturated solution is obtained).
- Allow to cool so that crystals of sodium chloride form; filter off the crystals and dry them between filter papers.
(a)(iii) Two other general methods of preparing soluble salts
- Reacting a dilute acid with excess insoluble base or metal oxide/carbonate, then filtering off the excess and crystallizing.
- Reacting a dilute acid with excess reactive metal, then filtering off the excess metal and crystallizing.
(b) Explanations
(i) Iron is more reactive (higher in the electrochemical series) than copper, so it displaces copper from copper(II) tetraoxosulphate(VI) solution. The reddish-brown copper metal is deposited on the iron sheet:
\[Fe_{(s)} + CuSO_{4(aq)} \to FeSO_{4(aq)} + Cu_{(s)}\](ii) The white gelatinous precipitate is aluminium hydroxide, \(Al(OH)_3\). Because aluminium hydroxide is amphoteric, it dissolves in excess sodium hydroxide to form a soluble aluminate (tetrahydroxoaluminate):
\[Al(OH)_3 + NaOH \to NaAl(OH)_4\](iii) Chlorine is a stronger oxidizing agent and oxidizes the pale-green iron(II) ions to yellow-brown iron(III) ions:
\[2FeCl_{2(aq)} + Cl_{2(g)} \to 2FeCl_{3(aq)}\](iv) Reaction of dilute hydrochloric acid with marble (calcium trioxocarbonate(IV)):
\[CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}\]Two industrial processes in which limestone is used as a raw material: manufacture of cement, and the extraction of iron in the blast furnace (as a flux). (Also acceptable: manufacture of glass and manufacture of lime.)
Answer Details
(a)(i) Two physical properties used as criteria for purity
- Melting point (a pure solid melts sharply at a fixed temperature).
- Boiling point (a pure liquid boils at a fixed temperature). Impurities lower the melting point and raise the boiling point, and both then occur over a range.
(a)(ii) Preparation of pure, dry sodium chloride by neutralization
- Using a pipette, place a known volume of dilute hydrochloric acid in a conical flask and add a few drops of an indicator (e.g. methyl orange).
- Titrate with sodium hydroxide solution from a burette until the indicator just changes colour (neutral point). Record the volume of alkali used.
- Repeat the experiment using the same volumes of acid and alkali but without the indicator, so that the salt is not contaminated by the dye.
- Evaporate the resulting neutral solution over a water bath to the point of crystallization (until a saturated solution is obtained).
- Allow to cool so that crystals of sodium chloride form; filter off the crystals and dry them between filter papers.
(a)(iii) Two other general methods of preparing soluble salts
- Reacting a dilute acid with excess insoluble base or metal oxide/carbonate, then filtering off the excess and crystallizing.
- Reacting a dilute acid with excess reactive metal, then filtering off the excess metal and crystallizing.
(b) Explanations
(i) Iron is more reactive (higher in the electrochemical series) than copper, so it displaces copper from copper(II) tetraoxosulphate(VI) solution. The reddish-brown copper metal is deposited on the iron sheet:
\[Fe_{(s)} + CuSO_{4(aq)} \to FeSO_{4(aq)} + Cu_{(s)}\](ii) The white gelatinous precipitate is aluminium hydroxide, \(Al(OH)_3\). Because aluminium hydroxide is amphoteric, it dissolves in excess sodium hydroxide to form a soluble aluminate (tetrahydroxoaluminate):
\[Al(OH)_3 + NaOH \to NaAl(OH)_4\](iii) Chlorine is a stronger oxidizing agent and oxidizes the pale-green iron(II) ions to yellow-brown iron(III) ions:
\[2FeCl_{2(aq)} + Cl_{2(g)} \to 2FeCl_{3(aq)}\](iv) Reaction of dilute hydrochloric acid with marble (calcium trioxocarbonate(IV)):
\[CaCO_{3(s)} + 2HCl_{(aq)} \to CaCl_{2(aq)} + H_2O_{(l)} + CO_{2(g)}\]Two industrial processes in which limestone is used as a raw material: manufacture of cement, and the extraction of iron in the blast furnace (as a flux). (Also acceptable: manufacture of glass and manufacture of lime.)
Question 7 Report
A hydrocarbon X which decolorizes bromine water but has no action on ammoniacal silver trioxonitrate (V) solution was found to have a molar mass of 58 g mol\(^{-1}\)
(a) Deduce the molecular formula of X. (H = 1, C = 12)
(b) Write the structures of two isomers of X.
Interpreting the clues
- X decolorises bromine water, so X is unsaturated (it contains a carbon-carbon double or triple bond and undergoes addition).
- X has no action on ammoniacal silver trioxonitrate(V) (Tollens' reagent gives no precipitate), so X is not a terminal alkyne (it has no \(\equiv C-H\) at the end of the chain). It is therefore an alkene, a diene, or an internal alkyne.
(a) Molecular formula
Test the unsaturated general formulae against the molar mass:
- Alkene \(C_nH_{2n}\): \(14n = 58 \Rightarrow n = 4.1\) (not a whole number).
- Alkyne / diene \(C_nH_{2n-2}\): \(14n - 2 = 58 \Rightarrow n = 4.3\) (not a whole number, and \(14n-2 = 54\) for \(n=4\)).
The only whole-number hydrocarbon fitting the unsaturated, non-terminal-alkyne description is \(C_4H_6\) (molar mass \(4\times12 + 6 = 54\ \text{g mol}^{-1}\)); the quoted value 58 appears to be a misprint for 54. Taking the consistent value:
\[\textbf{Molecular formula of X} = C_4H_6\](b) Two isomers of X (both unsaturated, neither a terminal alkyne)
- But-2-yne: \(CH_3-C\equiv C-CH_3\) (internal triple bond, so no reaction with ammoniacal \(AgNO_3\)).
- Buta-1,3-diene: \(CH_2=CH-CH=CH_2\) (two double bonds, decolorises bromine, no reaction with ammoniacal \(AgNO_3\)).
Answer Details
Interpreting the clues
- X decolorises bromine water, so X is unsaturated (it contains a carbon-carbon double or triple bond and undergoes addition).
- X has no action on ammoniacal silver trioxonitrate(V) (Tollens' reagent gives no precipitate), so X is not a terminal alkyne (it has no \(\equiv C-H\) at the end of the chain). It is therefore an alkene, a diene, or an internal alkyne.
(a) Molecular formula
Test the unsaturated general formulae against the molar mass:
- Alkene \(C_nH_{2n}\): \(14n = 58 \Rightarrow n = 4.1\) (not a whole number).
- Alkyne / diene \(C_nH_{2n-2}\): \(14n - 2 = 58 \Rightarrow n = 4.3\) (not a whole number, and \(14n-2 = 54\) for \(n=4\)).
The only whole-number hydrocarbon fitting the unsaturated, non-terminal-alkyne description is \(C_4H_6\) (molar mass \(4\times12 + 6 = 54\ \text{g mol}^{-1}\)); the quoted value 58 appears to be a misprint for 54. Taking the consistent value:
\[\textbf{Molecular formula of X} = C_4H_6\](b) Two isomers of X (both unsaturated, neither a terminal alkyne)
- But-2-yne: \(CH_3-C\equiv C-CH_3\) (internal triple bond, so no reaction with ammoniacal \(AgNO_3\)).
- Buta-1,3-diene: \(CH_2=CH-CH=CH_2\) (two double bonds, decolorises bromine, no reaction with ammoniacal \(AgNO_3\)).
Question 8 Report
(a) Give the formula indicating the relationship between entropy, free energy and enthalpy changes of a system.
(b) For each of the following, state whether entropy change is positive, negative or zero.
(i) H\(_2\)O\(_{(g)}\) -> H\(_2\)O\(_{(g)}\)
(ii) Cl\(_{2(g)}\) ---> 2CI\(_{(g)}\)
(iii) HCI\(_{(g)}\) -> HCl\(_{(g)}\)
(a) Relationship between free energy, enthalpy and entropy changes
The Gibbs-Helmholtz relationship is:
\[\Delta G = \Delta H - T\Delta S\]where \(\Delta G\) is the free energy change, \(\Delta H\) the enthalpy change, \(\Delta S\) the entropy change and \(T\) the absolute (Kelvin) temperature. A reaction is spontaneous (feasible) when \(\Delta G\) is negative.
(b) Sign of the entropy change
Entropy is a measure of disorder. Disorder increases when the number of gaseous molecules increases; it is unchanged when the state and number of particles are unchanged.
- (i) \(H_2O_{(g)} \to H_2O_{(g)}\): the substance stays in the same physical state with the same number of gaseous molecules, so the entropy change is zero.
- (ii) \(Cl_{2(g)} \to 2Cl_{(g)}\): one mole of gas becomes two moles of gas, so disorder increases and the entropy change is positive.
- (iii) \(HCl_{(g)} \to HCl_{(g)}\): no change in state or number of molecules, so the entropy change is zero.
Answer Details
(a) Relationship between free energy, enthalpy and entropy changes
The Gibbs-Helmholtz relationship is:
\[\Delta G = \Delta H - T\Delta S\]where \(\Delta G\) is the free energy change, \(\Delta H\) the enthalpy change, \(\Delta S\) the entropy change and \(T\) the absolute (Kelvin) temperature. A reaction is spontaneous (feasible) when \(\Delta G\) is negative.
(b) Sign of the entropy change
Entropy is a measure of disorder. Disorder increases when the number of gaseous molecules increases; it is unchanged when the state and number of particles are unchanged.
- (i) \(H_2O_{(g)} \to H_2O_{(g)}\): the substance stays in the same physical state with the same number of gaseous molecules, so the entropy change is zero.
- (ii) \(Cl_{2(g)} \to 2Cl_{(g)}\): one mole of gas becomes two moles of gas, so disorder increases and the entropy change is positive.
- (iii) \(HCl_{(g)} \to HCl_{(g)}\): no change in state or number of molecules, so the entropy change is zero.
Question 9 Report
(a) Name the type of solid structure possessed by:
(i) diamond;
(ii) iodine;
(iii) sodium chloride.
(b) Give:
(i) one alloy of tin:
(ii) a common reducing agent which is a compound of tin.
(a) Type of solid structure
- (i) Diamond: giant covalent (atomic/macromolecular) structure. Each carbon atom is covalently bonded tetrahedrally to four other carbon atoms.
- (ii) Iodine: simple molecular (molecular crystal) structure. Discrete \(I_2\) molecules are held together in the lattice by weak van der Waals forces.
- (iii) Sodium chloride: giant ionic (ionic crystal) structure. \(Na^+\) and \(Cl^-\) ions are held in a regular lattice by strong electrostatic (ionic) forces.
(b)
- (i) One alloy of tin: solder (an alloy of tin and lead). Other acceptable answers are bronze (tin and copper) or pewter (tin, antimony and copper).
- (ii) A common reducing agent that is a compound of tin: tin(II) chloride, \(SnCl_2\). The \(Sn^{2+}\) ion is readily oxidized to \(Sn^{4+}\), so it acts as a reducing agent.
Answer Details
(a) Type of solid structure
- (i) Diamond: giant covalent (atomic/macromolecular) structure. Each carbon atom is covalently bonded tetrahedrally to four other carbon atoms.
- (ii) Iodine: simple molecular (molecular crystal) structure. Discrete \(I_2\) molecules are held together in the lattice by weak van der Waals forces.
- (iii) Sodium chloride: giant ionic (ionic crystal) structure. \(Na^+\) and \(Cl^-\) ions are held in a regular lattice by strong electrostatic (ionic) forces.
(b)
- (i) One alloy of tin: solder (an alloy of tin and lead). Other acceptable answers are bronze (tin and copper) or pewter (tin, antimony and copper).
- (ii) A common reducing agent that is a compound of tin: tin(II) chloride, \(SnCl_2\). The \(Sn^{2+}\) ion is readily oxidized to \(Sn^{4+}\), so it acts as a reducing agent.
Question 10 Report
(a)(i) Define the term addition polymerization
(ii) What type of organic compounds undergo addition polymerization
(iii) List two factors which affect the strength of polymers
(b) The diagram below shows some reaction pathways involving ethanol
(i) Write the name and structural of the organic product X
(ii) State the reagent for the conversation indicated as step A.
(iii) What type of reaction will ethanol undergo CH\(_3\)CH\(_2\)COOH during the process of conversation indicated as step B?
(c)(i) Write three balanced equation for the complete combustion of ethanol in :date the volume of oxygen required at s.t.p for the complete combustion of ethanol. (H = 1, C = 12, O = 16, molar volume of gases at s.t.p. = 22.4 dm\(^3\))
(d)( i} State two substances produced when coal is heated in the absence of air
(ii) What name is given to the process in (d)(i) above?
(iii) State the importance of the non-volatile residue of the process named in (d)(iii) to the iron and steel industry.
(a)(i) Addition polymerization. It is the process in which many small unsaturated molecules (monomers containing C=C double bonds) join together to form a single large molecule (the polymer) without the loss of any small molecule.
(a)(ii) Type of compounds that undergo it: unsaturated organic compounds, that is alkenes (and other molecules containing a carbon-to-carbon double bond).
(a)(iii) Two factors affecting the strength of polymers:
- the chain length (relative molecular mass) of the polymer;
- the degree of branching / cross-linking between the chains (and the strength of intermolecular forces between chains).
(b) Reaction pathways of ethanol. Reading the diagram: ethanol (C2H5OH) reacts with propanoic acid (CH3CH2COOH) and an H+ catalyst to give X at the top; step A converts ethanol to CH3COOH; step B converts ethanol to C2H4.
(b)(i) Product X. Ethanol and propanoic acid undergo esterification, so X is ethyl propanoate.
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]
Structural formula of X: CH3–CH2–C(=O)–O–CH2–CH3
(b)(ii) Reagent for step A (ethanol \(\rightarrow\) ethanoic acid): acidified potassium dichromate(VI), K2Cr2O7/H+ (acidified KMnO4 is also acceptable), used as an oxidising agent with heat under reflux.
(b)(iii) Type of reaction in step B (ethanol \(\rightarrow\) ethene): a dehydration (elimination) reaction, in which a molecule of water is removed.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow[180^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \]
(c) Complete combustion of ethanol.
\[ \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]
From the equation, 1 mole of ethanol requires 3 moles of oxygen. At s.t.p. the volume of oxygen needed to burn one mole of ethanol completely is:
\[ V(\text{O}_2) = 3 \times 22.4 = 67.2\ \text{dm}^3 \]
(For a given mass \(m\) of ethanol, molar mass \(=46\), the volume is \(\dfrac{m}{46}\times 3\times 22.4\ \text{dm}^3\).)
(d)(i) Two substances produced when coal is heated in the absence of air: coke (solid residue) and coal gas; coal tar and ammoniacal liquor are also produced.
(d)(ii) Name of the process: destructive distillation of coal.
(d)(iii) Importance of the non-volatile residue (coke) to the iron and steel industry: coke is used in the blast furnace both as the fuel that supplies heat and as the reducing agent that reduces iron(III) oxide to iron:
\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]
(the carbon monoxide reducer is itself generated from the burning coke).
Answer Details
(a)(i) Addition polymerization. It is the process in which many small unsaturated molecules (monomers containing C=C double bonds) join together to form a single large molecule (the polymer) without the loss of any small molecule.
(a)(ii) Type of compounds that undergo it: unsaturated organic compounds, that is alkenes (and other molecules containing a carbon-to-carbon double bond).
(a)(iii) Two factors affecting the strength of polymers:
- the chain length (relative molecular mass) of the polymer;
- the degree of branching / cross-linking between the chains (and the strength of intermolecular forces between chains).
(b) Reaction pathways of ethanol. Reading the diagram: ethanol (C2H5OH) reacts with propanoic acid (CH3CH2COOH) and an H+ catalyst to give X at the top; step A converts ethanol to CH3COOH; step B converts ethanol to C2H4.
(b)(i) Product X. Ethanol and propanoic acid undergo esterification, so X is ethyl propanoate.
\[ \text{CH}_3\text{CH}_2\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \]
Structural formula of X: CH3–CH2–C(=O)–O–CH2–CH3
(b)(ii) Reagent for step A (ethanol \(\rightarrow\) ethanoic acid): acidified potassium dichromate(VI), K2Cr2O7/H+ (acidified KMnO4 is also acceptable), used as an oxidising agent with heat under reflux.
(b)(iii) Type of reaction in step B (ethanol \(\rightarrow\) ethene): a dehydration (elimination) reaction, in which a molecule of water is removed.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow[180^{\circ}\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O} \]
(c) Complete combustion of ethanol.
\[ \text{C}_2\text{H}_5\text{OH} + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 3\text{H}_2\text{O} \]
From the equation, 1 mole of ethanol requires 3 moles of oxygen. At s.t.p. the volume of oxygen needed to burn one mole of ethanol completely is:
\[ V(\text{O}_2) = 3 \times 22.4 = 67.2\ \text{dm}^3 \]
(For a given mass \(m\) of ethanol, molar mass \(=46\), the volume is \(\dfrac{m}{46}\times 3\times 22.4\ \text{dm}^3\).)
(d)(i) Two substances produced when coal is heated in the absence of air: coke (solid residue) and coal gas; coal tar and ammoniacal liquor are also produced.
(d)(ii) Name of the process: destructive distillation of coal.
(d)(iii) Importance of the non-volatile residue (coke) to the iron and steel industry: coke is used in the blast furnace both as the fuel that supplies heat and as the reducing agent that reduces iron(III) oxide to iron:
\[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]
(the carbon monoxide reducer is itself generated from the burning coke).
Question 11 Report
(a) Give one reason why a collision between reactants may not produce new species.
(b) Explain, illustrating with appropriate equation(s), why an aqueous solution of aluminium chloride is acidic.
(a) Why a collision between reactants may not produce new species
For a collision to be effective (fruitful) the colliding particles must have energy equal to or greater than the activation energy and must collide with the correct orientation. A collision fails to form products if the particles collide with too little energy (below the activation energy) or with an unfavourable orientation, so they simply bounce apart unchanged.
(b) Why aqueous aluminium chloride is acidic
The aluminium ion, \(Al^{3+}\), is very small and highly charged, so it strongly attracts and polarises the water molecules bonded to it. This causes hydrolysis (cationic hydrolysis): a bonded water molecule releases a hydrogen ion, so the solution contains an excess of \(H^+\) ions and is acidic.
Using the hydrated ion:
\[[Al(H_2O)_6]^{3+} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H^+\]or, written simply:
\[AlCl_3 + 3H_2O \rightleftharpoons Al(OH)_3 + 3HCl\]The liberated \(H^+\) (hydrochloric acid) makes the solution acidic.
Answer Details
(a) Why a collision between reactants may not produce new species
For a collision to be effective (fruitful) the colliding particles must have energy equal to or greater than the activation energy and must collide with the correct orientation. A collision fails to form products if the particles collide with too little energy (below the activation energy) or with an unfavourable orientation, so they simply bounce apart unchanged.
(b) Why aqueous aluminium chloride is acidic
The aluminium ion, \(Al^{3+}\), is very small and highly charged, so it strongly attracts and polarises the water molecules bonded to it. This causes hydrolysis (cationic hydrolysis): a bonded water molecule releases a hydrogen ion, so the solution contains an excess of \(H^+\) ions and is acidic.
Using the hydrated ion:
\[[Al(H_2O)_6]^{3+} \rightleftharpoons [Al(H_2O)_5(OH)]^{2+} + H^+\]or, written simply:
\[AlCl_3 + 3H_2O \rightleftharpoons Al(OH)_3 + 3HCl\]The liberated \(H^+\) (hydrochloric acid) makes the solution acidic.
Question 12 Report
(a) Arrange the first three members of the halogen family in their increasing order of electronegativity. Give the reason for your answer.
(b) State and explain what happens when chlorine reacts with starch iodide paper.
(a) Increasing order of electronegativity of the first three halogens
The first three members of the halogen family are fluorine, chlorine and bromine. In increasing order of electronegativity:
\[Br < Cl < F\]Reason: Electronegativity is the tendency of an atom to attract the shared (bonding) pair of electrons towards itself. Down the group the atomic size (number of shells) increases, so the bonding electrons are farther from the nucleus and are more shielded from it. The effective nuclear attraction for the shared electrons therefore decreases down the group. Since fluorine is the smallest atom its nucleus attracts electrons most strongly, while bromine, being the largest of the three, attracts them least. Hence electronegativity increases from bromine to chlorine to fluorine.
(b) Reaction of chlorine with starch-iodide paper
Observation: the starch-iodide paper turns blue-black.
Explanation: Chlorine is a stronger oxidizing agent (more electronegative) than iodine, so it displaces/oxidizes the iodide ions in the paper to liberate free iodine:
\[Cl_{2(g)} + 2KI_{(s)} \to 2KCl_{(s)} + I_{2(s)}\]The liberated iodine then reacts with the starch present in the paper to form the characteristic blue-black starch-iodine complex, confirming the presence of chlorine.
Answer Details
(a) Increasing order of electronegativity of the first three halogens
The first three members of the halogen family are fluorine, chlorine and bromine. In increasing order of electronegativity:
\[Br < Cl < F\]Reason: Electronegativity is the tendency of an atom to attract the shared (bonding) pair of electrons towards itself. Down the group the atomic size (number of shells) increases, so the bonding electrons are farther from the nucleus and are more shielded from it. The effective nuclear attraction for the shared electrons therefore decreases down the group. Since fluorine is the smallest atom its nucleus attracts electrons most strongly, while bromine, being the largest of the three, attracts them least. Hence electronegativity increases from bromine to chlorine to fluorine.
(b) Reaction of chlorine with starch-iodide paper
Observation: the starch-iodide paper turns blue-black.
Explanation: Chlorine is a stronger oxidizing agent (more electronegative) than iodine, so it displaces/oxidizes the iodide ions in the paper to liberate free iodine:
\[Cl_{2(g)} + 2KI_{(s)} \to 2KCl_{(s)} + I_{2(s)}\]The liberated iodine then reacts with the starch present in the paper to form the characteristic blue-black starch-iodine complex, confirming the presence of chlorine.
Question 13 Report
(a) Balance the nuclear equation below and hence identify Y.
\(^{238}_{234}U \to ^{234}_{92}Th + Y\)
(b) In.a tabular form, state two of the observations in the cathode ray experiment and the corresponding deductions.
(a) Balancing the nuclear equation
\[^{238}_{92}U \to\ ^{234}_{90}Th + Y\]Balancing the mass numbers: \(238 = 234 + A \Rightarrow A = 4\).
Balancing the atomic numbers: \(92 = 90 + Z \Rightarrow Z = 2\).
So \(Y\) has mass number 4 and atomic number 2, i.e. \(^{4}_{2}He\). Y is an alpha particle (a helium nucleus), and the process is alpha decay:
\[^{238}_{92}U \to\ ^{234}_{90}Th +\ ^{4}_{2}He\](b) Cathode-ray experiment: observations and deductions
| Observation | Deduction |
|---|---|
| The rays travel in straight lines and cast a sharp shadow of an object placed in their path | Cathode rays are made of fast-moving material particles |
| The rays are deflected towards the positive plate in an electric (or magnetic) field | The particles carry a negative charge (they are electrons) |
(A further acceptable pair: the rays turn a small paddle wheel placed in their path, showing the particles have mass and momentum.)
Answer Details
(a) Balancing the nuclear equation
\[^{238}_{92}U \to\ ^{234}_{90}Th + Y\]Balancing the mass numbers: \(238 = 234 + A \Rightarrow A = 4\).
Balancing the atomic numbers: \(92 = 90 + Z \Rightarrow Z = 2\).
So \(Y\) has mass number 4 and atomic number 2, i.e. \(^{4}_{2}He\). Y is an alpha particle (a helium nucleus), and the process is alpha decay:
\[^{238}_{92}U \to\ ^{234}_{90}Th +\ ^{4}_{2}He\](b) Cathode-ray experiment: observations and deductions
| Observation | Deduction |
|---|---|
| The rays travel in straight lines and cast a sharp shadow of an object placed in their path | Cathode rays are made of fast-moving material particles |
| The rays are deflected towards the positive plate in an electric (or magnetic) field | The particles carry a negative charge (they are electrons) |
(A further acceptable pair: the rays turn a small paddle wheel placed in their path, showing the particles have mass and momentum.)
Question 14 Report
Liquefied air is mainly a mixture of nitrogen and oxygen which can be separated into its components by fractional distillation (Boiling point of nitrogen is 196°C, Boiling point of oxygen is 174°C)
(a) Name the fraction which distills over first. Give the reason for your answer.
(b) Give another industrial application of fractional distillation as a separation technique.
(a) Fraction that distills over first
The fraction that distills over first is nitrogen.
Reason: Fractional distillation separates the components according to their boiling points, the component with the lower boiling point (the more volatile one) boiling off and distilling over first. Nitrogen has a lower boiling point (about \(-196\,^{\circ}C\)) than oxygen (about \(-183\,^{\circ}C\)). Therefore, as the liquefied air warms up, nitrogen reaches its boiling point first, vaporizes and distills over before oxygen.
(b) Another industrial application of fractional distillation
Fractional distillation is used in the refining of petroleum (crude oil), where crude oil is separated into useful fractions such as petrol, kerosene, diesel and lubricating oil according to their different boiling ranges. (It is also used to separate ethanol from a fermented ethanol-water mixture.)
Answer Details
(a) Fraction that distills over first
The fraction that distills over first is nitrogen.
Reason: Fractional distillation separates the components according to their boiling points, the component with the lower boiling point (the more volatile one) boiling off and distilling over first. Nitrogen has a lower boiling point (about \(-196\,^{\circ}C\)) than oxygen (about \(-183\,^{\circ}C\)). Therefore, as the liquefied air warms up, nitrogen reaches its boiling point first, vaporizes and distills over before oxygen.
(b) Another industrial application of fractional distillation
Fractional distillation is used in the refining of petroleum (crude oil), where crude oil is separated into useful fractions such as petrol, kerosene, diesel and lubricating oil according to their different boiling ranges. (It is also used to separate ethanol from a fermented ethanol-water mixture.)
