WAEC SSCE - Chemistry - 2002 (Essay)
Tambaya 1 Rahoto
(a)(i) State three methods of preparing salts, giving one example in each case of a salt so prepared.
(ii) What type of salt is each of the following? NaH\(_2\)PO\(_4\); (CH\(_3\)COO)\(_2\)Pb; KAI(SO\(_4\))\(_2\). 12H\(_2\)O.
(b)(i) Write an equation for the reaction between dilute HCI and a solution of AgNO\(_3\).
(ii) Explain why NaNO\(_3\) is preferred to AgNO\(_3\) in the preparation of oxygen by thermal decomposition of trioxonitrate (V) salts.
(iii) When silver wire was dipped into an aqueous solution of CuSO\(_4\), the wire remained intact but when the wire was replaced with zinc rod, the rod decreased in size. Give an explanation for this observation.
(c) When a sample of a crystalline salt X was exposed to air, there was a loss in mass.
(i) What phenomenon was exhibited by X?
(ii) Suggest two substances which X could be.
(iii) On heating 5.00 g of a fresh sample of X to constant mass, 1.80g was lost in the form of water vapour. Calculate the number of molecules of water of crystallization in one molecule of X. [H = 1.00; O = 16.00; Anhydrous form of X = 160 g mol\(^{-1}\)
(a)(i) Three methods of preparing salts
- Neutralisation of an acid by an alkali/base: \(HCl + NaOH \to NaCl + H_2O\) (gives sodium chloride).
- Action of an acid on a metal: \(Zn + H_2SO_4 \to ZnSO_4 + H_2\) (gives zinc tetraoxosulphate(VI)).
- Precipitation (double decomposition): \(AgNO_3 + NaCl \to AgCl + NaNO_3\) (gives silver chloride). Action of acid on a carbonate is also acceptable.
(a)(ii) Type of each salt
- \(NaH_2PO_4\): an acid salt (replaceable hydrogen remains).
- \((CH_3COO)_2Pb\): a normal salt (lead ethanoate).
- \(KAl(SO_4)_2\cdot12H_2O\): a double salt (potash alum).
(b)(i) \[HCl + AgNO_3 \to AgCl\downarrow + HNO_3\] (ionically \(Ag^+ + Cl^- \to AgCl\)).
(b)(ii) \(NaNO_3\) is preferred to \(AgNO_3\) because it decomposes simply to the nitrite and pure oxygen \((2NaNO_3 \to 2NaNO_2 + O_2)\), and it is cheap. \(AgNO_3\) decomposes to silver, oxygen and brown nitrogen(IV) oxide \((2AgNO_3 \to 2Ag + 2NO_2 + O_2)\), which contaminates the oxygen, and it is expensive.
(b)(iii) Silver is less reactive than copper, so it cannot displace copper and the wire stays intact. Zinc is more reactive than copper, so it displaces copper from the solution and itself dissolves, so the rod decreases in size: \[Zn + CuSO_4 \to ZnSO_4 + Cu\]
(c)(i) The phenomenon is efflorescence (loss of water of crystallisation to the air).
(c)(ii) Two possible substances: sodium trioxocarbonate(IV) decahydrate, \(Na_2CO_3\cdot10H_2O\), or sodium tetraoxosulphate(VI) decahydrate, \(Na_2SO_4\cdot10H_2O\).
(c)(iii) Mass of anhydrous salt \(= 5.00 - 1.80 = 3.20\ \text{g}\).
\[n(\text{anhydrous}) = \frac{3.20}{160} = 0.02\ \text{mol};\quad n(H_2O) = \frac{1.80}{18} = 0.10\ \text{mol}\]\[\frac{n(H_2O)}{n(\text{salt})} = \frac{0.10}{0.02} = 5\]There are 5 molecules of water of crystallisation in one molecule of X.
Bayanin Amsa
(a)(i) Three methods of preparing salts
- Neutralisation of an acid by an alkali/base: \(HCl + NaOH \to NaCl + H_2O\) (gives sodium chloride).
- Action of an acid on a metal: \(Zn + H_2SO_4 \to ZnSO_4 + H_2\) (gives zinc tetraoxosulphate(VI)).
- Precipitation (double decomposition): \(AgNO_3 + NaCl \to AgCl + NaNO_3\) (gives silver chloride). Action of acid on a carbonate is also acceptable.
(a)(ii) Type of each salt
- \(NaH_2PO_4\): an acid salt (replaceable hydrogen remains).
- \((CH_3COO)_2Pb\): a normal salt (lead ethanoate).
- \(KAl(SO_4)_2\cdot12H_2O\): a double salt (potash alum).
(b)(i) \[HCl + AgNO_3 \to AgCl\downarrow + HNO_3\] (ionically \(Ag^+ + Cl^- \to AgCl\)).
(b)(ii) \(NaNO_3\) is preferred to \(AgNO_3\) because it decomposes simply to the nitrite and pure oxygen \((2NaNO_3 \to 2NaNO_2 + O_2)\), and it is cheap. \(AgNO_3\) decomposes to silver, oxygen and brown nitrogen(IV) oxide \((2AgNO_3 \to 2Ag + 2NO_2 + O_2)\), which contaminates the oxygen, and it is expensive.
(b)(iii) Silver is less reactive than copper, so it cannot displace copper and the wire stays intact. Zinc is more reactive than copper, so it displaces copper from the solution and itself dissolves, so the rod decreases in size: \[Zn + CuSO_4 \to ZnSO_4 + Cu\]
(c)(i) The phenomenon is efflorescence (loss of water of crystallisation to the air).
(c)(ii) Two possible substances: sodium trioxocarbonate(IV) decahydrate, \(Na_2CO_3\cdot10H_2O\), or sodium tetraoxosulphate(VI) decahydrate, \(Na_2SO_4\cdot10H_2O\).
(c)(iii) Mass of anhydrous salt \(= 5.00 - 1.80 = 3.20\ \text{g}\).
\[n(\text{anhydrous}) = \frac{3.20}{160} = 0.02\ \text{mol};\quad n(H_2O) = \frac{1.80}{18} = 0.10\ \text{mol}\]\[\frac{n(H_2O)}{n(\text{salt})} = \frac{0.10}{0.02} = 5\]There are 5 molecules of water of crystallisation in one molecule of X.
Tambaya 2 Rahoto
(a) What term is used to describe each of the following processes?
(i) Alkaline hydrolysis of fats and oils;
(ii) The conversion of glucose into ethanol by enzymatic action;
(iii) Thermal decomposition of higher petroleum fractions into lower molecular mass hydrocarbons in the presence of catalyst.
(b)(i) Write the structure and IUPAC name for one alkanoic acid with the molecular formula C\(_4\)H\(_8\)0\(_2\).
(ii) Arrange the following compounds in order of increasing boiling point: Butane; Butanoic acid; Methylpropane.
(iii) Give an explanation for your answer in (b)(ii).
(c)(i) Ethanol was used for preparing a gas X which decolorized bromine water. Identify X and describe briefly its laboratory preparation.
(ii) Write an equation to show how ethanol reacts with sodium
(iii) Give the reagent and reaction conditions for the conversion of ethanol into C\(_2\)H\(_5\)COOC\(_2\)H\(_5\).
(d) State the type of recction involved in each of the conversions indicated below:
(i)C\(_6\)H\(_6\)C\(_6\)H\(_5\)CH\(_3\)
(ii) nC\(_2\)H\(_4\) \(\to\) (CH\(_2\) - CH\(_2\)),
(iii) CH\(_3\)CH\(_2\)CH(OH)CH\(_3\) -> CH\(_3\)CH\(_2\)CCH\(_3\)
(iv) (C\(_6\)H\(_{10}\)O\(_5\)) -> C\(_6\)H\(_{12}\)O\(_6\).
(iv) 
(a) Terms for the processes
- Alkaline hydrolysis of fats and oils: saponification.
- Conversion of glucose into ethanol by enzymes: fermentation.
- Thermal decomposition of higher petroleum fractions into lower ones over a catalyst: (catalytic) cracking.
(b)(i) Alkanoic acid with molecular formula C4H8O2: butanoic acid, CH3CH2CH2COOH.
(ii) Increasing boiling point: methylpropane < butane < butanoic acid.
(iii) Explanation: methylpropane and butane are isomers (C4H10). The branched methylpropane has a smaller surface area, so its van der Waals forces are weakest and its boiling point is lowest. Straight-chain butane has a larger contact area and stronger van der Waals forces, so a higher boiling point. Butanoic acid molecules are joined by strong hydrogen bonds through the -COOH group, giving the highest boiling point.
(c)(i) X is ethene (ethylene), C2H4, which decolorizes bromine water. Preparation: heat ethanol with excess concentrated tetraoxosulphate(VI) acid at about 170°C, which dehydrates the ethanol; alternatively pass ethanol vapour over heated aluminium oxide or broken porcelain. The gas is collected over water. \[\text{C}_2\text{H}_5\text{OH} \xrightarrow[170^\circ\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O}\]
(ii) Ethanol with sodium: \[2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\]
(iii) Conversion of ethanol into C2H5COOC2H5 (ethyl propanoate): warm the ethanol with propanoic acid (C2H5COOH) using a few drops of concentrated H2SO4 as catalyst (esterification). \[\text{C}_2\text{H}_5\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons} \text{C}_2\text{H}_5\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\]
(d) Type of reaction in each conversion
- C6H6 to C6H5CH3: substitution.
- nC2H4 to (CH2-CH2)n: addition polymerization.
- CH3CH2CH(OH)CH3 to CH3CH2COCH3: oxidation.
- (C6H10O5)n to C6H12O6: hydrolysis.
The reaction shown in the diagram is a cyclic diene combining with ethene across the double bonds to form a fused ring product, with no small molecule eliminated. This is an addition reaction (a cycloaddition).
Bayanin Amsa
(a) Terms for the processes
- Alkaline hydrolysis of fats and oils: saponification.
- Conversion of glucose into ethanol by enzymes: fermentation.
- Thermal decomposition of higher petroleum fractions into lower ones over a catalyst: (catalytic) cracking.
(b)(i) Alkanoic acid with molecular formula C4H8O2: butanoic acid, CH3CH2CH2COOH.
(ii) Increasing boiling point: methylpropane < butane < butanoic acid.
(iii) Explanation: methylpropane and butane are isomers (C4H10). The branched methylpropane has a smaller surface area, so its van der Waals forces are weakest and its boiling point is lowest. Straight-chain butane has a larger contact area and stronger van der Waals forces, so a higher boiling point. Butanoic acid molecules are joined by strong hydrogen bonds through the -COOH group, giving the highest boiling point.
(c)(i) X is ethene (ethylene), C2H4, which decolorizes bromine water. Preparation: heat ethanol with excess concentrated tetraoxosulphate(VI) acid at about 170°C, which dehydrates the ethanol; alternatively pass ethanol vapour over heated aluminium oxide or broken porcelain. The gas is collected over water. \[\text{C}_2\text{H}_5\text{OH} \xrightarrow[170^\circ\text{C}]{\text{conc. H}_2\text{SO}_4} \text{C}_2\text{H}_4 + \text{H}_2\text{O}\]
(ii) Ethanol with sodium: \[2\text{C}_2\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_2\text{H}_5\text{ONa} + \text{H}_2\]
(iii) Conversion of ethanol into C2H5COOC2H5 (ethyl propanoate): warm the ethanol with propanoic acid (C2H5COOH) using a few drops of concentrated H2SO4 as catalyst (esterification). \[\text{C}_2\text{H}_5\text{COOH} + \text{C}_2\text{H}_5\text{OH} \underset{\text{conc. H}_2\text{SO}_4}{\rightleftharpoons} \text{C}_2\text{H}_5\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}\]
(d) Type of reaction in each conversion
- C6H6 to C6H5CH3: substitution.
- nC2H4 to (CH2-CH2)n: addition polymerization.
- CH3CH2CH(OH)CH3 to CH3CH2COCH3: oxidation.
- (C6H10O5)n to C6H12O6: hydrolysis.
The reaction shown in the diagram is a cyclic diene combining with ethene across the double bonds to form a fused ring product, with no small molecule eliminated. This is an addition reaction (a cycloaddition).
Tambaya 3 Rahoto
(a) What is the shape of (i) p - orbital; (ii) a molecule of methane; (iii) a molecule of carbon (IV) oxide?
(b) Consider the following elements: Ne, S, CI, 0, Fe, Mg. State which of them
(i) exhibit(s) allotropy;
(ii) form(s) coloured ions;
(iii) is/are malleable;
(iv) consist(s) of molecules that are far apart at room temperature;
(v) form(s) hydrides by sharing electrons with hydrogen;
(vi) has/have complete outermost shell.
(c)(i) List three applications of radioactivity in different fields.
(ii) Explain clearly the difference between the following reactions involving electron loss from lead.
\(^{211} pb\) \(\to\) \(^{ 211}Bi\) + \(^0_{-1}\); Pb \(\to\) pb\(^{3+}\) _ 2e\(^-\)
(iii) Give one advantage and one disadvantage of nuclear power generation over the use of fossil fuels.
(a) Shapes
- (i) A p-orbital is dumb-bell (double-lobed) shaped.
- (ii) A molecule of methane, \(CH_4\), is tetrahedral (bond angle \(109.5^\circ\)).
- (iii) A molecule of carbon(IV) oxide, \(CO_2\), is linear (bond angle \(180^\circ\)).
(b) From \(Ne, S, Cl, O, Fe, Mg\):
- (i) Exhibit allotropy: S and O (sulphur: rhombic and monoclinic; oxygen: \(O_2\) and \(O_3\)).
- (ii) Form coloured ions: Fe (a transition metal).
- (iii) Malleable: Fe and Mg (metals).
- (iv) Molecules far apart at room temperature (gases): Ne, Cl and O.
- (v) Form hydrides by sharing electrons (covalent hydrides): S, Cl and O.
- (vi) Has a complete outermost shell: Ne (a noble gas).
(c)(i) Three applications of radioactivity: medicine (radiotherapy for cancer and radioactive tracers), agriculture (inducing mutations and preserving food), and archaeology/geology (radiocarbon dating). Detecting leaks and generating nuclear power are also acceptable.
(c)(ii) Difference between the two changes
\(^{211}Pb \to\ ^{211}Bi + \ ^{0}_{-1}e\) is a nuclear (radioactive) change: a beta particle (electron) is emitted from the nucleus, so a proton is formed and a new element (bismuth) results.
\(Pb \to Pb^{2+} + 2e^-\) is an ordinary chemical (ionisation) change: electrons are lost from the outermost shell, forming an ion of the same element (lead), with no change to the nucleus.
(c)(iii) Advantage of nuclear power over fossil fuels: a very small mass of fuel yields an enormous amount of energy and it produces no carbon(IV) oxide or other combustion gases. Disadvantage: it produces dangerous radioactive waste that is difficult to dispose of and there is a risk of harmful radiation leakage.
Bayanin Amsa
(a) Shapes
- (i) A p-orbital is dumb-bell (double-lobed) shaped.
- (ii) A molecule of methane, \(CH_4\), is tetrahedral (bond angle \(109.5^\circ\)).
- (iii) A molecule of carbon(IV) oxide, \(CO_2\), is linear (bond angle \(180^\circ\)).
(b) From \(Ne, S, Cl, O, Fe, Mg\):
- (i) Exhibit allotropy: S and O (sulphur: rhombic and monoclinic; oxygen: \(O_2\) and \(O_3\)).
- (ii) Form coloured ions: Fe (a transition metal).
- (iii) Malleable: Fe and Mg (metals).
- (iv) Molecules far apart at room temperature (gases): Ne, Cl and O.
- (v) Form hydrides by sharing electrons (covalent hydrides): S, Cl and O.
- (vi) Has a complete outermost shell: Ne (a noble gas).
(c)(i) Three applications of radioactivity: medicine (radiotherapy for cancer and radioactive tracers), agriculture (inducing mutations and preserving food), and archaeology/geology (radiocarbon dating). Detecting leaks and generating nuclear power are also acceptable.
(c)(ii) Difference between the two changes
\(^{211}Pb \to\ ^{211}Bi + \ ^{0}_{-1}e\) is a nuclear (radioactive) change: a beta particle (electron) is emitted from the nucleus, so a proton is formed and a new element (bismuth) results.
\(Pb \to Pb^{2+} + 2e^-\) is an ordinary chemical (ionisation) change: electrons are lost from the outermost shell, forming an ion of the same element (lead), with no change to the nucleus.
(c)(iii) Advantage of nuclear power over fossil fuels: a very small mass of fuel yields an enormous amount of energy and it produces no carbon(IV) oxide or other combustion gases. Disadvantage: it produces dangerous radioactive waste that is difficult to dispose of and there is a risk of harmful radiation leakage.
Tambaya 4 Rahoto
(a) Explain in terms of the kinetic theory why a tyre should not be overinflated.
(b)The following results were obtained at room temperature in an experiment to verify one of the gas laws using a glass syringe:
| Pressure (P) of air in syringe (atm) | Volume (V) of air in syringe (\(cm^3\)) | \(\frac{I}{V}\) |
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(i) Plot a graph of P against \(\frac{1}{v}\), using 1 cm to represent 0.01 atm on the vertical axis and 1cm to represent 0.02 unit on the horizontal axis.
(ii) Which of the gas laws is in agreement with the results?
(c) The flow chart below represents the stages involved in the manufacture of H\(_2\)SO\(_4\).
| S + O\(_2\) \(\to\) SO\(_2\) | SO\(_2\) +x \(\to\) SO\(_3\) | SO\(_3\) +Conc. H\(_2\)SO\(_4\) \(\to\) Y | Y +H\(_2\)O \(\to\) Conc H\(_2\)SO\(_4\) |
| stage I | stage II | stage III | stage IV |
(i) Name the process represented by the chart.
(ii) Identify reactant X and product Y.
(iii) What are the operating temperature and pressure at stage II?
(iv) Mention the stage which requires a catalyst and state the catalyst used.
(v) Give the reason why the SO\(_3\) produced in stage II is not dissolved directly in water to form the acid
(d) When K\(_4\)Cr\(_2\)C\(_7\) dissolves in water, the following equilibrium is established:
\(\mathrm{Cr_2O_{7(aq)}^{2-} + H_2O_{(l)} \to 2CrO_{4(aq)}^{2-} + 2H_{aq}}\)
(i) State the colour observed on adding a few drops of dilute H\(_2\)SO\(_4\) to the system.
(ii) Explain your answer in (d)(1).
(iii) What principle is applicable to this explanation?
(a) Why a tyre should not be overinflated (kinetic theory)
The air inside a tyre consists of molecules in rapid, random motion that continually collide with the inner walls; these collisions produce the pressure. Overinflating forces more molecules into the fixed volume, so the collisions become more frequent and the pressure rises. If the tyre becomes hot (from friction with the road or from sunshine), the molecules move even faster and strike the walls harder and more often, raising the pressure further. The pressure may then exceed what the tyre can withstand and it bursts.
(b) Verifying a gas law
The third column of the table is \( \dfrac{1}{V} \); it was obtained from the volume, for example \( \dfrac{1}{10.00} = 0.100 \) and \( \dfrac{1}{8.00} = 0.125 \).
| P / atm | V / cm\(^3\) | 1/V / cm\(^{-3}\) |
|---|---|---|
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(b)(i) The graph of \( P \) (vertical axis) against \( \dfrac{1}{V} \) (horizontal axis) is a straight line through the origin.
(b)(ii) Since \( P \) is directly proportional to \( \dfrac{1}{V} \), the results agree with Boyle's law (at constant temperature, \( PV = \text{constant} \)).
(c) Manufacture of \( H_2SO_4 \)
(i) The process is the Contact process.
(ii) Reactant \( X = \) oxygen (air), \( O_2 \); product \( Y = \) oleum (fuming sulphuric acid), \( H_2S_2O_7 \).
(iii) At stage II \( (2SO_2 + O_2 \rightleftharpoons 2SO_3) \) the operating temperature is about \( 450^{\circ}C \) and the pressure is about 1 to 2 atmospheres.
(iv) Stage II requires a catalyst; the catalyst is vanadium(V) oxide, \( V_2O_5 \).
(v) The \( SO_3 \) is not dissolved directly in water because the reaction is very exothermic and produces a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense; instead \( SO_3 \) is absorbed in concentrated \( H_2SO_4 \) to give oleum.
(d) Equilibrium: \( Cr_2O_{7(aq)}^{2-} \) (orange) \( + H_2O_{(l)} \rightleftharpoons 2CrO_{4(aq)}^{2-} \) (yellow) \( + 2H^+_{(aq)} \).
(i) On adding dilute \( H_2SO_4 \), the colour observed is orange.
(ii) The acid increases the concentration of \( H^+ \) ions; the equilibrium shifts to the left to remove the added \( H^+ \), forming more of the orange dichromate ion \( Cr_2O_7^{2-} \).
(iii) The principle applied is Le Chatelier's principle.
Bayanin Amsa
(a) Why a tyre should not be overinflated (kinetic theory)
The air inside a tyre consists of molecules in rapid, random motion that continually collide with the inner walls; these collisions produce the pressure. Overinflating forces more molecules into the fixed volume, so the collisions become more frequent and the pressure rises. If the tyre becomes hot (from friction with the road or from sunshine), the molecules move even faster and strike the walls harder and more often, raising the pressure further. The pressure may then exceed what the tyre can withstand and it bursts.
(b) Verifying a gas law
The third column of the table is \( \dfrac{1}{V} \); it was obtained from the volume, for example \( \dfrac{1}{10.00} = 0.100 \) and \( \dfrac{1}{8.00} = 0.125 \).
| P / atm | V / cm\(^3\) | 1/V / cm\(^{-3}\) |
|---|---|---|
| 0.100 | 10.00 | 0.100 |
| 0.125 | 8.00 | 0.125 |
| 0.150 | 6.60 | 0.150 |
| 0.175 | 5.60 | 0.179 |
| 0.200 | 4.80 | 0.208 |
| 0.225 | 4.40 | 0.227 |
(b)(i) The graph of \( P \) (vertical axis) against \( \dfrac{1}{V} \) (horizontal axis) is a straight line through the origin.
(b)(ii) Since \( P \) is directly proportional to \( \dfrac{1}{V} \), the results agree with Boyle's law (at constant temperature, \( PV = \text{constant} \)).
(c) Manufacture of \( H_2SO_4 \)
(i) The process is the Contact process.
(ii) Reactant \( X = \) oxygen (air), \( O_2 \); product \( Y = \) oleum (fuming sulphuric acid), \( H_2S_2O_7 \).
(iii) At stage II \( (2SO_2 + O_2 \rightleftharpoons 2SO_3) \) the operating temperature is about \( 450^{\circ}C \) and the pressure is about 1 to 2 atmospheres.
(iv) Stage II requires a catalyst; the catalyst is vanadium(V) oxide, \( V_2O_5 \).
(v) The \( SO_3 \) is not dissolved directly in water because the reaction is very exothermic and produces a dense, choking mist (fog) of fine sulphuric acid droplets that is difficult to condense; instead \( SO_3 \) is absorbed in concentrated \( H_2SO_4 \) to give oleum.
(d) Equilibrium: \( Cr_2O_{7(aq)}^{2-} \) (orange) \( + H_2O_{(l)} \rightleftharpoons 2CrO_{4(aq)}^{2-} \) (yellow) \( + 2H^+_{(aq)} \).
(i) On adding dilute \( H_2SO_4 \), the colour observed is orange.
(ii) The acid increases the concentration of \( H^+ \) ions; the equilibrium shifts to the left to remove the added \( H^+ \), forming more of the orange dichromate ion \( Cr_2O_7^{2-} \).
(iii) The principle applied is Le Chatelier's principle.
Tambaya 5 Rahoto
(a) Giving different examples, mention one metal in each case which produces hydrogen on reacting with
(i) dilute mineral acid
(ii) cold water;
(iii) steam;
(iv) hot, concentrated alkali.
(b) In an experiment, excess 0.50 mol dm\(^{-3}\) HCI was added to 1Og of granulated zinc in a beaker. Other conditions remaining constant, state how the reaction rate would be affected in each case, if the experiment was repeated using:
(i) 1.0 mol dm\(^{-3}\) HCI;
(ii) 8.0g of granulated zinc;
(iii) 10g of zinc dust;
(iv) a higher volume of 0.50 mol dm HCI;
(v) a reaction vessel dipped in crushed ice;
(vi) equal volumes of water and 0.50 mol dm\(^3\) HCI.
(c) Aluminium is extracted from its ore by electrolysis.
(i) Name the ore from which the metal is extracted.
(ii) State the role of molten cryolite in the extraction.
(iii) Describe in outline how the ore is purified before electrolysis
(iv) Calculate the current in amperes required to produce 18.0g of aluminium in 1.50 hours. [Al = 27.0; F = 96500C]
(d) Give the reason why
(i) aluminium, which is a reactive metal, is resistant to corrosion.
(ii) metals are generally good reducing agents.
(a) A metal that gives hydrogen with each reagent
- (i) Dilute mineral acid: zinc: \(Zn + 2HCl \to ZnCl_2 + H_2\).
- (ii) Cold water: sodium: \(2Na + 2H_2O \to 2NaOH + H_2\).
- (iii) Steam: iron: \(3Fe + 4H_2O \to Fe_3O_4 + 4H_2\).
- (iv) Hot concentrated alkali: aluminium (or zinc): \(2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2\).
(b) Effect on the rate of the zinc/HCl reaction (acid originally in excess)
- (i) Using \(1.0\ \text{mol dm}^{-3}\) HCl: higher concentration, so the rate increases.
- (ii) Using 8.0 g of the same granulated zinc: the acid is still in excess, so the rate is essentially unchanged (only less hydrogen is finally produced).
- (iii) Using 10 g of zinc dust: much larger surface area, so the rate increases markedly.
- (iv) Using a higher volume of the same \(0.50\ \text{mol dm}^{-3}\) HCl: concentration unchanged, so the rate is unchanged.
- (v) Vessel dipped in crushed ice: lower temperature, so the rate decreases.
- (vi) Equal volumes of water and acid (dilution to about \(0.25\ \text{mol dm}^{-3}\)): lower concentration, so the rate decreases.
(c) Extraction of aluminium
- (i) The ore is bauxite (impure \(Al_2O_3\)).
- (ii) Molten cryolite lowers the melting point of the alumina and dissolves it, improving conductivity and saving energy.
- (iii) Purification: dissolve the bauxite in hot concentrated \(NaOH\) (the amphoteric \(Al_2O_3\) dissolves as sodium aluminate), filter off insoluble impurities, precipitate \(Al(OH)_3\) by seeding/cooling, then heat it to give pure alumina.
- (iv) Current required: \(Al^{3+} + 3e^- \to Al\). \[n(Al) = \frac{18.0}{27.0} = 0.667\ \text{mol}\]\[Q = 0.667 \times 3 \times 96500 = 1.93\times10^{5}\ \text{C}\]\[t = 1.50 \times 3600 = 5400\ \text{s}\]\[I = \frac{Q}{t} = \frac{1.93\times10^{5}}{5400} = 35.7\ \text{A}\]
(d)(i) Although reactive, aluminium resists corrosion because it forms a thin, tough, adherent layer of aluminium oxide on its surface that seals the metal beneath from further attack.
(d)(ii) Metals are generally good reducing agents because they readily lose (donate) electrons, thereby reducing other species.
Bayanin Amsa
(a) A metal that gives hydrogen with each reagent
- (i) Dilute mineral acid: zinc: \(Zn + 2HCl \to ZnCl_2 + H_2\).
- (ii) Cold water: sodium: \(2Na + 2H_2O \to 2NaOH + H_2\).
- (iii) Steam: iron: \(3Fe + 4H_2O \to Fe_3O_4 + 4H_2\).
- (iv) Hot concentrated alkali: aluminium (or zinc): \(2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2\).
(b) Effect on the rate of the zinc/HCl reaction (acid originally in excess)
- (i) Using \(1.0\ \text{mol dm}^{-3}\) HCl: higher concentration, so the rate increases.
- (ii) Using 8.0 g of the same granulated zinc: the acid is still in excess, so the rate is essentially unchanged (only less hydrogen is finally produced).
- (iii) Using 10 g of zinc dust: much larger surface area, so the rate increases markedly.
- (iv) Using a higher volume of the same \(0.50\ \text{mol dm}^{-3}\) HCl: concentration unchanged, so the rate is unchanged.
- (v) Vessel dipped in crushed ice: lower temperature, so the rate decreases.
- (vi) Equal volumes of water and acid (dilution to about \(0.25\ \text{mol dm}^{-3}\)): lower concentration, so the rate decreases.
(c) Extraction of aluminium
- (i) The ore is bauxite (impure \(Al_2O_3\)).
- (ii) Molten cryolite lowers the melting point of the alumina and dissolves it, improving conductivity and saving energy.
- (iii) Purification: dissolve the bauxite in hot concentrated \(NaOH\) (the amphoteric \(Al_2O_3\) dissolves as sodium aluminate), filter off insoluble impurities, precipitate \(Al(OH)_3\) by seeding/cooling, then heat it to give pure alumina.
- (iv) Current required: \(Al^{3+} + 3e^- \to Al\). \[n(Al) = \frac{18.0}{27.0} = 0.667\ \text{mol}\]\[Q = 0.667 \times 3 \times 96500 = 1.93\times10^{5}\ \text{C}\]\[t = 1.50 \times 3600 = 5400\ \text{s}\]\[I = \frac{Q}{t} = \frac{1.93\times10^{5}}{5400} = 35.7\ \text{A}\]
(d)(i) Although reactive, aluminium resists corrosion because it forms a thin, tough, adherent layer of aluminium oxide on its surface that seals the metal beneath from further attack.
(d)(ii) Metals are generally good reducing agents because they readily lose (donate) electrons, thereby reducing other species.
Tambaya 6 Rahoto
a)(i) Give two uses of ammonia.
(ii) Name the process by which ammoniacal liquor can be obtained from coal and list two other products of the reaction
(iii) What type of reaction is involved in the conversion of ammoniacal liquor to (NH\(_4\))\(_2\)SO\(_4\) by dilute H\(_2\)SO\(_4\)?
(iv) Sketch and label an energy profile diagram to show the effect of presence of Pt/Rh on the reaction represented by the following equation: 4NH\(_3\) + 5O\(_2\) \(\to\) 6H\(_2\)O + 4NO; \(\Delta\)H = —907 kJmol\(^1\)
(b) Rock salt is an impure form of sodium chloride.
(i) Outline a suitable procedure for preparing a pure sample of sodium chloride from rock salt.
(ii) State two methods that can be used to prepare chlorine from rock salt. Write an appropriate equation in each case.
(c) Lead pigments were used in a water colour painting which turned black after prolonged exposure to an air pollutant. The original colour was restored by using H\(_2\)O\(_2\) which converted the black substance to a simple, white lead (II) salt.
(i) Which pollutant turned the painting black?
(ii) Write the formula of the black substance
(iii) What is the white salt?
(iv) State the role of H\(_2\)O in the restoration process.
(a)(i) Two uses of ammonia are:
- Manufacture of fertilizers such as ammonium sulphate and ammonium nitrate.
- Manufacture of trioxonitrate(V) acid, i.e. nitric acid.
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
- Coke
- Coal tar
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
- Crush the rock salt and dissolve it in water.
- Filter the mixture to remove insoluble impurities such as sand.
- Evaporate the filtrate until it is nearly saturated.
- Allow the solution to cool so that sodium chloride crystals form.
- Filter, wash the crystals with a little cold distilled water and dry them.
(b)(ii) Two methods of preparing chlorine from rock salt are:
-
Electrolysis of concentrated sodium chloride solution (brine):
\(2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electrolysis}} 2NaOH(aq) + H_2(g) + Cl_2(g)\) -
Oxidation of sodium chloride with manganese(IV) oxide and concentrated tetraoxosulphate(VI) acid:
\(2NaCl + MnO_2 + 2H_2SO_4 \rightarrow Cl_2 + MnSO_4 + Na_2SO_4 + 2H_2O\)
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
Bayanin Amsa
(a)(i) Two uses of ammonia are:
- Manufacture of fertilizers such as ammonium sulphate and ammonium nitrate.
- Manufacture of trioxonitrate(V) acid, i.e. nitric acid.
(a)(ii) Ammoniacal liquor is obtained by the destructive distillation of coal.
Two other products are:
- Coke
- Coal tar
Coal gas is also produced.
(a)(iii) The reaction is a neutralisation (acid-base) reaction.
\(2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4\)
(a)(iv) Energy profile diagram for the reaction:
The reaction is exothermic because the products are at a lower energy level than the reactants. Pt/Rh lowers the activation energy but does not change the value of \(\Delta H\).
(b)(i) Procedure for preparing pure sodium chloride from rock salt:
- Crush the rock salt and dissolve it in water.
- Filter the mixture to remove insoluble impurities such as sand.
- Evaporate the filtrate until it is nearly saturated.
- Allow the solution to cool so that sodium chloride crystals form.
- Filter, wash the crystals with a little cold distilled water and dry them.
(b)(ii) Two methods of preparing chlorine from rock salt are:
-
Electrolysis of concentrated sodium chloride solution (brine):
\(2NaCl(aq) + 2H_2O(l) \xrightarrow{\text{electrolysis}} 2NaOH(aq) + H_2(g) + Cl_2(g)\) -
Oxidation of sodium chloride with manganese(IV) oxide and concentrated tetraoxosulphate(VI) acid:
\(2NaCl + MnO_2 + 2H_2SO_4 \rightarrow Cl_2 + MnSO_4 + Na_2SO_4 + 2H_2O\)
(c)(i) The air pollutant is hydrogen sulphide, \(H_2S\).
(c)(ii) The black substance is lead(II) sulphide, \(PbS\).
(c)(iii) The white salt is lead(II) tetraoxosulphate(VI), \(PbSO_4\).
(c)(iv) Hydrogen peroxide, \(H_2O_2\), acts as an oxidising agent. It oxidises black lead(II) sulphide to white lead(II) sulphate:
\(PbS + 4H_2O_2 \rightarrow PbSO_4 + 4H_2O\)
