WAEC SSCE - Chemistry - 1992 (Essay)
Pergunta 1 Relatório
(a) Name one metal in each case Which:
(i) exists as a liquid at room temperature,
(ii) can be found in nature in the uncombined state
(b) X and Y are 0.5 mol. dm\(^{-3}\) freshly prepared aqueous solutions of two salts of iron
(II) With excess sodium hydroxide solution, X gave a dirty green precipitate which was not obtained in the case of Y.
(a) Naming metals
- (i) Liquid at room temperature: mercury (Hg).
- (ii) Found in nature in the uncombined (native) state: gold (Au) (silver or platinum are also acceptable).
(b) The two iron(II) solutions X and Y
With excess sodium hydroxide, a genuine iron(II) salt gives a dirty green precipitate of iron(II) hydroxide: \[Fe^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Fe(OH)_{2(s)}\] Since X gave this dirty green precipitate, X is the true iron(II) salt (Fe2+). Y did not give the dirty green precipitate; this shows that the iron(II) in Y had been oxidized by air to iron(III), so with sodium hydroxide it gives instead a reddish-brown precipitate of iron(III) hydroxide, Fe(OH)3. Iron(II) hydroxide itself is unstable and slowly turns reddish-brown on standing as it is oxidized to iron(III) hydroxide.
Detalhes da Resposta
(a) Naming metals
- (i) Liquid at room temperature: mercury (Hg).
- (ii) Found in nature in the uncombined (native) state: gold (Au) (silver or platinum are also acceptable).
(b) The two iron(II) solutions X and Y
With excess sodium hydroxide, a genuine iron(II) salt gives a dirty green precipitate of iron(II) hydroxide: \[Fe^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Fe(OH)_{2(s)}\] Since X gave this dirty green precipitate, X is the true iron(II) salt (Fe2+). Y did not give the dirty green precipitate; this shows that the iron(II) in Y had been oxidized by air to iron(III), so with sodium hydroxide it gives instead a reddish-brown precipitate of iron(III) hydroxide, Fe(OH)3. Iron(II) hydroxide itself is unstable and slowly turns reddish-brown on standing as it is oxidized to iron(III) hydroxide.
Pergunta 2 Relatório
(a)(i) State Graham's law of diffusion.
(ii) Consider the reaction represented by the following equation:
\(\mathrm{N_2O_{4(g)}}\) 
\(2\mathrm{NO_{2(g)}}\)
| Night yellow | dark brown |
State what would happen to the vapour density of \(\mathrm{N_2O_4}\) as the temperature of the system is increased. If the system is cooled, would the gases become lighter or darker in colour? Explain your answer in each case.
(b) Explain the following observations:
(i) an inflated balloon that was left in the sun. burst after some time;
(ii) a pure sample of a liquid did not have a constant boiling point at the top and at the base of a high mountain
(c)(i) List two gaseous reducing agents
(ii) Write one equation each to illustrate the reducing property of the gases you listed in (c)(i) above.
(a)(i) Graham's law of diffusion.
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[ r \propto \frac{1}{\sqrt{\rho}} \qquad\text{or}\qquad \frac{r_1}{r_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{M_2}{M_1}} \]
(a)(ii) The equilibrium.
The diagram shows a reversible reaction, heat driving it forward and cooling driving it back:
\[ \text{N}_2\text{O}_{4(g)} \;\underset{\text{cool}}{\overset{\text{heat}}{\rightleftharpoons}}\; 2\text{NO}_{2(g)} \]
(pale yellow \(\rightleftharpoons\) dark brown)
Effect on vapour density as temperature rises: The forward dissociation is endothermic, so raising the temperature shifts the position of equilibrium to the right. One mole of N2O4 (molar mass 92) breaks into two moles of NO2 (molar mass 46 each), so the number of gas particles increases while the total mass is unchanged. The average molar mass of the mixture therefore falls. Since vapour density \( = \dfrac{M}{2} \), the vapour density of N2O4 decreases as the temperature is increased.
Effect of cooling on colour: On cooling, the exothermic reverse reaction is favoured, so the equilibrium shifts back towards N2O4. The dark brown NO2 is converted into pale (almost colourless) N2O4, so the gases become lighter in colour.
(b)(i) Balloon bursting in the sun. Heat from the sun raises the temperature of the trapped air. The gas molecules gain kinetic energy and move faster, striking the walls more often and more forcefully; the gas also tends to expand. At fixed volume this raises the internal pressure. When the pressure exceeds the elastic limit of the rubber, the balloon bursts.
(b)(ii) No constant boiling point up a mountain. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure. Atmospheric pressure falls with altitude, so at the top of the high mountain the pressure is lower and the liquid boils at a lower temperature, whereas at the base the pressure is higher and the boiling point is higher. The boiling point of a pure liquid is therefore fixed only for a fixed pressure, not for a changing one.
(c)(i) Two gaseous reducing agents: hydrogen (H2) and carbon monoxide (CO).
(c)(ii) Equations showing the reducing property:
Hydrogen reduces copper(II) oxide to copper:
\[ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \]
Carbon monoxide reduces copper(II) oxide (or iron(III) oxide) to the metal:
\[ \text{CuO} + \text{CO} \rightarrow \text{Cu} + \text{CO}_2 \]
Detalhes da Resposta
(a)(i) Graham's law of diffusion.
At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its density (or of its molar mass).
\[ r \propto \frac{1}{\sqrt{\rho}} \qquad\text{or}\qquad \frac{r_1}{r_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{M_2}{M_1}} \]
(a)(ii) The equilibrium.
The diagram shows a reversible reaction, heat driving it forward and cooling driving it back:
\[ \text{N}_2\text{O}_{4(g)} \;\underset{\text{cool}}{\overset{\text{heat}}{\rightleftharpoons}}\; 2\text{NO}_{2(g)} \]
(pale yellow \(\rightleftharpoons\) dark brown)
Effect on vapour density as temperature rises: The forward dissociation is endothermic, so raising the temperature shifts the position of equilibrium to the right. One mole of N2O4 (molar mass 92) breaks into two moles of NO2 (molar mass 46 each), so the number of gas particles increases while the total mass is unchanged. The average molar mass of the mixture therefore falls. Since vapour density \( = \dfrac{M}{2} \), the vapour density of N2O4 decreases as the temperature is increased.
Effect of cooling on colour: On cooling, the exothermic reverse reaction is favoured, so the equilibrium shifts back towards N2O4. The dark brown NO2 is converted into pale (almost colourless) N2O4, so the gases become lighter in colour.
(b)(i) Balloon bursting in the sun. Heat from the sun raises the temperature of the trapped air. The gas molecules gain kinetic energy and move faster, striking the walls more often and more forcefully; the gas also tends to expand. At fixed volume this raises the internal pressure. When the pressure exceeds the elastic limit of the rubber, the balloon bursts.
(b)(ii) No constant boiling point up a mountain. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure. Atmospheric pressure falls with altitude, so at the top of the high mountain the pressure is lower and the liquid boils at a lower temperature, whereas at the base the pressure is higher and the boiling point is higher. The boiling point of a pure liquid is therefore fixed only for a fixed pressure, not for a changing one.
(c)(i) Two gaseous reducing agents: hydrogen (H2) and carbon monoxide (CO).
(c)(ii) Equations showing the reducing property:
Hydrogen reduces copper(II) oxide to copper:
\[ \text{CuO} + \text{H}_2 \rightarrow \text{Cu} + \text{H}_2\text{O} \]
Carbon monoxide reduces copper(II) oxide (or iron(III) oxide) to the metal:
\[ \text{CuO} + \text{CO} \rightarrow \text{Cu} + \text{CO}_2 \]
Pergunta 3 Relatório
(a) Give one example of naturally-occurring acids
(b) If 0.5 mole of a mono-alkanoic acid weighs 44g, determine the molecular formula and the name of the acid. (H = 1, C = 12, O = 16)
(a) A naturally-occurring acid
Ethanoic acid (found in vinegar). Other acceptable examples are methanoic acid (in ant stings), citric acid (in citrus fruits) and lactic acid (in sour milk).
(b) Molecular formula and name of the acid
\[\text{molar mass} = \frac{44}{0.5} = 88\,g\,mol^{-1}\]
A mono-alkanoic (monocarboxylic) acid has the general formula CnH2nO2, whose molar mass is 14n + 32.
\[14n + 32 = 88 \Rightarrow 14n = 56 \Rightarrow n = 4\]
Molecular formula = C4H8O2 (that is, C3H7COOH).
Name of the acid: butanoic acid.
Detalhes da Resposta
(a) A naturally-occurring acid
Ethanoic acid (found in vinegar). Other acceptable examples are methanoic acid (in ant stings), citric acid (in citrus fruits) and lactic acid (in sour milk).
(b) Molecular formula and name of the acid
\[\text{molar mass} = \frac{44}{0.5} = 88\,g\,mol^{-1}\]
A mono-alkanoic (monocarboxylic) acid has the general formula CnH2nO2, whose molar mass is 14n + 32.
\[14n + 32 = 88 \Rightarrow 14n = 56 \Rightarrow n = 4\]
Molecular formula = C4H8O2 (that is, C3H7COOH).
Name of the acid: butanoic acid.
Pergunta 4 Relatório
(a) State the method of collecting gases which are denser than air.
(b) Name two gases that can be used to perform the fountain experiment in the laboratory. State the physical property which makes it suitable for the experiment.
(a) Collecting gases denser than air
Gases that are denser than air are collected by upward delivery (downward displacement of air): the delivery tube is passed to the bottom of an upright gas jar, and the denser gas sinks and fills the jar from the base upwards, pushing the air out.
(b) Gases for the fountain experiment
Two suitable gases are ammonia (NH3) and hydrogen chloride (HCl).
The physical property that makes them suitable is their very high solubility in water. When a little water enters the flask of gas, the gas dissolves almost instantly, causing a large drop in pressure inside the flask, so more water is forced up rapidly to form a fountain.
Detalhes da Resposta
(a) Collecting gases denser than air
Gases that are denser than air are collected by upward delivery (downward displacement of air): the delivery tube is passed to the bottom of an upright gas jar, and the denser gas sinks and fills the jar from the base upwards, pushing the air out.
(b) Gases for the fountain experiment
Two suitable gases are ammonia (NH3) and hydrogen chloride (HCl).
The physical property that makes them suitable is their very high solubility in water. When a little water enters the flask of gas, the gas dissolves almost instantly, causing a large drop in pressure inside the flask, so more water is forced up rapidly to form a fountain.
Pergunta 5 Relatório
(a) Name: (i) one structural isomer of glucose.
(ii) the process by which starch is converted to glucose.
(b) The open-chain structure of glucose is shown below.
(a) State the functional groups present in the structure.
(ii) Which of the functional groups would react with warm Fehling's solution?
(a)(i) Structural isomer of glucose.
Fructose. Both glucose and fructose share the molecular formula \(C_6H_{12}O_6\) but differ in structure: glucose is an aldohexose (carries an aldehyde, \(-CHO\), group) while fructose is a ketohexose (carries a ketone, \(>C=O\), group). Having the same molecular formula but a different arrangement of atoms makes them structural isomers.
(a)(ii) Process converting starch to glucose.
Hydrolysis (either acid hydrolysis using dilute acid, or enzymatic hydrolysis using the enzyme amylase/diastase). Water molecules split the long starch chain at its glycosidic linkages to release individual glucose units.
(b)(i) Functional groups present in the structure.
Reading the open-chain structure shown, the top carbon carries \(H-C=O\), and each of the remaining carbons carries an \(-O-H\) group (four secondary \(-OH\) groups plus a terminal \(-CH_2OH\)). Therefore the functional groups are:
- the aldehyde group, \(-CHO\) (one, at the top of the chain);
- the hydroxyl group, \(-OH\) (alcohol groups, several along the chain).
(b)(ii) Group that reacts with warm Fehling's solution.
The aldehyde group, \(-CHO\). Warm Fehling's solution (a blue \(Cu^{2+}\) complex) is reduced by the aldehyde group; the copper is reduced from \(Cu^{2+}\) to \(Cu^{+}\), depositing a brick-red precipitate of copper(I) oxide, \(Cu_2O\), while the aldehyde is oxidised to a carboxylic acid group. This is why glucose is classed as a reducing sugar. The hydroxyl groups do not give this test.
Detalhes da Resposta
(a)(i) Structural isomer of glucose.
Fructose. Both glucose and fructose share the molecular formula \(C_6H_{12}O_6\) but differ in structure: glucose is an aldohexose (carries an aldehyde, \(-CHO\), group) while fructose is a ketohexose (carries a ketone, \(>C=O\), group). Having the same molecular formula but a different arrangement of atoms makes them structural isomers.
(a)(ii) Process converting starch to glucose.
Hydrolysis (either acid hydrolysis using dilute acid, or enzymatic hydrolysis using the enzyme amylase/diastase). Water molecules split the long starch chain at its glycosidic linkages to release individual glucose units.
(b)(i) Functional groups present in the structure.
Reading the open-chain structure shown, the top carbon carries \(H-C=O\), and each of the remaining carbons carries an \(-O-H\) group (four secondary \(-OH\) groups plus a terminal \(-CH_2OH\)). Therefore the functional groups are:
- the aldehyde group, \(-CHO\) (one, at the top of the chain);
- the hydroxyl group, \(-OH\) (alcohol groups, several along the chain).
(b)(ii) Group that reacts with warm Fehling's solution.
The aldehyde group, \(-CHO\). Warm Fehling's solution (a blue \(Cu^{2+}\) complex) is reduced by the aldehyde group; the copper is reduced from \(Cu^{2+}\) to \(Cu^{+}\), depositing a brick-red precipitate of copper(I) oxide, \(Cu_2O\), while the aldehyde is oxidised to a carboxylic acid group. This is why glucose is classed as a reducing sugar. The hydroxyl groups do not give this test.
Pergunta 6 Relatório
(a)(i) List two uses of aluminium and state how each use is related to the properties of the element.
(ii) State the reason why aluminium oxide is said to be amphoteric.
(b) Calcium is extracted by the electrolysis of fused calcium chloride containing about one-sixth of its mass of calcium fluoride.
(i) Sketch and label the cell used for the extraction.
(ii) Write equations for the reactions at the electrodes.
(iii) State the role of the calcium fluoride in the extraction.
(c) W, X, Y and Z represent four metals which have the following properties: W does not react with cold water but it liberates hydrogen from steam; X is one of the products formed when its trioxonitrate (V) decomposes on strong heating; Z forms the oxide when heated in air and it displaces W from an aqueous solutions of a salt of W; Y tarnishes rapidly exposure and reacts vigorously with cold water. Use the information provided to deduce the order of reactivity of the metals.
(a)(i) Uses of aluminium and related properties
- Cooking utensils: aluminium is a good conductor of heat, is light, and resists attack by food because of its protective oxide coating.
- Overhead electric cables: aluminium is a good conductor of electricity and has low density, making the cables light.
(a)(ii) Aluminium oxide is amphoteric because it reacts with both acids and bases.
With an acid:
\[Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\]
With a base:
\[Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAl(OH)_4\]
(b)(i) Electrolytic cell for extracting calcium
(b)(ii) Electrode reactions
At the cathode:
\[Ca^{2+} + 2e^- \rightarrow Ca\]
At the anode:
\[2Cl^- \rightarrow Cl_2 + 2e^-\]
(b)(iii) Calcium fluoride lowers the melting point of calcium chloride. This allows electrolysis to take place at a lower temperature and hence reduces energy required.
(c) The order of reactivity, starting with the least reactive metal, is:
\[\boxed{X < W < Z < Y}\]
Equivalently, in decreasing order of reactivity: \(Y > Z > W > X\).
- Y reacts vigorously with cold water and is therefore the most reactive.
- Z displaces W from a solution of a salt of W, so Z is more reactive than W.
- W reacts with steam but not with cold water.
- X is produced when its trioxonitrate(V) is strongly heated, showing that it is the least reactive.
Detalhes da Resposta
(a)(i) Uses of aluminium and related properties
- Cooking utensils: aluminium is a good conductor of heat, is light, and resists attack by food because of its protective oxide coating.
- Overhead electric cables: aluminium is a good conductor of electricity and has low density, making the cables light.
(a)(ii) Aluminium oxide is amphoteric because it reacts with both acids and bases.
With an acid:
\[Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O\]
With a base:
\[Al_2O_3 + 2NaOH + 3H_2O \rightarrow 2NaAl(OH)_4\]
(b)(i) Electrolytic cell for extracting calcium
(b)(ii) Electrode reactions
At the cathode:
\[Ca^{2+} + 2e^- \rightarrow Ca\]
At the anode:
\[2Cl^- \rightarrow Cl_2 + 2e^-\]
(b)(iii) Calcium fluoride lowers the melting point of calcium chloride. This allows electrolysis to take place at a lower temperature and hence reduces energy required.
(c) The order of reactivity, starting with the least reactive metal, is:
\[\boxed{X < W < Z < Y}\]
Equivalently, in decreasing order of reactivity: \(Y > Z > W > X\).
- Y reacts vigorously with cold water and is therefore the most reactive.
- Z displaces W from a solution of a salt of W, so Z is more reactive than W.
- W reacts with steam but not with cold water.
- X is produced when its trioxonitrate(V) is strongly heated, showing that it is the least reactive.
Pergunta 7 Relatório
(a) Give the electrolyte of a named secondary electrochemical cell.
(b) Consider the cell represented as shown below: \(Cu_{(s)}Cu^{2+}_{(aq)}/Zn^{2+}_{(aq)}/Zn{(s)}\)
(i) What does the vertical double stroke represent?
(ii) Which of the metals forms the cathode?
(iii) Name the electrode to which each half-cell should be coupled if the oxidation potential of the half-cells are to be determined.
(a) Electrolyte of a named secondary cell
In the lead-acid accumulator, the electrolyte is dilute tetraoxosulphate(VI) acid (dilute sulphuric acid, H2SO4). (In the nickel-cadmium cell the electrolyte is potassium hydroxide solution.)
(b) The cell
\[Cu_{(s)} | Cu^{2+}_{(aq)} \parallel Zn^{2+}_{(aq)} | Zn_{(s)}\]
- (i) The vertical double stroke represents the salt bridge (the boundary between the two half-cells).
- (ii) Copper forms the cathode, because it is the less reactive metal, so reduction (gain of electrons) occurs at the copper electrode.
- (iii) To determine the oxidation (electrode) potential of each half-cell, each is coupled to the standard hydrogen electrode (SHE), which is taken as the reference with a potential of zero.
Detalhes da Resposta
(a) Electrolyte of a named secondary cell
In the lead-acid accumulator, the electrolyte is dilute tetraoxosulphate(VI) acid (dilute sulphuric acid, H2SO4). (In the nickel-cadmium cell the electrolyte is potassium hydroxide solution.)
(b) The cell
\[Cu_{(s)} | Cu^{2+}_{(aq)} \parallel Zn^{2+}_{(aq)} | Zn_{(s)}\]
- (i) The vertical double stroke represents the salt bridge (the boundary between the two half-cells).
- (ii) Copper forms the cathode, because it is the less reactive metal, so reduction (gain of electrons) occurs at the copper electrode.
- (iii) To determine the oxidation (electrode) potential of each half-cell, each is coupled to the standard hydrogen electrode (SHE), which is taken as the reference with a potential of zero.
Pergunta 8 Relatório
(a) What is meant by the term acid salt? Give one example.
(b) State the reason why an all-glass apparatus must be used for the laboratory preparation of concentrated trioxonitrate (V) acid.
(a) Acid salt
An acid salt is a salt formed by the partial replacement of the ionizable (replaceable) hydrogen atoms of a polybasic acid by a metal or ammonium ion, so it still contains replaceable hydrogen. Example: sodium hydrogen tetraoxosulphate(VI), NaHSO4 (sodium hydrogen carbonate, NaHCO3, is also acceptable).
(b) Why an all-glass apparatus is used
Concentrated trioxonitrate(V) acid, and its hot vapour, is a strong, corrosive oxidizing agent that attacks and corrodes rubber, cork and metal connections. Glass is chemically unreactive towards the acid, so an all-glass apparatus prevents corrosion, leakage and contamination of the acid.
Detalhes da Resposta
(a) Acid salt
An acid salt is a salt formed by the partial replacement of the ionizable (replaceable) hydrogen atoms of a polybasic acid by a metal or ammonium ion, so it still contains replaceable hydrogen. Example: sodium hydrogen tetraoxosulphate(VI), NaHSO4 (sodium hydrogen carbonate, NaHCO3, is also acceptable).
(b) Why an all-glass apparatus is used
Concentrated trioxonitrate(V) acid, and its hot vapour, is a strong, corrosive oxidizing agent that attacks and corrodes rubber, cork and metal connections. Glass is chemically unreactive towards the acid, so an all-glass apparatus prevents corrosion, leakage and contamination of the acid.
Pergunta 9 Relatório
(a) Name the crystalline allotrope of sulphur that is stable at room temperature.
(b)(i) Give one example of a fuel that contains significant amount of sulphur as an impurity.
(ii) State one environmental disadvantage of using a fuel that has high sulphur content
(a) Crystalline allotrope stable at room temperature
Rhombic sulphur (alpha-sulphur), which is stable below about 96 °C.
(b)(i) A fuel containing significant sulphur as an impurity
Coal (crude oil / petroleum and diesel are also acceptable).
(ii) One environmental disadvantage
Burning such a fuel releases sulphur(IV) oxide (SO2), which pollutes the air and reacts with rain water to form acid rain. Acid rain damages buildings, plants, soils and aquatic life, and the gas also causes respiratory problems.
Detalhes da Resposta
(a) Crystalline allotrope stable at room temperature
Rhombic sulphur (alpha-sulphur), which is stable below about 96 °C.
(b)(i) A fuel containing significant sulphur as an impurity
Coal (crude oil / petroleum and diesel are also acceptable).
(ii) One environmental disadvantage
Burning such a fuel releases sulphur(IV) oxide (SO2), which pollutes the air and reacts with rain water to form acid rain. Acid rain damages buildings, plants, soils and aquatic life, and the gas also causes respiratory problems.
Pergunta 10 Relatório
(a) Three elements A, B and C have atomic numbers 8, 11 and 12 respectively.
(i) Write the formula of the compound formed by the chemical combination of A and B.
(ii) State which of the three elements belong(s) to the s-block of the Periodic Table. Give reason for your answer:
(b) List the component elements of bleaching powder.
The elements are A (atomic number 8, oxygen), B (atomic number 11, sodium) and C (atomic number 12, magnesium).
(a)(i) Formula of the compound of A and B
Sodium (B) forms Na+ and oxygen (A) forms O2-, so the compound is sodium oxide, Na2O.
(ii) Elements in the s-block
B (sodium, 1s22s22p63s1) and C (magnesium, 1s22s22p63s2) belong to the s-block, because their highest-energy (valence) electron enters an s-orbital. Oxygen (A) is a p-block element (its last electron enters a 2p orbital).
(b) Component elements of bleaching powder
Bleaching powder is calcium oxochlorate(I) chloride, CaOCl2. Its component elements are calcium, chlorine and oxygen.
Detalhes da Resposta
The elements are A (atomic number 8, oxygen), B (atomic number 11, sodium) and C (atomic number 12, magnesium).
(a)(i) Formula of the compound of A and B
Sodium (B) forms Na+ and oxygen (A) forms O2-, so the compound is sodium oxide, Na2O.
(ii) Elements in the s-block
B (sodium, 1s22s22p63s1) and C (magnesium, 1s22s22p63s2) belong to the s-block, because their highest-energy (valence) electron enters an s-orbital. Oxygen (A) is a p-block element (its last electron enters a 2p orbital).
(b) Component elements of bleaching powder
Bleaching powder is calcium oxochlorate(I) chloride, CaOCl2. Its component elements are calcium, chlorine and oxygen.
Pergunta 11 Relatório
(a)(i) State three differences between electrovalent compound and covalent compound.
(ii) Name the type of chemical bonding involved in the formation of ammonium ion from ammonia.
(b)(i) Name the quantum numbers which define an electron within an atom.
(ii) State the orbital in which the fifth electron of an atom is most likely to be found. Sketch the shape of the orbital.
(iii) State the period and the group to which the element boron belongs in the Periodic Table.
(c)(i) What is meant by the entropy of a chemical system?
(ii) Calculate the free energy change for a given reaction at 300 K using the following data obtained for the reaction: \(\Delta\) = -710KJ mol\(^{-1}\); \(\Delta\)S = 0.15 KJ mol\(^{-1}\)K\(^{-1}\)
(iii) From your evaluation in (c)(i) above, state whether the reaction is spontaneous or not at the given temperature. Give reason for your answer.
(a)(i) Differences between electrovalent and covalent compounds
| Electrovalent (ionic) compounds | Covalent compounds |
|---|---|
| They consist of oppositely charged ions held by strong electrostatic forces. | They consist of discrete molecules formed by sharing electrons. |
| They conduct electricity when molten or when dissolved in water because their ions are mobile. | They generally do not conduct electricity because they contain no mobile ions. |
| They usually have high melting and boiling points and are commonly soluble in water. | They usually have relatively low melting and boiling points and are commonly soluble in organic solvents. |
(a)(ii) The bond formed when ammonia accepts a hydrogen ion to form ammonium ion is a coordinate (dative covalent) bond.
\[\mathrm{NH_3 + H^+ \longrightarrow NH_4^+}\]
The nitrogen atom donates its lone pair of electrons to \(\mathrm{H^+}\).
(b)(i) The four quantum numbers are:
- principal quantum number, \(n\);
- azimuthal or angular momentum quantum number, \(l\);
- magnetic quantum number, \(m_l\);
- spin quantum number, \(m_s\).
(b)(ii) The electronic arrangement of an atom with five electrons is \(1s^2\,2s^2\,2p^1\). Therefore, the fifth electron is in a 2p orbital. A 2p orbital is dumb-bell shaped.
(b)(iii) Boron, \(1s^2\,2s^2\,2p^1\), is in Period 2 and Group 13 (Group IIIA) of the Periodic Table.
(c)(i) Entropy is a measure of the degree of disorderliness or randomness of a chemical system.
(c)(ii)
\[\Delta G=\Delta H-T\Delta S\]
\[\Delta G=-710-\left(300\times0.15\right)\]
\[\Delta G=-710-45=-755\ \text{kJ mol}^{-1}\]
(c)(iii) The reaction is spontaneous at \(300\,\text{K}\), because \(\Delta G\) is negative.
Detalhes da Resposta
(a)(i) Differences between electrovalent and covalent compounds
| Electrovalent (ionic) compounds | Covalent compounds |
|---|---|
| They consist of oppositely charged ions held by strong electrostatic forces. | They consist of discrete molecules formed by sharing electrons. |
| They conduct electricity when molten or when dissolved in water because their ions are mobile. | They generally do not conduct electricity because they contain no mobile ions. |
| They usually have high melting and boiling points and are commonly soluble in water. | They usually have relatively low melting and boiling points and are commonly soluble in organic solvents. |
(a)(ii) The bond formed when ammonia accepts a hydrogen ion to form ammonium ion is a coordinate (dative covalent) bond.
\[\mathrm{NH_3 + H^+ \longrightarrow NH_4^+}\]
The nitrogen atom donates its lone pair of electrons to \(\mathrm{H^+}\).
(b)(i) The four quantum numbers are:
- principal quantum number, \(n\);
- azimuthal or angular momentum quantum number, \(l\);
- magnetic quantum number, \(m_l\);
- spin quantum number, \(m_s\).
(b)(ii) The electronic arrangement of an atom with five electrons is \(1s^2\,2s^2\,2p^1\). Therefore, the fifth electron is in a 2p orbital. A 2p orbital is dumb-bell shaped.
(b)(iii) Boron, \(1s^2\,2s^2\,2p^1\), is in Period 2 and Group 13 (Group IIIA) of the Periodic Table.
(c)(i) Entropy is a measure of the degree of disorderliness or randomness of a chemical system.
(c)(ii)
\[\Delta G=\Delta H-T\Delta S\]
\[\Delta G=-710-\left(300\times0.15\right)\]
\[\Delta G=-710-45=-755\ \text{kJ mol}^{-1}\]
(c)(iii) The reaction is spontaneous at \(300\,\text{K}\), because \(\Delta G\) is negative.
Pergunta 12 Relatório
(a) Give the products of the following reactions:
(i) hydrolysis of simple proteins.
(ii) alkaline hydrolysis of fats and oils.
(b) A combustion tube was packed with small pieces of broken clay pot and the tube maintained at a temperature of 750K. When the vapour of decane was passed into the tube, the main products included a gaseous hydrocarbon X.
(i) Name the process involved in the reaction. Give its industrial application.
(ii) State the function of the pieces of broken pot in the experiment.
(iii) Give one chemical test to distinguish between X and methane.
(iv) Draw a labelled diagram for the laboratory preparation of X.
(c)(i) State what would be observed if a piece of sodium was added to 10cm\(^3\) of propanol in a beaker. Write an equation for the reaction.
(ii) Give the main product formed when excess acidified potassium heptaoxodichromate (VI) reacts with each of the following: propan-1- ol: propan - 2 -of; State the type of process involved in the reactions.
(a)
- Hydrolysis of simple proteins produces amino acids.
- Alkaline hydrolysis of fats and oils produces soap (the sodium or potassium salts of fatty acids) and glycerol.
(b)
- The process is cracking. It is used industrially to convert large, less useful and less demanded hydrocarbon fractions into smaller, more useful fractions such as petrol and alkenes.
- The broken pieces of clay pot prevent local overheating and provide a medium for more uniform heating in the combustion tube.
- Pass each gas through bromine water, or expose it to bromine vapour. Hydrocarbon X decolourises the reddish-brown bromine water because it is an unsaturated hydrocarbon (an alkene), whereas methane produces no change.
- Laboratory preparation of ethene
Ethene is prepared by dehydrating ethanol with concentrated tetraoxosulphate(VI) acid at about 170°C. The ethene formed is collected over water.
Equation:
\[\mathrm{CH_3CH_2OH\ \xrightarrow[170^\circ C]{conc.\ H_2SO_4}\ CH_2{=}CH_2 + H_2O}\]
(c)
- There is effervescence as sodium gradually dissolves. Hydrogen gas is evolved and it gives a pop sound with a lighted splint. Sodium propoxide is formed.
\[\mathrm{2CH_3CH_2CH_2OH + 2Na \rightarrow 2CH_3CH_2CH_2ONa + H_2}\] - With excess acidified potassium heptaoxodichromate(VI):
- Propan-1-ol is oxidised to propanoic acid, \(\mathrm{CH_3CH_2COOH}\).
- Propan-2-ol is oxidised to propanone, \(\mathrm{CH_3COCH_3}\).
Detalhes da Resposta
(a)
- Hydrolysis of simple proteins produces amino acids.
- Alkaline hydrolysis of fats and oils produces soap (the sodium or potassium salts of fatty acids) and glycerol.
(b)
- The process is cracking. It is used industrially to convert large, less useful and less demanded hydrocarbon fractions into smaller, more useful fractions such as petrol and alkenes.
- The broken pieces of clay pot prevent local overheating and provide a medium for more uniform heating in the combustion tube.
- Pass each gas through bromine water, or expose it to bromine vapour. Hydrocarbon X decolourises the reddish-brown bromine water because it is an unsaturated hydrocarbon (an alkene), whereas methane produces no change.
- Laboratory preparation of ethene
Ethene is prepared by dehydrating ethanol with concentrated tetraoxosulphate(VI) acid at about 170°C. The ethene formed is collected over water.
Equation:
\[\mathrm{CH_3CH_2OH\ \xrightarrow[170^\circ C]{conc.\ H_2SO_4}\ CH_2{=}CH_2 + H_2O}\]
(c)
- There is effervescence as sodium gradually dissolves. Hydrogen gas is evolved and it gives a pop sound with a lighted splint. Sodium propoxide is formed.
\[\mathrm{2CH_3CH_2CH_2OH + 2Na \rightarrow 2CH_3CH_2CH_2ONa + H_2}\] - With excess acidified potassium heptaoxodichromate(VI):
- Propan-1-ol is oxidised to propanoic acid, \(\mathrm{CH_3CH_2COOH}\).
- Propan-2-ol is oxidised to propanone, \(\mathrm{CH_3COCH_3}\).
Pergunta 13 Relatório
(a) State the main ore of tin.
(b) Write an equation for the reaction involved in the smelting of purified tin ore
(c) List two alloys of tin
(a) Main ore of tin
Cassiterite, which is tin(IV) oxide, SnO2.
(b) Smelting of purified tin ore
The oxide is reduced by heating with carbon (coke) in a furnace: \[SnO_2 + 2C \rightarrow Sn + 2CO\]
(c) Two alloys of tin
- Solder (tin and lead).
- Bronze (tin and copper). (Pewter and type metal are also tin alloys.)
Detalhes da Resposta
(a) Main ore of tin
Cassiterite, which is tin(IV) oxide, SnO2.
(b) Smelting of purified tin ore
The oxide is reduced by heating with carbon (coke) in a furnace: \[SnO_2 + 2C \rightarrow Sn + 2CO\]
(c) Two alloys of tin
- Solder (tin and lead).
- Bronze (tin and copper). (Pewter and type metal are also tin alloys.)
Pergunta 14 Relatório
Magnesium ribbon reacts with dilute hydrochloric acid at room temperature.
(a) State three ways by which the reaction can be made to proceed faster
(b) Write an equation for the reaction.
(a) Three ways to make the reaction faster
- Increase the temperature of the acid.
- Increase the concentration of the hydrochloric acid.
- Increase the surface area of the magnesium (use powdered or finely cut magnesium instead of a whole ribbon).
(b) Equation
\[Mg_{(s)} + 2HCl_{(aq)} \rightarrow MgCl_{2(aq)} + H_{2(g)}\]
Detalhes da Resposta
(a) Three ways to make the reaction faster
- Increase the temperature of the acid.
- Increase the concentration of the hydrochloric acid.
- Increase the surface area of the magnesium (use powdered or finely cut magnesium instead of a whole ribbon).
(b) Equation
\[Mg_{(s)} + 2HCl_{(aq)} \rightarrow MgCl_{2(aq)} + H_{2(g)}\]
