WAEC SSCE - Chemistry - 2012 (Essay)
Question 1 Rapport
(a) (i) Define nuclear fission.
(ii) A certain natural decay series starts with \(^{238}_{92}U\) and ends with \(^{230}_{90}Th\). Each step involves the loss of an alpha or a beta particle. Using the given information, deduce how many alpha beta particpes were emitted.
(b) Consider the equilibrium reaction represented by the following equation:
\(\mathrm{A_2(g) + 3B_2(g) \rightleftharpoons 2AB_3(g)}\); \(\Delta H = +x\mathrm{kJmol}^{-1}\)
Explain briefly the effect of each of the following changes on the equilibrium composition: (i) increase in concentrat of B; (ii) decrease in pressure of the system; (iii) addition of catalyst.
(c) The lattice energies of three sodium halides are as follows:
| Compound | NaF | NaBr | Nal |
| Lattice energy/\(\mathrm{kJmol}^{-1}\) | 890 | 719 | 670 |
Explain briefly the trend.
(d) State the property exhibited by nitrogen (IV) oxide in each of the following reactions:
(i) \(\mathrm{4Cu + 2NO_2 \to 4CuO + N_2}\);
(ii) \(\mathrm{H_2O + 2NO_2 \to HNO_3 + HNO_2}\).
(e) Iron is manufacturcd in a blast furnace using iron ore (\(\mathrm{Fe_2O_3}\)), coke and limestone. Write the equation for the reaction(s) at the: (i) top of the furnace; (ii) middle of the furnace; (iii) bottom of the furnace.
(f) (i) Name two products of destructive distillation of coal. (ii) Give one use of each product in (f)(i).
(a)(i) Nuclear fission
The splitting of a heavy atomic nucleus into two or more lighter nuclei of comparable mass, accompanied by the release of neutrons and a very large amount of energy, usually when the nucleus is struck by a neutron.
(a)(ii) Alpha and beta particles emitted from \(^{238}_{92}\)U to \(^{230}_{90}\)Th
Change in mass number: \(238 - 230 = 8\). Only alpha emission changes the mass number (by 4 each):
\[ \text{number of }\alpha = \frac{8}{4} = 2 \]Change in atomic number: 2 alphas alone would give \(92 - (2\times 2) = 88\), but the product has Z = 90. Each beta emission raises Z by 1:
\[ 88 + b = 90 \Rightarrow b = 2 \]So 2 alpha particles and 2 beta particles were emitted.
(b) Equilibrium: A\(_2\)(g) + 3B\(_2\)(g) \(\rightleftharpoons\) 2AB\(_3\)(g); \(\Delta H = +x\) kJ mol\(^{-1}\) (endothermic)
- (i) Increase in concentration of B: equilibrium shifts to the right (forward) to use up the added B, giving more AB\(_3\).
- (ii) Decrease in pressure: the system shifts toward the side with the greater number of gas moles. The left has 4 moles and the right 2 moles, so equilibrium shifts to the left (backward), forming more A\(_2\) and B\(_2\).
- (iii) Addition of a catalyst: no effect on the equilibrium composition; it only speeds up the forward and backward reactions equally, so equilibrium is reached sooner.
(c) Trend in lattice energy of the sodium halides
| Compound | NaF | NaBr | NaI |
|---|---|---|---|
| Lattice energy/kJ mol\(^{-1}\) | 890 | 719 | 670 |
From F\(^-\) to Br\(^-\) to I\(^-\) the ionic radius of the halide increases, so the distance between the Na\(^+\) and the halide ion grows. Lattice energy is inversely related to this interionic distance, so the electrostatic attraction weakens and the lattice energy falls: NaF > NaBr > NaI.
(d) Property of nitrogen(IV) oxide
- (i) 4Cu + 2NO\(_2\) \(\rightarrow\) 4CuO + N\(_2\): nitrogen is reduced from +4 to 0, so NO\(_2\) acts as an oxidising agent.
- (ii) H\(_2\)O + 2NO\(_2\) \(\rightarrow\) HNO\(_3\) + HNO\(_2\): nitrogen goes to +5 (in HNO\(_3\)) and +3 (in HNO\(_2\)), so NO\(_2\) undergoes disproportionation (it is simultaneously oxidised and reduced).
(e) Reactions in the blast furnace
- (i) Top (cooler zone) - reduction of the ore by carbon(II) oxide: \[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]
- (ii) Middle - decomposition of limestone and formation of carbon(II) oxide: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \qquad \text{CO}_2 + \text{C} \rightarrow 2\text{CO} \]
- (iii) Bottom (hottest zone, at the tuyeres) - combustion of coke: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]
(Formation of slag, CaO + SiO\(_2\) \(\rightarrow\) CaSiO\(_3\), also occurs in the lower middle region.)
(f)(i) Two products of the destructive distillation of coal: coke and coal tar (coal gas and ammoniacal liquor are also produced).
(f)(ii) One use of each:
- Coke: used as a fuel and as the reducing agent in the extraction of iron in the blast furnace.
- Coal tar: used as a raw material for making dyes, drugs and antiseptics (and for surfacing roads).
Détails de la réponse
(a)(i) Nuclear fission
The splitting of a heavy atomic nucleus into two or more lighter nuclei of comparable mass, accompanied by the release of neutrons and a very large amount of energy, usually when the nucleus is struck by a neutron.
(a)(ii) Alpha and beta particles emitted from \(^{238}_{92}\)U to \(^{230}_{90}\)Th
Change in mass number: \(238 - 230 = 8\). Only alpha emission changes the mass number (by 4 each):
\[ \text{number of }\alpha = \frac{8}{4} = 2 \]Change in atomic number: 2 alphas alone would give \(92 - (2\times 2) = 88\), but the product has Z = 90. Each beta emission raises Z by 1:
\[ 88 + b = 90 \Rightarrow b = 2 \]So 2 alpha particles and 2 beta particles were emitted.
(b) Equilibrium: A\(_2\)(g) + 3B\(_2\)(g) \(\rightleftharpoons\) 2AB\(_3\)(g); \(\Delta H = +x\) kJ mol\(^{-1}\) (endothermic)
- (i) Increase in concentration of B: equilibrium shifts to the right (forward) to use up the added B, giving more AB\(_3\).
- (ii) Decrease in pressure: the system shifts toward the side with the greater number of gas moles. The left has 4 moles and the right 2 moles, so equilibrium shifts to the left (backward), forming more A\(_2\) and B\(_2\).
- (iii) Addition of a catalyst: no effect on the equilibrium composition; it only speeds up the forward and backward reactions equally, so equilibrium is reached sooner.
(c) Trend in lattice energy of the sodium halides
| Compound | NaF | NaBr | NaI |
|---|---|---|---|
| Lattice energy/kJ mol\(^{-1}\) | 890 | 719 | 670 |
From F\(^-\) to Br\(^-\) to I\(^-\) the ionic radius of the halide increases, so the distance between the Na\(^+\) and the halide ion grows. Lattice energy is inversely related to this interionic distance, so the electrostatic attraction weakens and the lattice energy falls: NaF > NaBr > NaI.
(d) Property of nitrogen(IV) oxide
- (i) 4Cu + 2NO\(_2\) \(\rightarrow\) 4CuO + N\(_2\): nitrogen is reduced from +4 to 0, so NO\(_2\) acts as an oxidising agent.
- (ii) H\(_2\)O + 2NO\(_2\) \(\rightarrow\) HNO\(_3\) + HNO\(_2\): nitrogen goes to +5 (in HNO\(_3\)) and +3 (in HNO\(_2\)), so NO\(_2\) undergoes disproportionation (it is simultaneously oxidised and reduced).
(e) Reactions in the blast furnace
- (i) Top (cooler zone) - reduction of the ore by carbon(II) oxide: \[ \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2 \]
- (ii) Middle - decomposition of limestone and formation of carbon(II) oxide: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \qquad \text{CO}_2 + \text{C} \rightarrow 2\text{CO} \]
- (iii) Bottom (hottest zone, at the tuyeres) - combustion of coke: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]
(Formation of slag, CaO + SiO\(_2\) \(\rightarrow\) CaSiO\(_3\), also occurs in the lower middle region.)
(f)(i) Two products of the destructive distillation of coal: coke and coal tar (coal gas and ammoniacal liquor are also produced).
(f)(ii) One use of each:
- Coke: used as a fuel and as the reducing agent in the extraction of iron in the blast furnace.
- Coal tar: used as a raw material for making dyes, drugs and antiseptics (and for surfacing roads).
Question 2 Rapport
(a) (i) What is diffusion?
(ii) State Charles' law:
(iii) Sketch a graph to illustrate Charles' law.
(iv) A given mass of a gas occupied 150 cm3 at 27 °C and a pressure of \(1.013 \times 10^{5}\ \mathrm{Nm}^{-2}\). Calculate the temperature at which its volume will be doubled at the same pressure.
(v) Arrange the three states of matter in order of increasing: (i) kinetic energy; (ii) forces of cohesion.
(b) (i) State Le Chatelier's principle. (ii) A metal M forms two oxides containing 11.1% and 20.0% of oxygen. Show that these figures agree with the law of multiple proportion.
(c) The table below shows the physical properties of substances A, B and C.
| Substance | Melting point/°C | Boiling point/°C | Solubility in water at 25°C |
| A | 30 | 117 | Insolube |
| B | 31 | 160 | Insoluble |
| C | 1200 | 1200 | Insoluble |
(i) If A and B are miscible when melted and B and C react when heated, describe how a mixture of A, B, and C could be separated.
(ii) When 25.25g of the mixture A, B and C was separated, 7.52 g of A and 8.48 g of B were recovered. Assuming i there was no loss of components during separation, calculate the percentage by mass of C in the mixture
(a)(i) Diffusion
The spontaneous movement of particles of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.
(a)(ii) Charles' law
The volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, provided the pressure remains constant.
(a)(iii) Sketch
A straight-line graph of volume (y-axis) against absolute temperature in kelvin (x-axis) that passes through the origin (0 K, 0 volume).
(a)(iv) Temperature at which the volume doubles (constant pressure)
\(V_1 = 150\ \text{cm}^3\), \(T_1 = 27 + 273 = 300\ \text{K}\), \(V_2 = 300\ \text{cm}^3\).
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow T_2 = \frac{V_2 T_1}{V_1} = \frac{300 \times 300}{150} = 600\ \text{K} = 327\ ^{o}\text{C} \](a)(v) Order of increasing:
- Kinetic energy: solid < liquid < gas.
- Forces of cohesion: gas < liquid < solid.
(b)(i) Le Chatelier's principle
When a system in equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to oppose (reduce) that change.
(b)(ii) Law of multiple proportions for the two oxides of M
| Oxide | % oxygen | % metal M | Mass of O per 1 g of M |
|---|---|---|---|
| 1 | 11.1 | 88.9 | 11.1 / 88.9 = 0.125 |
| 2 | 20.0 | 80.0 | 20.0 / 80.0 = 0.250 |
The masses of oxygen that combine with a fixed mass of M are in the simple whole-number ratio 2 : 1, which agrees with the law of multiple proportions.
(c) Substances A, B and C
| Substance | m.p./\(^{o}\)C | b.p./\(^{o}\)C | Solubility in water (25 \(^{o}\)C) |
|---|---|---|---|
| A | 30 | 117 | Insoluble |
| B | 31 | 160 | Insoluble |
| C | 1200 | 1200 | Insoluble |
(c)(i) Separation scheme
- Warm the mixture gently to about 35 \(^{o}\)C. A and B (m.p. 30 and 31 \(^{o}\)C) both melt and, being miscible, form one liquid; C (m.p. 1200 \(^{o}\)C) stays solid. Do not overheat, or B would react with C.
- Filter the warm mixture: solid C is retained on the filter paper while the molten A + B mixture passes through. Wash and dry C.
- Separate A and B from the liquid by fractional distillation. A distils over first at 117 \(^{o}\)C and is collected; B distils at 160 \(^{o}\)C and is collected separately.
(c)(ii) Percentage by mass of C
\[ \text{Mass of C} = 25.25 - 7.52 - 8.48 = 9.25\ \text{g} \] \[ \%\,\text{C} = \frac{9.25}{25.25}\times 100 = 36.6\% \]Détails de la réponse
(a)(i) Diffusion
The spontaneous movement of particles of a substance from a region of higher concentration to a region of lower concentration until they are evenly spread.
(a)(ii) Charles' law
The volume of a fixed mass of gas is directly proportional to its absolute (Kelvin) temperature, provided the pressure remains constant.
(a)(iii) Sketch
A straight-line graph of volume (y-axis) against absolute temperature in kelvin (x-axis) that passes through the origin (0 K, 0 volume).
(a)(iv) Temperature at which the volume doubles (constant pressure)
\(V_1 = 150\ \text{cm}^3\), \(T_1 = 27 + 273 = 300\ \text{K}\), \(V_2 = 300\ \text{cm}^3\).
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow T_2 = \frac{V_2 T_1}{V_1} = \frac{300 \times 300}{150} = 600\ \text{K} = 327\ ^{o}\text{C} \](a)(v) Order of increasing:
- Kinetic energy: solid < liquid < gas.
- Forces of cohesion: gas < liquid < solid.
(b)(i) Le Chatelier's principle
When a system in equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that tends to oppose (reduce) that change.
(b)(ii) Law of multiple proportions for the two oxides of M
| Oxide | % oxygen | % metal M | Mass of O per 1 g of M |
|---|---|---|---|
| 1 | 11.1 | 88.9 | 11.1 / 88.9 = 0.125 |
| 2 | 20.0 | 80.0 | 20.0 / 80.0 = 0.250 |
The masses of oxygen that combine with a fixed mass of M are in the simple whole-number ratio 2 : 1, which agrees with the law of multiple proportions.
(c) Substances A, B and C
| Substance | m.p./\(^{o}\)C | b.p./\(^{o}\)C | Solubility in water (25 \(^{o}\)C) |
|---|---|---|---|
| A | 30 | 117 | Insoluble |
| B | 31 | 160 | Insoluble |
| C | 1200 | 1200 | Insoluble |
(c)(i) Separation scheme
- Warm the mixture gently to about 35 \(^{o}\)C. A and B (m.p. 30 and 31 \(^{o}\)C) both melt and, being miscible, form one liquid; C (m.p. 1200 \(^{o}\)C) stays solid. Do not overheat, or B would react with C.
- Filter the warm mixture: solid C is retained on the filter paper while the molten A + B mixture passes through. Wash and dry C.
- Separate A and B from the liquid by fractional distillation. A distils over first at 117 \(^{o}\)C and is collected; B distils at 160 \(^{o}\)C and is collected separately.
(c)(ii) Percentage by mass of C
\[ \text{Mass of C} = 25.25 - 7.52 - 8.48 = 9.25\ \text{g} \] \[ \%\,\text{C} = \frac{9.25}{25.25}\times 100 = 36.6\% \]Question 3 Rapport
(a) (i) What is the structure of the atom as proposed by Rutherford?
(ii) Distinguish between the atomic number and the mass number of an element.
(iii) Explain briefly why the relative atomic mass of chlorine is not a whole number.
(b) (i) What is meant by first ionization energy?
(ii) List three properties of electrovalent compounds.
(iii) Consider the following pairs of elements: I. \(_9\)F and \(_{17}\)Cl; \(_{12}\)Mg and \(_{20}\)Ca. Explain briefly why the elements in each pair have similar chemical properties.
(c) Explain briefly the following terms using an appropriate example in each case.
(i) homologous series; (ii) heterolytic fission.
(d) State the indicator(s) which could be used to determine the end-point of the following titrations:
(i) dilute hydrochloric acid against sodium hydroxide solution;
(ii) dilute hydrochloric acid against ammonium hydroxide solution;
(iii) ethanoic acid against sodium hydroxide solution.
(e) A solid chloride E which sublimed on heating reacted with an alkali F to give a choking gas G. G turned moist red litmus paper blue. identify E, F and G.
(a)(i) Rutherford's atomic structure: the atom has a very small, dense, positively charged nucleus at the centre where nearly all the mass is concentrated, with the negatively charged electrons moving in the mostly empty space around the nucleus.
(a)(ii) Atomic number vs mass number: the atomic number is the number of protons in the nucleus of an atom; the mass number is the total number of protons and neutrons in the nucleus.
(a)(iii) Chlorine exists as two isotopes, 35Cl and 37Cl, in the fixed ratio 3:1. The relative atomic mass is the weighted average of these isotopic masses, \(\frac{(35\times3)+(37\times1)}{4}=35.5\), which is not a whole number.
(b)(i) First ionization energy: the minimum energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous singly charged positive ions.
(b)(ii) Three properties of electrovalent compounds:
- They have high melting and boiling points.
- They are usually soluble in water but insoluble in organic solvents.
- They conduct electricity when molten or in aqueous solution.
(b)(iii) Each pair has the same number of electrons in the outermost shell (F and Cl each have 7; Mg and Ca each have 2). Since chemical properties depend on the number of valence electrons, members of the same group have similar chemical properties.
(c)(i) Homologous series: a family of organic compounds with the same general formula in which successive members differ by a CH2 unit and show a gradual change in physical properties, e.g. the alkanes (CH4, C2H6, C3H8...).
(c)(ii) Heterolytic fission: the breaking of a covalent bond in which both shared electrons go to one of the atoms, producing oppositely charged ions, e.g. \(\text{HCl} \to \text{H}^+ + \text{Cl}^-\).
(d) Indicators:
- (i) HCl vs NaOH (strong acid/strong base): methyl orange or phenolphthalein.
- (ii) HCl vs NH4OH (strong acid/weak base): methyl orange.
- (iii) Ethanoic acid vs NaOH (weak acid/strong base): phenolphthalein.
(e) E = ammonium chloride (NH4Cl, a solid chloride that sublimes); F = an alkali such as sodium hydroxide (NaOH); G = ammonia (NH3), the choking gas that turns moist red litmus blue.
\[ \text{NH}_4\text{Cl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O} + \text{NH}_3 \]Détails de la réponse
(a)(i) Rutherford's atomic structure: the atom has a very small, dense, positively charged nucleus at the centre where nearly all the mass is concentrated, with the negatively charged electrons moving in the mostly empty space around the nucleus.
(a)(ii) Atomic number vs mass number: the atomic number is the number of protons in the nucleus of an atom; the mass number is the total number of protons and neutrons in the nucleus.
(a)(iii) Chlorine exists as two isotopes, 35Cl and 37Cl, in the fixed ratio 3:1. The relative atomic mass is the weighted average of these isotopic masses, \(\frac{(35\times3)+(37\times1)}{4}=35.5\), which is not a whole number.
(b)(i) First ionization energy: the minimum energy required to remove one mole of electrons from one mole of gaseous atoms to form one mole of gaseous singly charged positive ions.
(b)(ii) Three properties of electrovalent compounds:
- They have high melting and boiling points.
- They are usually soluble in water but insoluble in organic solvents.
- They conduct electricity when molten or in aqueous solution.
(b)(iii) Each pair has the same number of electrons in the outermost shell (F and Cl each have 7; Mg and Ca each have 2). Since chemical properties depend on the number of valence electrons, members of the same group have similar chemical properties.
(c)(i) Homologous series: a family of organic compounds with the same general formula in which successive members differ by a CH2 unit and show a gradual change in physical properties, e.g. the alkanes (CH4, C2H6, C3H8...).
(c)(ii) Heterolytic fission: the breaking of a covalent bond in which both shared electrons go to one of the atoms, producing oppositely charged ions, e.g. \(\text{HCl} \to \text{H}^+ + \text{Cl}^-\).
(d) Indicators:
- (i) HCl vs NaOH (strong acid/strong base): methyl orange or phenolphthalein.
- (ii) HCl vs NH4OH (strong acid/weak base): methyl orange.
- (iii) Ethanoic acid vs NaOH (weak acid/strong base): phenolphthalein.
(e) E = ammonium chloride (NH4Cl, a solid chloride that sublimes); F = an alkali such as sodium hydroxide (NaOH); G = ammonia (NH3), the choking gas that turns moist red litmus blue.
\[ \text{NH}_4\text{Cl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O} + \text{NH}_3 \]Question 4 Rapport
(a) Define each of the following terms: I. biotechnology; II. biogas.
(ii) State two applications of biotechnology
(b) (i) Describe briefly the production of ethanol from sugar cane juice.
(ii) State the by-product of the process in (b)(i)
(iii) Mention two uses of the by-product.
(iv) Ethanol can be produced from both cane sugar and petroleum. Explain briefly why the ethanol from cane sugar is renewable but that from petroleum is non-renewable.
(c) Distinguish between heavy chemicals and fine chemicals. Give one example of each chemical.
(d) Arrange the following gases in increasing order of deviation from ideal gas behaviour: HCl; O\(_2\); Cl\(_2\). Give reason(s) for your answer.
(a)(i) I. Biotechnology: the use of living organisms (or their components, such as enzymes or micro-organisms) to make useful products or to carry out useful processes for man. II. Biogas: a mixture of gases (mainly methane and carbon(IV) oxide) produced by the anaerobic bacterial decomposition of organic matter, used as a fuel.
(a)(ii) Applications of biotechnology: production of alcoholic drinks/ethanol by fermentation; production of antibiotics and drugs; production of biogas; making of bread, yoghurt and cheese (any two).
(b)(i) Ethanol from sugar cane juice: the sugar cane juice, which contains sucrose, is diluted and yeast is added. The enzyme invertase (sucrase) in yeast hydrolyses sucrose to glucose and fructose, then zymase ferments these sugars to ethanol at about 30-40 °C in the absence of air:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 \]The ethanol is then concentrated by fractional distillation.
(b)(ii) By-product: carbon(IV) oxide (CO2).
(b)(iii) Uses of the by-product: making fizzy (carbonated) drinks; used as a fire extinguisher; solid CO2 (dry ice) as a refrigerant.
(b)(iv) Ethanol from cane sugar is renewable because sugar cane is a crop that can be replanted and regrown within a short time, so the source is continuously replaceable. Ethanol from petroleum is non-renewable because petroleum is a fossil fuel formed over millions of years and cannot be replaced once used up.
(c) Heavy vs fine chemicals: heavy chemicals are produced in very large quantities, are relatively cheap and impure, e.g. tetraoxosulphate(VI) acid (H2SO4) or sodium trioxocarbonate(IV). Fine chemicals are produced in small quantities, are of high purity and are expensive, e.g. drugs/pharmaceuticals or dyes.
(d) Increasing order of deviation from ideal behaviour: O2 < Cl2 < HCl. Reason: deviation from ideal behaviour increases with the strength of intermolecular forces and with molecular size/polarity. O2 is small and non-polar (least deviation); Cl2 is larger and more polarisable; HCl is polar with strong dipole-dipole attractions, so it deviates most.
Détails de la réponse
(a)(i) I. Biotechnology: the use of living organisms (or their components, such as enzymes or micro-organisms) to make useful products or to carry out useful processes for man. II. Biogas: a mixture of gases (mainly methane and carbon(IV) oxide) produced by the anaerobic bacterial decomposition of organic matter, used as a fuel.
(a)(ii) Applications of biotechnology: production of alcoholic drinks/ethanol by fermentation; production of antibiotics and drugs; production of biogas; making of bread, yoghurt and cheese (any two).
(b)(i) Ethanol from sugar cane juice: the sugar cane juice, which contains sucrose, is diluted and yeast is added. The enzyme invertase (sucrase) in yeast hydrolyses sucrose to glucose and fructose, then zymase ferments these sugars to ethanol at about 30-40 °C in the absence of air:
\[ \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{zymase}} 2\text{C}_2\text{H}_5\text{OH} + 2\text{CO}_2 \]The ethanol is then concentrated by fractional distillation.
(b)(ii) By-product: carbon(IV) oxide (CO2).
(b)(iii) Uses of the by-product: making fizzy (carbonated) drinks; used as a fire extinguisher; solid CO2 (dry ice) as a refrigerant.
(b)(iv) Ethanol from cane sugar is renewable because sugar cane is a crop that can be replanted and regrown within a short time, so the source is continuously replaceable. Ethanol from petroleum is non-renewable because petroleum is a fossil fuel formed over millions of years and cannot be replaced once used up.
(c) Heavy vs fine chemicals: heavy chemicals are produced in very large quantities, are relatively cheap and impure, e.g. tetraoxosulphate(VI) acid (H2SO4) or sodium trioxocarbonate(IV). Fine chemicals are produced in small quantities, are of high purity and are expensive, e.g. drugs/pharmaceuticals or dyes.
(d) Increasing order of deviation from ideal behaviour: O2 < Cl2 < HCl. Reason: deviation from ideal behaviour increases with the strength of intermolecular forces and with molecular size/polarity. O2 is small and non-polar (least deviation); Cl2 is larger and more polarisable; HCl is polar with strong dipole-dipole attractions, so it deviates most.
Question 5 Rapport
(a) (i) Define standard electrode potential.
(ii) State two factors that affect the value of standard electrode potential.
(iii) Give two uses of the values of standard electrode potential.
(iv) Draw and label a diagram for an electrochemic cell made up of Cu\(^{2+}\)/Cu; E° = +0.34 V Zn\(^{2+}\)/Zn; E° = -0.76 V
(v) Calculate the e.m.f. of the cell in (a)(iv) above.
(b) (i) In terms of electron transfer, define: I. oxidation; II. oxidizing agent.
(ii) Balance the following redox reaction: MnO\(^{-4}\) + l\(^{-}\) \(\to\) l\(_{2}\) + Mn\(^{2+}\)
(c) Classiify each of the following oxides as basic, amphoteric, acidic or neutral: (i) Carbon (II) oxide; (ii) Sulphu (IV) oxide; (iii) Aluminium oxide; (iv) Lithium oxide.
(d) What is hydrogen bonding?
(a)(i) The standard electrode potential, E°, is the potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions: temperature 298 K, ion concentration 1.0 mol dm−3, and gas pressure 1 atm.
(ii) Two factors which affect electrode potential are:
- concentration of ions in the electrolyte;
- temperature.
(iii) Standard electrode potentials are used to:
- predict the direction or feasibility of a redox reaction;
- calculate the e.m.f. of an electrochemical cell.
(iv) A correctly labelled electrochemical cell is shown below.
Cell notation: \[\text{Zn(s)}\;|\;\text{Zn}^{2+}\text{(aq, 1 mol dm}^{-3}\text{)}\;||\;\text{Cu}^{2+}\text{(aq, 1 mol dm}^{-3}\text{)}\;|\;\text{Cu(s)}\] Zinc is the anode (negative electrode) where oxidation occurs, while copper is the cathode (positive electrode) where reduction occurs.
(v)
(b)(i)
- Oxidation is the loss of electrons.
- An oxidizing agent is a substance that accepts electrons and causes another substance to be oxidized.
(ii) In acidic solution, the balanced ionic equation is:
(c)
| Oxide | Classification |
|---|---|
| Carbon(II) oxide, CO | Neutral |
| Sulphur(IV) oxide, SO2 | Acidic |
| Aluminium oxide, Al2O3 | Amphoteric |
| Lithium oxide, Li2O | Basic |
(d) Hydrogen bonding is the strong intermolecular attraction between a hydrogen atom covalently bonded to nitrogen, oxygen, or fluorine in one molecule and a lone pair of electrons on nitrogen, oxygen, or fluorine in a neighbouring molecule.
Détails de la réponse
(a)(i) The standard electrode potential, E°, is the potential of a half-cell measured relative to the standard hydrogen electrode under standard conditions: temperature 298 K, ion concentration 1.0 mol dm−3, and gas pressure 1 atm.
(ii) Two factors which affect electrode potential are:
- concentration of ions in the electrolyte;
- temperature.
(iii) Standard electrode potentials are used to:
- predict the direction or feasibility of a redox reaction;
- calculate the e.m.f. of an electrochemical cell.
(iv) A correctly labelled electrochemical cell is shown below.
Cell notation: \[\text{Zn(s)}\;|\;\text{Zn}^{2+}\text{(aq, 1 mol dm}^{-3}\text{)}\;||\;\text{Cu}^{2+}\text{(aq, 1 mol dm}^{-3}\text{)}\;|\;\text{Cu(s)}\] Zinc is the anode (negative electrode) where oxidation occurs, while copper is the cathode (positive electrode) where reduction occurs.
(v)
(b)(i)
- Oxidation is the loss of electrons.
- An oxidizing agent is a substance that accepts electrons and causes another substance to be oxidized.
(ii) In acidic solution, the balanced ionic equation is:
(c)
| Oxide | Classification |
|---|---|
| Carbon(II) oxide, CO | Neutral |
| Sulphur(IV) oxide, SO2 | Acidic |
| Aluminium oxide, Al2O3 | Amphoteric |
| Lithium oxide, Li2O | Basic |
(d) Hydrogen bonding is the strong intermolecular attraction between a hydrogen atom covalently bonded to nitrogen, oxygen, or fluorine in one molecule and a lone pair of electrons on nitrogen, oxygen, or fluorine in a neighbouring molecule.
Question 6 Rapport
(a) (i) What is a structural isomer?
(ii) Write all the structural isomeric alkanols with the molecular formula C\(_4\)H\(_{10}\)O.
(iii) Which of the isomers from (a)(ii) above does not react easily on heating with acidified K\(_2\)Cr\(_2\)O\(_7\)?
(b) Chlorine reacted with excessentane in the presence of light. Chloropentane and a gas which fumes on cont with air were produced.
(i) Write an equation for the reaction.
(ii) Draw the structure of the major product.
(iii) What is the role of light in the reaction?
(iv) If a mixture of pentane and the major product is heated, which compounc would distil off first? Give a reason for your answer.
(v) Write the formula of the main product that would have be formed if but -1-ene has been used instead of pentane.
(c) Give the name and structural formula of the product which would be formed by hydration of each of the followinc compounds:
(i) CH\(_3\)CH(CH\(_3\))CH=CH\(_2\); (ii) CH\(_2\)=CHCOOH.
(d) (i) Write the structure of the amino acid, CH\(_3\)CH(NH\(_2\))COOH in: I. acidic medium; II. alkaline medium.
(ii) On analysis, an ammonium salt cf an alkanoic acid gave 60.5% carbon and 6.5% hydrogen. If 0.309 g of the salt yielded 0.0313 g of nitrogen, determine the empirical formula cf the salt. [H = 1.00; C =12.0; N =14.0; O = 16.0]
(a)(i) Structural isomer: one of two or more compounds having the same molecular formula but different structural arrangements (different order of bonding) of their atoms.
(a)(ii) Alkanols of C4H10O (the four alcohols):
- Butan-1-ol: CH3CH2CH2CH2OH
- Butan-2-ol: CH3CH2CH(OH)CH3
- 2-methylpropan-1-ol: (CH3)2CHCH2OH
- 2-methylpropan-2-ol: (CH3)3COH
(a)(iii) 2-methylpropan-2-ol (the tertiary alcohol) does not react easily with acidified K2Cr2O7 because a tertiary alcohol has no hydrogen atom on the carbon bearing the OH group and so resists oxidation.
(b) Chlorine with excess pentane in light
(i) \(\text{C}_5\text{H}_{12} + \text{Cl}_2 \xrightarrow{\text{light}} \text{C}_5\text{H}_{11}\text{Cl} + \text{HCl}\)
(ii) Major product (1-chloropentane): CH3CH2CH2CH2CH2Cl.
(iii) Light supplies the energy that homolytically splits the Cl2 molecule into chlorine free radicals, initiating the substitution (free-radical) reaction.
(iv) Pentane would distil off first because it has the lower boiling point; 1-chloropentane has a higher relative molecular mass and stronger dipole/van der Waals forces, hence a higher boiling point.
(v) With but-1-ene the reaction is addition, giving 1,2-dichlorobutane, CH2ClCHClCH2CH3.
(c) Hydration products
- (i) CH3CH(CH3)CH=CH2 adds water (Markovnikov) to give 3-methylbutan-2-ol, CH3CH(CH3)CH(OH)CH3.
- (ii) CH2=CHCOOH gives 3-hydroxypropanoic acid, HOCH2CH2COOH.
(d)(i) Amino acid CH3CH(NH2)COOH:
- I. Acidic medium (amino group protonated): CH3CH(NH3+)COOH.
- II. Alkaline medium (acid deprotonated): CH3CH(NH2)COO-.
(d)(ii) Empirical formula of the ammonium salt
%N \(= \dfrac{0.0313}{0.309}\times100 = 10.1\%\). Then %O \(= 100 - 60.5 - 6.5 - 10.1 = 22.9\%\).
| Element | % | moles (%/Ar) | ratio (/0.72) |
|---|---|---|---|
| C | 60.5 | 5.04 | 7 |
| H | 6.5 | 6.50 | 9 |
| N | 10.1 | 0.72 | 1 |
| O | 22.9 | 1.43 | 2 |
Empirical formula: C7H9NO2 (consistent with an ammonium salt, RCOONH4).
Détails de la réponse
(a)(i) Structural isomer: one of two or more compounds having the same molecular formula but different structural arrangements (different order of bonding) of their atoms.
(a)(ii) Alkanols of C4H10O (the four alcohols):
- Butan-1-ol: CH3CH2CH2CH2OH
- Butan-2-ol: CH3CH2CH(OH)CH3
- 2-methylpropan-1-ol: (CH3)2CHCH2OH
- 2-methylpropan-2-ol: (CH3)3COH
(a)(iii) 2-methylpropan-2-ol (the tertiary alcohol) does not react easily with acidified K2Cr2O7 because a tertiary alcohol has no hydrogen atom on the carbon bearing the OH group and so resists oxidation.
(b) Chlorine with excess pentane in light
(i) \(\text{C}_5\text{H}_{12} + \text{Cl}_2 \xrightarrow{\text{light}} \text{C}_5\text{H}_{11}\text{Cl} + \text{HCl}\)
(ii) Major product (1-chloropentane): CH3CH2CH2CH2CH2Cl.
(iii) Light supplies the energy that homolytically splits the Cl2 molecule into chlorine free radicals, initiating the substitution (free-radical) reaction.
(iv) Pentane would distil off first because it has the lower boiling point; 1-chloropentane has a higher relative molecular mass and stronger dipole/van der Waals forces, hence a higher boiling point.
(v) With but-1-ene the reaction is addition, giving 1,2-dichlorobutane, CH2ClCHClCH2CH3.
(c) Hydration products
- (i) CH3CH(CH3)CH=CH2 adds water (Markovnikov) to give 3-methylbutan-2-ol, CH3CH(CH3)CH(OH)CH3.
- (ii) CH2=CHCOOH gives 3-hydroxypropanoic acid, HOCH2CH2COOH.
(d)(i) Amino acid CH3CH(NH2)COOH:
- I. Acidic medium (amino group protonated): CH3CH(NH3+)COOH.
- II. Alkaline medium (acid deprotonated): CH3CH(NH2)COO-.
(d)(ii) Empirical formula of the ammonium salt
%N \(= \dfrac{0.0313}{0.309}\times100 = 10.1\%\). Then %O \(= 100 - 60.5 - 6.5 - 10.1 = 22.9\%\).
| Element | % | moles (%/Ar) | ratio (/0.72) |
|---|---|---|---|
| C | 60.5 | 5.04 | 7 |
| H | 6.5 | 6.50 | 9 |
| N | 10.1 | 0.72 | 1 |
| O | 22.9 | 1.43 | 2 |
Empirical formula: C7H9NO2 (consistent with an ammonium salt, RCOONH4).
