Chemistry (9-1) 0971 | Paper 4 Mock 01 | Theory (Extended)
Question 1 Report
PET is the polyester used to make transparent drinks bottles. It is made from ethane-1,2-diol (Fig. 12.1) and benzene-1,4-dicarboxylic acid (Fig. 12.2).
Fig. 12.1:
Fig. 12.2:
(a) Name the class of compound to which each monomer belongs. [2]
(b) Draw the structure of the ester linkage formed when the -OH group of one monomer reacts with the -COOH group of the other. [3]
(c) Name the small molecule eliminated and state how many of these molecules are lost each time one ester linkage is formed. [2]
(d) Describe how a long PET molecule is built up from these two monomers. [3]
(e) When PET is hydrolysed, what are the products? [2]
(f) State two reasons why recycling PET bottles is better than sending them to landfill. [2]
PET is a polyester built by condensation of a diol (ethane-1,2-diol, HO-CH2-CH2-OH) and a dicarboxylic acid (benzene-1,4-dicarboxylic acid). The question tests the ester linkage, the loss of water, and how a chain grows from difunctional monomers.
(a) Class of each monomer [2]. Ethane-1,2-diol is an alcohol (specifically a diol, two -OH groups) [1]; benzene-1,4-dicarboxylic acid is a carboxylic acid (a dicarboxylic acid, two -COOH groups) [1].
(b) Ester linkage [3]. When an -OH reacts with a -COOH, the -OH of the acid and the -H of the alcohol leave as water, leaving the ester group -C(=O)-O-. The displayed linkage is:
Marks: carbon double-bonded to oxygen (C=O) [1]; that carbon single-bonded to a second oxygen (C-O) [1]; the linkage correctly joining the acid residue on one side to the alcohol residue on the other [1].
(c) Small molecule eliminated [2]. Water, H2O [1]; exactly one molecule of water is lost for every one ester linkage formed [1].
(d) Building a long PET molecule [3]. Each diol carries two -OH groups and each acid carries two -COOH groups [1]. An -OH reacts with a -COOH to form an ester linkage and release a molecule of water [1]. Because both ends of every monomer are reactive, the chain keeps growing alternately -acid-diol-acid-diol- into a long polymer [1].
(e) Hydrolysis products [2]. Hydrolysis reverses the polymerisation: water splits every ester linkage, regenerating the two starting monomers, ethane-1,2-diol [1] and benzene-1,4-dicarboxylic acid [1].
(f) Two reasons to recycle rather than landfill [2]. Any two of: it saves crude oil (the raw material for the plastic); it saves energy; it reduces the volume of non-biodegradable waste and litter in landfill; it reduces carbon dioxide emissions from making new plastic [1] [1].
Exam tip: the ester group is C=O and C-O on the SAME carbon; drawing the two oxygens on different carbons loses the linkage mark.
Answer Details
PET is a polyester built by condensation of a diol (ethane-1,2-diol, HO-CH2-CH2-OH) and a dicarboxylic acid (benzene-1,4-dicarboxylic acid). The question tests the ester linkage, the loss of water, and how a chain grows from difunctional monomers.
(a) Class of each monomer [2]. Ethane-1,2-diol is an alcohol (specifically a diol, two -OH groups) [1]; benzene-1,4-dicarboxylic acid is a carboxylic acid (a dicarboxylic acid, two -COOH groups) [1].
(b) Ester linkage [3]. When an -OH reacts with a -COOH, the -OH of the acid and the -H of the alcohol leave as water, leaving the ester group -C(=O)-O-. The displayed linkage is:
Marks: carbon double-bonded to oxygen (C=O) [1]; that carbon single-bonded to a second oxygen (C-O) [1]; the linkage correctly joining the acid residue on one side to the alcohol residue on the other [1].
(c) Small molecule eliminated [2]. Water, H2O [1]; exactly one molecule of water is lost for every one ester linkage formed [1].
(d) Building a long PET molecule [3]. Each diol carries two -OH groups and each acid carries two -COOH groups [1]. An -OH reacts with a -COOH to form an ester linkage and release a molecule of water [1]. Because both ends of every monomer are reactive, the chain keeps growing alternately -acid-diol-acid-diol- into a long polymer [1].
(e) Hydrolysis products [2]. Hydrolysis reverses the polymerisation: water splits every ester linkage, regenerating the two starting monomers, ethane-1,2-diol [1] and benzene-1,4-dicarboxylic acid [1].
(f) Two reasons to recycle rather than landfill [2]. Any two of: it saves crude oil (the raw material for the plastic); it saves energy; it reduces the volume of non-biodegradable waste and litter in landfill; it reduces carbon dioxide emissions from making new plastic [1] [1].
Exam tip: the ester group is C=O and C-O on the SAME carbon; drawing the two oxygens on different carbons loses the linkage mark.
Question 2 Report
Displacement reactions can be used to place the halogens in order of reactivity. A few drops of each halogen solution are added to solutions of potassium halides. The results are recorded in Table 11.1, where 'no reaction' means no colour change was seen.
| chlorine water | bromine water | iodine solution | |
|---|---|---|---|
| potassium chloride solution | no reaction | no reaction | no reaction |
| potassium bromide solution | … | no reaction | no reaction |
| potassium iodide solution | … | … | no reaction |
(a) Complete the three shaded boxes, stating in each case whether a reaction occurs and, if so, the colour change seen. [3]
(b) Use the results to place chlorine, bromine and iodine in order of decreasing reactivity. [1]
(c) Explain why a more reactive halogen is able to displace a less reactive halogen from a solution of its salt. Refer to electron gain in your answer. [3]
(d) Write the ionic equation, with state symbols, for the reaction between bromine and potassium iodide solution. [2]
(e) In the reaction in (d), state which species is reduced and explain your choice in terms of electrons. [2]
(f) Predict whether astatine (below iodine in Group VII) would displace iodine from potassium iodide solution, and justify your prediction. [2]
Reactivity in Group VII decreases down the group, so chlorine is the most reactive of these three and iodine the least. A halogen can only displace another halogen that is below it (less reactive) in the group.
(a) The completed results, with the colour change seen when a halogen is displaced:
| chlorine water | bromine water | iodine solution | |
|---|---|---|---|
| potassium chloride | no reaction | no reaction | no reaction |
| potassium bromide | reaction: colourless to orange/brown | no reaction | no reaction |
| potassium iodide | reaction: colourless to brown | reaction: orange to brown | no reaction |
Marks: chlorine with potassium bromide, colourless to orange/brown (bromine set free) [1]; chlorine with potassium iodide, colourless to brown (iodine set free) [1]; bromine with potassium iodide, orange to brown (iodine set free) [1].
(b) Order of decreasing reactivity: chlorine, then bromine, then iodine [1].
(c) A halogen reacts by gaining an electron to become a halide ion. A more reactive halogen gains electrons more readily [1], so it pulls the electron away from the ion of the less reactive halogen [1]; the more reactive halogen becomes a halide ion while the less reactive halogen is set free as neutral molecules, that is, it is displaced [1].
(d) Ionic equation with state symbols [species 1; balancing and states 1]:
\[ \text{Br}_2(aq) + 2\text{I}^{-}(aq) \rightarrow 2\text{Br}^{-}(aq) + \text{I}_2(aq) \](e) Bromine, \(\text{Br}_2\), is reduced [1], because each bromine atom gains an electron to become \(\text{Br}^{-}\), and gain of electrons is reduction [1]. (The iodide ions are oxidised, losing electrons to form \(\text{I}_2\).)
(f) Astatine lies below iodine, so it is the least reactive halogen here and gains electrons less readily than iodine. It would therefore not displace iodine [1] from potassium iodide solution [1].
Answer Details
Reactivity in Group VII decreases down the group, so chlorine is the most reactive of these three and iodine the least. A halogen can only displace another halogen that is below it (less reactive) in the group.
(a) The completed results, with the colour change seen when a halogen is displaced:
| chlorine water | bromine water | iodine solution | |
|---|---|---|---|
| potassium chloride | no reaction | no reaction | no reaction |
| potassium bromide | reaction: colourless to orange/brown | no reaction | no reaction |
| potassium iodide | reaction: colourless to brown | reaction: orange to brown | no reaction |
Marks: chlorine with potassium bromide, colourless to orange/brown (bromine set free) [1]; chlorine with potassium iodide, colourless to brown (iodine set free) [1]; bromine with potassium iodide, orange to brown (iodine set free) [1].
(b) Order of decreasing reactivity: chlorine, then bromine, then iodine [1].
(c) A halogen reacts by gaining an electron to become a halide ion. A more reactive halogen gains electrons more readily [1], so it pulls the electron away from the ion of the less reactive halogen [1]; the more reactive halogen becomes a halide ion while the less reactive halogen is set free as neutral molecules, that is, it is displaced [1].
(d) Ionic equation with state symbols [species 1; balancing and states 1]:
\[ \text{Br}_2(aq) + 2\text{I}^{-}(aq) \rightarrow 2\text{Br}^{-}(aq) + \text{I}_2(aq) \](e) Bromine, \(\text{Br}_2\), is reduced [1], because each bromine atom gains an electron to become \(\text{Br}^{-}\), and gain of electrons is reduction [1]. (The iodide ions are oxidised, losing electrons to form \(\text{I}_2\).)
(f) Astatine lies below iodine, so it is the least reactive halogen here and gains electrons less readily than iodine. It would therefore not displace iodine [1] from potassium iodide solution [1].
Question 3 Report
Fig. 7.1 shows the apparatus a student sets up to obtain a cooling curve for a pure solid that melts a little above room temperature. A boiling tube containing the molten substance and a thermometer is supported so that it cools in the surrounding air.
(a) Describe how the student would use this apparatus to obtain the readings needed to plot a cooling curve. [3]
(b) State the two quantities that must be measured during the experiment. [2]
(c) Explain, in terms of the particles, what happens to the substance as it cools from a liquid to a solid. [3]
(d) The solid is first melted by standing the boiling tube in a beaker of hot water rather than heating it directly in a Bunsen flame. Explain one advantage of using a water bath. [2]
(e) Suggest two improvements that would make the recorded temperatures more reliable. [3]
Labelled answer diagram:
This tests the practical method for obtaining a cooling curve and the reasoning behind good technique.
(a) Method [3]: first melt the substance so the thermometer bulb is fully covered by the liquid [1]; then allow it to cool and record the temperature at regular fixed time intervals (for example every minute) using a stopwatch [1]; continue until the substance has fully solidified / the temperature has fallen well below the freezing point, then plot temperature against time [1]. Regular timed readings are what let you draw a smooth curve and spot the flat freezing section.
(b) The two quantities measured are the temperature of the substance [1] and the time [1] - these become the y-axis and x-axis of the cooling curve.
(c) As it cools from liquid to solid [3]: the liquid particles lose kinetic energy and move more slowly [1]; the attractive forces pull the particles together into a regular fixed arrangement [1]; the particles can then only vibrate about fixed positions, so a solid forms [1].
(d) Using a water bath rather than a direct flame: the heating is more even and gentle, so the substance does not overheat above its melting point [1]; and it is safer because the substance is flammable and there is no naked flame near it (it also avoids local overheating) [1].
(e) Improvements for more reliable temperatures (any two, plus a valid third) [3]: stir the liquid so the temperature is even throughout [1]; use a thermometer with finer / smaller divisions or read at eye level to avoid parallax error [1]; and repeat the experiment and average, or take readings at more frequent intervals [1].
Answer Details
Labelled answer diagram:
This tests the practical method for obtaining a cooling curve and the reasoning behind good technique.
(a) Method [3]: first melt the substance so the thermometer bulb is fully covered by the liquid [1]; then allow it to cool and record the temperature at regular fixed time intervals (for example every minute) using a stopwatch [1]; continue until the substance has fully solidified / the temperature has fallen well below the freezing point, then plot temperature against time [1]. Regular timed readings are what let you draw a smooth curve and spot the flat freezing section.
(b) The two quantities measured are the temperature of the substance [1] and the time [1] - these become the y-axis and x-axis of the cooling curve.
(c) As it cools from liquid to solid [3]: the liquid particles lose kinetic energy and move more slowly [1]; the attractive forces pull the particles together into a regular fixed arrangement [1]; the particles can then only vibrate about fixed positions, so a solid forms [1].
(d) Using a water bath rather than a direct flame: the heating is more even and gentle, so the substance does not overheat above its melting point [1]; and it is safer because the substance is flammable and there is no naked flame near it (it also avoids local overheating) [1].
(e) Improvements for more reliable temperatures (any two, plus a valid third) [3]: stir the liquid so the temperature is even throughout [1]; use a thermometer with finer / smaller divisions or read at eye level to avoid parallax error [1]; and repeat the experiment and average, or take readings at more frequent intervals [1].
Question 4 Report
Ethene reacts readily with bromine at room temperature. Fig. 4.1 shows the displayed formula of ethene, the molecule that reacts.
Fig. 4.1
(a) Bromine can be added as bromine water. Describe what you would see when ethene is bubbled through bromine water. [2]
(b) State the type of reaction that occurs between ethene and bromine. [1]
(c) Name the product formed when ethene reacts with bromine. [1]
(d) Give the molecular formula of this product. [1]
(e) Draw the displayed formula of the product, showing every atom and every bond. [2]
(f) State whether the product is saturated or unsaturated, and explain your answer. [2]
(g) State what the fast reaction of ethene with bromine tells you about the ethene molecule. [1]
This question is about the addition reaction between ethene and bromine, testing the observation, reaction type, product, its formula and displayed structure, and what the reaction reveals about ethene.
(a) The orange (yellow-brown) colour of the bromine water [1] disappears; the mixture turns colourless, that is the bromine water is decolourised [1]. This colour change is the standard test for an unsaturated compound.
(b) The reaction is an addition reaction [1]: the bromine adds across the double bond and nothing is given off.
(c) The product is 1,2-dibromoethane (dibromoethane) [1].
(d) Its molecular formula is \( C_2H_4Br_2 \) [1]: the two bromine atoms simply add to the ethene, \( C_2H_4 \).
(e) The displayed formula shows the C=C double bond opened to a single bond, with one bromine atom added to each carbon and every bond drawn in full:
One mark for the correct single C-C bond with a bromine on each carbon, one mark for the correct four hydrogen atoms with all bonds shown [2].
(f) The product is saturated [1]: the C=C double bond has opened, leaving only single bonds, so no more atoms can add on [1].
(g) The fast reaction tells you that ethene is unsaturated, that is it contains a reactive carbon-carbon double bond that readily takes part in addition [1].
Answer Details
This question is about the addition reaction between ethene and bromine, testing the observation, reaction type, product, its formula and displayed structure, and what the reaction reveals about ethene.
(a) The orange (yellow-brown) colour of the bromine water [1] disappears; the mixture turns colourless, that is the bromine water is decolourised [1]. This colour change is the standard test for an unsaturated compound.
(b) The reaction is an addition reaction [1]: the bromine adds across the double bond and nothing is given off.
(c) The product is 1,2-dibromoethane (dibromoethane) [1].
(d) Its molecular formula is \( C_2H_4Br_2 \) [1]: the two bromine atoms simply add to the ethene, \( C_2H_4 \).
(e) The displayed formula shows the C=C double bond opened to a single bond, with one bromine atom added to each carbon and every bond drawn in full:
One mark for the correct single C-C bond with a bromine on each carbon, one mark for the correct four hydrogen atoms with all bonds shown [2].
(f) The product is saturated [1]: the C=C double bond has opened, leaving only single bonds, so no more atoms can add on [1].
(g) The fast reaction tells you that ethene is unsaturated, that is it contains a reactive carbon-carbon double bond that readily takes part in addition [1].
Question 5 Report
This question is about copper and about the arrangement of elements in the Periodic Table.
Table 18.1 gives information about the two isotopes of copper.
| Isotope of copper | Nucleon number | Number of protons | Number of neutrons | Abundance / % |
|---|---|---|---|---|
| copper-63 | 63 | 29 | … | 70 |
| copper-65 | 65 | 29 | … | 30 |
(a) Complete the shaded 'Number of neutrons' column of Table 18.1. [2]
(b) Explain, in terms of subatomic particles, why the two isotopes are both atoms of copper yet have different masses. [2]
(c) Calculate the relative atomic mass of copper from the abundance data. Show all your working. [3]
(d) Copper is a transition element. State two properties of copper, other than its density, that are typical of a transition metal but not of a Group I metal. [2]
(e) Fig. 18.1 shows the electron arrangement of a different element, Z, in the same period as copper.
(i) Deduce the proton number, group and period of element Z. [3]
(ii) Element Z reacts vigorously with cold water. Write a balanced symbol equation for the reaction, giving the hydroxide of Z (formula ZOH) and a gas. [2]
(iii) Explain why element Z is more reactive than the element directly above it in the same group. [2]
(f) Copper forms ions Cu+ and Cu2+. State the term for this behaviour and give the formulae of the two oxides these ions form with O2−. [3]
(g) State, giving a reason, whether copper or element Z is the better conductor of heat, or whether they would be similar. [1]
(h) Both copper and element Z are metals. State one feature of their structure that they share. [1]
(a) Neutrons = nucleon number \(-\) protons. Copper-63: \(63 - 29 = 34\) neutrons [1]; copper-65: \(65 - 29 = 36\) neutrons [1].
(b) Both isotopes have the same number of protons, 29, so both are copper [1]; they differ in neutrons (34 and 36), giving different nucleon numbers and therefore different masses [1]. Chemical identity is fixed by proton number, not by mass.
(c) Relative atomic mass is the weighted mean [method 1]:
\[ A_r = \frac{(63\times 70) + (65\times 30)}{100} = \frac{4410 + 1950}{100} = \frac{6360}{100} \quad [1] \] \[ = 63.6 \quad [1] \](d) Any two transition-metal properties, other than density: forms coloured compounds; shows variable valency (\(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\)); acts as a catalyst; has a high melting point [2]. A Group I metal shows none of these (white compounds, a single 1+ charge, low melting point).
(e) Element Z has \(2 + 8 + 8 + 1 = 19\) electrons.
(i) Proton number 19 [1]; Group I [1] (1 outer electron); Period 4 [1] (4 occupied shells).
(ii) A Group I metal reacts with cold water to give its hydroxide and hydrogen gas [products 1; balanced 1]:
\[ 2\text{Z} + 2\text{H}_2\text{O} \rightarrow 2\text{ZOH} + \text{H}_2 \](iii) Z is more reactive than the element directly above it because it has one more electron shell, so its outer electron is further from the nucleus and more shielded [1]; the weaker nuclear pull means that outer electron is lost more easily, and losing that electron is how a Group I metal reacts [1].
(f) Forming both \(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\) is variable valency (variable oxidation state) [1]. With the oxide ion \(\text{O}^{2-}\) the charges balance to give \(\text{Cu}_2\text{O}\) [1] (two 1+ ions per oxide) and \(\text{CuO}\) [1] (one 2+ ion per oxide).
(g) They would be similar; both are good conductors of heat [1], because both are metals with delocalised electrons free to move and transfer thermal energy.
(h) They share a giant metallic lattice of positive ions in a sea of delocalised electrons [1].
Answer Details
(a) Neutrons = nucleon number \(-\) protons. Copper-63: \(63 - 29 = 34\) neutrons [1]; copper-65: \(65 - 29 = 36\) neutrons [1].
(b) Both isotopes have the same number of protons, 29, so both are copper [1]; they differ in neutrons (34 and 36), giving different nucleon numbers and therefore different masses [1]. Chemical identity is fixed by proton number, not by mass.
(c) Relative atomic mass is the weighted mean [method 1]:
\[ A_r = \frac{(63\times 70) + (65\times 30)}{100} = \frac{4410 + 1950}{100} = \frac{6360}{100} \quad [1] \] \[ = 63.6 \quad [1] \](d) Any two transition-metal properties, other than density: forms coloured compounds; shows variable valency (\(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\)); acts as a catalyst; has a high melting point [2]. A Group I metal shows none of these (white compounds, a single 1+ charge, low melting point).
(e) Element Z has \(2 + 8 + 8 + 1 = 19\) electrons.
(i) Proton number 19 [1]; Group I [1] (1 outer electron); Period 4 [1] (4 occupied shells).
(ii) A Group I metal reacts with cold water to give its hydroxide and hydrogen gas [products 1; balanced 1]:
\[ 2\text{Z} + 2\text{H}_2\text{O} \rightarrow 2\text{ZOH} + \text{H}_2 \](iii) Z is more reactive than the element directly above it because it has one more electron shell, so its outer electron is further from the nucleus and more shielded [1]; the weaker nuclear pull means that outer electron is lost more easily, and losing that electron is how a Group I metal reacts [1].
(f) Forming both \(\text{Cu}^{+}\) and \(\text{Cu}^{2+}\) is variable valency (variable oxidation state) [1]. With the oxide ion \(\text{O}^{2-}\) the charges balance to give \(\text{Cu}_2\text{O}\) [1] (two 1+ ions per oxide) and \(\text{CuO}\) [1] (one 2+ ion per oxide).
(g) They would be similar; both are good conductors of heat [1], because both are metals with delocalised electrons free to move and transfer thermal energy.
(h) They share a giant metallic lattice of positive ions in a sea of delocalised electrons [1].
Question 6 Report
Chlorine gas is bubbled through a colourless solution of potassium bromide.
(a) State the colour change seen in the solution during the reaction. [1]
(b) Name this type of reaction. [1]
(c) Write the ionic equation for the reaction between chlorine and bromide ions. [2]
(d) Explain why chlorine is able to react with potassium bromide in this way. [1]
This question tests the halogen displacement rule: a more reactive halogen removes a less reactive halogen from a solution of its ions. Chlorine is above bromine in Group VII, so it is the more reactive of the two.
(a) The potassium bromide solution starts colourless. As chlorine displaces bromine, free bromine (Br2) dissolves in the water and colours it, so the colour change is colourless to orange / yellow-brown [1]. The orange-brown colour is the visible sign that a halogen has been set free.
(b) One element (bromine) is pushed out of its compound by a more reactive element (chlorine), so this is a displacement reaction [1]. Because electrons are also transferred, redox is accepted.
(c) The balanced ionic equation, showing only the species that change, is:
Cl2 + 2Br- → 2Cl- + Br2 [1 for correct species; 1 for balancing with charges = 2]
Check: 2 Cl and 2 Br on each side, and total charge is 2- on both sides. The potassium ions (K+) are spectator ions, so they are left out of the ionic equation.
(d) Chlorine reacts this way because it is more reactive than bromine, so it is able to take the electrons from the bromide ions and displace bromine from solution [1]. Reactivity falls down Group VII, so the halogen that is higher up always displaces the one below it.
Exam tip: for the ionic equation, cancel the spectator ion (K+) and always check the charges balance, not just the atoms.
Answer Details
This question tests the halogen displacement rule: a more reactive halogen removes a less reactive halogen from a solution of its ions. Chlorine is above bromine in Group VII, so it is the more reactive of the two.
(a) The potassium bromide solution starts colourless. As chlorine displaces bromine, free bromine (Br2) dissolves in the water and colours it, so the colour change is colourless to orange / yellow-brown [1]. The orange-brown colour is the visible sign that a halogen has been set free.
(b) One element (bromine) is pushed out of its compound by a more reactive element (chlorine), so this is a displacement reaction [1]. Because electrons are also transferred, redox is accepted.
(c) The balanced ionic equation, showing only the species that change, is:
Cl2 + 2Br- → 2Cl- + Br2 [1 for correct species; 1 for balancing with charges = 2]
Check: 2 Cl and 2 Br on each side, and total charge is 2- on both sides. The potassium ions (K+) are spectator ions, so they are left out of the ionic equation.
(d) Chlorine reacts this way because it is more reactive than bromine, so it is able to take the electrons from the bromide ions and displace bromine from solution [1]. Reactivity falls down Group VII, so the halogen that is higher up always displaces the one below it.
Exam tip: for the ionic equation, cancel the spectator ion (K+) and always check the charges balance, not just the atoms.
