Chemistry 0620 | Paper 4 Mock 01 | Theory (Extended)
Question 1 Report
Fig. 1.1 represents the electron arrangement of a neutral atom of element J. Each dot is one electron.
(a) State the number of electrons in the outer (highest) energy level of atom J. [1]
(b) Use Fig. 1.1 to deduce the group of the Periodic Table to which J belongs. [1]
(c) State how many occupied electron shells J has, and hence give its period number. [2]
(d) An atom of J contains 18 neutrons. Calculate its nucleon number. [1]
What this tests: reading group and period from an electron diagram, and working out nucleon number.
The dots fill the shells 2, then 8, then 7 outward from the centre.
(a) The outer (third) shell holds 7 electrons [1].
(b) The number of outer-shell electrons gives the group number, so J is in Group VII [1].
(c) There are 3 occupied shells [1], and the number of occupied shells gives the period, so J is in Period 3 [1].
(d) The proton number equals the total number of electrons: \(2+8+7 = 17\). Nucleon number = protons + neutrons = \(17 + 18 = 35\) [1]. (These values identify J as chlorine.)
Answer Details
What this tests: reading group and period from an electron diagram, and working out nucleon number.
The dots fill the shells 2, then 8, then 7 outward from the centre.
(a) The outer (third) shell holds 7 electrons [1].
(b) The number of outer-shell electrons gives the group number, so J is in Group VII [1].
(c) There are 3 occupied shells [1], and the number of occupied shells gives the period, so J is in Period 3 [1].
(d) The proton number equals the total number of electrons: \(2+8+7 = 17\). Nucleon number = protons + neutrons = \(17 + 18 = 35\) [1]. (These values identify J as chlorine.)
Question 2 Report
Sulfuric acid is manufactured by the Contact process, outlined in Fig. 14.1.
Fig. 14.1
(a) Write balanced equations for the three main reaction stages: burning sulfur in air; the oxidation of sulfur dioxide; and the formation of oleum. [3]
(b) For the sulfur dioxide oxidation stage, state the temperature and the pressure used, and give one reason for the choice of each. [4]
(c) Name the catalyst used in this stage. [1]
(d) Explain why sulfur trioxide is dissolved in concentrated sulfuric acid rather than directly in water. [2]
(e) Calculate the maximum mass of sulfuric acid that could be made from 64 tonnes of sulfur dioxide. (Ar: S = 32, O = 16, H = 1) [3]
(f) Give two uses of sulfuric acid. [2]
This question tests the Contact process for making sulfuric acid: the three reaction stages, the conditions and their reasons, and a mass calculation.
(a) The three main stages, each balanced [1] each:
Burning sulfur: \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)
Oxidation of sulfur dioxide: \( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \)
Formation of oleum: \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)
(b) For the SO2 oxidation stage the temperature is about 450°C [1]: the forward reaction is exothermic, so a lower temperature would give a higher yield, but 450°C keeps the rate fast enough, making it a compromise [1]. The pressure is only about 2 atm [1]: the yield is already about 98% at low pressure, so a higher, more expensive pressure is not worthwhile [1].
(c) The catalyst is vanadium(V) oxide, V2O5 [1].
(d) Adding SO3 straight into water produces a large amount of heat and a dangerous, choking mist (fog) of sulfuric acid that is hard to condense [1]. Dissolving SO3 in concentrated sulfuric acid to form oleum, then carefully adding water, is safe and controlled [1].
(e) Work in tonne-moles. The relative formula masses are SO2 = 32 + (2 × 16) = 64 and H2SO4 = (2 × 1) + 32 + (4 × 16) = 98.
\[ n(\text{SO}_2) = \frac{64}{64} = 1 \text{ tonne-mol} \]The stages carry through one mole of sulfur atoms, so 1 mol SO2 gives 1 mol H2SO4 [1]. Using M(H2SO4) = 98 [1]:
\[ \text{mass} = 1 \times 98 = 98 \text{ tonnes} \; [1] \](f) Any two uses of sulfuric acid: making fertilisers, detergents, paints, use as car battery acid, or making other chemicals [1][1].
Exam reminder: in tonne-mole calculations you can keep working in tonnes as long as you divide by the relative formula mass; the mole ratio from the balanced equations is what links reactant to product.
Answer Details
This question tests the Contact process for making sulfuric acid: the three reaction stages, the conditions and their reasons, and a mass calculation.
(a) The three main stages, each balanced [1] each:
Burning sulfur: \( \text{S} + \text{O}_2 \rightarrow \text{SO}_2 \)
Oxidation of sulfur dioxide: \( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \)
Formation of oleum: \( \text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \)
(b) For the SO2 oxidation stage the temperature is about 450°C [1]: the forward reaction is exothermic, so a lower temperature would give a higher yield, but 450°C keeps the rate fast enough, making it a compromise [1]. The pressure is only about 2 atm [1]: the yield is already about 98% at low pressure, so a higher, more expensive pressure is not worthwhile [1].
(c) The catalyst is vanadium(V) oxide, V2O5 [1].
(d) Adding SO3 straight into water produces a large amount of heat and a dangerous, choking mist (fog) of sulfuric acid that is hard to condense [1]. Dissolving SO3 in concentrated sulfuric acid to form oleum, then carefully adding water, is safe and controlled [1].
(e) Work in tonne-moles. The relative formula masses are SO2 = 32 + (2 × 16) = 64 and H2SO4 = (2 × 1) + 32 + (4 × 16) = 98.
\[ n(\text{SO}_2) = \frac{64}{64} = 1 \text{ tonne-mol} \]The stages carry through one mole of sulfur atoms, so 1 mol SO2 gives 1 mol H2SO4 [1]. Using M(H2SO4) = 98 [1]:
\[ \text{mass} = 1 \times 98 = 98 \text{ tonnes} \; [1] \](f) Any two uses of sulfuric acid: making fertilisers, detergents, paints, use as car battery acid, or making other chemicals [1][1].
Exam reminder: in tonne-mole calculations you can keep working in tonnes as long as you divide by the relative formula mass; the mole ratio from the balanced equations is what links reactant to product.
Question 3 Report
An atom of an element is represented using nuclide notation. Fig. 1.1 shows the nuclide notation for one atom of this element.
Fig. 1.1
(a) State what is meant by the term isotopes. [2]
(b) Use Fig. 1.1 to deduce the number of neutrons in this atom. [1]
(c) State the number of electrons in a neutral atom of this element. [1]
(d) A different isotope of the same element has a nucleon number of 35. State the number of neutrons in one atom of this isotope. [1]
(e) State one physical property that is different for the two isotopes. [1]
What this tests: the definition of isotopes and using nuclide notation to find neutrons and electrons.
(a) Isotopes [2] Isotopes are atoms of the same element [1] that have the same number of protons but different numbers of neutrons (same proton number, different nucleon numbers) [1].
(b) Neutrons [1] In nuclide notation the top number (37) is the nucleon number and the bottom number (17) is the proton number, so neutrons \(= 37 - 17 = 20\) [1].
(c) Electrons [1] A neutral atom has equal numbers of protons and electrons, so there are 17 electrons [1].
(d) Other isotope [1] With nucleon number 35, neutrons \(= 35 - 17 = 18\) [1].
(e) Different physical property [1] The two isotopes have different masses (accept different density or different rate of diffusion) [1]. Their chemical properties are identical because they have the same electron arrangement.
Answer Details
What this tests: the definition of isotopes and using nuclide notation to find neutrons and electrons.
(a) Isotopes [2] Isotopes are atoms of the same element [1] that have the same number of protons but different numbers of neutrons (same proton number, different nucleon numbers) [1].
(b) Neutrons [1] In nuclide notation the top number (37) is the nucleon number and the bottom number (17) is the proton number, so neutrons \(= 37 - 17 = 20\) [1].
(c) Electrons [1] A neutral atom has equal numbers of protons and electrons, so there are 17 electrons [1].
(d) Other isotope [1] With nucleon number 35, neutrons \(= 35 - 17 = 18\) [1].
(e) Different physical property [1] The two isotopes have different masses (accept different density or different rate of diffusion) [1]. Their chemical properties are identical because they have the same electron arrangement.
Question 4 Report
Fig. 12.1 shows a blast furnace divided into three zones, A, B and C, at different heights.
(a) Name the three solid raw materials added at the top of the furnace. [3]
(b) In zone C, coke burns in the hot air. Write the balanced equation and state whether the reaction is exothermic or endothermic. [2]
(c) In zone B, carbon monoxide is formed. Describe how it is formed and write the balanced equation. [2]
(d) In zone A, the ore is reduced. Write the balanced equation for the reduction of Fe2O3 by carbon monoxide. [2]
(e) Explain, in terms of oxygen, why the reaction in part (d) is described as a reduction. [1]
(f) Limestone decomposes in the furnace. Write the equation and name this type of reaction. [2]
(g) The calcium oxide produced reacts with silicon dioxide impurity. Write the equation and name the product formed. [2]
(h) State two ways in which operating a blast furnace can harm the environment. [2]
This question follows the reactions inside a blast furnace zone by zone, testing raw materials, equations, redox, and environmental effects.
(a) The three solid raw materials added at the top are: iron ore (haematite, \( \mathrm{Fe_2O_3} \)) [1]; coke (carbon) [1]; and limestone (calcium carbonate) [1].
(b) In zone C the coke burns in the hot air blasted in:
\[ C + O_2 \rightarrow CO_2 \]This reaction is exothermic [1] (equation [1]), and it supplies the heat that drives the furnace.
(c) In zone B the carbon dioxide rising up the furnace meets more hot coke and is reduced to carbon monoxide [1]:
\[ CO_2 + C \rightarrow 2CO \]The equation scores [1]. Carbon monoxide is the gas that actually reduces the ore.
(d) In zone A the carbon monoxide reduces the iron ore to molten iron:
\[ Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \]One mark for formulae, one for balancing [2].
(e) It is a reduction because the iron(III) oxide loses its oxygen [1] to the carbon monoxide.
(f) The limestone decomposes when heated:
\[ CaCO_3 \rightarrow CaO + CO_2 \]This is called thermal decomposition [1] (equation [1]).
(g) The calcium oxide removes the sandy silicon dioxide impurity:
\[ CaO + SiO_2 \rightarrow CaSiO_3 \]The product is calcium silicate, which runs off as molten slag [1] (equation [1]).
(h) Any two environmental harms, one mark each up to [2]: carbon dioxide released causes climate change / global warming; carbon monoxide produced is a toxic gas; sulfur dioxide from impurities causes acid rain; heat and dust pollution; loss of land and habitats through mining the ore.
Answer Details
This question follows the reactions inside a blast furnace zone by zone, testing raw materials, equations, redox, and environmental effects.
(a) The three solid raw materials added at the top are: iron ore (haematite, \( \mathrm{Fe_2O_3} \)) [1]; coke (carbon) [1]; and limestone (calcium carbonate) [1].
(b) In zone C the coke burns in the hot air blasted in:
\[ C + O_2 \rightarrow CO_2 \]This reaction is exothermic [1] (equation [1]), and it supplies the heat that drives the furnace.
(c) In zone B the carbon dioxide rising up the furnace meets more hot coke and is reduced to carbon monoxide [1]:
\[ CO_2 + C \rightarrow 2CO \]The equation scores [1]. Carbon monoxide is the gas that actually reduces the ore.
(d) In zone A the carbon monoxide reduces the iron ore to molten iron:
\[ Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2 \]One mark for formulae, one for balancing [2].
(e) It is a reduction because the iron(III) oxide loses its oxygen [1] to the carbon monoxide.
(f) The limestone decomposes when heated:
\[ CaCO_3 \rightarrow CaO + CO_2 \]This is called thermal decomposition [1] (equation [1]).
(g) The calcium oxide removes the sandy silicon dioxide impurity:
\[ CaO + SiO_2 \rightarrow CaSiO_3 \]The product is calcium silicate, which runs off as molten slag [1] (equation [1]).
(h) Any two environmental harms, one mark each up to [2]: carbon dioxide released causes climate change / global warming; carbon monoxide produced is a toxic gas; sulfur dioxide from impurities causes acid rain; heat and dust pollution; loss of land and habitats through mining the ore.
Question 5 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 6 Report
Fig. 7.1
Fig. 7.1 shows the apparatus used to titrate a solution of sodium carbonate against dilute nitric acid.
(a) State how the burette reading should be taken so that a parallax error is avoided. [1]
(b) State why a conical flask is used to hold the sodium carbonate solution rather than an open beaker. [1]
(c) Methyl orange is used as the indicator. State the colour change seen in the flask as the acid is added to the sodium carbonate solution up to the end-point. [2]
(d) A pipette delivers 20.0 cm3 of sodium carbonate solution, concentration 0.0500 mol/dm3, into the flask. The average titre of nitric acid is 16.00 cm3. The equation is:
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
(i) Calculate the amount, in mol, of sodium carbonate used. [1]
(ii) Calculate the amount, in mol, of nitric acid used. [1]
(iii) Calculate the concentration of the nitric acid in mol/dm3. [2]
(e) Calculate the concentration of the nitric acid in g/dm3. [2]
(f) Describe a test, and its positive result, to show that the gas given off is carbon dioxide. [2]
(g) Name the salt formed in this titration. [1]
Sodium carbonate is titrated against nitric acid. The mole ratio from the equation is 1:2, and the question also tests practical technique, indicator colour, converting mol/dm3 to g/dm3, and the test for carbon dioxide.
(a) Avoiding parallax error [1]. Read the scale with your eye level with the bottom of the meniscus [1], so the line of sight is horizontal.
(b) Why a conical flask, not a beaker [1]. A conical flask can be swirled vigorously to mix the reagents without any solution splashing out, so no liquid is lost [1].
(c) Methyl orange colour change [2]. As acid is added to the alkaline carbonate the indicator changes from yellow [1] to orange/red at the end-point [1].
(d)(i) Moles of sodium carbonate [1]. \( n(\text{Na}_2\text{CO}_3) = c \times V = 0.0500 \times \dfrac{20.0}{1000} = 0.00100\ \text{mol} \) [1].
(d)(ii) Moles of nitric acid [1]. The equation shows 1 Na2CO3 reacts with 2 HNO3, so \( n(\text{HNO}_3) = 2 \times 0.00100 = 0.00200\ \text{mol} \) [1].
(d)(iii) Concentration of nitric acid [2]. This is in the 16.00 cm3 titre, so \( c = \dfrac{n}{V} = \dfrac{0.00200}{16.00/1000} = 0.125\ \text{mol/dm}^3 \) [2].
(e) Concentration in g/dm3 [2]. First the molar mass: \( M_r(\text{HNO}_3) = 1 + 14 + (3 \times 16) = 63 \) [1]. Then \( 0.125\ \text{mol/dm}^3 \times 63\ \text{g/mol} = 7.875\ \text{g/dm}^3 \) [1].
(f) Test for carbon dioxide [2]. Bubble the gas through limewater (calcium hydroxide solution) [1]; a positive result is the limewater turning milky/cloudy as a white precipitate of calcium carbonate forms [1].
(g) Salt formed [1]. Sodium nitrate (NaNO3) [1].
Exam tip: to change mol/dm3 into g/dm3, multiply by the molar mass; to go back, divide.
Answer Details
Sodium carbonate is titrated against nitric acid. The mole ratio from the equation is 1:2, and the question also tests practical technique, indicator colour, converting mol/dm3 to g/dm3, and the test for carbon dioxide.
(a) Avoiding parallax error [1]. Read the scale with your eye level with the bottom of the meniscus [1], so the line of sight is horizontal.
(b) Why a conical flask, not a beaker [1]. A conical flask can be swirled vigorously to mix the reagents without any solution splashing out, so no liquid is lost [1].
(c) Methyl orange colour change [2]. As acid is added to the alkaline carbonate the indicator changes from yellow [1] to orange/red at the end-point [1].
(d)(i) Moles of sodium carbonate [1]. \( n(\text{Na}_2\text{CO}_3) = c \times V = 0.0500 \times \dfrac{20.0}{1000} = 0.00100\ \text{mol} \) [1].
(d)(ii) Moles of nitric acid [1]. The equation shows 1 Na2CO3 reacts with 2 HNO3, so \( n(\text{HNO}_3) = 2 \times 0.00100 = 0.00200\ \text{mol} \) [1].
(d)(iii) Concentration of nitric acid [2]. This is in the 16.00 cm3 titre, so \( c = \dfrac{n}{V} = \dfrac{0.00200}{16.00/1000} = 0.125\ \text{mol/dm}^3 \) [2].
(e) Concentration in g/dm3 [2]. First the molar mass: \( M_r(\text{HNO}_3) = 1 + 14 + (3 \times 16) = 63 \) [1]. Then \( 0.125\ \text{mol/dm}^3 \times 63\ \text{g/mol} = 7.875\ \text{g/dm}^3 \) [1].
(f) Test for carbon dioxide [2]. Bubble the gas through limewater (calcium hydroxide solution) [1]; a positive result is the limewater turning milky/cloudy as a white precipitate of calcium carbonate forms [1].
(g) Salt formed [1]. Sodium nitrate (NaNO3) [1].
Exam tip: to change mol/dm3 into g/dm3, multiply by the molar mass; to go back, divide.
