4ch1-2c-que-20230614

Question 1

Use the Periodic Table to help you answer this question.

(a) (i) Name the element with atomic number 14.
(ii) Name the element in Group 2 and Period 3.
(iii) Name an element that is a liquid at room temperature.
(iv) Give the electronic configuration of an atom of phosphorus.
(v) Give the formula of sodium sulfide.

(b) Explain, in terms of electron configuration, why neon is unreactive.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• Topic 1(d): The Periodic Table — parts (a)(i), (a)(ii), (a)(iii)
• Topic 1(c): Atomic structure — part (a)(iv)
• Topic 1(e): Chemical formulae, equations and calculations — part (a)(v)
• Topic 1(d): The Periodic Table — part (b)

▶️ Answer/Explanation

Solution

(a)(i) silicon
ALLOW Si

(a)(ii) magnesium
ALLOW Mg

(a)(iii) bromine
ALLOW mercury / ALLOW Br / Br₂ / REJECT bromide

(a)(iv) \( 2,8,5 \) or \( 2.8.5 \)
ACCEPT diagram showing electron configuration

(a)(v) \( \text{Na}_2\text{S} \)
ALLOW Na⁺₂S²⁻

(b) An explanation that links the following two points:
• It has a full outer shell / 8 electrons in outer shell / (electron configuration) \( 2.8 \) (1 mark)
• (So) it does not need to lose or gain (or share) electrons / is stable (1 mark)

Question 2

This question is about gases in the atmosphere.

(a) Which of these gases has the lowest percentage by volume in the atmosphere?

A) argon
B) carbon dioxide
C) nitrogen
D) oxygen

(b) When copper(II) carbonate is heated, copper(II) oxide and carbon dioxide form.

(i) What is the name of this type of reaction?

A) addition
B) decomposition
C) oxidation
D) substitution

(ii) Which colour change occurs when copper(II) carbonate is heated?

A) blue to black
B) blue to orange
C) green to black
D) green to orange

(iii) Give the chemical equation for this reaction.

(c) A student uses this apparatus to find the percentage of oxygen in a sample of air.

The student leaves the apparatus until there is no further change in volume of gas in the syringe.

These are the student’s results:

Calculate the percentage of oxygen in the sample of air. Give your answer to two significant figures.

(d) Explain why an increasing amount of carbon dioxide in the atmosphere is likely to cause a problem for the environment.

Most-appropriate topic codes (Edexcel IGCSE Chemistry):

• 2(c): Gases in the atmosphere — parts (a), (c), (d)
• 2(d): Reactivity series — part (b)(ii), (b)(iii)
• 3(b): Rates of reaction — part (b)(i)
• 1(e): Chemical formulae, equations and calculations — part (c)

▶️ Answer/Explanation

Solution

(a) B (carbon dioxide)
Carbon dioxide makes up only about 0.04% of the atmosphere, which is lower than argon (~0.93%), nitrogen (~78%), and oxygen (~21%).

(b)(i) B (decomposition)
A single compound (copper(II) carbonate) breaks down into two simpler substances (copper(II) oxide and carbon dioxide) upon heating.

(b)(ii) C (green to black)
Copper(II) carbonate is a green solid. When heated, it decomposes to form copper(II) oxide, which is a black solid.

(b)(iii) \(\mathrm{CuCO_3 \rightarrow CuO + CO_2}\)
The thermal decomposition reaction produces one mole each of solid copper(II) oxide and gaseous carbon dioxide.

(c) 19%
Detailed calculation: The oxygen in the air reacts, causing the gas volume to decrease. The volume of oxygen used is the initial syringe volume minus the final: \(100 – 27 = 73\ \text{cm}^3\). The total initial air volume is the flask volume plus the syringe volume: \(280 + 100 = 380\ \text{cm}^3\). Percentage oxygen = \((73 / 380) \times 100 = 19.2\% \approx 19\%\) (to 2 s.f.).

(d) Carbon dioxide is a greenhouse gas. It absorbs and re-radiates infrared radiation (heat) from the Earth’s surface, trapping it in the atmosphere. This enhanced greenhouse effect leads to global warming, which causes climate change, melting of polar ice caps, rising sea levels, and more extreme weather events.

Question 3

This question is about alcohols.

Ethanol can be manufactured using two different methods.

hydration of ethene
fermentation of glucose

This is the equation for hydration.

\[\mathrm{C_2H_4 + H_2O\rightarrow C_2H_5OH}\]

(a) Complete the equation for fermentation.

\[\mathrm{C_6H_{12}O_6\rightarrow \hspace{2cm} + \hspace{2cm}}\]

(b) The table gives some information about the two methods.

(i) Complete the table by giving the missing information.
(ii) Explain one advantage and one disadvantage of using fermentation rather than hydration to produce ethanol.
You should use information from the table to help your answer.

(c) Explain why fermentation needs to occur in the absence of air.

(d) Propanol has this percentage composition by mass.

\[{\mathsf{C}}=60.0\%\qquad{\mathsf{H}}=13.3\%\qquad{\mathsf{O}}=26.7\%\]

(i) Show by calculation that the empirical formula of propanol is \(\mathsf{C}_3\mathsf{H}_8\mathsf{O}\).
(ii) Draw the displayed formula of propanol.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 4(e) Alcohols: Manufacture of ethanol — parts (a), (b), (c)
• 1(e) Chemical formulae, equations and calculations: Empirical formula — part (d)(i)
• 4(a) Introduction to organic chemistry: Displayed formulae — part (d)(ii)

▶️ Answer/Explanation

Solution

(a) \(\mathrm{C_6H_{12}O_6\rightarrow 2C_2H_5OH + 2CO_2}\)

(b)(i) Table completed: Hydration catalyst = phosphoric acid; Fermentation temperature ≈ 30°C; Hydration pressure = 60-70 atm.

(b)(ii) Advantage: Fermentation uses a lower temperature/pressure or a renewable resource (glucose). Disadvantage: It is slower, less efficient, or produces an impure product requiring purification.

(c) Air contains oxygen, which would oxidise the ethanol produced to ethanoic acid. Fermentation needs anaerobic conditions to ensure ethanol is formed.

(d)(i) Assume 100g sample. Moles: C = 60.0/12 = 5.0; H = 13.3/1 = 13.3; O = 26.7/16 = 1.67. Divide by smallest (1.67): C=3, H=8, O=1. So empirical formula is C₃H₈O.

(d)(ii) Displayed formula of propan-1-ol:

(O-H bond must be shown).

Detailed Reasoning: For (a), fermentation of glucose by yeast produces ethanol and carbon dioxide. For (b)(i), the conditions are from standard knowledge. For (b)(ii), compare the two processes using the table. For (c), oxygen reacts with ethanol. For (d)(i), calculate moles from percentages and find the simplest whole number ratio.

Question 4

A student uses a titration to find the concentration of potassium hydroxide solution.

Student’s method:

Measure 25.0 cm³ of the potassium hydroxide solution into a conical flask.
Add a few drops of methyl orange indicator to the conical flask.
Fill a burette with dilute sulfuric acid and record the initial burette reading.
Place the conical flask on a white tile.
Add acid from the burette to the mixture in the conical flask, swirling continuously.
When the indicator changes colour at the end point, record the final burette reading.
Repeat the titration to obtain concordant results.

(a) Name the most suitable piece of apparatus to measure out 25.0 cm³ of potassium hydroxide solution.

(b) Complete the table below to show the colour of methyl orange in potassium hydroxide solution and in dilute sulfuric acid.

(c) Explain why the student places the conical flask on a white tile.

(d) Explain why the student swirls the flask continuously.

(e) State what is meant by the term concordant results.

(f) The student finds that \(15.00 \, \text{cm}^3\) of sulfuric acid of concentration \(0.180 \, \text{mol/dm}^3\) reacts with \(25.0 \, \text{cm}^3\) of potassium hydroxide solution.

The equation for the reaction is:

\[ 2\text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O} \]

Calculate the concentration of the potassium hydroxide solution.

(g) The ionic equation for the reaction between an acid and an alkali is:

\[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \]

Explain why the \(\text{OH}^-\) ion is a proton acceptor in this reaction.

Most-appropriate topic codes (Pearson Edexcel International GCSE Chemistry 4CH1):

• 2(f): Acids, alkalis and titrations — parts (a), (b), (c), (d), (e), (f), (g)
• 2(g): Acids, bases and salt preparations — part (g) proton transfer theory
• 1(e): Chemical formulae, equations and calculations — part (f) reacting mass and concentration calculations
• Appendix 5: Command word taxonomy — terms “explain”, “state”, “calculate”

▶️ Answer/Explanation

Solution

(a) Pipette (with pipette filler)

(b)

Solution	Colour of methyl orange
potassium hydroxide solution	yellow
dilute sulfuric acid	red

(c) To see the colour change (of the indicator) more clearly at the end point.

(d) To mix the solutions thoroughly / to ensure even reaction between acid and alkali.

(e) Titre volumes / results within \( \pm 0.2 \, \text{cm}^3 \) of each other.

(f)

M1: Moles of \(\text{H}_2\text{SO}_4 = 0.01500 \times 0.180 = 0.00270 \, \text{mol}\)

M2: Moles of \(\text{KOH} = 0.00270 \times 2 = 0.00540 \, \text{mol}\)

M3: Concentration of \(\text{KOH} = \frac{0.00540}{0.0250} = 0.216 \, \text{mol/dm}^3\)

Answer: \(0.216 \, \text{mol/dm}^3\)

(g) An explanation linking the following points:

An \(\text{H}^+\) ion is a proton.
The \(\text{OH}^-\) ion reacts with / accepts / bonds with the \(\text{H}^+\) ion to form water.

Therefore, the \(\text{OH}^-\) ion is a proton acceptor.

Question 6

When copper(II) sulfate solution is electrolysed, copper forms at the negative electrode.
A student uses this apparatus to investigate the electrolysis of copper(II) sulfate solution.

(a) Describe how the student could test a sample of copper(II) sulfate solution to show that it contains copper(II) ions.

(b) Describe how copper metal forms at the negative electrode.

(c) State the appearance of the copper that forms on the negative electrode.

Oxygen forms at the positive electrode.

(d) Give a test for oxygen.

(e) Complete the half-equation for the formation of oxygen at the positive electrode.

\( \text{……} \rightarrow \text{……} + \text{……} + \text{……} \)

(f) State why the formation of oxygen at the positive electrode is an oxidation reaction.

Most-appropriate topic codes (Edexcel IGCSE Chemistry 4CH1):

• 1(i) Electrolysis: — parts (b), (c), (e), (f)
• 2(h) Chemical tests: — parts (a), (d)

▶️ Answer/Explanation

Solution

(a)
• M1: Add sodium hydroxide (solution) / potassium hydroxide / ammonia (solution).
• M2: A blue precipitate forms.

OR
• M1: Perform a flame test.
• M2: A blue-green flame is observed.

(b)
• Copper ions \( \text{(Cu}^{2+}) \) are positively charged / are cations.
• They are attracted to / move towards the negative electrode (cathode).
• At the cathode, each \( \text{Cu}^{2+} \) ion gains two electrons \( \text{(2e}^- \text{)} \).
• This reduction forms copper atoms: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \).

(c)
• Pink solid / deposit / coating.
• Accept: orange-brown / red-brown.

(d)
• Insert a glowing splint into the gas.
• The splint relights.

(e)
\( 4\text{OH}^- \rightarrow 2\text{H}_2\text{O} + \text{O}_2 + 4\text{e}^- \)
Also accept: \( 2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + \text{O}_2 + 4\text{e}^- \)

(f)
• Because electrons are lost (by the hydroxide ions or water molecules).

Question 6

Ethanoic acid reacts with methanol to form an ester.

The equation shows the displayed formulae for the reactants and product:

(a) Name a suitable catalyst to increase the rate of this reaction.

(b) State how you would know that an ester has formed.

(c) Name this ester.

The table shows the number of bonds in the reactants and the number of bonds in the products.

(d) State which two bonds need to be broken in the reactants.

(e) Explain why the enthalpy change in this reaction is approximately \(0 \, \text{kJ/mol}\).

Most-appropriate topic codes (Edexcel IGCSE Chemistry – 4CH1):

• 4(g): Esters — Parts (a), (b), (c), ester formation, properties, and naming.
• 3(a): Energetics — Part (e), enthalpy change and bond energy calculations.
• 1(g): Covalent Bonding / 4(a): Introduction to Organic Chemistry — Part (d), identifying specific bonds broken/formed in a reaction.
• 3(b): Rates of Reaction — Part (a), use of a catalyst.

▶️ Answer/Explanation

Solution

(a) Sulfuric acid / \(H_2SO_4\)
Other strong acids (e.g., HCl, \(H_3PO_4\)) are also acceptable.

(b) A distinctive / sweet / fruity smell is produced.

(c) Methyl ethanoate

(d) C—O and O—H
(The C—O in the acid’s —COOH group and the O—H in the alcohol’s —OH group are broken to form the ester linkage and water).

(e) An explanation that links the following two points:
• M1: The same (two) bonds (i.e., C—O and O—H) are broken and (the same type of bonds are) formed.
• M2: The energy needed to break these bonds is (approximately) equal to the energy released when the new bonds form, resulting in an overall enthalpy change close to zero.

Question 7

Methane reacts with steam to form carbon monoxide and hydrogen.

(a) Write the equation for the reaction.

\[ \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g) \quad \Delta H = +206 \, \text{kJ/mol} \]

(b) Explain why carbon monoxide is poisonous to humans.

(c) The reaction conditions for this reaction are a temperature of \(700^\circ \text{C}\) and a pressure of 5 atmospheres.

(i) The temperature of the reaction mixture is reduced to \(600^\circ \text{C}\), but the pressure is kept at 5 atmospheres. Explain the effect on the yield of hydrogen at equilibrium.
(ii) The pressure of the reaction mixture is reduced to 4 atmospheres, but the temperature is kept at \(700^\circ \text{C}\). Explain the effect on the yield of hydrogen at equilibrium.

(d) Calculate the volume, in \(\text{dm}^3\), of methane gas at rtp needed to produce 6.6 tonnes of hydrogen gas.

Molar volume = \(24 \, \text{dm}^3\)
1 tonne = \(10^6 \, \text{g}\)

Give your answer in standard form.

Most-appropriate topic codes (Edexcel International GCSE Chemistry 4CH1):

• 3(c): Reversible reactions and equilibria — parts (c)(i) and (c)(ii)
• 1(e): Chemical formulae, equations and calculations — part (d)
• 1(g): Covalent bonding / Organic chemistry (combustion products) — part (b)
• 4(b): Crude oil / Combustion — part (b) link to carbon monoxide poisoning
• Specification Points: 3.22C, 1.35C, 4.13

▶️ Answer/Explanation

Solution

(a) Equation given in the question:
\[ \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + 3\text{H}_2(g) \quad \Delta H = +206 \, \text{kJ/mol} \]

(b) • Carbon monoxide reduces the capacity of blood to transport oxygen round the body.
• It binds with haemoglobin (forming carboxyhaemoglobin), preventing oxygen transport.

(c)(i) An explanation that links the following two points:
• Yield of hydrogen decreases.
• Because the forward reaction is endothermic (\( \Delta H = +206 \, \text{kJ/mol}\)), so lowering the temperature shifts the equilibrium to the left (reactants side).

(c)(ii) An explanation that links the following two points:
• Yield of hydrogen increases.
• Because there are fewer moles of gas on the left-hand side (2 mol) than on the right-hand side (4 mol), so reducing the pressure shifts the equilibrium to the side with more moles (right/products side).

(d) Calculation steps:

M1: Amount of \( \text{H}_2 \) = \( \frac{6.6 \times 10^6 \, \text{g}}{2 \, \text{g/mol}} = 3.3 \times 10^6 \, \text{mol} \)

M2: From the equation, 1 mol \( \text{CH}_4 \) produces 3 mol \( \text{H}_2 \).
Amount of \( \text{CH}_4 \) = \( \frac{3.3 \times 10^6}{3} = 1.1 \times 10^6 \, \text{mol} \)

M3: Volume of \( \text{CH}_4 \) at rtp = \( 1.1 \times 10^6 \times 24 = 26,400,000 \, \text{dm}^3 \)

M4: In standard form = \( 2.64 \times 10^7 \, \text{dm}^3 \)

Final Answer: \( 2.6 \times 10^7 \, \text{dm}^3 \) (to 2 significant figures).

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