An original Thinka practice paper modelled on the structure and difficulty of the Nov 2023 (V1) Cambridge IGCSE Chemistry (0620) paper. Not affiliated with or reproduced from Cambridge.
Section Question 1
Identify chemical substances from a provided list based on given chemical and physical descriptions.
Answer the following question using only the substances from the list. Each substance may be used once, more than once or not at all.
Give the name of the substance that is a gas used to provide an inert atmosphere in balloons.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Helium is a noble gas in Group VIII of the Periodic Table. It is extremely unreactive (inert) and has a very low density, making it ideal and safe for filling balloons.
Answer the following question using only the substances from the list. Each substance may be used once, more than once or not at all.
Give the name of the substance that is the main source of iron in a blast furnace.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Hematite is the chief ore of iron, and its main chemical constituent is iron(III) oxide (\(\text{Fe}_2\text{O}_3\)), which is reduced to molten iron in the blast furnace.
Answer the following question using only the substances from the list. Each substance may be used once, more than once or not at all.
Give the name of the substance that contains a halide ion which reacts with acidified silver nitrate to form a yellow precipitate.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Potassium iodide contains iodide ions (\(\text{I}^-\)). When acidified aqueous silver nitrate is added, silver iodide (\(\text{AgI}\)) forms as a yellow precipitate.
Answer the following question using only the substances from the list. Each substance may be used once, more than once or not at all.
Give the name of the substance that reacts with dilute hydrochloric acid to produce carbon dioxide gas.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Carbonates react with dilute acids to produce carbon dioxide, water, and a salt. Calcium carbonate (\(\text{CaCO}_3\)) reacts with dilute hydrochloric acid to liberate carbon dioxide gas.
Marking scheme
calcium carbonate [1]
Ready to test yourself?
Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.
Solve questions on atomic structure, isotopes, extraction of metals (aluminium), physical properties, and dot-and-cross diagrams.
6 Question · 24 marks
Question 1 · structured
2 marks
Oxygen has three naturally occurring isotopes, \( ^{16}\text{O} \), \( ^{17}\text{O} \) and \( ^{18}\text{O} \).
Complete the table to show the number of protons, neutrons and electrons in one atom of the isotope \( ^{18}\text{O} \).
particle number in one atom of \( ^{18}\text{O} \)
protons ..................
neutrons ..................
electrons ..................
Show answer & marking schemeHide answer & marking scheme
Worked solution
Oxygen has an atomic number of 8 (found in the Periodic Table), which means any atom of oxygen contains 8 protons. In a neutral atom, the number of electrons is equal to the number of protons, which is 8. The mass number of this isotope is 18. The number of neutrons is calculated as: mass number \( - \) atomic number \( = 18 - 8 = 10 \).
Gallium, Ga, has a relative atomic mass of 69.72. Naturally occurring gallium consists of only two isotopes, \(^{69}\text{Ga}\) and \(^{71}\text{Ga}\).
Calculate the percentage abundance of the \(^{71}\text{Ga}\) isotope in naturally occurring gallium. Show your working.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let the percentage abundance of \(^{71}\text{Ga}\) be \(x\%\).
The percentage abundance of \(^{69}\text{Ga}\) is therefore \((100 - x)\%\).
Therefore, the percentage abundance of the \(^{71}\text{Ga}\) isotope is 36%.
Marking scheme
M1: Setting up a correct algebraic equation representing the relative atomic mass calculation (1 mark) - e.g. \(69.72 = \frac{71x + 69(100 - x)}{100}\) or \(69.72 = 71y + 69(1 - y)\)
M2: Correct rearrangement/simplification of the equation (1 mark) - e.g. \(2x = 72\) or \(2y = 0.72\)
M3: Correct final answer of 36 (%) (1 mark) - Accept: 36 or 36%
Question 3 · theory
3 marks
Gallium, Ga, has a relative atomic mass of 69.72. Naturally occurring gallium consists of only two isotopes, \(^{69}\text{Ga}\) and \(^{71}\text{Ga}\).
Calculate the percentage abundance of the \(^{71}\text{Ga}\) isotope in naturally occurring gallium. Show your working.
Show answer & marking schemeHide answer & marking scheme
Worked solution
Let the percentage abundance of \(^{71}\text{Ga}\) be \(x\%\).
The percentage abundance of \(^{69}\text{Ga}\) is therefore \((100 - x)\%\).
Therefore, the percentage abundance of the \(^{71}\text{Ga}\) isotope is 36%.
Marking scheme
M1: Setting up a correct algebraic equation representing the relative atomic mass calculation (1 mark) - e.g. \(69.72 = \frac{71x + 69(100 - x)}{100}\) or \(69.72 = 71y + 69(1 - y)\)
M2: Correct rearrangement/simplification of the equation (1 mark) - e.g. \(2x = 72\) or \(2y = 0.72\)
M3: Correct final answer of 36 (%) (1 mark) - Accept: 36 or 36%
Question 4 · theory
7 marks
Magnesium is a metal extracted by the electrolysis of molten magnesium chloride, \(\text{MgCl}_2\).
(a) State two physical properties of magnesium that make it useful in the construction of aircraft alloys. [2]
(b) Write the ionic half-equation, including state symbols, for the reaction occurring at the negative electrode (cathode) during this electrolysis. [2]
(c) Explain why magnesium metal is not produced if an aqueous solution of magnesium chloride is electrolysed instead of the molten salt. [3]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Magnesium is chosen for aerospace applications because of its low density (lightweight) and good strength when mixed with other metals to form alloys.
(b) At the cathode (negative electrode), magnesium ions gain electrons (reduction) to form magnesium atoms: \(\text{Mg}^{2+}(\text{l}) + 2\text{e}^- \rightarrow \text{Mg}(\text{l})\)
(c) In an aqueous solution of magnesium chloride, both magnesium ions (\(\text{Mg}^{2+}\)) and hydrogen ions (\(\text{H}^+\)) from the water are present at the cathode. Since magnesium is higher in the reactivity series (more reactive) than hydrogen, hydrogen ions are more easily reduced (discharged) at the cathode, yielding hydrogen gas (\(\text{H}_2\)) rather than magnesium metal.
Marking scheme
(a) [Max 2 marks] - 1 mark for low density / lightweight (REJECT: 'light' on its own). - 1 mark for high strength-to-weight ratio / strong / malleable.
(b) [Max 2 marks] - 1 mark for correct species and balancing: \(\text{Mg}^{2+} + 2\text{e}^- \rightarrow \text{Mg}\). - 1 mark for correct state symbols: \(\text{Mg}^{2+}(\text{l})\) and \(\text{Mg}(\text{l})\) [accept \(\text{Mg}(\text{s})\)].
(c) [Max 3 marks] - 1 mark for stating that water/aqueous solution contains hydrogen ions / \(\text{H}^+\). - 1 mark for stating that magnesium is more reactive than hydrogen / hydrogen is lower in the reactivity series. - 1 mark for stating that hydrogen ions are preferentially discharged / hydrogen gas is produced.
Question 5 · theory
4 marks
Aluminium is a highly useful metal extracted from bauxite.
(a) During extraction, purified aluminium oxide is dissolved in molten cryolite. Give two reasons for using molten cryolite in this process. [2]
(b) Identify two physical characteristics of aluminium that make it suitable for the manufacture of aircraft bodies. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Pure aluminium oxide has a very high melting point (around 2000 °C). Dissolving it in molten cryolite lowers the melting point of the mixture to about 950 °C, which significantly reduces the energy required and the operating costs. Cryolite also acts as a solvent and improves the electrical conductivity of the electrolyte.
(b) Aluminium is highly suitable for aircraft bodies because of its low density (which makes the aircraft lightweight and fuel-efficient) and its high strength (especially when alloyed/combined with other metals) to withstand flight stresses.
Marking scheme
(a) 1 mark for each point (max 2 marks): - Lowers the melting point / operating temperature of the electrolyte [1] - Acts as a solvent (to dissolve aluminium oxide) [1] - Increases / improves electrical conductivity [1]
(b) 1 mark for each physical property (max 2 marks): - Low density / lightweight [1] - Strong / high strength (when alloyed) [1] - Malleable / easy to shape [1] Note: Reject 'resistant to corrosion' as this is a chemical property.
Question 6 · theory
5 marks
Magnesium and oxygen are elements in the Periodic Table.
(a) Magnesium has three naturally occurring isotopes, $^{24}\text{Mg}$, $^{25}\text{Mg}$ and $^{26}\text{Mg}$. State the meaning of the term *isotopes*. [1]
(b) Magnesium burns in oxygen with a bright white flame to form the ionic compound magnesium oxide, $\text{MgO}$. Write the symbol equation for this reaction. [2]
(c) During this reaction, magnesium atoms react with oxygen atoms to form ions. Deduce the electronic configuration and charge of both ions in magnesium oxide:
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons.
(b) Magnesium ($\text{Mg}$) reacts with oxygen diatomic molecules ($\text{O}_2$) to produce magnesium oxide ($\text{MgO}$): $$2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$
(c) A magnesium atom has an electronic configuration of 2,8,2. It loses 2 valence electrons to form a stable octet, resulting in a magnesium ion ($\text{Mg}^{2+}$) with configuration 2,8 and a charge of 2+. An oxygen atom has an electronic configuration of 2,6. It gains the 2 electrons from magnesium to form a stable octet, resulting in an oxide ion ($\text{O}^{2-}$) with configuration 2,8 and a charge of 2-.
Marking scheme
(a) M1: Atoms of the same element with the same number of protons / proton number but different numbers of neutrons / neutron number. [1]
(c)(i) M1: Magnesium ion electronic configuration 2,8 AND charge 2+ (or +2). [1] (c)(ii) M2: Oxide ion electronic configuration 2,8 AND charge 2- (or -2). [1]
Section Question 3
Analyse displacement reactions, group trends, halogens properties, and complete balanced symbol and ionic equations.
13 Question · 78 marks
Question 1 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 2 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 3 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 4 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 5 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 6 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 7 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 8 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 9 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 10 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 11 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 12 · metal_displacement_and_equations
6 marks
Order of reactivity can be determined by displacement reactions.
(a) A student investigates the reactivities of four metals: chromium ($\text{Cr}$), copper ($\text{Cu}$), iron ($\text{Fe}$), and nickel ($\text{Ni}$), by adding each metal separately to aqueous solutions of the other metal nitrates.
*(Note: Shaded or empty diagonal cells indicate where a metal is added to its own nitrate solution.)*
(i) Complete Table 3.1 by writing a tick ($\boldsymbol{\checkmark}$) or a cross ($\boldsymbol{X}$) in the two blank spaces, and place the four metals in their order of reactivity with the most reactive first.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) - Chromium ($\text{Cr}$) is more reactive than nickel, so it reacts with nickel(II) nitrate (write $\boldsymbol{\checkmark}$ in the first column, last row). - Iron ($\text{Fe}$) is less reactive than chromium, so it does not react with chromium(III) nitrate (write $\boldsymbol{X}$ in the third column, first row). - Reactivity order: Chromium (most reactive) > Iron > Nickel > Copper (least reactive).
(ii) Iron reacts with copper(II) nitrate to form iron(II) nitrate and copper: $$\text{Fe} + \text{Cu(NO}_3)_2 \rightarrow \text{Fe(NO}_3)_2 + \text{Cu}$$
(b) Group VII reactivity decreases down the group: $\text{F} > \text{Cl} > \text{Br} > \text{I} > \text{At}$. The element that lies between bromine and astatine is iodine ($\text{I}$ or $\text{I}_2$).
Marking scheme
(a) (i) - M1: Both table spaces completed correctly ($\boldsymbol{\checkmark}$ for chromium with nickel(II) nitrate AND $\boldsymbol{X}$ for iron with chromium(III) nitrate) [1] - M2: All four metals in correct reactivity order (chromium, iron, nickel, copper) [2] *(Allow 1 mark if only one pair is reversed)*
(ii) - M1: Correct formulas for all reactants and products ($\text{Fe}$, $\text{Cu(NO}_3)_2$, $\text{Fe(NO}_3)_2$, $\text{Cu}$) [1] - M2: Correctly balanced equation [1]
(b) - M1: Iodine / $\text{I}$ / $\text{I}_2$ [1]
Question 13 · free_response
6 marks
A student investigates the reactivity of the halogens by mixing aqueous solutions of the elements with aqueous solutions of halide salts.
(a) Aqueous chlorine is added to a solution of potassium iodide. Describe the colour change observed. from .............................................................. to ........................................................... [2]
(b) Aqueous bromine is added to a solution of sodium iodide. Write the balanced ionic equation for this displacement reaction. Include state symbols. ................................................................................................................................................. [3]
(c) State the trend in the reactivity of the Group VII elements as the group is descended. ................................................................................................................................................. [1]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) When chlorine displacing iodine from potassium iodide solution occurs, iodine is liberated. The starting solution of potassium iodide is colourless, and the liberated iodine turns the solution brown. from: colourless to: brown (or orange-brown/yellow-brown)
(b) Bromine reacts with sodium iodide to form sodium bromide and iodine. The ionic equation only includes the species that change state or oxidation number. Diatomic bromine, \(Br_2(aq)\), and iodide ions, \(2I^-(aq)\), react to form bromide ions, \(2Br^-(aq)\), and diatomic iodine, \(I_2(aq)\). Equation: \(\text{Br}_2\text{(aq)} + 2\text{I}^-\text{(aq)} \rightarrow 2\text{Br}^-\text{(aq)} + \text{I}_2\text{(aq)}\)
(c) Going down Group VII (from fluorine to astatine), the reactivity of the halogens decreases.
Marking scheme
(a) - from: colourless [1] - to: brown / orange-brown / yellow-brown [1] (Reject: green, red, clear)
(b) - \(Br_2\) and \(I^-\). as reactants, and \(Br^-\). and \(I_2\). as products [1] - correctly balanced: \(\text{Br}_2 + 2\text{I}^- \rightarrow 2\text{Br}^- + \text{I}_2\) [1] - all state symbols correct and shown as \((aq)\) [1]
(c) - reactivity decreases (down the group) [1]
Section Question 4
Answer questions on rate of reaction, gas tests, collision theory, rate curve sketching, and multi-step molar concentration calculations.
2 Question · 12 marks
Question 1 · gas_tests_and_rate_theory
6 marks
A student investigates the rate of reaction between magnesium ribbon and dilute hydrochloric acid.
(b) The rate of reaction decreases as the reaction proceeds. Explain, using collision theory, why the rate of reaction decreases. [2]
(c) Calculate the concentration of the dilute hydrochloric acid in $\text{g/dm}^3$ if $50.0\text{ cm}^3$ of this acid reacts completely with excess magnesium to produce $0.0030\text{ mol}$ of hydrogen gas. Show your working. [3]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Test: Use a lighted splint. Observation: Squeaky pop / 'pop' sound.
(b) As the reaction proceeds, the concentration of the acid decreases because reactant particles are used up. This reduces the number of reactant particles per unit volume, leading to a decrease in the frequency of collisions (fewer successful collisions per second).
(c) Step 1: Determine the moles of $\text{HCl}$ that react. According to the equation, $1\text{ mol of } \text{H}_2$ is produced from $2\text{ mol of } \text{HCl}$. $$\text{Moles of } \text{HCl} = 0.0030 \times 2 = 0.0060\text{ mol}$$
Step 2: Calculate the concentration in $\text{mol/dm}^3$. $$\text{Volume in } \text{dm}^3 = \frac{50.0}{1000} = 0.0500\text{ dm}^3$$ $$\text{Concentration in } \text{mol/dm}^3 = \frac{0.0060}{0.0500} = 0.12\text{ mol/dm}^3$$
Step 3: Calculate the concentration in $\text{g/dm}^3$. $$M_r(\text{HCl}) = 1.0 + 35.5 = 36.5\text{ g/mol}$$ $$\text{Concentration in } \text{g/dm}^3 = 0.12 \times 36.5 = 4.38\text{ g/dm}^3$$
(b) [2 marks] - M1: Concentration of acid / reactant particles decreases (as they are used up) [1] - M2: Frequency of collisions decreases / fewer successful collisions per unit time [1]
(c) [3 marks] - M1: Moles of $\text{HCl} = 0.0060\text{ mol}$ [1] - M2: Concentration of $\text{HCl} = 0.12\text{ mol/dm}^3$ [1] - M3: Concentration in $\text{g/dm}^3 = 4.38\text{ g/dm}^3$ [1] (Allow ECF from M1 or M2)
Question 2 · extended_theory
6 marks
Excess zinc is added to $25.0\text{ cm}^3$ of dilute sulfuric acid. The total volume of hydrogen gas collected at r.t.p. is $120\text{ cm}^3$.
Calculate the concentration of the dilute sulfuric acid in $\text{g/dm}^3$ using the following steps.
- Calculate the number of moles of hydrogen gas produced. - Determine the number of moles of sulfuric acid which reacted. - Calculate the concentration of the sulfuric acid in $\text{mol/dm}^3$. - Calculate the concentration of the sulfuric acid in $\text{g/dm}^3$.
Show answer & marking schemeHide answer & marking scheme
Worked solution
- **Moles of $\text{H}_2$ produced:** $$\text{moles} = \frac{\text{volume in cm}^3}{24000\text{ cm}^3/\text{mol}} = \frac{120}{24000} = 0.0050\text{ mol}$$
- **Moles of $\text{H}_2\text{SO}_4$ reacted:** From the balanced equation, $1\text{ mol}$ of $\text{H}_2\text{SO}_4$ produces $1\text{ mol}$ of $\text{H}_2$. $$\text{moles of } \text{H}_2\text{SO}_4 = 0.0050\text{ mol}$$
- **Concentration of $\text{H}_2\text{SO}_4$ in $\text{mol/dm}^3$:** $$\text{concentration} = \frac{\text{moles} \times 1000}{\text{volume in cm}^3} = \frac{0.0050 \times 1000}{25.0} = 0.20\text{ mol/dm}^3$$
- **Concentration of $\text{H}_2\text{SO}_4$ in $\text{g/dm}^3$:** First, determine the relative molecular mass ($M_r$) of $\text{H}_2\text{SO}_4$: $$M_r = 2(1.0) + 32.1 + 4(16.0) = 98.1\text{ (or } 98\text{)}$$ $$\text{concentration in g/dm}^3 = \text{concentration in mol/dm}^3 \times M_r = 0.20 \times 98.1 = 19.62\text{ g/dm}^3\text{ (or } 19.6\text{ g/dm}^3\text{ using } M_r = 98\text{)}$$
Marking scheme
- **M1**: Moles of $\text{H}_2 = \frac{120}{24000} = 0.005(0)\text{ mol}$ [1] - **M2**: Moles of $\text{H}_2\text{SO}_4 = 0.005(0)\text{ mol}$ (Accept error carried forward from M1) [1] - **M3**: Correct working shown for concentration calculation: $\frac{\text{M2} \times 1000}{25.0}$ [1] - **M4**: Concentration in $\text{mol/dm}^3 = 0.20$ [1] - **M5**: Relative molecular mass ($M_r$) of $\text{H}_2\text{SO}_4 = 98$ or $98.1$ [1] - **M6**: Concentration in $\text{g/dm}^3 = 19.6$ or $19.62$ (Accept error carried forward from M4 and M5) [1]
Section Question 5
Explain equilibrium features, predict the shift of equilibrium based on Le Chatelier's principle, and discuss greenhouse gases and environmental chemistry.
3 Question · 16 marks
Question 1 · equilibrium_and_le_chatelier
8 marks
Carbon dioxide reacts with hot carbon in a closed container to form carbon monoxide. The reaction is reversible and reaches dynamic equilibrium.
(a) State one characteristic of a chemical reaction at dynamic equilibrium. [1]
(b) Describe and explain the effect on the position of equilibrium when:
(i) the overall pressure of the system is increased. [2]
(ii) the temperature of the system is reduced. [2]
(c) Carbon dioxide is a greenhouse gas that contributes to climate change.
(i) Identify one greenhouse gas other than carbon dioxide. [1]
(ii) Explain how the presence of greenhouse gases in the atmosphere leads to an increase in global temperatures. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) Any one of the following:
The rate of the forward reaction equals the rate of the reverse reaction.
The concentrations of reactants and products remain constant in a closed system.
(b) (i)
Effect: The position of equilibrium shifts to the left / towards the reactants (producing more carbon dioxide).
Explanation: There is 1 mole of gas on the left-hand side (\(\text{CO}_2\text{(g)}\)) and 2 moles of gas on the right-hand side (\(2\text{CO(g)}\)). Since solid carbon (\(\text{C(s)}\)) does not affect gas pressure, increasing the pressure shifts the equilibrium to the side with fewer moles of gas to decrease the pressure.
(b) (ii)
Effect: The position of equilibrium shifts to the left / towards the reactants.
Explanation: The forward reaction is endothermic (\(\Delta H\) is positive). Reducing the temperature shifts the equilibrium in the exothermic direction (to the left) to release heat and oppose the temperature decrease.
(c) (i) Any one of:
Methane (\(\text{CH}_4\))
Water vapour (\(\text{H}_2\text{O}\))
Nitrous oxide (\(\text{N}_2\text{O}\))
(c) (ii)
Greenhouse gases absorb infrared / thermal radiation emitted or reflected from the Earth's surface.
This absorbed energy is re-radiated in all directions, including back towards the Earth's surface, trapping heat in the atmosphere.
Marking scheme
(a) [1 mark]
• Rate of forward reaction equals rate of reverse reaction OR concentrations of reactants and products remain constant.
(b) (i) [2 marks]
• Equilibrium shifts to the left / reactants [1]
• Fewer moles of gas on the reactant side / left-hand side (1 mole of gas vs 2 moles of gas) [1]
(b) (ii) [2 marks]
• Equilibrium shifts to the left / reactants [1]
• The forward reaction is endothermic (or reverse reaction is exothermic) [1]
(c) (i) [1 mark]
• Methane / water vapour / nitrous oxide [1]
Reject: carbon monoxide / oxygen / nitrogen
(c) (ii) [2 marks]
• Greenhouse gases absorb thermal / infrared energy emitted/reflected from the Earth's surface [1]
• They re-emit this thermal energy back towards the Earth, trapping heat [1]
Question 2 · structured
4 marks
Carbon dioxide reacts with hydrogen to produce methane and steam in a reversible reaction:
The reaction reaches dynamic equilibrium in a closed system.
(a) State and explain the effect on the position of equilibrium when the temperature is increased. [2]
(b) Both carbon dioxide and methane are greenhouse gases. Explain, in terms of thermal energy, how greenhouse gases lead to global warming. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) According to Le Chatelier's principle, increasing the temperature favors the endothermic reaction direction to absorb the added heat. Since the forward reaction is exothermic ($\Delta H = -165\text{ kJ/mol}$), the reverse reaction is endothermic. Therefore, the position of equilibrium shifts to the left, towards the reactants.
(b) Solar radiation warms the Earth's surface, which then re-emits thermal energy as longer-wavelength infrared radiation. Greenhouse gases in the atmosphere absorb this re-emitted infrared radiation, trapping the thermal energy and preventing it from escaping into space, thereby causing global warming.
Marking scheme
**(a)** * **M1**: (position of equilibrium) shifts to the left / towards the reactants / in the reverse direction [1] * **M2**: (because the forward) reaction is exothermic / reverse reaction is endothermic [1]
**(b)** * **M1**: Earth's surface re-emits thermal / infrared radiation (originally received from the sun) [1] * **M2**: Greenhouse gases absorb this re-emitted thermal / infrared radiation (reducing its loss into space / trapping it) [1]
Question 3 · structured
4 marks
Carbon dioxide reacts with hydrogen to produce methane and steam in a reversible reaction:
The reaction reaches dynamic equilibrium in a closed system.
(a) State and explain the effect on the position of equilibrium when the temperature is increased. [2]
(b) Both carbon dioxide and methane are greenhouse gases. Explain, in terms of thermal energy, how greenhouse gases lead to global warming. [2]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) According to Le Chatelier's principle, increasing the temperature favors the endothermic reaction direction to absorb the added heat. Since the forward reaction is exothermic ($\Delta H = -165\text{ kJ/mol}$), the reverse reaction is endothermic. Therefore, the position of equilibrium shifts to the left, towards the reactants.
(b) Solar radiation warms the Earth's surface, which then re-emits thermal energy as longer-wavelength infrared radiation. Greenhouse gases in the atmosphere absorb this re-emitted infrared radiation, trapping the thermal energy and preventing it from escaping into space, thereby causing global warming.
Marking scheme
**(a)** * **M1**: (position of equilibrium) shifts to the left / towards the reactants / in the reverse direction [1] * **M2**: (because the forward) reaction is exothermic / reverse reaction is endothermic [1]
**(b)** * **M1**: Earth's surface re-emits thermal / infrared radiation (originally received from the sun) [1] * **M2**: Greenhouse gases absorb this re-emitted thermal / infrared radiation (reducing its loss into space / trapping it) [1]
Section Question 6
Solve problems on organic chemistry involving fermentation, catalytic hydration of alkenes, acid-base properties, oxidation of alcohols, and salt formation.
Propan-1-ol, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}\), is an alcohol that can be manufactured industrially and undergoes various chemical reactions. (a) Propan-1-ol is manufactured by the catalytic addition of steam to an alkene. (i) Name the alkene used in this reaction. [1] (ii) State the two reaction conditions of temperature and pressure used in this process. [2] (iii) Name the catalyst used in this process. [1] (b) Propan-1-ol can be oxidised to form a carboxylic acid. (i) Name the carboxylic acid formed when propan-1-ol is oxidised. [1] (ii) Acidified potassium manganate(VII) can be used as the oxidising agent. State the colour change seen during this reaction. [2] (c) Ethanol can be manufactured by a different method called fermentation. State one advantage of manufacturing ethanol by fermentation compared to the catalytic addition of steam to ethene. [1]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a)(i) The catalytic addition of steam to propene produces propan-1-ol. (a)(ii) The reaction requires a temperature of 300°C and a pressure of 60 atm (or 6000 kPa). (a)(iii) Phosphoric acid (\(\text{H}_3\text{PO}_4\)) is the catalyst used in the industrial hydration of alkenes. (b)(i) Primary alcohols like propan-1-ol are oxidised to form carboxylic acids with the same number of carbon atoms; in this case, propanoic acid is produced. (b)(ii) Acidified potassium manganate(VII) is an oxidising agent which changes color from purple to colourless as it is reduced. (c) Fermentation uses renewable resources (such as sugar cane or starch), whereas the catalytic addition of steam to ethene relies on non-renewable crude oil for the source of ethene.
6 Propan-1-ol is a liquid alcohol that can be produced through the catalytic hydration of propene.
(a) (i) Identify the specific catalyst required when propene reacts with steam. [1] (ii) Explain why this chemical process is classified as an addition reaction. [1]
(b) When heated with acidified potassium manganate(VII), propan-1-ol is oxidised to propanoic acid. (i) Describe the colour change that occurs during this oxidation process. [1] (ii) State the type of chemical change propan-1-ol undergoes, and define this change in terms of oxygen transfer. [1]
(c) Propanoic acid is a weak acid. (i) Explain what is meant by the term weak in the context of acids. [1] (ii) Define an acid using the concept of proton transfer. [1]
Show answer & marking schemeHide answer & marking scheme
Worked solution
(a) (i) The catalyst used in the hydration of alkenes to form alcohols is phosphoric acid, \(\text{H}_3\text{PO}_4\). (ii) It is an addition reaction because two reactant molecules (propene and steam) combine to form a single product (propan-1-ol) and no other products are made. (b) (i) Acidified potassium manganate(VII) is purple. Upon reacting with and oxidising the alcohol, it is reduced to colourless manganese(II) ions, so the colour changes from purple to colourless. (ii) The alcohol undergoes oxidation, which is defined as the gain of oxygen by a substance. (c) (i) A weak acid is one that only partially dissociates or ionises in aqueous solution to release hydrogen ions. (ii) By Brønsted-Lowry definition, an acid is a substance that acts as a proton (\(\text{H}^+\)) donor.
(a)(ii) Only one product is formed / two molecules combine to form a single molecule [1]
(b)(i) (from) purple to colourless / decolourised [1] Reject: clear (instead of colourless)
(b)(ii) oxidation AND gain of oxygen [1]
(c)(i) partial dissociation / partial ionisation (in aqueous solution) [1] Reject: does not dissolve fully
(c)(ii) proton donor / donor of \(\text{H}^+\) [1]
Question 3 · structural
3 marks
Propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$, is a weak acid that reacts with metal carbonates. (a) Propanoic acid reacts with solid calcium carbonate. (i) State the name of the salt formed in this reaction. (ii) State one observation, other than a change in temperature, that shows a reaction is occurring. (iii) Give the chemical formula of the salt formed in this reaction.
Show answer & marking schemeHide answer & marking scheme
Worked solution
(i) The reaction between propanoic acid and calcium carbonate produces the salt calcium propanoate, carbon dioxide gas, and water. (ii) The production of carbon dioxide gas results in visible effervescence (bubbles/fizzing), and the solid calcium carbonate will dissolve as it reacts to form the soluble salt. (iii) Calcium forms $\text{Ca}^{2+}$ ions, and the propanoate ion is $\text{CH}_3\text{CH}_2\text{COO}^-$. To balance the charges, two propanoate ions combine with one calcium ion, giving the formula $(\text{CH}_3\text{CH}_2\text{COO})_2\text{Ca}$.
thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.