Question
This question is about alkanes and alkenes.
(a) Short-chain alkanes and alkenes can be formed from long-chain alkanes in a chemical reaction.
(i) Name the type of chemical reaction which forms short‑chain alkanes and alkenes from long‑chain alkanes.[1]
(ii) Decane has 10 carbon atoms. It forms ethane and ethene as the only products in this type of chemical reaction.
Write the chemical equation for this reaction.[3]
(b) Ethane reacts with chlorine at room temperature to form chloroethane, C2H5Cl, and one other product.
(i) Name the other product formed.[1]
(ii) State the condition needed for this reaction to take place.[1]
(c) Ethene reacts with chlorine at room temperature to form dichloroethane, C2H4Cl2.
C2H4 + Cl2 → C2H4Cl2
(i) State why this is an addition reaction.[1]
(ii) The chemical equation for this reaction can be represented as shown.
The energy change for the reaction is –180kJ/mol.
Use the bond energies in the table to calculate the bond energy of a C–Cl bond, in kJ/mol.
Use the following steps.
step 1 Calculate the energy needed to break bonds.
energy needed to break bonds = kJ
step 2 Use your answer in step 1 and the energy change for the reaction to determine the energy released when bonds are formed.
energy released when bonds form = kJ
step 3 Use your answer in step 2 and bond energy values to determine the energy of a C–Cl bond.
bond energy of a C–Cl bond = kJ/mol
[4][Total: 11]
Answer/Explanation
Ans:
5(a)(i) cracking
5(a)(ii) C10H22 → 4C2H4 + C2H6
C10H22 as only reactant
formulae of ethene and ethane as only products
correct equation
5(b)(i) hydrogen chloride
5(b)(ii) ultraviolet light
5(c)(i) (only) one product is formed
5(c)(ii) M1 Bond energy in breaking bonds
= [(4 × 410) + 610 + 240] = 2490 (kJ / mol)
M2 Use of total E change to find bond energy of C2H4Cl2
= M1 + 180 = 2490 + 180 = 2670 (kJ / mol)
M3 Determination of total C–Cl bond energy
= M2 – [(4 × 410) + 350] = 2670 – 1990 = 680 (kJ / mol)
M4 Determination of each C–Cl bond energy
= M3 / 2 = 680 / 2 = 340 (kJ / mol)
Question
Chalcopyrite, \(FeCuS_2\), is used in the manufacture of sulfuric acid in the Contact process.
(a) In the first stage of the process, chalcopyrite reacts with oxygen in the air to produce
sulfur dioxide, \(SO_2\), iron(III) oxide and copper(II) oxide.
Complete the chemical equation for the reaction of \(FeCuS_2\) with oxygen.
\(4FeCuS_2 + 13O_2\) → …………….. + …………….. + ……………..
(b) Sulfur dioxide is then converted to sulfur trioxide.
\(2SO_2 + O_2 \leftrightarrow 2SO_3\)
The reaction is exothermic. It is also an equilibrium.
(i) State two features of an equilibrium.
1 …………………………………………………………………………………………………………………………
2 …………………………………………………………………………………………………………………………
(ii) State the temperature and pressure used in this reaction.
Include units.
● temperature …………………………………………………………………………………………………..
● pressure ………………………………………………………………………………………………………..
(iii) Name the catalyst used.
(iv) Explain why a catalyst is used.
(v) Describe and explain, in terms of equilibrium, what happens when the temperature is
increased.
(c) Concentrated sulfuric acid is a dehydrating agent.
When glucose is dehydrated, carbon and one other product are formed.
Complete the equation to show the dehydration of glucose, \(C_6H_{12}O_6\).
\(C_6H_{12}O_6\) → ………..C + …………………
Answer/Explanation
Answer:
(a) \((4FeCuS 13O_2) → 2Fe_2O_3 + 4CuO + 8SO_2\)
\(Fe_2O_3\) and CuO as a product (1)
Equation fully correct (1)
(b) (i) M1 rate of forward reaction = rate of reverse reaction (1)
M2 concentration of reactants and products are constant (1)
(ii) M1 450 °C (1)
M2 1-2 atm (1)
(iii) vanadium(V) oxide
(iv) increase rate of reaction
(v) M1 equilibrium shifts to left hand side (1)
M2 forward reaction is exothermic (1)
(c) \(C_6H_{12}O_6 → 6C + 6H_2O\)
\(H_2O\) (1)
balance (1)
Question
(a) Dilute sulfuric acid is electrolysed using the apparatus shown in the diagram.
(i) State what is meant by the term electrolysis.[2]
(ii) Explain why inert electrodes are used.[1]
(iii) Name the products formed at each electrode.
negative electrode……………………………………………..
positive electrode…………………………………………..[2]
(iv) Write an ionic half-equation for the reaction at the negative electrode.[2]
(b) Sulfuric acid is manufactured using the Contact process. This manufacture involves four stages.
(i) Stage 1 involves the combustion of sulfur to form sulfur dioxide.
Write the chemical equation for stage 1.[1]
(ii) The equation for stage 2 is shown.
$
2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})
$
The reaction can reach equilibrium.
Explain what is meant by the term equilibrium.[2]
(iii) The energy level diagram for the forward reaction in stage 2 is shown.
[2]
Explain what the diagram shows about the energy changes in the forward reaction.
(c) In stage 3 sulfur trioxide, $\mathrm{SO}_3$, is converted to oleum, $\mathrm{H}_2 \mathrm{~S}_2 \mathrm{O}_7$.
In stage 4 oleum reacts to form sulfuric acid, $\mathrm{H}_2 \mathrm{SO}_4$.
State what oleum reacts with in stage 4 .[1]
(d) A sample of sulfuric acid, $\mathrm{H}_2 \mathrm{SO}_4$, has a concentration of $0.75 \mathrm{~mol} / \mathrm{dm}^3$.
Calculate the concentration of sulfuric acid in $\mathrm{g} / \mathrm{dm}^3$.
$\mathrm{g} / \mathrm{dm}^3[2]$[Total: 15]
▶️Answer/Explanation
Ans:
5(a)(i) breakdown by (the passage of) electricity (1)
of an ionic compound in molten/aqueous (state) (1)
5(a)(ii) they do not react 1
5(a)(iii) negative electrode: hydrogen (gas) (1)
positive electrode: oxygen (gas) (1)
5(a)(iv)
$\mathrm{H}^{+}+\mathrm{e}(-)$ as the only species on the left (1) equation fully correct (1)
$
2 \mathrm{H}^{+}+2 \mathrm{e}(-) \rightarrow \mathrm{H}_2 \text { (scores 2) }
$
5(b)(i) $\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2$
5(b)(ii) rate of forward reaction is equal to rate of reverse reaction (1)
constant concentration (of reactants and products) (1)
5(b)(iii) exothermic / heat / energy is released / surroundings warm up
products have lower energy than reactants / ORA
5(c) water / H2O
5(d) (Mr =) 98
(0.75× 98 =) 73.5