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.
(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.
(b) Ethane reacts with chlorine at room temperature to form chloroethane, C2H5Cl, and one other product.
(i) Name the other product formed.
(ii) State the condition needed for this reaction to take place.
(c) Ethene reacts with chlorine at room temperature to form dichloroethane, C2H4Cl2.
C2H4 + Cl2 → C2H4Cl2
(i) State why this is an addition reaction.
(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.
step 2 – Use your answer in step 1 and the energy change for the reaction to determine the energy released when bonds are formed.
step 3 – Use your answer in step 2 and bond energy values to determine the energy of a C–Cl bond.
▶️ Answer/Explanation
(a)(i) Ans: cracking
Cracking breaks long-chain hydrocarbons into shorter alkanes and alkenes using heat and a catalyst.
(a)(ii) Ans: C10H22 → 4C2H4 + C2H6
Decane (C10H22) cracks to form 4 ethene molecules and 1 ethane molecule, balancing carbon and hydrogen atoms.
(b)(i) Ans: hydrogen chloride
Chlorination of ethane substitutes one hydrogen with chlorine, producing chloroethane and HCl.
(b)(ii) Ans: ultraviolet light
UV light provides energy to break Cl–Cl bonds, initiating free-radical substitution.
(c)(i) Ans: Only one product is formed.
Addition reactions combine two molecules (ethene + chlorine) into a single product (dichloroethane).
(c)(ii) Ans: 340 kJ/mol
Step 1: Energy to break bonds = (4 × C–H) + (C=C) + (Cl–Cl) = 4×410 + 610 + 240 = 2490 kJ/mol.
Step 2: Energy released = 2490 + 180 (exothermic) = 2670 kJ/mol.
Step 3: Subtract C–H and C–C bond energies: 2670 – (4×410 + 350) = 680 kJ/mol for two C–Cl bonds → 340 kJ/mol per bond.
(a) Chalcopyrite, \(FeCuS_2\), is used in the manufacture of sulfuric acid in the Contact process.
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.
(ii) State the temperature and pressure used in this reaction.
Include units.
(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
(a) Ans: \(4FeCuS_2 + 13O_2 → 2Fe_2O_3 + 4CuO + 8SO_2\)
Balancing the equation: 4 chalcopyrite molecules produce 2 iron(III) oxide, 4 copper(II) oxide, and 8 sulfur dioxide molecules to conserve all atoms.
(b)(i) Ans:
1. Forward and reverse reactions occur at equal rates
2. Concentrations of reactants and products remain constant
These are the two fundamental characteristics of dynamic equilibrium.
(b)(ii) Ans:
● Temperature: 450°C
● Pressure: 1-2 atm
These conditions provide optimal yield while considering reaction rate and energy costs.
(b)(iii) Ans: Vanadium(V) oxide (\(V_2O_5\))
This is the industrial catalyst used in the Contact process.
(b)(iv) Ans: To increase the reaction rate without being consumed
The catalyst provides an alternative pathway with lower activation energy, speeding up both forward and reverse reactions equally.
(b)(v) Ans:
The equilibrium shifts left (towards \(SO_2\) and \(O_2\)) because the forward reaction is exothermic. According to Le Chatelier’s Principle, increasing temperature favors the endothermic direction.
(c) Ans: \(C_6H_{12}O_6 → 6C + 6H_2O\)
Dehydration removes water molecules from glucose, leaving elemental carbon. The balanced equation shows 6 carbon atoms and 6 water molecules formed per glucose molecule.