Home / iGCSE Chemistry 0620 Extended – 11.5 Alkenes- Exam Style Questions Paper 4

iGCSE Chemistry 0620 Extended - 11.5 Alkenes- Exam Style Questions Paper 4- New Syllabus

Question

Ethene, $\mathrm{C_2H_4}$, is the first member of a family of similar compounds which contains the alkene functional group.
(a) State the term for a family of similar compounds which contain the same functional group.
(b) Determine the difference in relative molecular mass between $\mathrm{C_2H_4}$ and the next member in this family.
(c) Write the symbol equation for the complete combustion of $\mathrm{C_2H_4}$.
(d) $\mathrm{C_2H_4(g)}$ reacts with steam to form ethanol.
$\mathrm{C_2H_4(g)} + \mathrm{H_2O(g)} \rightleftharpoons \mathrm{C_2H_5OH(g)}$, $\Delta H = -45\mathrm{kJ/mol}$.
The process happens in a closed system and the reaction reaches an equilibrium.
The conditions for this process are 300 °C and 60 atm pressure. $\mathrm{H_2PO_4(g)}$ is used as a catalyst.
(i) Complete Table 6.1 to show the effect, if any, on the concentration of $\mathrm{C_2H_4(g)}$ when changes are applied (words: increases, decreases, no change).

(ii) Explain, in terms of collision theory, why the rate of the forward reaction increases if the temperature increases.
(e) Compound B has the displayed formula shown in Fig. 6.1.

(i) Deduce the molecular formula of compound B.
(ii) State why compound B is unsaturated.
(iii) Draw the structure of one repeat unit of the polymer formed when compound B undergoes addition polymerisation.
(iv) Explain why 1 mol of compound B reacts with 2 mol of sodium hydroxide, $\mathrm{NaOH}$.
(v) Calculate the volume, in $\mathrm{cm}^3$, of $0.250\mathrm{mol/dm^3}$ $\mathrm{NaOH}$ that reacts with $0.100\mathrm{mol}$ of compound B.

Most-appropriate topic codes (Cambridge IGCSE Chemistry 0620):

• Topic 11.1 — Formulae, functional groups and terminology (Parts (a), (b), (e)(i), (e)(ii))
• Topic 11.5 — Alkenes (Part (c))
• Topic 11.6 — Alcohols, Topic 6.3 — Reversible reactions and equilibrium (Parts (d)(i), (d)(ii))
• Topic 11.8 — Polymers (Part (e)(iii))
• Topic 11.7 — Carboxylic acids (Parts (e)(iv), (e)(v))

▶️ Answer/Explanation

(a) Homologous series

(b) $14$
Each successive member of a homologous series differs by a $\mathrm{-CH_2-}$ unit, which has a relative mass of $12 + 2(1) = 14$.

(c) $\mathrm{C_2H_4} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 2\mathrm{H_2O}$
Complete combustion of hydrocarbons yields carbon dioxide and water.

(d)(i) Temperature is decreased: decreases. Some $\mathrm{C_2H_5OH(g)}$ is removed: decreases. Pressure is increased: decreases. A more effective catalyst is used: no change.

(d)(ii) As temperature increases, the kinetic energy of particles increases. This leads to a higher frequency of collisions between particles. Crucially, a higher percentage/proportion/fraction of collisions have energy greater than or equal to the activation energy.

(e)(i) $\mathrm{C_8H_6O_4}$

(e)(ii) It has a carbon-carbon bond which is not a single bond (a carbon-carbon double bond).
Unsaturated compounds contain double or triple carbon-carbon bonds.

(e)(iii) The repeat unit is drawn with a single $\mathrm{C\text{–}C}$ backbone bond and continuation bonds, with the correct substituents attached.

(e)(iv) Compound B has two carboxylic acid groups.
Each $\mathrm{-COOH}$ group is acidic and can donate one proton, reacting with one mole of $\mathrm{NaOH}$ in a neutralisation reaction.

(e)(v) $800\text{ cm}^3$
Moles of $\mathrm{NaOH}$ reacting $= 0.100 \times 2 = 0.200\text{ mol}$. Volume $= \frac{\text{moles}}{\text{concentration}} = \frac{0.200}{0.250} = 0.80\text{ dm}^3 = 800\text{ cm}^3$.

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