Phosphorus, sulfur and chlorine can all react with oxygen to form oxides.
- (a)(i)Phosphorus reacts with an excess of oxygen to form phosphorus(V) oxide. Write an equation to show the reaction of phosphorus with excess oxygen.[1]
- (a)(ii)Describe the reaction of phosphorus(V) oxide with water.[2]
- (a)(iii)State the structure and bonding of solid phosphorus(V) oxide.[1]
- (b)(i)The two most common oxides of sulfur are SO2 and SO3. When SO₂ dissolves in water, a small proportion of it reacts with water to form a weak Brønsted-Lowry acid. Explain the meaning of the term weak Brønsted-Lowry acid.[2]
- (b)(ii)Write the equation for the reaction of SO2 with water.[1]
- (b)(iii)SO2 reacts with NO2 in the atmosphere to form SO3 and NO. NO is then oxidised in air to form NO2. SO2 + NO2 → SO3 + NO; 2NO + O2 → 2NO2; State the role of NO2 in this two-stage process.[1]
- (c)(i)Emissions of SO2 from coal-fired power stations can be reduced by mixing the coal with powdered limestone. Limestone is heated to form CaO in reaction 1. This then reacts with SO2 and O₂ to form CaSO4 in reaction 2. reaction 1: CaCO3(s) → CaO(s) + CO₂(s); reaction 2: CaO(s) + SO2(g) + ½O2(g) → CaSO4(s); State the type of reaction occurring in reaction 1.[1]
- (c)(ii)Use the data to calculate the enthalpy change of reaction 2. compound | ΔH/kJ mol⁻¹ ---|--- CaO(s) | -635 SO2(g) | -297 CaSO4(s) | -1434 enthalpy change of reaction 2 = ................................... kJ mol⁻¹[2]
- (d)(i)Chlorine forms several oxides, including Cl₂O, ClO2 and Cl₂O6. Draw a 'dot-and-cross' diagram of Cl₂O. Show outer-shell electrons only.[1]
- (d)(ii)ClO2 can be prepared by reacting NaClO2 with Cl2. Write the oxidation state of chlorine in each species in the boxes provided. 2NaClO2 + Cl2 → 2ClO2 + 2NaCl; oxidation state of chlorine: [+3] [ ] [ ] [ ][1]
- (d)(iii)Cl₂O6(g) is produced by the reaction of ClO2(g) with O3(g). 2ClO2(g) + 2O3(g) ⇌ Cl₂O6(g) + 2O2(g) ΔH = −216 kJ mol⁻¹. The reaction takes place at 500K and 100kPa. State and explain the effect on the yield of Cl₂O6(g) when the experiment is carried out: • at 1000K and 100 kPa • at 500 K and 500 kPa.[4]
- (e)(i)Element E is a Period 5 element. E reacts with oxygen to form an insoluble white oxide that has a melting point of 1910°C. The oxide of E conducts electricity only when liquid. Suggest the type of bonding shown by the oxide of E. Explain your answer.[2]
- (e)(ii)E also reacts readily with Cl2(g) to form a white solid that reacts exothermically with water. The resulting solution reacts with aqueous silver nitrate to form a white precipitate that dissolves in dilute ammonia. Suggest the type of bonding shown by the chloride of E. Explain your answer.[2]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2020). The mark or mark type is shown in brackets.
- [1] P4 + 5O2 → P4O10
- [1] reacts vigorously
- [1] solid disappears / colourless solution forms
- [1] hydrolysis
- [1] exothermic
- [1] acid(ic) (solution)
- [1] steamy / misty fumes
- [1] Simple and covalent OR molecular and covalent
- [1] M1: proton / H+ donor
- [1] M2: partially dissociates (in solution)
- [1] SO2 + H2O → H2SO3
- [1] (homogeneous) catalyst
- [1] thermal decomposition
- [1] M1: ΔH = –1434 – (–635 + –297)
- [1] M2: = –502 (kJ mol⁻¹)
- [1] Cl2O dot-and-cross diagram with correct outer shell electrons
- [1] oxidation states: Cl2 = 0, ClO2 = +4, NaCl = -1
- [1] M1: (at 1000 K and 100 kPa) (yield) decreases
- [1] M2: reaction is exothermic AND equilibrium moves left
- [1] M3: (at 500 K and 500 kPa) (yield) increases
- [1] M4: fewer moles (of gas) on right-hand side AND equilibrium moves right
- [1] M1: ionic
- [1] M2: ions only able / free to move / free to conduct (when liquid / molten)
- [1] M1: covalent
- [1] M2: hydrolysed (by water)
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