9701_s18_qp_21 - revisedeck

1

Sulfuric acid is manufactured by the Contact process. One stage in this process is the conversion of sulfur dioxide into sulfur trioxide in the presence of a heterogeneous catalyst of vanadium(oxide, V2O5. 2 O=S=O+ O=O2 O=S=OΔH = –196 kJ mol–1 = O State the effect of a catalyst on a reaction. Explain how a catalyst causes this effect. State the meaning of the term heterogeneous as applied to catalysts. Some bond energies are given. bond bond energy / kJ mol–1 S=O (in SO2) O=O Use the data, and the enthalpy change for the conversion of sulfurdioxide into sulfurtrioxide, to calculate a value for the S=O bond energy in SO3. S=O bond energy in SO3 = kJ mol–1 The Contact process is usually carried out at a temperature of about 400 °C and a pressure just above atmospheric pressure. Using a higher or lower temperature and pressure would affect both the rate of production of sulfurtrioxide and the yield of sulfurtrioxide. A reaction pathway diagram for both the catalysed and uncatalysed reactions between SO2 and O2 is shown. A B C D E energy progress of reaction 2SO2 + O2 2SO3 The letters A–E represent energy changes. Complete the table by stating which letter, A–E, represents the energy change described. energy change letter the energy change for the production of SO3 the activation energy for the production of SO3 in the absence of a catalyst the activation energy for the first step in the decomposition of SO3 in the presence of a catalyst The equation for this stage of the Contact Process is shown. 2SO2+ O22SO3ΔH = –196 kJ mol–1 State and explain the effect of increasing temperature on the rate of production of SO3. State and explain the effect of increasing temperature on the yield of SO3. The SO3 produced is converted to sulfuricacid in two stages. In the first stage the SO3 is reacted with concentrated sulfuricacid to produce oleum, H2S2O7. The oleum is then reacted with water to form sulfuric acid. Suggest an equation for the reaction of oleum, H2S2O7, with water to form sulfuricacid. SO2 reacts with water to form sulfurous acid. Sulfurous acid is a weak Brønsted‑Lowry acid, while sulfuric acid is a strong Brønsted‑Lowry acid. Complete the ‘dot-and-cross’ diagram to show the bonding in a molecule of SO2. Show outer electrons only. State the meaning of the term strong Brønsted‑Lowry acid. Write an equation to show the acid-base behaviour of sulfuricacid with water. Include state symbols.

12. Nitrogen and sulfur → 12.1. Nitrogen and sulfur

8. Reaction kinetics → 8.3. Homogeneous and heterogeneous catalysts

Open

2

Crude oil is a complex mixture of hydrocarbon molecules. The hydrocarbon molecules in crude oil are separated by fractional distillation. Fractional distillation is used because the different hydrocarbon molecules in crude oil have different boiling points. Explain why the hydrocarbon molecules in crude oil have different boiling points. Some of the hydrocarbon molecules obtained from crude oil are processed further by cracking. Suggest why some hydrocarbon molecules are processed further by cracking. Cracking one mole of dodecane, C12H26, produces two moles of ethene and one mole of another hydrocarbon molecule. Write the equation for this cracking reaction. The ethene can be used in the production of poly. Give the full name of the process used to produce polyfrom ethene. Give two reasons why polyshould be reused or recycled rather than just thrown away. Part of a polymer chain, produced by the same type of process as poly, is shown. CH2 CH3 C COOCH3 CH2 CH3 C COOCH3 CH2 CH3 C COOCH3 Give the displayed formula of the monomer used to produce this polymer.

14. Hydrocarbons → 14.1. Alkanes

14. Hydrocarbons → 14.2. Alkenes

20. Polymerisation → 20.1. Addition polymerisation

Open

3

The elements in the third period exhibit periodicity in both their chemical and physical properties. A graph of the atomic and ionic radii across the third period is shown. 0.25 0.20 0.15 0.10 0.05 0.00 atomic or ionic radius / nm atoms and ions Na Na+ Mg Mg2+ Al Al 3+ Si Si4+ P3– P S2– S Cl – Cl = atomic radius / nm = ionic radius / nm Explain the decrease in atomic radius across the third period. Explain why, for sodium to silicon, the ionic radii are less than the atomic radii. Explain why, for phosphorus to chlorine, the ionic radii are greater than the atomic radii. The first ionisation energies of the elements across the third period show a general increase. Aluminium and sulfur do not follow this general trend. Explain why aluminium has a lower first ionisation energy than magnesium. Explain why sulfur has a lower first ionisation energy than phosphorus. The elements in the third period, from sodium to silicon, can react with chlorine to form chlorides. State and explain the pattern of change of oxidation number which occurs to both chlorine and the different Period3 elements when they react together. Give the equations to show the reactions of sodiumchloride and silicon(chloride when separately added to water. sodiumchloride silicon(chloride Complete the table to describe the structure and bonding in sodium chloride and silicon(chloride. structure bonding sodiumchloride silicon(chloride

9. The Periodic Table: chemical periodicity → 9.2. Periodicity of chemical properties of the elements in Period 3

Open

4

X is CH3CH(OH)CH2CH3. The reaction between X and alkaline aqueous iodine produces a yellow precipitate. Give the name of the compound formed as a yellow precipitate in this reaction. Give the name of X. There are three structural isomers of X that are alcohols. Draw the structures of these three isomers of X. Two reactions of X are shown. CH3CH(OH)CH2CH3 CH3COCH2CH3 C4H8 reaction 2 reaction 1 X Identify the type of reaction involved in reaction1. Identify the reagents for reaction1. Reaction2 can be carried out by passing the vapour of X over hot aluminiumoxide. The product of reaction2, C4H8, is actually a mixture of three isomers. Give the full names of the three isomers formed by reaction2. The reaction of methylpropene, (CH3)2CCH2, with hydrogenbromide, HBr, produces a mixture of two halogenoalkanes. One of the halogenoalkanes, 2‑bromo‑2‑methylpropane, is formed as the major product while 1‑bromo‑2‑methylpropane is formed in small quantities. Complete the mechanism to show the reaction of methylpropene with HBr to form the major product. Include the structure of the intermediate and all necessary charges, dipoles, lone pairs and curly arrows. The structure of 2‑bromo‑2‑methylpropane is not required. CH3 H3C C H H C H Br 2-bromo-2-methylpropane Explain why 2-bromo-2-methylpropane is the major product of this reaction.

15. Halogen compounds → 15.1. Halogenoalkanes

14. Hydrocarbons → 14.2. Alkenes

14. Hydrocarbons → 14.1. Alkanes

Open