9701_m22_qp_22 - revisedeck

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Fig.1.1 shows how first ionisation energies vary across Period2. ionisation energy Li Be B C element N O F Ne A Construct an equation to represent the first ionisation energy of oxygen. Include state symbols. State and explain the general trend in first ionisation energies across Period2. Explain why ionisation energy A in Fig. 1.1 does not follow the general trend in first ionisation energies across Period2. ElementE is in Period3 of the Periodic Table. The first eight ionisation energy values of E are shown in Table1.1. Table 1.1 ionisation 1st 2nd 3rd 4th 5th 6th 7th 8th ionisation energy / kJ mol–1 11 600 14 800 18 400 23 400 27 500 Deduce the full electronic configuration of E. Explain your answer. full electronic configuration of E = explanation

1. Atomic structure → 1.4. Ionisation energy

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Some oxides of elements in Period3 are shown. Na2O Al 2O3 P4O6 P4O10 SO2 SO3 Na reacts with O2 to form Na2O. Na is the reducing agent in this reaction. Define reducing agent. Write an equation for the reaction of Na2O with water. Al 2O3 is an amphoteric oxide found in bauxite. State what is meant by amphoteric. Al 2O3 is purified from bauxite in several steps. The first step involves heating Al 2O3 with an excess of NaOH. A colourless solution forms. Write an equation for this reaction. Al 2O3 is used as a catalyst in the dehydration of alcohols. State the effect of using Al 2O3 as a catalyst in the dehydration of alcohols. Use the Boltzmann distribution in Fig.2.1 to help explain your answer. energy number of molecules P4O6 is a white solid that has a melting point of 24 °C. Solid P4O6 reacts with water to form H3PO3. Deduce the type of structure and bonding shown by P4O6. Explain your answer. Determine the oxidation number of P in H3PO3. When P4O6is heated with oxygen it forms P4O10. P4O6+ 2O2→ P4O10∆Hr = –1372 kJ mol–1 The enthalpy change of formation, ∆Hf, of P4O10is –3012 kJ mol–1. Calculate the enthalpy change of formation, ∆Hf, of P4O6. ∆Hf of P4O6= kJ mol–1 Write an equation for the reaction of P4O10 with water. SO2 and SO3 are found in the atmosphere. The oxidation of SO2 to SO3 in the atmosphere is catalysed by NO2. The first step of the catalytic oxidation is shown in equation1. equation 1 SO2+ NO2SO3+ NOConstruct an equation to show how NO2 is regenerated in the catalytic oxidation of SO2. NO2 can also react with unburned hydrocarbons to form photochemical smog. State the product of this reaction that contributes to photochemical smog. Fig.2.2 shows how the temperature of the atmosphere varies with height from the ground. –120 –100 –80 –60 –40 temperature / C –20 height / km The equilibrium reaction in equation1 has ∆Hr = –168 kJ mol–1. Suggest how the position of this equilibrium differs at a height of 20 km compared with a height of 50 km from the ground. Explain your answer.

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

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The hydrogen halides HCl, HBr and HI are all colourless gases at room temperature. The hydrogen halides can be formed by reacting the halogens with hydrogen. Describe and explain the relative reactivity of the halogens down the group when they react with hydrogen to form HCl, HBr and HI. HCl is a product of several different reactions. Some of these are shown in Fig.3.1. SiCl 4 NaCl HCl reaction 1 H2O reaction 2 concentrated H2SO4 Write an equation for reaction1. In reaction2, NaCl reacts with concentrated H2SO4 to form HCl and NaHSO4. When NaBr reacts with concentrated H2SO4, the products include Br2 and SO2. Identify the typeof reaction that occur in each case by completing Table3.1. Explain the difference in these reactions. Table 3.1 reactants typeof reaction NaCl and concentrated H2SO4 NaBr and concentrated H2SO4 explanation When heated with a Bunsen burner, HCl does not decompose, whereas HI forms H2 and I2. Explain the difference in the effect of heating on HCl and HI. The hydrogen halides dissolve in water to form strong Brønsted–Lowry acids. The concentration of a strong acid can be determined by titration. State what is meant by strong Brønsted–Lowry acid. On Fig.3.2, sketch the pH titration curves produced when: ● 0.1 mol dm–3 NaOHis added to 25 cm3 of 0.1 mol dm–3 HBr, to excess ● 0.1 mol dm–3 NH3is added to 25 cm3 of 0.1 mol dm–3 HBr, to excess. pH volume of NaOH / cm3 reaction of NaOHand HBr14 pH volume of NH3 / cm3 reaction of NH3and HBr0 10 15 20 25 30 35 40 45 10 15 20 25 30 35 40 45 HBr reacts with propene to form two bromoalkanes, CH3CH2CH2Br and (CH3)2CHBr. Complete the diagram to show the mechanism of the reaction of HBr and propene to form the major organic product. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate. Draw the structures of the intermediate and the major organic product. H Br C C C H H H H H H Explain why the two bromoalkanes are not produced in equal amounts by this reaction. The reaction of CH3CH2CH2Br and NaOH is different depending on whether water or ethanol is used as a solvent. Complete Table 3.2 to identify the organic and inorganic products of the reaction of CH3CH2CH2Br and NaOH in each solvent. Table 3.2 solvent organic productinorganic productwater ethanol

7. Equilibria → 7.2. Brønsted–Lowry theory of acids and bases

11. Group 17 → 11.2. The chemical properties of the halogen elements and the hydrogen halides

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Compounds J and K are found in plant oils. OH O C C J CHO H3C CH3 H C C K H CH3 COOH Complete Table4.1 to state what you would observe when J reacts with the reagents listed. Table 4.1 reagent observation with J 2,4-dinitrophenylhydrazine (2,4-DNPH) Tollens’ reagent sodium metal J has two optical isomers. Draw the three-dimensional structures of the two optical isomers of J. K is used to make the addition polymer Perspex®. A synthesis of Perspex® is shown in Fig.4.2. H C C K H CH3 COOH H C C Perspex® M H CH3 COOCH3 reactant L reaction 1 reaction 2 Identify L. State the conditions required for reaction1. L = conditions = Draw one repeat unit of the addition polymer Perspex®. Use information from Table4.2 to suggest how the infrared spectra of M and Perspex® would differ. Explain your answer. Table 4.2 bond functional group containing the bond characteristic infrared absorption range (in wavenumbers) / cm–1 C–O hydroxy, ester 1040–1300 C=C aromatic compound, alkene 1500–1680 C=O amide carbonyl, carboxyl ester 1640–1690 1670–1740 1710–1750 C≡N nitrile 2200–2250 C–H alkane 2850–3100 N–H amine, amide 3300–3500 O–H carboxyl hydroxy 2500–3000 3200–3650 K can be made from propanone in the three-step synthesis shown in Fig.4.3. H C C K H CH3 COOH C O propanone H3C CH3 C C H3C CH3 C H3C CH3 COOH OH OH N step 1 step 2 step 3 Complete Table4.3 to identify the reagentused and the type of reaction in each step. Table 4.3 step reagenttype of reaction Al 2O3

14. Hydrocarbons → 14.2. Alkenes

15. Halogen compounds → 15.1. Halogenoalkanes

14. Hydrocarbons → 14.1. Alkanes

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