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1 Candidate Name Centre Number 2 Candidate Number GCE A level 335/01 CHEMISTRY CH5 A.M. THURSDAY, 19 June hour 40 minutes JD*(S ) 4 B 5 ADDITIONAL MATERIALS TOTAL MARK In addition to this examination paper, you will need: a calculator; an 8 page answer book; a copy of the Periodic Table supplied by WJEC. Refer to it for any relative atomic masses you require. INSTRUCTIONS TO CANDIDATES Write your name, centre number and candidate number in the spaces at the top of this page. Section A Answer all the questions in the spaces provided. Section B Answer both questions in Section B in a separate answer book which should then be placed inside this question-and-answer book. Candidates are advised to allocate their time appropriately between Section A (35 marks) and Section B (40 marks). INFORMATION FOR CANDIDATES The number of marks is given in brackets at the end of each question or part-question. The maximum mark for this paper is 75. FOR EXAMINER S USE ONLY Section Question Mark Your answers must be relevant and must make full use of the information given to be awarded full marks for a question. You are reminded that marking will take into account the Quality of Written Communication used in all your written answers A 1 2 3

2 2 Examiner only SECTION A Answer all the questions in the spaces provided. 1. (a) The thermal decomposition of ozone is shown in the equation below. 2O 3 (g) 3O 2 (g) Kinetic studies have shown that the reaction is second order with respect to ozone. (i) Write the rate equation for the reaction and use it to explain the term order of reaction. [2] (ii) The value of the rate constant at 298 K is dm 3 mol 1 s 1. If the concentration of ozone is mol dm 3, calculate the rate of reaction at 298 K and state its units. [2] (iii) In the stratosphere, chlorine radicals act as catalysts and speed up the decomposition of ozone. Explain how catalysts increase the rate of reaction. [2] (iv) Give one problem caused as a consequence of ozone depletion. [1]

3 3 (b) Radioactive decay shows first-order kinetics. (i) The graph below shows the decay of a sample of radioactive iodine 131 I. 100 Examiner only Relative concentration of 131 I time / days I. Use the graph to find the half-life of 131 I. [1] II. Using the tangents drawn at relative concentrations of 131 I of 60 and 30, or otherwise, explain how the graph shows that the radioactive decay is first-order. [2] (ii) Archaeologists can determine the age of dead organic matter by measuring the proportion of radioactive carbon-14 present. Carbon-14 emits β-particles and has a half-life of 5570 years. I. Calculate the age of a piece of wood found to contain th as much 14 8 C as living material. [1] 1 II. Write an equation to show the radioactive decay of 14 C by β emission. [1] Total [12] Turn over.

4 4 Examiner only 2. (a) Iodine and chlorine both react with aqueous sodium thiosulphate, but in different ways. (i) Complete equation A for the reaction of iodine with thiosulphate ions. [1] Equation A: I 2 + 2S 2 O (ii) When chlorine is passed into an aqueous solution containing thiosulphate ions, the products are sulphate ions (SO 4 2 ), chloride ions (Cl ) and hydrogen ions (H + ). Write the half equations for: I. the reduction of chlorine to chloride ions; [1] II. the oxidation of thiosulphate ions in water to sulphate ions and hydrogen ions. [1] Half equation B: III. Hence, write the overall equation for this reaction. [1] (iii) I. State the initial and final oxidation numbers of sulphur in both equations A and B by completing the table below. [1] Equation A Half equation B Initial oxidation number of sulphur Final oxidation number of sulphur II. Explain how these values show that chlorine is a stronger oxidising agent than iodine. [1] Explanation

5 5 Examiner only (b) Chlorine can react with aqueous sodium hydroxide to form sodium chloride, sodium chlorate(i) and sodium chlorate(v). Give one large scale use for sodium chloride sodium chlorate(v) [1] (c) Sodium chlorate(i) is used in bleach. The concentration of sodium chlorate(i) in domestic bleach can be found by reacting the bleach with an acidified iodide solution to form iodine and then titrating with a thiosulphate solution. (i) Name a suitable indicator for this reaction. [1] (ii) The equation for this reaction is given below. ClO (aq) + 2H + (aq) + 2I (aq) I 2 (aq) + Cl (aq) + H 2 O(l) A 25 0 cm 3 sample of domestic bleach was diluted to 250 cm 3 in a volumetric flask cm 3 of this solution was added to an excess of acidified potassium iodide and the iodine produced reacted with 20 4 cm 3 of aqueous sodium thiosulphate of concentration mol dm 3. Use the equation above, together with equation A in (a)(i), to calculate the concentration of sodium chlorate(i) in the original bleach sample to three significant figures. [4] mol dm 3 Turn over.

6 6 Examiner only (d) Describe how you could distinguish between separate aqueous solutions containing chloride ions and iodide ions. [2] Total [14] 3. In living systems, enzymes that enable biochemical reactions to take place can function only within a narrow ph range. An important buffer system in human beings is the carbonate buffer, consisting of carbonic acid and its conjugate base, the hydrogencarbonate ion. H 2 CO 3 (aq) H + (aq) + HCO 3 (aq) (K a for carbonic acid = mol dm 3 at body temperature) (a) Write an expression for the dissociation constant, K a, for the acid. [1] (b) Define the term ph. [1] (c) Calculate the ph of blood in a person, given that at body temperature: [3] [H 2 CO 3 ] = mol dm 3 [HCO 3 ] = mol dm (d) State the purpose of a buffer solution. [1]

7 7 Examiner only (e) Explain how an aqueous solution of ethanoic acid and sodium ethanoate can act as a buffer solution when a small amount of acid or alkali is separately added to it. [3] Total [9] Total Section A [35] Turn over.

8 8 SECTION B Answer both questions in this section in the separate answer book provided. 4. Copper and iron are typical transition metals. (a) A characteristic of these metals is the ability to form coloured ions. (i) (ii) Give the electronic configuration of copper(i) ions, Cu +, and state why copper(i) compounds are not usually coloured. [2] Copper(II) ions can form the following coloured complexes: [Cu(H 2 O) 6 ] 2+ and [CuCl 4 ] 2 I. State the shape and colour for each complex. [4] II. Describe the bonding in copper(ii) complexes. [2] (b) (i) Define the term standard molar enthalpy change of formation, H f. [1] φ (ii) The enthalpy of formation of copper(ii) fluoride, CuF 2, can be determined indirectly using a Born-Haber cycle. Use the data given below to calculate the enthalpy of formation of copper(ii) fluoride in kj mol 1. [4] Process H / kj mol 1 φ Cu(s) Cu(g) F 2 (g) F(g) 79 Cu(g) Cu + (g) + e 745 Cu + (g) Cu 2+ (g) + e 1960 F(g) + e F (g) 348 Cu 2+ (g) + 2F (g) CuF 2 (s) 3037

9 9 (c) Another transition metal characteristic is the ability to show variable oxidation states. The diagram below shows part of the apparatus that was used to measure the standard electrode potential of the Fe 3+ (aq)/fe 2+ (aq) half cell. The standard electrode potential of the half cell was +0 77V. V hydrogen, 1 atm. Fe 3+ /Fe 2+ (aq) (i) A Name a solution that must be placed in beaker A and state its concentration. [1] B (ii) Name a metal that could be used as an electrode in beaker B. [1] (iii) Name the part of the cell that is missing and state its purpose. [1] (iv) State, giving a reason, the direction in which the electrons flow along the wire through the voltmeter. [1] (v) The standard electrode potential for the Cu 2+ (aq)/cu(s) electrode is V. If the hydrogen electrode is replaced by the Cu 2+ (aq)/cu(s) electrode, calculate the new reading on the voltmeter, V. [1] (d) Some standard electrode potentials, E, are given below. φ I 2 (s) + e _ Br 2 (l) + e _ Cl 2 (g) + e _ System I (aq) Br (aq) Cl (aq) E / volts φ Using the information from the table, state which of the halides will reduce Fe 3+ to Fe 2+. Give a reason for your answer. [2] Total [20] Turn over.

10 10 5. Two important trends in inorganic chemistry are that: metallic character increases down a group; metallic character decreases across a period. (a) (b) (c) (d) Show how the first trend is true by describing the structure and bonding of carbon (diamond) and of lead. [5] Group IV elements can show oxidation states of II and IV in their compounds. State, giving examples, how the relative stability of these oxidation states changes as the group is descended and give a reason for this trend. [3] Explain why the reactivity of Group I metals increases as the group is descended. [2] State what would be observed if sodium hydride were added to water and predict the ph of the resulting solution. [2] (e) State the trend in the bonding of the chlorides across Period 3. [1] (f) (g) Write equations to show what happens when sodium chloride and phosphorus(v) chloride are separately added to water. Predict a value for the ph of the resulting solutions and explain how your answer confirms the trend given in part (e). [5] Name an oxide of one of the elements of Period 2 or 3 which would dissolve in water to give an acidic solution and give the formula of the salt which would form if the acidic solution were added to aqueous sodium hydroxide. [2] Total [20] Total Section B [40]

11 CHEMISTRY CH5 Section A 1. (a) (i) Rate = k[o 3 ] 2 (1) Order (2) is the power to which the concentration (of O 3 ) is raised (1) [2] (ii) Rate = (0.023) 2 (1) = mol dm -3 s -1 (1) [2] (Mark consequentially to (i)) (iii) Catalysts provide a different route (1) of lower activation energy (1) [2] (iv) Causes skin cancer / damages eyesight / lets in (more) harmful UV radiation [1] (b) (i) I. 8 days [1] II. Tangent at 6 days = 5.5 Tangent at 12 days = 2.75 (1) As concentration doubles, rate doubles. Therefore first order (1) (Or compares at least two half-life values (1) to show that half life is constant. (1)) [2] (ii) I = years [1] II. 14 C 14 N + β [1] Total [12] 23

12 2. (a) (i) I 2 + 2S 2 O 3 2 2I + S 4 O 6 2 [1] (ii) I. Cl 2 + 2e 2Cl [1] II. S 2 O H 2 O 2SO H + + 8e [1] III. 4Cl 2 + S 2 O H 2 O 8Cl + 2SO H + (iii) I. Equation A: S +2 to +2.5 (½) II. Equation B: S +2 to +6 (½) [1] Chlorine produces greater oxidation number (change) in sulphur (therefore is more powerful oxidising agent) [1] (b) Sodium chloride in food industry, sodium chlorate(v) as weedkiller [1] (c) (i) Starch solution [1] [1] (ii) Moles S2O = = 1.88 (1) Moles I 2 = Moles ClO = (1) (Moles ClO in 250 cm 3 solution = ) Moles ClO in original bleach = (1) Concentration ClO = = mol dm 3 (1) [4] (d) Add HNO 3 (aq) followed by AgNO 3 (aq) (½ each) (1) White precipitate shows chloride, yellow precipitate shows iodide (½ each) (1) [2] (Accept adding chlorine / bromine water no change if chloride brown solution if iodide) Total [14] 24

13 3. (a) K a + [HCO3 ][H ] = [H CO ] 2 3 [1] (b) ph = log[h + ] [1] [H ] (c) = (1) [H + ] = (1) ph = log = 7.4 (1) [3] (d) Keeps ph constant (when small amounts of acid or alkali are added). [1] (e) Solution contains a large amount of CH 3 COOH and CH 3 COO ions. (Accept correct equations) (1) When an acid is added, the CH 3 COO ions react with the H + ions, removing them from solution and keeping the ph constant. (1) When an alkali is added the CH 3 COOH reacts with the OH ions, removing them from solution and keeping the ph constant. (1) [3] (Accept answer in terms of H + ions reacting with OH ions) Total [9] Total Section A [35] 25

14 Section B 4. (a) (i) 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 (1) Cu + has full 3d orbitals. (1) [2] (ii) I. [Cu(H 2 O) 6 ] 2+ octahedral (1) blue (1) [CuCl 4 ] 2 tetrahedral (1) yellow (accept green) [4] II. Complexes consist of a Cu 2+ ion surrounded by ligands (1) that form coordinate bonds with it. (1) [2] (b) (i) Enthalpy change when 1 mole of compound is formed from its elements in their standard states / under standard conditions. [1] (ii) H f = H at Cu + I.E. Cu + H at F 2 + E.A. F + H lat form CuF 2 (1) Doubling value for forming 2F and 2F (1) (These marks can be obtained from Born-Haber cycle) H f CuF 2 = (1) H f CuF 2 = _ 531 kj mol -1 [4] (c) (i) HCl, 1 mol dm -3 [1] (Accept H + ) (ii) Platinum [1] (iii) (iv) Salt bridge. It allows charged species to flow between the solutions / allows conductivity without mixing solutions / completes the circuit [1] From beaker A to beaker B because the electrons flow to the positive electrode. [1] (v) emf = = 0.43 V [1] (d) Iodide (1) Only one with more negative / less positive standard potential than Fe 3+, Fe 2+ half-cell (1) [2] (2 nd mark can be obtained from calculation values) Total [20] 26

15 Q. 5(a)(i) is to be assessed for the quality of Written Communication. Candidates must satisfy the following criteria to be awarded the marks designated as [L]. Selection of a form and style of writing which is appropriate to this question. Organisation of the relevant information clearly and coherently. Use of legible text with accurate spelling, grammar and punctuation. 5. (a) Diamond shows giant covalent (macromolecular)structure. (1) In diamond each carbon atom is joined to four others. The four outer electrons are all involved in bonding (1) giving tetrahedral structure. (1) In lead there is a giant metallic structure (1) with a regular array of metal ions surrounded by a sea of delocalised (valence) electrons. (1) The bonding force is the attraction between the positive ions and the delocalised electrons. (1) Therefore carbon at the top of the group is a non metal, while lead at the bottom of the group is a metal. (1) [5] (Accept any 5 points) (b) Oxidation state II becomes more stable (1) CO 2 more stable than CO (½), but PbO more stable than PbO 2 (½) Inert pair effect the two s electrons become more stable as the group is descended. (1) [3] (c) (They react by losing an electron.) As group is descended it is easier to lose an electron / ionisation energy decreases (1) since effective nuclear charge decreases / ionisation energy decreases / there is less attraction to the nucleus due to increased shielding (accept further distance from the nucleus.) (1) [2] (d) White solid disappearing / colourless solution / bubbles (1) ph = (1) [2] (e) From ionic to covalent / becomes more covalent across period 3. [1] (f) NaCl Na + (aq) + Cl (aq) (1) ph = 7 (1) PCl 5 + 4H 2 O H 3 PO 4 + 5HCl (1) ph = 0 3 (1) Ionic chlorides dissolve in water, covalent chlorides react with water (1) [5] (g) e.g., Carbon dioxide / phosphorus(v) oxide / sulphur dioxide (accept formula) (1) Na 2 CO 3 / Na 3 PO 4 / Na 2 SO 3 (1) [2] Total [20] Total Section B [40] 27

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