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Stretch and Challenge What is Stretch and Challenge? What it is: the application of knowledge and/or problem solving rather than just recalling data demonstrating your breadth of knowledge (including facts learned during your AS course) – not the narrow testing of one specification topic being examined by more open-ended questions which may have no single correct answer applying your scientific knowledge to new contexts you haven't met before part of your normal A2 examination paper. What it isn’t: additional content to learn lots of extra essay questions – although some Stretch and Challenge questions may be tested by extended prose questions being examined by questions that are flagged: Stretch and Challenge only for those students aiming for an A* grade – your examination paper contains questions that will test you on your knowledge of chemistry and, whatever grade you are aiming for, you need to answer all the questions to the best of your ability. Why should I be concerned about Stretch and Challenge? It's about learning new skills and ways of thinking Stretch and Challenge should help improve your problem solving and thinking skills. You will probably be required to apply your knowledge in new contexts as well as make links across the various topics you've encountered throughout your AS and A2 studies. Improving your marks The skills Stretch and Challenge help develop could mean the difference between one grade and another, especially if you're aiming for an A or A* grade – although these skills should be helpful in attaining whatever grade you are aiming for. Useful beyond A level The Stretch and Challenge skills should also be useful if you want to carry on with your studies in further education or want to start a career. Universities which use entrance exams or interviews will be looking for students who can show more than just rote learning – the thinking skills demonstrated through © Pearson Education Ltd 2009 This document may have been altered from the original 1 Stretch and Challenge help to prepare you for these situations. Similarly, most businesses will be looking for people who have good problem-solving skills and the ability to apply knowledge in different contexts – so practising such skills during your A2 course will stand you in good stead whatever you choose to do. How should I prepare for Stretch and Challenge questions? Stretch and Challenge questions are just part of your examination paper, so all the usual advice for revision applies. You need to know the content of the course as laid out by the specification. However, to do well in Stretch and Challenge you also need to practise thinking in different ways rather than just learning the content by rote. Don't limit yourself to revising topics in isolation Try to spot links between modules and use strategies such as mind maps to connect up different topics. You could use the textbook spreads which introduce each chapter for examples. Read around the subject Use different sources, for instance, newspapers, magazines (e.g. New Scientist), websites (e.g. NewScientist.com – the link for which can be found in on this CD). Your goal isn't to learn lots of additional information, but to be aware of how many of the topics or ways of scientific thinking are relevant to everyday life. It will also help you recognise that different views on the same topic exist. Ask your teacher for advice This could be about the best practice questions to use for Stretch and Challenge – including past examination questions, specimen examination questions or other materials specific to your school. Look at the examples in this document and try answering the questions. Think carefully about your exam technique Allow yourself enough time to answer all your questions and be prepared to have a go – you’re sure to get some marks! Bear in mind you need to do as well as you can on the whole paper. What will Stretch and Challenge questions look like? Stretch and Challenge questions can crop up anywhere in an A2 examination paper, although they will not be flagged as such. However, the examples below should help you to spot the Stretch and Challenge opportunities. Stretch and Challenge questions can account for around 5–10% of the marks in your examination paper, so they could make the difference between one grade © Pearson Education Ltd 2009 This document may have been altered from the original 2 and another. Sometimes, a Stretch and Challenge question could be the final part of a long question consisting of several parts, while at other times it will be a single question. It is quite likely that Stretch and Challenge questions will require some sort of extended prose answer, but could equally ask you to draw a diagram, calculate something or solve a problem. Stretch and Challenge questions often use the command words explain, discuss, evaluate, or suggest. Sample questions Below is a set of sample questions. However, the nature of Stretch and Challenge questions means they may include material from other topics. They are not past exam questions nor are they written by the examination board, they are designed to give you a flavour of the type of question you may be asked. As with most questions, even if you do not achieve full marks you can usually pick up several marks using a careful approach. The answers have been laid out to show any potential marking points – i.e. if the question carries six marks, the corresponding answer will have six points to it; if the question has been allocated eight marks then the corresponding answer will have eights points, etc. Look carefully through each question. There are many opportunities to gain marks and don’t be put off by the slightly more complex wording of the question. © Pearson Education Ltd 2009 This document may have been altered from the original 3 A2 Unit F324: Rings, polymers and analysis Module 1 – Rings, acids and amines Question 1 Civetone can be extracted from the anal gland of the civet cat. It was used as an ingredient in expensive perfumes. The skeletal formula for civetone is shown in Figure 1. Figure 1 Civetone (a) What is the molecular formula of civetone? [1] (b) Predict three reactions for civetone. For each reaction give the: o reagents and conditions o skeletal formula of the organic product. [6] (c) Cyclohex-3-en-1-one contains the same functional groups as civetone. Cyclohex-3en-1-one has the carbon-carbon double bond in the Z-orientation. Suggest why the carbon-carbon double bond cannot exist in the E-orientation. [2] (d) Ammonia reacts with carbonyl compounds to form imines. R2C=O + NH3 R2C=NH + H2O This reaction involves a nucleophilic addition followed by an elimination. The reaction is catalysed by the presence of hydrogen ions. (i) Draw the displayed formula of the imine R2C=NH. Show on the formula all of the bond angles. [1] (ii) Use the curly–arrow model to suggest a mechanism for the nucleophilic addition stage of the reaction. Show any relevant dipole and lone pair. [4] (iii) 2-methylbutanal reacts with methylamine. Draw the organic structure of the product from this reaction. [2] (iv) Explain why dimethylamine will not react in a similar way with carbonyl compounds. [1] [Total: 17] © Pearson Education Ltd 2009 This document may have been altered from the original 4 Sample answer: (a) C17H30O (b) NaBH4/ethanol solvent gives secondary alcohol Br2(aq)/dibromo derivative HBr(g)/room temperature gives bromo derivative (c) The carbon–carbon double bond cannot exist in the E-orientation as the C–C bonds would be as far apart from each other as possible, so the ring would be under too much strain and the ends of the carbon chain would be too far away from each other to form a ring. (d) (i) All bond angles are 120°. © Pearson Education Ltd 2009 This document may have been altered from the original 5 (ii) (iii) CH3CH2CH(CH3)CH=NCH3 (iv) It is impossible to have the elimination stage because there is no proton to be lost. Examiner’s comments: (a) Be careful not to forget the hydrogen atoms, which are not shown in the skeletal formula. (b) Stretch and Challenge questions such as this allow you to demonstrate knowledge and understanding from different parts of the specification. Firstly, you need to identify the functional groups present in the molecule and then decide the most appropriate reactions to describe. Since you have the choice of which reactions to describe, it is best to choose the most straightforward ones, such as reduction or addition. An alternative answer could be to describe the more complex reaction with 2,4-dinitrophenylhydrazine to make 2,4-dinitrophenylhydrazone. (c) In a cyclohexane ring the bond angles are approximately 109 °; with an E double bond it is impossible for the molecule to distort to this bond angle. (d) This illustrates another aspect of Stretch and Challenge questions – i.e. you must use your knowledge of organic chemistry to answer questions about a novel type of reaction. Here, you need to apply your knowledge regarding the chemistry of carbonyl compounds to this new type of reaction. For example, using © Pearson Education Ltd 2009 This document may have been altered from the original 6 your answer from part (i) the bond angles will be similar to those found in carbonyl compounds or alkenes. From part (ii), the mechanism is very similar to that for the reaction of NaBH4 with carbonyl compounds. It also relies on your familiarity with the nucleophilic nature of amines. In part (ii), remember a curly arrow shows the movement of an electron pair either from a bond or a lone pair. You must show the correct use of the curly arrow starting from the lone pair on the nitrogen atom to attack the electron deficient carbon atom. It is also important to include the partial charges on C and O. The next step shows the addition of a proton to the electron-rich nitrogen atom. There is no need to use the curly arrow model for the proton transfer step. (iii) (iv) A displayed formula could also be drawn. If you look at the mechanism in part (ii), the last stage involves loss of a proton. Since the nitrogen atom does not have a proton to lose, the reaction cannot happen. Question 2 Research chemists synthesise many compounds to test as possible anticancer drugs. The structure of one of these compounds is shown in Figure 2. Figure 2 Compound A (a) The primary amine attached to the benzene ring is less basic than the other –NH2 group in the molecule. The secondary amine group is more basic than either of the the –NH2 groups. Explain these differences in basicity. [3] (b) The research chemists dissolve compound A in aqueous sodium hydroxide. They then react the solution with aqueous bromine. Complete the structure shown in Figure 3 to show the product formed. [2] © Pearson Education Ltd 2009 This document may have been altered from the original 7 Figure 3 (c) The research chemists add compound A to excess dilute nitric acid. Compound A dissolves to form a solution containing ions. Complete the structure shown in Figure 3 to show the organic ion formed. [2] (d) The research chemists add compound A to an excess of potassium dichromate(VI) solution and dilute sulfuric acid. The mixture is refluxed for one hour. Complete the structure shown in Figure 3 to show the product formed. [1] [Total: 8] Sample answer: (a) In the –NH2 group attached to the benzene ring, the lone pair is donated into the πsystem of the benzene ring and so is less available to be donated to a proton. Secondary amine has two electron releasing groups attached, which makes the lone pair more available and so more basic. © Pearson Education Ltd 2009 This document may have been altered from the original 8 (b) (c) (d) Examiner’s comments: This is another question where you have to apply your knowledge and understanding regarding the chemistry of several functional groups. In part (a), it would have been better to ring the structure but the answer clearly indicates it is the secondary amine group that is the most basic. Remember that amines are basic because they can donate a lone pair of electrons to accept a proton. In parts (b) and (c), you need to apply your knowledge about the reactions of phenylamine and phenols in order to answer the question. Remember to look at the mark allocation; even with a Stretch and Challenge question this will help you organise your answer. Since parts (b) and (c) are worth two marks each, it is likely that there may be reactions at more than one site in the molecule. In part (c) you could also give products that involved electrophilic substitution. The phenol could react with dilute nitric acid to substitute hydrogens for a nitro group. The electrophilic substitution could occur on positions 2, 4 and 6 relative to the –OH group. © Pearson Education Ltd 2009 This document may have been altered from the original 9 In part (d), you should recognise that acidified potassium dichromate is an oxidising agent. Module 2 – Polymers and synthesis Question 1 Figure 4 shows the molecular structure for the molecule atropine. Atropine is present in the plant commonly called deadly nightshade and is toxic to humans. However, it can also act as one of the few antidotes to nerve gases such as sarin. Figure 4 Atropine (a) Deduce the empirical formula for atropine. [1] (b) Calculate the percentage by mass of nitrogen present in atropine. [1] (c) Atropine contains several functional groups. (i) Describe, with the aid of an equation, three different reactions for atropine. Each reaction should involve a different functional group. (ii) Draw the structure of the products from each of these reactions. [9] (d) Indicate with an asterisk * every chiral carbon atom that is present in a molecule of atropine. [1] (e) Synthetic atropine can be prepared by chemists. (i) Suggest one difference between synthetic atropine and natural atropine (ii) The last stage in one synthetic route to produce atropine involves two compounds reacting to make an ester. Suggest the identity of the two compounds. [1] [2] [Total: 15] Sample answer: (a) C17H23O3N (b) 4.84% © Pearson Education Ltd 2009 This document may have been altered from the original 10 (c) (i) and (ii) Esterification: RCH2OH + 2[O] RCOOH + H2O Bromination: C6H5R + Br2 C6H4RBr + HBr This reaction needs the presence of a halogen carrier such as AlBr3. o Amine will react in presence of HCl(aq) to make: R3N + HCl R3NH+ + Cl– © Pearson Education Ltd 2009 This document may have been altered from the original 11 (d) (e) (i) Synthetic atropine is a mixture of optical isomers but natural atropine will be just one optical isomer. (ii) Examiner’s comments: (a) Remember that an empirical formula is the simplest mole ratio of each element in the compound. (b) An error-carried-forward mark would be available from an incorrect empirical formula. (c) Three different reactions can be taken from the chemistry of esters, alcohols, amines or benzene rings. In each case, you must name the functional group and the reagents for the reaction. You must also write the correct equations and products. Stretch and Challenge questions often allow you to demonstrate knowledge and understanding from different parts of the specification. Again, you first need to identify the functional groups present in the molecule and then decide the most appropriate reactions to describe. As you can choose which reactions to describe, it is best to choose the most straightforward ones, such as oxidation of the alcohol. You also need to write the equation – you can use the notation rather than writing out the full structure, e.g. RCH2OH for the alcohol reaction, R’COOR for the ester, RR’NCH3 for the amine, and for the benzene ring just show an R-group attached to the benzene ring. © Pearson Education Ltd 2009 This document may have been altered from the original 12 Other possible reactions you can have described are o Esters reacting with hot NaOH(aq) to make: o Alcohols reacting with Na or with PCl5 o The benzene ring reacting with HNO3/H2SO4 o Finally, remember not to describe three reactions of the same functional group. Question 2 Table 1 shows some amino acids and their isoelectric points. Table 1 Amino acid Structure Isoelectric point Glycine 6.1 Glutamic acid 3.1 © Pearson Education Ltd 2009 This document may have been altered from the original 13 Lysine 10.0 Proline 6.3 (a) (i) (ii) Explain why proline is an α-amino acid. Proline has two stereoisomers. Draw these two stereoisomers, clearly showing the 3D relationship between the two isomers. [1] [2] (b) Draw the structure of glycine at a pH of 6.1. [1] (c) Draw the structure of lysine at a pH of 6.1. [1] (d) Draw the structure of glutamic acid at a pH of 6.1. [1] (e) Proline is refluxed with hot sodium hydroxide. Draw the structure of the organic product. [1] (f) Proline and lysine react together to form two different dipeptides. Draw the structures of these two dipeptides. (g) [2] Glutamic acid is refluxed with excess ethanol and concentrated sulfuric acid catalyst. Write an equation for this reaction. [2] [Total: 11] Sample answer: (a) (i) Although it is a secondary amine rather than a primary amine the amine group is on the number 2 carbon. © Pearson Education Ltd 2009 This document may have been altered from the original 14 (ii) (b) (c) (d) © Pearson Education Ltd 2009 This document may have been altered from the original 15 (e) (f) (g) Examiner’s comments: This question involves the application of your knowledge and understanding of the chemistry of amino acids. The Stretch and Challenge aspect is that one of the amino acids chosen is less familiar than the other three and includes a ring structure. (a) In part (i), remember α-amino acids are 2-aminocarboxylic acids. Make certain that you show the three-dimensional arrangement in part (ii); it is easier to show the bonds of the ring in the plane of the paper. (b), (c) and (d) The isoelectric point is the pH at which the amino acid exists as a zwitterion. Above this pH the amino acid will be negatively charged and below this pH the amino acid will be positively charged. You also need to consider the functional groups on the side chain and how they would react with acids or bases. (e) This is just an acid–base reaction to make a salt. © Pearson Education Ltd 2009 This document may have been altered from the original 16 (f) Remember to form the peptide linkage using the amino group on carbon number 2 and not the one on the side chain. (g) The reagents suggest esterification, so it is the carboxyl group on the side chain that reacts. Module 3 – Analysis Question 1 A research chemist analysed the percentage composition by mass of the drug, compound A. He found the percentage composition to be 59.90% carbon, 4.44% hydrogen and 35.6% oxygen. The research chemist also analysed the drug using three different spectroscopic techniques. Look at the three different spectra for compound A. Figure 5 NMR spectrum for compound A When D2O was added to the sample and the NMR spectrum repeated, the signal at δ = 11.2 ppm disappeared. © Pearson Education Ltd 2009 This document may have been altered from the original 17 Figure 6 Mass spectrum for compound A Figure 7 IR spectrum of compound A © Pearson Education Ltd 2009 This document may have been altered from the original 18 Use the given information and the three different spectra from Figures 5–7 to identify, as far as you can, compound A. Explain your reasoning carefully. [Total: 14] Sample answer: (a) Empirical formula calculation Mole ratio: C:H:O is 4.99:4.44:2.23 2.24:2.00:1.00 Empirical formula is C9H8O4 Molecular formula is C9H8O4 as NMR spectrum indicates eight protons and m/z from mass spectrum indicates Mr is 180. (b) NMR spectrum Contains six different types of protons. δ = 2.3 ppm indicates a –CH3 attached to a carbonyl group or benzene ring. δ = between 7.5 to 8.2 ppm indicates a C–H attached to a benzene ring. δ = 11.2 ppm indicates a proton in a carboxylic acid – confirmed by loss of signal in D2O. (c) Mass spectrum Molecular ion is m/z = 180 So Mr is 180. m/z = 43 could indicate CH3CO– or C3H7 e.g. CH3CH2CH2– or CH3CHCH3 present in molecule. (d) IR spectrum Broad absorption between 2500 to 3000 cm–1 indicates O–H stretching in carboxylic acid. Absorptions at 1680 and 1750 cm–1 suggest C=O stretching in two carbonyl groups. Strong absorptions 1200 and 1300 cm–1 (possibly) indicate C–O stretching. Presence of C–O and C=O suggest presence of an ester. (e) Compound A’s structure Examiner’s comments: Although this Stretch and Challenge question is not structured, you must make certain that you analyse all the data you have been given and that your final suggested structure is © Pearson Education Ltd 2009 This document may have been altered from the original 19 consistent with all of the given information. It is better to analyse each piece of data separately, however, you must also use any information you have already deduced from other analysis. When you have analysed each piece of information, you must bring all this data together when suggesting the final structure for the compound. It is not possible to decide the substitution pattern in the benzene ring. © Pearson Education Ltd 2009 This document may have been altered from the original 20 A2 Unit F325: Equilibria, energetics and elements Module 1 – Rates, equilibrium and pH Question 1 A student wants to determine the solubility of calcium hydroxide in water. She decides to use a titration method. Stage 1 A sample of 5.00 g calcium hydroxide Ca(OH)2(s) was added to a conical flask. Stage 2 About 100 cm3 of distilled water was added to the conical flask and the flask with its contents was stoppered. The flask was then shaken for five minutes and left for a further five hours. Stage 3 The contents of the conical flask were filtered into a clean, dry conical flask. The filtrate obtained was saturated calcium hydroxide. Stage 4 A 25.0 cm3 sample of the saturated calcium hydroxide was titrated against 0.0750 mol dm–3 hydrochloric acid. In Stage 2, an equilibrium between solid calcium hydroxide and its constituent aqueous ions is established. Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) In Stage 4, the titre was 22.40 cm3. (a) Calculate the solubility in g dm–3 of the saturated calcium hydroxide. (b) The student decides to repeat the experiment but makes a mistake in Stage 2. She adds about 100 cm3 of 1.00 mol dm–3 sodium hydroxide instead of water. She finds that the solubility of the calcium hydroxide is much smaller than before. Explain why. [1] (c) The student puts 20.0 cm3 of 0.005 mol dm–3 calcium hydroxide into a conical flask. She places a pH probe connected to a datalogger into the calcium hydroxide. She then slowly adds 0.010 mol dm–3 hydrochloric acid using a burette. Eventually, she adds 40.0 cm3 of hydrochloric acid. Sketch on the graph shown in Figure 8 how the pH of the solution in the conical flask changes during the experiment. [4] © Pearson Education Ltd 2009 This document may have been altered from the original [4] 21 Figure 8 [Total: 9] Sample answer: (a) Moles of HCl = 1.68 × 10–3 Moles of Ca(OH)2 = 8.40 × 10–4 [Ca(OH)2] = 3.36 × 10–2 mol dm–3 Solubility = 2.49 g dm–3 (b) To minimise increase in [OH–(aq], the equilibrium shifts to the left so that [Ca2+(aq)] is lowered. (c) The [OH–(aq)] is 0.020 mol dm–3, therefore pOH is 2.00 and pH is 12.00. At the end of the titration there will be excess acid. The concentration of the [H +(aq) cannot be greater than 0.010 mol dm–3 so the pH cannot be lower than 2.00. Two moles of HCl reacts with one mole of Ca(OH)2 but double the concentration of acid is needed so that neutralisation occurs when 20.0 cm3 of acid is added. So, the graph will start at pH 12.0 and will have a vertical section at 20 cm3 that covers a range of 6 pH units and will end at just above pH 2.0. © Pearson Education Ltd 2009 This document may have been altered from the original 22 Examiner’s comments: In part (a) the calculation is unstructured and this is typical of Stretch and Challenge questions. Part (b) is an application of le Chatelier’s principle. In part (c) the mark allocation suggests that you need to do more than just a sketch graph. The starting pH, finishing pH and the volume of hydrochloric acid needed to just neutralise the aqueous calcium hydroxide need to be calculated. Question 2 This question is about reaction kinetics of gas phase reactions. (a) The following data was obtained for the reaction between hydrogen and nitrogen(II) oxide. 2H2(g) + 2NO(g) 2H2O(g) + N2(g) Table 2 [NO(g)]/mol dm–3 [H2(g)] /mol dm–3 initial rate of reaction mol dm–3 s–1 0.70 1.40 1.40 0.39 0.39 0.78 0.19 0.76 1.52 Suggest, with reasons, a mechanism for the reaction that is consistent with the data. [4] © Pearson Education Ltd 2009 This document may have been altered from the original 23 (b) Ozone is made in the stratosphere when oxygen atoms react with oxygen molecules. The concentration of oxygen atoms in the stratosphere is estimated to be 3.0 × 10–14 mol dm–3 and the concentration of oxygen molecules, 1.3 × 10–14 mol dm–3. O(g) + O2(g) O3(g) This reaction is first order with respect to both oxygen atoms and oxygen molecules. The rate constant k for the reaction is 3.9 × 10–5 dm3 mol–1 s–1. Calculate the number of ozone molecules made in one dm 3 of the stratosphere every second. [2] [Total: 6] Sample answer: (a) The reaction is first order with respect to H2 because, as concentration of H2 doubles, the rate doubles. Reaction is second order with respect to NO because, as the concentration of NO doubles, the rate quadruples. Therefore, rate = k[H2][NO]2 (2) This means that the slowest step involves one molecule of H2 reacting with two molecules of NO. Slow step H2 + 2NO H2O + N2 + O Fast step H2 + O H2O (b) Rate = 3.9 × 10–5 × (3.0 × 10–14) (1.3 × 10–14) = 1.521 × 10–32 mol dm–3 s–1 Examiner’s comments: Although parts (a) and (b) of the question are unstructured, you must make certain that you structure your answer. In part (a), if the question had been structured you would have had to work out the orders of reaction and/or the rate equation and use this to decide the particles involved in the rate-determining step. Finally, the overall equation needs to be considered to suggest a mechanism. All of these things should be shown for a complete answer to the question. In part (b), write down the rate equation and substitute in the values given. Question 3 Buffer solutions minimise pH changes when small amounts of acids or alkalis are added. (a) Calculate the pH of a buffer solution that is 0.100 mol dm–3 with respect to sodium methanoate and 0.200 mol dm–3 with respect to methanoic acid. The Ka of methanoic acid is 1.8 × 10–4 mol dm–3. © Pearson Education Ltd 2009 This document may have been altered from the original [3] 24 (b) What changes to the concentrations of the sodium methanoate and the methanoic acid would you need to make to decrease the pH of the buffer by one unit? [1] [Total: 4] Sample answer: (a) [HCOOH(aq)]eqm ≈ 0.100 mol dm–3 and [HCOO–(aq)]eqm ≈ 0.200 mol dm–3 Ka = [HCOO–(aq)]eqm[H+(aq)] [HCOOH(aq)]eqm + So [H (aq)] = 9.0 × 10–5 mol dm–3 pH = 4.05 (b) [HCOOH(aq)] = 1.00 mol dm–3 and [HCOO–(aq)] = 0.200 mol dm–3 or [HCOOH(aq)] = 0.100 mol dm–3 and [HCOO–(aq)] = 0.0200 mol dm–3 Examiner’s comments: Part (a) is another unstructured question typical for a Stretch and Challenge question. Remember that the final step is to use pH = –log10[H+(aq)]. In part (b), a decrease of one unit of pH means a change of a factor of ten in the [H +]. Question 4 Dinitrogen(IV) oxide is in equilibrium with nitrogen dioxide. N2O4(g) 2NO2(g) 0.500 mol of N2O4 was allowed to react and reach equilibrium in a 5.00 dm3 sealed container. At equilibrium, there were 0.375 mol of NO2 present. (a) Calculate the numerical value for the equilibrium constant. [3] (b) Explain the effect on the position of equilibrium and the equilibrium constant if the volume of the sealed container was suddenly reduced but the temperature remained constant. [2] [Total: 5] Sample answer: (a) If 0.375 moles of NO2 is made then 0.1875 moles of N2O4 must have reacted So at equilibrium, N2O4 = 0.3125 moles Therefore, [N2O4]eqm = 0.0625 mol dm–3 and [NO2]eqm = 0.0075 mol dm–3 Kc = [NO2(g)] 2eqm [N2O4(g)] eqm Kc = 9.00 × 10–4 mol dm–3 © Pearson Education Ltd 2009 This document may have been altered from the original 25 (b) Position of equilibrium moves to the left but equilibrium constant stays the same. Examiner’s comments: In part (a), it is easier to use moles rather than the equilibrium concentrations; however, the equilibrium constant Kc is based on the equilibrium concentrations in mol dm–3. In part (b), remember that the equilibrium constant will only change if the temperature changes. Module 2 – Energy Question 1 Magnesium oxide is a giant ionic solid. (a) Use the following data in Table 3 to calculate the first ionisation energy of magnesium. Table 3 Process Enthalpy change/kJ mol–1 Lattice enthalpy for magnesium oxide –3889 Enthalpy of atomisation for oxygen +248 Enthalpy of atomisation for magnesium +150 Second ionisation energy of magnesium +1450 First electron affinity for oxygen –141 Second electron affinity for oxygen +798 Enthalpy change of formation of magnesium oxide –602 [4] (b) (c) Explain the difference in enthalpy change between the first and second electron affinity of oxygen. [2] Magnesium nitrate decomposes when heated to form magnesium oxide, nitrogen dioxide and oxygen. 2Mg(NO3)2(s) 2MgO(s) + 4NO2(g) + O2(g) Use oxidation numbers to show that this reaction involves both oxidation and reduction. (d) [2] Magnesium carbonate does not decompose at room temperature but will decompose if heated strongly with a Bunsen burner. (i) Using the data in Table 4, calculate the free energy change of reaction ΔG in © Pearson Education Ltd 2009 This document may have been altered from the original 26 kJ mol–1 for the decomposition of magnesium carbonate at 300 K (room temperature). MgCO3(s) MgO(s) + CO2(g) Table 4 Substance Enthalpy change of formation/kJ mol–1 Entropy/J K–1 mol–1 MgCO3(s) –1113 66 MgO(s) –602 27 NO2(g) +33 240 [5] (ii) (e) Explain why magnesium carbonate does not decompose at room temperature. [1] Use the data in Table 5 to predict if lithium chloride or sodium chloride has a more exothermic enthalpy change of solution. Table 5 Process Enthalpy change/kJ mol–1 Lattice enthalpy for lithium chloride –848 Lattice enthalpy for sodium chloride –776 Enthalpy change of hydration for chloride ion –381 Enthalpy change of hydration for lithium ion –499 Enthalpy change of hydration for sodium ion –390 [2] [Total: 16] Sample answer: (a) ∆Hle = –∆H1ea –∆H2ea –∆H1ie –∆H2ie –∆HatMg –∆HatO +∆Hf so ∆1ie = –∆Hle –∆H1ea –∆H2ea –∆H2ie –∆HatMg –∆HatO +∆Hf = +3889 +141 –798 –1450 –150 –248 –602 = +782 kJ mol–1 (b) First electron affinity is exothermic because the electron gained is attracted by the positive nucleus. Second electron affinity is endothermic because the electron gained is repelled by the overall negative charge of the O– ion. (c) N goes from +5 to +4, which is reduction. O goes from –2 to 0, which is oxidation. © Pearson Education Ltd 2009 This document may have been altered from the original 27 (d) (i) (ii) (e) ∆H = +1113 +(-603) + (+33) = +544 kJ mol–1 ∆S = -66 +(27)+(240) = +201 J K–1 mol–1 ∆G = ∆H − T∆S = +544000 – (300 × +201) J K–1 mol–1 ∆G = +483.7 kJ mol–1 This is not a spontaneous reaction because of the positive free energy change. LiCl ∆Hsoln = +848 + (–381) + (–499) = –32 kJ mol–1 NaCl ∆Hsoln = +776 + (–381) + (–390) = +5 kJ mol–1, so LiCl has a more exothermic enthalpy change of solution. Examiner’s comments: For part (a), instead of using a normal Born–Haber cycle to calculate the lattice enthalpy you will need to calculate the first ionisation energy of magnesium. Drawing a Born–Haber cycle will help you answer the question. The answer to part (b) requires knowledge and understanding from more than one part of the specification. It is best to define both terms and then tackle the explanation. In part (c), it is best to write the oxidation numbers underneath each symbol in the equation. The calculation in part (d) is unstructured and you must remember to ensure that the correct units are used in the equation that links free energy with the enthalpy change and the entropy change. Entropy data is normally given in J K–1 mol–1 so that either it must be converted to kJ K–1 mol–1 or the enthalpy change converted to J mol–1. Question 2 This question is about redox reactions. (a) The half-equation for the reduction of iodate(V) ions is shown below. IO3–(aq) + 6H+(aq) + 6e– I–(aq) + 3H2O(l) Iodate(V) ions oxidise iodide ions in the presence of acid. Construct the ionic equation for this oxidation. (b) [2] Look at the standard electrode potentials shown in Table 6. Table 6 Process Zn2+(aq) + 2e– Standard electrode potential/V Zn(s) –0.76 © Pearson Education Ltd 2009 This document may have been altered from the original 28 VO3–(aq) + 4H+(aq) e– VO2+(aq) + 2H2O(l) +1.00 VO2+ + 2H+(aq) + e– ⇌ V3+ + H2O(l) +1.00 V3+(aq) + e– V2+(aq) –0.26 V2+(aq) + e– V(s) –1.20 Excess zinc powder is added to acidified aqueous ammonium vanadate(V), NH 4VO3. Use the information in Table 6 to predict the final products of the reaction. Include equations in your answer. [5] (c) Aqueous ferrate(VI) ions FeO42– decompose in the presence of hydrogen ions, forming iron(III) ions, oxygen and water. Construct the ionic equation for this reaction. [2] [Total: 9] Sample answer: (a) 2I–(aq) I2(aq) + 2e– So adding 6I–(aq) 3I2(aq) + 6e– To IO3–(aq) + 6H+(aq) + 6e– I–(aq) + 3H2O(l) Gives IO3–(aq) + 6H+(aq) + 6e– + 6I–(aq) I–(aq) + 3H2O(l) + 3I2(aq) + 6e– Cancelling out this will give IO3–(aq) + 6H+(aq) + 5I–(aq) 3I2(aq) + 3H2O(l) (b) 2VO3–(aq) + Zn(aq) + 8H+(aq) 2VO2+(aq) + Zn2+(aq) + 4H2O(l) 2VO2+(aq) + Zn(aq) + 4H+(aq) 2V3+(aq) + Zn2+(aq) + 2H2O(l) 2V3+(aq) + Zn(aq) 2V2+(aq) + + Zn2+(aq) The more positive the electrode potential the more likely the reduction so VO 3–(aq), VO2+(aq) and V3+(aq) are more likely to be reduced than Zn2+(aq) Zn2+(aq) is more likely to be reduced than V2+(aq) So that VO3–(aq) should be reduced to V2+(aq). (c) 2FeO42–(aq) + 10H+(aq) 5H2O(l) + 1½O2(g) + 2Fe3+(aq) Examiner’s comments: For part (a), you must be able to recall the equation for the reduction of iodine and, when the two half-equations are combined, it is important to cancel out species that are included on both the left-hand and right-hand sides of the overall equation. Part (b) involves applying your knowledge about electrode potentials should allow you to make the required prediction. Remember, you must show your reasoning and not just the answer. The equation for part (c) is very difficult to balance – start by trying to balance the charges. One mark is available for getting the reactants and products correct, i.e. FeO42–(aq) + H+(aq) H2O(l) + O2(g) + Fe3+(aq) © Pearson Education Ltd 2009 This document may have been altered from the original 29 Module 3 – Transition elements Question 1 A student has isolated some crystals of a compound. The student knows that the compound has the formula (C2O4)2KxHy.zH2O. The values of x, y and z can be determined by titration. Stage 1 Stage 2 Stage 3 Water is added to a known mass of the compound to make 250 cm3 of solution B. A 10.0 cm3 sample of solution B is titrated against 0.100 mol dm–3 sodium hydroxide solution. A 25.0 cm3 sample of solution B is added to a conical flask. This sample has 25.0 cm3 of 1 mol dm–3 sulfuric acid added to it. The flask is heated to 60 °C and the hot contents of the flask are titrated against 0.0200 mol dm–3 potassium manganate(VII). In Stage 1, the compound dissolves. In Stage 2, hydrogen ions react with hydroxide ions. In Stage 3, ethanedioate ions are oxidised to form carbon dioxide. 5C2O42–(aq) + 16H+(aq) + 2MnO4–(aq) 2Mn2+(aq) + 10CO2(g) + 8H2O(l) (a) In Stage 3, the flask is heated before heating. Suggest why this is done. [1] (b) In Stage 2, the titre was 9.55 cm3. Calculate the concentration of hydrogen ions in solution B. [2] (c) In Stage 1, the student used 2.017 g of the compound (C2O4)2KxHy.zH2O. In Stage 3, the titre was 31.75 cm3. Calculate the values of x, y and z. [5] [Total: 8] Sample answer: (a) So the reaction is fast enough to do a titration (b) Moles of H+ = 2.38 × 10–3 (c) Moles MnO4– = 6.35 × 10–4 Moles C2O42– = 2½ × moles MnO4– = 1.59 × 10–3 Each mole of the compound provides two moles of C2O42–, so moles of compound = 7.94 × 10–4. M = mass ÷ moles = 254 y = 2.38 × 10–3 ÷ 7.94 × 10–4 = 3 x=1 z = (254 – 48 – 128 – 3 – 39) ÷18 = 2 © Pearson Education Ltd 2009 This document may have been altered from the original 30 Examiner’s comments: In part (a), you must make a sensible suggestion. Part (b) is a straightforward calculation. The calculation for part (c) is unstructured and relies on you using the relationships between molar mass, mass and moles, and between moles, volume and concentration. It is important that you organise your answer and indicate at each stage the quantity that you are calculating. Error-carried-forward marks would be available, but only if it is clear how the error had been carried forward. Question 2 Copper reacts with excess hot, concentrated sulfuric(VI) acid to form a blue solution A, a colourless gas B and a liquid C. The gas B contains 50.0% by mass of oxygen. When excess aqueous ammonia was added slowly to the reaction mixture containing solution A, an exothermic reaction initially took place forming salt D. As more aqueous ammonia was added, a blue precipitate E was formed and finally a dark blue solution F was formed. Concentrated hydrochloric acid was added to a small sample of solution F and a green solution G was formed. Identify compounds A to G showing your reasoning. Write equations for all the reactions that take place. [Total: 13] Sample answer: 2H2SO4 + Cu CuSO4 + 2H2O + SO2 A is CuSO4. B is SO2 because, using the percentage data, the mole ratio S:O is 1.56:3.13. C is H2O. 2NH3 + H2SO4 (NH4)2SO4 D is (NH4)2SO4 because sulfuric acid is neutralised by ammonia. Cu2+(aq) + 2OH–(aq) Cu(OH)2(s) [Cu(H2O)6]2+ + 4NH3 [Cu(NH3)4(H2O)2]2+ + 4H2O Or Cu(OH)2 + 4NH3 + 2H2O [Cu(NH3)4(H2O)2]2+ + 2OH– E is Cu(OH)2 because copper(II) ions initially form a blue precipitate with aqueous ammonia. This precipitate re-dissolves in excess aqueous ammonia to form the complex ion F, which is [Cu(NH3)4(H2O)2]2+. NH3 + HCl NH4Cl [Cu(NH3)4(H2O)2]2+ + 4Cl– [CuCl4]2– + 4NH3 + 2H2O(l) The excess ammonia is neutralised by hydrochloric acid and then a ligand substitution takes place to make G, which is CuCl42–. Examiner’s comments: This unstructured Stretch and Challenge question focuses on the chemistry of copper compounds. It is better to organise your answer and focus on each of the sets of reactions © Pearson Education Ltd 2009 This document may have been altered from the original 31 described. The use of the percentage composition by mass enables you to work out that sulfuric acid is reduced to sulfur dioxide. You should remember the colours of the copper complex ions mentioned in the question and this helps to identify the unknowns. Question 3 Iron forms many complex ions. Table 7 shows the stability constant Kstab for two complexes of iron(III). Table 7 Complex Stability constant Colour of complex [Fe(CN)6]3– 4.08 × 1052 dark blue 102 blood red [Fe(H2O)5(SCN)]2+ 9.17 × (a) Write an expression for the stability constant of [Fe(H2O)5(SCN)]2+. [1] (b) An aqueous solution containing [Fe(CN)6]3– and SCN– is mixed together. Use the stability constants to predict what you would observe. [2] [Total: 3] Sample answer: (a) Kstab = (b) {[Fe(H2O)5(SCN)]2+(aq)} [Fe3+(aq)] [SCN-(aq] The solution would stay dark blue. The Kstab for [Fe(CN)6]3– is much larger than that for the other complex [Fe(H2O)5(SCN)]2+, therefore the equilibrium will be entirely on the side of [Fe(CN)6]3–. Examiner’s comments: In part (a), curly brackets have been used instead of square brackets to avoid confusion with the use of square brackets in complex ions. Water is not shown in this expression since the primary role of water is as a solvent and its concentration does not change. Kstab is just an equilibrium constant but it allows you to compare the relative stabilities of complex ions. In part (b), since [Fe(CN)6]3– has such a large Kstab, it is a more stable complex than [Fe(H2O)5(SCN)]2+. © Pearson Education Ltd 2009 This document may have been altered from the original 32 And finally... The three key pieces of advice to remember if you want to aim for those Stretch and Challenge marks are: Aim to answer all the questions you need to on the examination paper as well as you can. Look for questions which invite you to make links across the different but familiar topics. Don’t be afraid of questions which appear to be about something you haven't studied – look hard, because it’s probably a Stretch and Challenge question which requires you to use your existing knowledge but in a new context. © Pearson Education Ltd 2009 This document may have been altered from the original 33