Write your name here Surname Other names Pearson Edexcel Level 3 GCE Centre Number Candidate Number Chemistry Advanced Subsidiary Paper 2: Core Organic and Physical Chemistry Sample Assessment Materials for first teaching September 2015 Time: 1 hour 30 minutes You must have: Data Booklet Scientific calculator, ruler Paper Reference 8CH0/02 Total Marks Instructions Use black ink or ball-point pen. Fill in the boxes at the top of this page with your name, centre number and candidate number. Answer all questions. Answer the questions in the spaces provided there may be more space than you need. Information The total mark for this paper is 80. The marks for each question are shown in brackets use this as a guide as to how much time to spend on each question. You may use a scientific calculator. For questions marked with an *, marks will be awarded for your ability to structure your answer logically showing the points that you make are related or follow on from each other where appropriate. Advice Read each question carefully before you start to answer it. Try to answer every question. Check your answers if you have time at the end. Show all your working in calculations and include units where appropriate. Turn over S47552A 2015 Pearson Education Ltd. 1/1/1/1/1/1/1/ *s47552a0124* 53
Answer ALL questions. Write your answers in the spaces provided. Some questions must be answered with a cross in a box. If you change your mind about an answer, put a line through the box and then mark your new answer with a cross. 1 Methylpropane is an alkane. (a) Draw the displayed formula of methylpropane. (b) Give the empirical formula of methylpropane. (c) A partially balanced equation for the complete combustion of butane is: C 4 H 10 + xo 2 4CO 2 + 5H 2 O The number of moles of oxygen, x, needed to balance this equation is A 4.5 B 6.5 C 9 D 13 2 54 *S47552A0224*
(d) Incomplete combustion of butane produces a mixture of products. Which substance is not produced during the incomplete combustion of butane? A C B CO C H 2 D H 2 O (e) A student is given a bottle, containing an alkane, with an incomplete label. he ane The student realises that the alkane could be hexane or heptane and wants to identify it. The student injects a sample of the liquid, with a mass of 0.235 g, into a sealed gas syringe. The syringe is placed in an oven at 425 K and the liquid vaporises. The volume of the gas produced is 83 cm 3 at a pressure of 100 kpa (100 000 N m 2 ). Identify the alkane. Justify your answer by the use of a calculation. (4) (Total for Question 1 = 8 marks) *S47552A0324* 55 3
2 The equations for three reactions are: reaction 1 C(s) + O 2 (g) CO 2 (g) r H d = 394 kj mol 1 reaction 2 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) r H d = 890 kj mol 1 reaction 3 2H 2 (g) + O 2 (g) 2H 2 O(l) r H d = 572 kj mol 1 (a) Give two reasons why all three reactions are classified as examples of combustion. (2) (b) Only reaction 1 represents a standard enthalpy change of formation. Give reasons why reactions 2 and 3 do not. (2) 4 56 *S47552A0424*
(c) Which physical property should be kept constant when measuring an enthalpy change? A concentration B C D pressure temperature volume (d) A reaction occurs under standard conditions. Which of these represents a possible standard condition? A 1 g cm 3 B 4.18 J g 1 K 1 C 24 dm 3 D 298 K (Total for Question 2 = 6 marks) *S47552A0524* 57 5
3 Ammonia is used in the manufacture of nitric acid. The equation for one step in this manufacturing process is: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) r H = 900 kj mol 1 *(a) A manufacturer carries out this reaction at a temperature of 1200 K and a pressure of 10 atm. A scientist proposes that a temperature of 1000 K should be used at the same pressure. Evaluate the effects of making this change on the rate and yield of this reaction. (6) 6 58 *S47552A0624*
(b) When this reaction is used in industry, the catalyst is an alloy of platinum and rhodium. The diagram shows the reaction profile for the uncatalysed reaction. enthalpy 4NH 3 + 5O 2 4NO + 6H 2 O progress of reaction (i) On the diagram, draw the reaction profile for the catalysed reaction. (ii) Label the diagram to show the enthalpy change, r H the activation energy, E a for the catalysed reaction. (c) Write the expression for the equilibrium constant, K c, for this reaction. (2) (Total for Question 3 = 10 marks) *S47552A0724* 59 7
4 This question is about the thermal decomposition of ammonia. This reaction is catalysed by platinum and is represented by the equation: 2NH 3 (g) N 2 (g) + 3H 2 (g) r H d = + 92 kj mol 1 The diagram shows a sketch of the Maxwell-Boltzmann curve for the distribution of molecular energies for a fixed amount of ammonia gas at a given temperature. E a represents the activation energy of the uncatalysed reaction. fraction of molecules with energy, E E a (no catalyst) energy, E (a) (i) On the diagram, draw a vertical line to represent the activation energy of the catalysed reaction. Label this line E a (with catalyst). (ii) Use the diagram to explain why the use of a catalyst increases the rate of decomposition of ammonia. (3) 8 60 *S47552A0824*
(b) The table shows the bond enthalpies for the N N and H H bonds. Bond Bond enthalpy / kj mol 1 N N + 944 H H + 436 Use this data, together with the standard enthalpy change of reaction, r H d, for the decomposition of ammonia, to calculate a value, in kj mol 1, for the mean bond enthalpy of the N H bond in ammonia. (3) (Total for Question 4 = 7 marks) *S47552A0924* 61 9
5 Some reactions involving copper(ii) sulfate are investigated. The diagram shows the apparatus used to measure the temperature change when anhydrous copper(ii) sulfate, CuSO 4 (s), is dissolved in water. thermometer plastic cup water The chemical change can be represented by this equation: The method is: CuSO 4 (s) + aq CuSO 4 (aq) Pour some water into the plastic cup and record the steady temperature Add some anhydrous copper(ii) sulfate and stir until it has all dissolved Record the maximum temperature reached in the reaction These results are recorded. Mass of water used 50.0 g Initial temperature of water 18.2 C Final temperature of water 25.4 C Mass of anhydrous copper(ii) sulfate used 4.70 g 10 62 *S47552A01024*
(a) Calculate the standard enthalpy change, in kj mol 1, for the reaction, giving your answer to an appropriate number of significant figures. (5) *S47552A01124* 63 11
(b) The method used in this experiment leads to an inaccurate value for the temperature change. Describe an alternative method to find a more accurate value for the temperature change. (3) (c) The experiment is repeated using the same method but with hydrated copper(ii) sulfate, CuSO 4.5H 2 O(s), as well as with anhydrous copper(ii) sulfate. The same method is used to calculate the standard enthalpy changes for these reactions: CuSO 4.5H 2 O(s) + aq CuSO 4 (aq) r H d = +8.5 kj mol 1 1 CuSO 4 (s) + aq CuSO 4 (aq) r H d = 41 kj mol 1 2 Calculate the standard enthalpy change for this reaction: CuSO 4 (s) + aq CuSO 4.5H 2 O(s) (2) 12 64 *S47552A01224*
(d) The experimental value in part (c) of r H d 2 = 41 kj mol 1 for the reaction involving anhydrous copper(ii) sulfate was different from a data book value. The anhydrous copper(ii) sulfate used in this experiment was pale blue, rather than white. Explain how this observation partly accounts for the difference between the experimental value and the data book value. (3) (Total for Question 5 = 13 marks) *S47552A01324* 65 13
6 A student investigates the rates of hydrolysis of three different halogenoalkanes. The student uses this method. 5 cm 3 of ethanol is put into each of three test tubes and five drops of a different halogenoalkane is added to each. The test tubes are placed into a water bath at 50 C. 5 cm 3 of aqueous silver nitrate is put into each of three clean test tubes in the water bath. When the solutions have reached the temperature of the water bath, 5 cm 3 of the silver nitrate solution is mixed with each test tube containing a halogenoalkane in ethanol. A stop clock is started when the solutions are mixed. The time taken for a precipitate to appear in each test tube is recorded in a table. Halogenoalkane 1-chlorobutane 1-bromobutane 1-iodobutane Time taken for precipitate to appear 20 minutes and 50 seconds 9 minutes and 15 seconds 5 seconds (a) (i) State the colour of the precipitate formed in the 1-iodobutane test tube. (ii) Calculate the relative rates of reaction for each of the three halogenoalkanes. (2) 14 66 *S47552A01424*
(iii) Explain the relative rates of hydrolysis of 1-chlorobutane and 1-iodobutane in this experiment. (3) (b) The student saw, on a website, a comparison of the rates of hydrolysis of these bromoalkanes: F CH 3 CH 2 CH 2 CH 2 Br G CH 3 CH 2 CHBrCH 3 H (CH 3 ) 3 CBr (i) The correct order for the rates of hydrolysis of these bromoalkanes, showing the fastest first, is: A F G H B C D G F H H F G H G F (ii) Identify, with a reason, which of F, G and H is a tertiary bromoalkane. (2) (iii) A student said that bromoalkane F had the name 4-bromobutane. Give a reason why this name is incorrect. (Total for Question 6 = 10 marks) *S47552A01524* 67 15
7 But-1-ene and but-2-ene are unsaturated hydrocarbons with the molecular formula C 4 H 8. Each one reacts with a few drops of bromine in an addition reaction. (a) Give the colour change that occurs in these reactions. (b) (i) Name the type of mechanism for these reactions. (ii) Draw the mechanism for the reaction between but-2-ene and bromine. (4) 16 68 *S47552A01624*
(c) But-1-ene reacts with hydrogen bromide to form two different products. (i) Analysis of one of the products showed that it contains 34.9% carbon and 6.60% hydrogen by mass. Calculate the empirical formula of this product. (3) (ii) Give the name of the major product of this reaction. (d) The carbon-carbon double bond in but-2-ene can be represented by C C. A disadvantage of this representation is that it shows the two covalent bonds to be the same. However, there are two different types of covalent bond ( and ) in but-2-ene. Compare and contrast these two different types of bonds. (3) *S47552A01724* 69 17
(e) Draw the skeletal formula of trans-but-2-ene. (Total for Question 7 = 14 marks) 70 18 *S47552A01824*
8 This question is about some isomeric alcohols with molecular formula C 5 H 12 O. (a) The table shows some information about three of these alcohols. Give the structural formula of each of these alcohols. (3) Alcohol Description Structural formula P the straight-chain primary alcohol Q the secondary alcohol with a branched carbon chain R the alcohol not oxidised by potassium dichromate(vi) in dilute sulfuric acid (b) T is another different isomeric alcohol with the molecular formula C 5 H 12 O. T is a secondary alcohol. The most prominent peak in the mass spectrum of T occurred at m/z = 45, with another peak at m/z = 43 Deduce the structure of T. Justify your answer by identifying the structural formula of each fragment ion. (3) *S47552A01924* 71 19
Some of the alcohols were heated with potassium dichromate(vi) and sulfuric acid. The organic compounds were separated from the reaction mixtures and purified. The infrared spectra of two of these organic compounds are shown. 100 Compound Y Transmittance (%) 50 0 4000 3000 2000 1500 1000 500 Wavenumber (cm 1 ) 100 Compound Z Transmittance (%) 50 0 4000 3000 2000 1500 1000 500 Wavenumber (cm 1 ) 72 20 *S47552A02024*
(c) (i) Deduce, using the table of absorptions from the data booklet, the type of compound responsible for each spectrum. Include in your answer references to wavenumbers and their corresponding bonds. (2) (ii) One of the two compounds, Y and Z, of mass 2.74 g is completely burned in oxygen. Carbon dioxide and water are the only two products. The masses of carbon dioxide and water formed are 5.89 g and 2.44 g respectively. Show by calculation that this data is consistent with a formula of C 5 H 10 O 2 for this compound. (4) (Total for Question 8 = 12 marks) TOTAL FOR PAPER = 80 MARKS *S47552A02124* 73 21
BLANK PAGE 74 22 *S47552A02224*
BLANK PAGE *S47552A02324* 75 23
76 24 *S47552A02424*
CHEMISTRY AS PAPER 2 MARKSCHEME Question Number 1(a) Answer Additional Guidance Mark Every atom and every bond must be shown 1 1(b) C2H5 1 1(c) B 1 1(d) C 1 1(e) re-arrangement of pv = nrt equation conversion of volume and substitution evaluation of number of moles of alkane calculation of Mr, using number of moles and mass of alkane to identify it Correct answer, no working scores 4 marks Example of calculation n = pv RT = 100 000 x (83 x 10-6 ) 8.31 x 425 4 = 0.00235 (moles) Mr = 0.235 / 0.00235 = 99.99 or 100 i.e. this is heptane (Total for Question 1 = 8 marks) 77
Question Number 2(a) An answer that makes reference to the following points: Answer Additional Guidance Mark 2 (all equations) show oxygen/o2 as a reactant/on the left (all) ΔrH o values are negative /(all) reactions are exothermic 2(b) An answer that makes reference to the following points: 2 reaction 2 is not, because there is more than 1 product Accept because methane/ch4 is not an element reaction 3 is not, because 2 moles of product are formed 2(c) B 1 2(d) D 1 (Total for Question 2 = 6 marks) 78
Question Number *3(a) This question assesses a student s ability to show a coherent and logically structured answer with linkages and fullysustained reasoning. Answer Additional Guidance Mark Marks are awarded for indicative content and for how the answer is structured and shows lines of reasoning. The following table shows how the marks should be awarded for indicative content. Number of Number of indicative marks marking awarded for points seen indicative in answer marking points 6 4 5 4 3 3 2 2 1 1 0 0 Guidance on how the mark scheme should be applied: The mark for indicative content should be added to the mark for lines of reasoning. For example, an answer with five indicative marking points that is partially structured with some linkages and lines of reasoning, scores 4 marks (3 marks for indicative content and 1 mark for partial structure and some linkages and lines of reasoning). If there are no linkages between points, the same five indicative marking points would yield an overall score of 3 marks (3 marks for indicative content and no marks for linkages). 6 79
Question Number *3(a) cont. Answer Additional Guidance Mark The following table shows how the marks should be awarded for structure and lines of reasoning. Answer shows a coherent and logical structure with linkages and fully sustained lines of reasoning demonstrated throughout Answer is partially structured with some linkages and lines of reasoning Answer has no linkages between points and is unstructured Number of marks awarded for structure of answer and sustained line of reasoning 2 1 0 Accept slows down the reaction Indicative content: temperature decrease lowers the rate of the reaction because there are fewer molecules/particles with E Ea and therefore there are fewer successful collisions per second temperature decrease increases the yield (of the product) because the (forward) reaction is exothermic lower rate and increased yield are opposing factors and it is not possible to tell which has greater effect on overall yield in a given time. Accept shifts the position of equilibrium to the right/in forward direction 80
Question Number 3(b)(i) 3(b)(ii) 3(c) Answer Additional Guidance Mark 1 curve starting at reactants level, ending at products level, and peaking lower than original curve ecf from candidate s curve 2 ΔrH shown as the approximately vertical distance between reactants and products Ea shown as the approximately vertical distance between reactants and peak of drawn curve for catalysed reaction Kc = [!"! ]! [!!!(!)]! [!!!! ]! [!! (!)]! State symbols not essential 1 (Total for Question 3 = 10 marks) 81
Question Number 4(a)(i) Answer Additional Guidance Mark Line needs to be to the left of Ea (no catalyst) but to the right of the hump in the curve 1 Shading not required but can be included and used in (a)(ii) 4(a)(ii) An explanation that makes reference to the following points: 3 the area under the curve to the right of the activation energy represents the fraction of molecules that have E Ea Accept description that involves shaded areas under the curve the area under the curve to the right of Ea (with catalyst) is greater than the area under the curve to the right of Ea (no catalyst) Accept more molecules have enough energy to react (therefore) there are more successful collisions (per second per unit vol.) 82
Question Number 4(b) calculation of total bonds formed Answer Additional Guidance Mark Correct answer, no working scores 3 marks 3 re-arrangement of ΔrH expression Example of calculation evaluation of final answer (bonds formed) = E(N N) + 3E(H H) = 944 + (3 x 436) = 2252 (kj mol 1 ) ΔrH o = bonds broken (bonds formed), so (bonds formed) = bonds broken ΔrH o 6E(N H) = 2252 (-92) E(N H) = (+) 391 (kj mol 1 ) Accept 390.7 (Total for Question 4 = 7 marks) 83
Question Number 5(a) calculation of temperature change and substitution into Q = mcδt Answer Additional Guidance Mark conversion to kj calculation of amount of copper(ii) sulfate Allow ecf for steps in calculation; including for final answer dependent on rounding in steps of the calculation. Correct answer with sign, to 3.s.f and no working scores 5 marks 5 use of ΔH =!! Mark consequentially on incorrect temperature change correct answer with sign and to 3. s.f Example of calculation Q = 50.0 x 4.18 x 7.2 (J) = 1504.8 (J) Q =!"#$.!!""" = 1.5048 (kj) moles of copper(ii) sulfate = 4.70 159.6 = 0.029449 mol ΔH =!! = 1.5048 0.029449 = ( 51.099) = 51.1 (kj mol 1 ) 84
Question Number 5(b) A description that makes reference to the following points: Answer Additional Guidance Mark 3 record temperature at timed intervals (after adding the solid) plot these temperature values (on a graph) extrapolate the line of best fit to the time of adding the solid 5(c) use of Hess cycle or expression e.g. ΔrH = ΔrH2 ΔrH1 or ΔrH = ΔHsolution(anhydrous) ΔHsolution(hydrated) Correct answer with sign and no working scores 2 marks 2 correct final answer with sign ( 49.5 kj mol 1 ) 5(d) An explanation that makes reference to the following points: 3 (the pale blue colour shows that) anhydrous copper (II) sulfate is partially hydrated therefore less anhydrous copper (II) sulfate reacts with water / some of the (blue) solid reacts endothermically therefore less thermal energy transferred / heat energy released during experiment or (experimental value) has a smaller negative value (Total for Question 5 = 13 marks) 85
Question Number Answer Additional Guidance Mark 6(a)(i) (pale) yellow 1 6(a)(ii) conversion to seconds and use of!!"#$!"!"# Example of calculation!!"#$ :!!!! :!! 2 divide by lowest to get whole number ratio or 8.00 x 10 4 : 1.80 x 10 3 : 0.20 6(a)(iii) An explanation that makes reference to the following points: the C Cl bond is shorter than the C I bond / the Cl atom is smaller than the I atom 1 : 2.25 : 250 Allow reverse argument throughout 3 therefore, the C Cl bond is stronger than the C I bond therefore, the C Cl bond needs more energy to break, and so the reaction is slower 86
Question Number Answer Additional Guidance Mark 6(b)(i) D 1 6(b)(ii) H/ (CH3)3 CBr 2 because C of C Br is joined to 3 alkyl groups/joined to 3 carbon atoms 6(b)(iii) number prefix used should be as low as possible / bromine atom is on the first carbon atom of the chain 1 (Total for Question 6 = 10 marks) 87
Question Number Answer Additional Guidance Mark 7(a) (red-)brown to colourless Allow orange for red-brown Do not allow clear for colourless 1 7(b)(i) electrophilic (addition) 1 7(b)(ii) 4 partial charges on Br atoms curly arrow from C=C bond to ( +) Br atom AND curly arrow from Br Br bond to ( ) Br atom structure of carbocation with + charge Accept cyclic bromonium ion with + charge shown in centre of ring curly arrow from Br to C with + charge AND structure of 1, 2-dibromobutane product 88
Question Number 7(c)(i) calculation of % by mass of Br Answer Additional Guidance Mark Example of calculation 3 evaluation of number of moles of C, H, and Br %Br = 100 34.9 6.60 = 58.5% final empirical formula based on whole number ratio C!".!!"!.!" = 2.91 H!!".! = 6.60 Br!".! = 0.732 2.91 = 3.98 6.60 = 9.02 0.732 = 1 0.732 0.732 0.732 i.e. C4H9Br 7(c)(ii) 2 bromobutane 1 89
Question Number 7(d) An answer that includes at least one similarity and one difference to gain maximum marks: Answer Additional Guidance Mark 3 Similarities: both involve the overlap of two (atomic) orbitals (each containing a single electron) both involve an electrostatic attraction between a bonding pair of electrons and two nuclei Accept electrostatic attraction between a shared pair of electrons and two nuclei Differences: the bond represents electron density/electrons between the two carbon atoms/nuclei/end-on overlapping of atomic orbitals Accept end-to-end/head-on/axial overlapping the bond represents electron density/electrons above and below the bond/ sideways overlapping of orbitals the σσ bond is formed by the overlap of a single lobe from one (atomic) orbital with a single lobe from the another (atomic) orbital the ππ bond is formed by the overlap of two lobes from one (atomic) orbital with two lobes of another (atomic) orbital 90
Question Number 7(e) Answer Additional Guidance Mark Do not accept a structural or displayed formula. 1 (Total for Question 7 = 14 marks) 91
Question Number 8(a) P CH3CH2CH2CH2CH2OH Answer Additional Guidance Mark Accept displayed formulae 3 Q (CH3)2CHCH(OH)CH3 R (CH3)2C(OH)CH2CH3 8(b) m/z = 45 is due to CH3CH(OH) + Penalise the lack of a plus sign once only 3 m/z = 43 is due to CH3CH2CH2 + therefore the structure is CH3CH2CH2CH(OH)CH3 (CH3)2CHCH(OH)CH3 not allowed as it is Q 8(c)(i) An answer that makes reference to the following points: Y is a ketone because the spectrum contains an absorption for C=O (carbonyl group) 1700 1720 cm -1 Absence of bonds or absence of wavenumber ranges maximum 1 Allow values within the ranges 2 Z is a carboxylic acid because the spectrum contains absorptions for C=O 1725-1700 cm -1 AND O H (acids) 3300 2500 cm -1 92
Question Number 8(c)(ii) calculation of mass of carbon from mass of CO2 Answer Additional Guidance Mark calculation of mass of hydrogen from mass of H2O Example of calculation Mass of carbon = (!" x 5.89) = 1.606 g!! 4 subtraction to find mass of O, and evaluation of number of moles of C, H and O Mass of hydrogen = (!!" x 2.44) = 0.271 g Mass of oxygen = 2.74 1.606 0.271 confirm whole number ratio = 0.863 g C!.!"!!" = 0.134 H!.!"#! = 0.271 O!.!"#!" = 0.054 Ratio = 2.48 : 5.02 : 1 = 5 : 10 : 2 (Total for Question 8 = 12 marks) 93