Edexcel Chemistry AS Paper 2 Unofficial Markscheme 09/06/2017

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timothysimons
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Can everyone comment the answers they remember so an unofficial markscheme can be put together?
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saffah
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Molecular formula: C6H12
Empherical formula: CH2
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Tortellini69
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Everyone in my year found this test brutal, even the ones who had been getting A's all year struggled to finish the test,
The only answer I remember was the mass of the substance in 250cm cubes and I got 9.5
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Tortellini69
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-Empiracle formula ch2 molecular formula c6h12
-6 marker- raising the temp deceased the yield of the wanted product, increasing the pressure increased the yield of the wanted product
-When it asked for the structure of isomers i wasn't sure if it wanted displayed formula or structural so I did structural formula.
-for the enthalpy change of combustion I got around -2700kjmol-1
-for functional groups I got cc double bond(wasn't sure) and OH
- atom economy I got around 63%
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timothysimons
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(Original post by Tortellini69)
-Empiracle formula ch2 molecular formula c6h12
-6 marker- raising the temp deceased the yield of the wanted product, increasing the pressure increased the yield of the wanted product
-When it asked for the structure of isomers i wasn't sure if it wanted displayed formula or structural so I did structural formula.
-for the enthalpy change of combustion I got around -2700kjmol-1
-for functional groups I got cc double bond(wasn't sure) and OH
- atom economy I got around 63%
theres so much i screwed up on, especially the practical question nd titration question at the end, kmt i really hope the grade boundaries are like 40% for a D otherwise i have no chance of passing
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ahmed111111
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do you guys remember some of the specific questions that came up or topics if so can i have a list? thank you
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James1121111
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Do you remember all the topics that came up if so could you create a list so I know what to revise for thanks
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timothysimons
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(Original post by James1121111)
Do you remember all the topics that came up if so could you create a list so I know what to revise for thanks
equilibrium
calculations
practical skills
kinetics
energetics
infrared/mass spec
mechanisms
tests for functional groups
hydrogen bonding, shapes of molecules was a few marks
There was a test for double bonds and alcohols
and for kinetics there was enthalpy profile diagrams
and there was a kj/mol question
the practical thing they gave the method you had to answer q's on it
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Student1899
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I took this exam in June 2017 and based on marks given by the examiner, my 'unofficial' answers for this paper are:
--------------------------------------------------------------------------------------
Paper 8CH0/02 taken on 9 June 2017:
1. A flouromethane
2. C CO2
3.
(a) A + O2 ----> CO2 + H2O
3.143g + 1.284g
mass C = 12/44 * 3.143 = 0.854 C ratio 08.54/12 = 0.0714 => 1
mass H = 2/18 * 1.284 = 0.143 H ratio 0.143/1 = 0.143 => 2
emiprical ratio CH2, RFM = 84, so molecular formula is C6H12
(b)
(i) Use q = m * c * delta T, delta T = 29.5 - 21.3 = 8.2 C
q = 250 * 4.18 * 8.2 = 8569J
mass of A = 112.990-112.732 = 0.258g, moles of A = 0.258/84.0 = 0.003071
delta H = 8.569/0.003071 = 2789.9 KJ/mol = 2790 KJ/mol to 3 SF
(ii) It is thermally conductive so less heat will be absorbed than a glass beaker.
So more heat will be transferred to H2O = more reliable results.
4.
(a) Molecular formula of geraninol is C10H18O
Molar Mass = 154 g/mol
(b)
(i) The last peak m+ peak shows the molar mass at 154
(ii) m/z = 69 could be a molecule with C5H9
............--CH3
....CH2-CH=C..........
............--CH3
(c) First functional group - Alcohol OH group (bond O-H) is shown from 3600-3200
Second functional group - Alkene (C=C) bond shown from 1680-1620
(d) First functional group - Heat with Potassium Dichromate, if it turns from
orange to green = contains primary or secondary alcohol
Second functional group - Add to Bromine water, if it decolourises from
pale orange = contains Alkene
(e) Two molecules Isoprene contains 5 carbons and geranimol contains 10 carbons.
So you need two Isoprene molecules to have correct number of carbons.
(f)
...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... H.. Br.... Br H

...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... H.. Br.... H..Br........
......
...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... Br..H......H..Br

...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... Br..H.... Br..H
5.
(a) The volume that 1 moles of a gas occupies at room temperature and pressure = 24 dm3
(b)
(i) Percentage uncertainty = error/measurement * 100%
Uncertainty in Volume = 0.5 * 2/50 * 100
Uncertainty in Mass = 0.0005 * 2/107.563 + 0.0005 * 2/107.655
Total Uncertainty = 1.002%
(ii) Uncertainty in Volume becomes 0.5 * 2/100 * 100, so uncertainty is halved
Uncertainty in Mass = 0.0005 * 2/214 approx, so uncertainty is halved
(iii) mass of 50 cm3 of gas = 107.655 - 107.563 = 0.092g,
moles of gas = 50/24000 = 0.00208 moles
Molar Mass = 0.092/0.00208 = 4.423 g/mol
(iv) If there was a leak then the pressure in the syringe would be lower. By pulling
the syringe out and releasing it, the plunger will settle back to the correct
level. If gas had leaked the new volume will be less than 50cm3. The leak would
lower the mass of the gas and hence the calculated molar mass.
(c)
As pV=nRT, if T is reduced then the number of moles of gas would increase. The
measured mass of gas would increase and the calculated molar mass would also
increase.
(d) If it was not, there would be H2O molecules in the air, which would take up
volume and so affect results = less wanted gas held in gas syringe
6.
(a)
The production of sulfur trioxide is exothermic, so according to Le Chatelier's
principle, if you raise the temperature the system would counteract this by
favouring the endothermic change. So although the increased temperature
increases the rate at which equilibrium is reached, but it lowers the
percentage yield. The lower yield means less product will be made, which
isn't economically optimal.

Higher pressure favours the reaction that produces fewer molecules as fewr
molecules produce lower pressure. According to Le Chatelier's principle,
more sulfur trioxide would be produced and percentage yield would increase.
Higher pressure also brings molecules closer together so the rate of
reaction increases. However, high pressure require very strong pipes,
containment vessels and lots of energy, so is expensive.
(b)
(i)
Label y-axis Energy and x-axis Reaction progress. Draw high horizontal
line with label 2SO2+O2 for reactants. Draw curve with a high peak
labeled A and coming down to lower horizontal line with label SO3 for
products. Draw second curve with lower peak and add label B for catalysed
reaction.
(ii)
Ea = Activation energy - Upward arrow from higher horizontal line to peak
of curve labeled B
Delta H = Enthalpy change - Downward arrow from higher horizontal line to
lower horizontal line
(c)
(i).......... [SO3]^2
......Kc = -------------
............[SO2]^2[O2]
(ii) B dm3/mol
7.
(a)
(i) B Z,Y,X X=Cl, Y=Br, X=I
(ii) The halogenoalkanes are insoluble in water. Using ethanol ensures that the
halogenoalkane dissolves so that it can react with molecules in the silver
nitrate solution.
(iii) It ensures that the halogenoalkane, ethanol solutions have reached the
temperature of the water bath.
(iv) Halide ions are formed for primary halogenoalkane as a reaction between
halogenoalkane and water in the solvent.
R-Hal + H2O --> R-OH + H+ + Hal-
A tertiary halogenoalkane can ionise by itself
R-Hal -- R+ + Hal-
(v) Ag+ (aq) + Cl- (aq) --> AgCl (s)
(b)
(i) Label y-axis Concentration of 1-bromo-2-methylpropane (mol/dm3) with
scale 0.01, 0.02, 0.03 etc. Label x-axis Time (s) with scale 50, 100,
150 etc. Mark each reading with x and draw the downward curve.
ii) Draw a tangent to the curve at 100s and then a right angled triangle.
The rate of reaction is the gradient of the triangle.
(0.043 - 0.01)/(140 - 50) = 0.033/90 = 0.0003667 mol/dm3 s
(c)
(i) D nucleophile
................................ ............H.. CH3 H
................................ ............|.. |.. |
..................H.. CH3 H............OH---C---C---C--H
...... &-<------&+|.. |.. |................ |.. |.. |
...... Br---------C---C---C--H.. --->...... H.. H.. H
................^ |.. |.. |..........
................| H.. H.. H..............+..Br-
.............. OH-

(iii) D substitution
8.
(a)
.............................. OH..H.. OH
.............................. |.. |.. |
...... propane-1,3-diol....H---C---C---C---H..
.............................. |.. |.. |
.............................. H.. H.. H

.............................. OH..OH..H
.............................. |.. |.. |
...... propane-1,2-diol....H---C---C---C---H
.............................. |.. |.. |
.............................. H.. H.. H
(b)
(i) A sodium dichromate(VI) + sulfuric acid
(ii) B orange to green
(c)
(i)
H2X + 2NaOH --> Na2X + 2H2O
m = c * v c=0.4 mol/dm3, v = 18.45/1000
= 0.4 * 18.45/1000 = 0.00738 moles
Ratio of propanedioic acid is half of sodium hydroxide
moles of propanedioic acid = 0.5 * 0.00738 = 0.00369 moles
(ii)
moles = mass/relative atomic mass (RAM), RAM = 104.0
mass = moles * RAM = 0.00369 * 104 = 0.38376g in 10cm3
mass in 250 cm3 = 25 * 0.38376 = 9.594g
(ii)
mass of propane-1,3-diol that reacted = 9.594 * 76.0/104.0 = 7.011
% yield = 7.011/15.2 * 100 = 46.1%
(iii) Not all reactants react to make productPaper 8CH0/02 taken on 9 June 2017:
1. A flouromethane
2. C CO2
3.
(a) A + O2 ----> CO2 + H2O
3.143g + 1.284g
mass C = 12/44 * 3.143 = 0.854 C ratio 08.54/12 = 0.0714 => 1
mass H = 2/18 * 1.284 = 0.143 H ratio 0.143/1 = 0.143 => 2
emiprical ratio CH2, RFM = 84, so molecular formula is C6H12
(b)
(i) Use q = m * c * delta T, delta T = 29.5 - 21.3 = 8.2 C
q = 250 * 4.18 * 8.2 = 8569J
mass of A = 112.990-112.732 = 0.258g, moles of A = 0.258/84.0 = 0.003071
delta H = 8.569/0.003071 = 2789.9 KJ/mol = 2790 KJ/mol to 3 SF
(ii) It is thermally conductive so less heat will be absorbed than a glass beaker.
So more heat will be transferred to H2O = more reliable results.
4.
(a) Molecular formula of geraninol is C10H18O
Molar Mass = 154 g/mol
(b)
(i) The last peak m+ peak shows the molar mass at 154
(ii) m/z = 69 could be a molecule with C5H9
............--CH3
....CH2-CH=C..........
............--CH3
(c) First functional group - Alcohol OH group (bond O-H) is shown from 3600-3200
Second functional group - Alkene (C=C) bond shown from 1680-1620
(d) First functional group - Heat with Potassium Dichromate, if it turns from
orange to green = contains primary or secondary alcohol
Second functional group - Add to Bromine water, if it decolourises from
pale orange = contains Alkene
(e) Two molecules Isoprene contains 5 carbons and geranimol contains 10 carbons.
So you need two Isoprene molecules to have correct number of carbons.
(f)
...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... H.. Br.... Br H

...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... H.. Br.... H..Br........
......
...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... Br..H......H..Br

...... H.. C--H3..H..H
...... |.. |......|..|
.. H---C---C------C--C--H
...... |.. |......|..|
...... Br..H.... Br..H
5.
(a) The volume that 1 moles of a gas occupies at room temperature and pressure = 24 dm3
(b)
(i) Percentage uncertainty = error/measurement * 100%
Uncertainty in Volume = 0.5 * 2/50 * 100
Uncertainty in Mass = 0.0005 * 2/107.563 + 0.0005 * 2/107.655
Total Uncertainty = 1.002%
(ii) Uncertainty in Volume becomes 0.5 * 2/100 * 100, so uncertainty is halved
Uncertainty in Mass = 0.0005 * 2/214 approx, so uncertainty is halved
(iii) mass of 50 cm3 of gas = 107.655 - 107.563 = 0.092g,
moles of gas = 50/24000 = 0.00208 moles
Molar Mass = 0.092/0.00208 = 4.423 g/mol
(iv) If there was a leak then the pressure in the syringe would be lower. By pulling
the syringe out and releasing it, the plunger will settle back to the correct
level. If gas had leaked the new volume will be less than 50cm3. The leak would
lower the mass of the gas and hence the calculated molar mass.
(c)
As pV=nRT, if T is reduced then the number of moles of gas would increase. The
measured mass of gas would increase and the calculated molar mass would also
increase.
(d) If it was not, there would be H2O molecules in the air, which would take up
volume and so affect results = less wanted gas held in gas syringe
6.
(a)
The production of sulfur trioxide is exothermic, so according to Le Chatelier's
principle, if you raise the temperature the system would counteract this by
favouring the endothermic change. So although the increased temperature
increases the rate at which equilibrium is reached, but it lowers the
percentage yield. The lower yield means less product will be made, which
isn't economically optimal.

Higher pressure favours the reaction that produces fewer molecules as fewr
molecules produce lower pressure. According to Le Chatelier's principle,
more sulfur trioxide would be produced and percentage yield would increase.
Higher pressure also brings molecules closer together so the rate of
reaction increases. However, high pressure require very strong pipes,
containment vessels and lots of energy, so is expensive.
(b)
(i)
Label y-axis Energy and x-axis Reaction progress. Draw high horizontal
line with label 2SO2+O2 for reactants. Draw curve with a high peak
labeled A and coming down to lower horizontal line with label SO3 for
products. Draw second curve with lower peak and add label B for catalysed
reaction.
(ii)
Ea = Activation energy - Upward arrow from higher horizontal line to peak
of curve labeled B
Delta H = Enthalpy change - Downward arrow from higher horizontal line to
lower horizontal line
(c)
(i).......... [SO3]^2
......Kc = -------------
............[SO2]^2[O2]
(ii) B dm3/mol
7.
(a)
(i) B Z,Y,X X=Cl, Y=Br, X=I
(ii) The halogenoalkanes are insoluble in water. Using ethanol ensures that the
halogenoalkane dissolves so that it can react with molecules in the silver
nitrate solution.
(iii) It ensures that the halogenoalkane, ethanol solutions have reached the
temperature of the water bath.
(iv) Halide ions are formed for primary halogenoalkane as a reaction between
halogenoalkane and water in the solvent.
R-Hal + H2O --> R-OH + H+ + Hal-
A tertiary halogenoalkane can ionise by itself
R-Hal -- R+ + Hal-
(v) Ag+ (aq) + Cl- (aq) --> AgCl (s)
(b)
(i) Label y-axis Concentration of 1-bromo-2-methylpropane (mol/dm3) with
scale 0.01, 0.02, 0.03 etc. Label x-axis Time (s) with scale 50, 100,
150 etc. Mark each reading with x and draw the downward curve.
ii) Draw a tangent to the curve at 100s and then a right angled triangle.
The rate of reaction is the gradient of the triangle.
(0.043 - 0.01)/(140 - 50) = 0.033/90 = 0.0003667 mol/dm3 s
(c)
(i) D nucleophile
................................ ............H.. CH3 H
................................ ............|.. |.. |
..................H.. CH3 H............OH---C---C---C--H
...... &-<------&+|.. |.. |................ |.. |.. |
...... Br---------C---C---C--H.. --->...... H.. H.. H
................^ |.. |.. |..........
................| H.. H.. H..............+..Br-
.............. OH-

(iii) D substitution
8.
(a)
.............................. OH..H.. OH
.............................. |.. |.. |
...... propane-1,3-diol....H---C---C---C---H..
.............................. |.. |.. |
.............................. H.. H.. H

.............................. OH..OH..H
.............................. |.. |.. |
...... propane-1,2-diol....H---C---C---C---H
.............................. |.. |.. |
.............................. H.. H.. H
(b)
(i) A sodium dichromate(VI) + sulfuric acid
(ii) B orange to green
(c)
(i)
H2X + 2NaOH --> Na2X + 2H2O
m = c * v c=0.4 mol/dm3, v = 18.45/1000
= 0.4 * 18.45/1000 = 0.00738 moles
Ratio of propanedioic acid is half of sodium hydroxide
moles of propanedioic acid = 0.5 * 0.00738 = 0.00369 moles
(ii)
moles = mass/relative atomic mass (RAM), RAM = 104.0
mass = moles * RAM = 0.00369 * 104 = 0.38376g in 10cm3
mass in 250 cm3 = 25 * 0.38376 = 9.594g
(ii)
mass of propane-1,3-diol that reacted = 9.594 * 76.0/104.0 = 7.011
% yield = 7.011/15.2 * 100 = 46.1%
(iii) Not all reactants react to make product


--------------------------------------------------------------------------------------
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Student1899
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I have also posted unofficial answers for Paper 1 (Paper 8CH0/01 - 9 June 2017) in following Student Room thread

https://www.thestudentroom.co.uk/sho...4741892&page=2
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