The thread is a shambles so just put some solutions in order. If anyone has corrections / knows marks allocated fire away. Cheers. Edit: * THANKS to all who contributed from chem4/5 thread.
Q1 Periodicity (12 marks)
Covalent
P
P4O10 + 6H2O -> H3PO4
(3 marks)
Ionic
Na
2Na + 2H2O --> 2NaOH + H2
(4 marks)
Al2O3
Ionic
Reacts with acids and bases
3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
(5 marks)
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Q2 BH Cycle (8 marks)
Thermodynamics Calcs ****
2K(s) + ½O2(g)
2K(g) + ½O2(g)
2K(g) + O(g)
(3 marks)
Enthalpy of Lattice dissociation = +2328kJ.mol-1
(3 marks)
K+ ions are bigger than Na+ ions, same charge so lower charge density
Weaker attraction between K+ and O^(2-) than Na+ and O^(2-)
(2 marks)
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Q3 Enthalpy of solution (6 marks)
MgCl2 + aq --> Mg^2+(aq)+ 2Cl-(aq) (1 mark)
Solubility will increase with temperature. The question stated itself that solubility is defined as the amount of solute available to dissolve in a solvent ( in this case water ). Hence, by the reverse reaction due to exothermic enthalpy of solution, more magnesium chloride will be produced, hence increasing its solubility. (3 marks)
Enthalpy of solution = -155 kJ.mol-1 (2 marks)
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Q4 Electrochemistry (11 marks)
Weakest oxidising agent is Zn2+ (1 mark)
Standard conditions to give EMF of -0.76:
100kPa/298K/zero current/1M all solutions (2 marks)
Cell giving EMF of 0.48
Zn|Zn2+||Co2+|Co (2 marks)
SO2 + V2O5 -> SO3 + V2O4
V2O4 + 0.5 O2 -> V2O5 (2 marks)
Reactants need purifying to prevent catalyst poisioning (1 mark)
E(Au+/Au) > E(O2/ H2O) So Gold not oxidised by O2- Alternatively calculate EMF ofthe cell to get -0.45V which proves reaction not feasible (3 marks)
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Q5 Hydrogen cells (9 marks)
O2 + 4e- + 2H2O ---> 4OH- Postive electrode
H2 + 2OH- ---> 2H2O + 2e- Negative electrode
Overall: O2 + 2H2 ---> 2H2O
(3 marks)
Increase pressure increase EMF (2 marks)
Graph = straight horizontal line (1 mark)
No effect on EMF for increase in surface area of Pt (1 mark)
Environmental advantage – CO2 not produced, greenhouse gas, global warming (1 mark)
Why H fuel cells may not be Carbon neutral?
H2 obtained by electrolysis of H2O
May use energy from combustion of fossil fuels
(1 mark)
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Q6 Contact process/enthalpy calcs (14 marks)
Enthalpy change -196 (3 marks)
Entropy -189 (2 marks)
What does sign show about the product compared to the reactants? Products more ordered/less disordered (1 mark)
DeltaG = -135 kJ.mol-1
Delta G is less than/= 0 therefore feasible (3 marks)
Reactants pure to prevent catalyst poisioning (1 mark)
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Q7 Chromium/inorganic RXNs (12 marks)
CrCl3 + 6H2O -> 3Cl- + [Cr(H2O]6]3+ (1 mark)
P = [Cr(H2O)3(OH)3]
NaOH / any OH-
[Cr(H20)6]2+ + 3OH --> [Cr(H2O)3(OH)3]+ 6H20(3 marks)
Q = CO2
Na2CO3 / any CO3^(2-)
2[Cr(H20)6]2+ + 3CO3^(2-) --> 2[Cr(H2O)3(OH)3]+ 3CO2 + 3H2O
(3 marks)
RXN 4 = [Cr(OH)6]3-
Excess NaOH / Excess of any OH-
[Cr(H20)6]2+ + 6OH- --> [Cr(OH)6]3-+6H2O(3 marks)
Zn + HCL
Blue(2 marks)
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Q8 Cobalt Chem (12 marks)
Co Atom is 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7
Co2+ 1s2 2s2p6 3s2p6d7
Characteristic features: complex formation, coloured compounds, var. ox state. (5marks)
[Co(h2o)6]2+ + 3NH2CH2CH2NH2 --> [Co(NH2CH2CH2NH2)3]2+ + 6H2O
4 mol reactant to 7 mol product therefore deltaS positive
DeltaH roughly 0 as same type and number of bond broken
So delta G is negative and rxn is feasible
Structure of complex ion: Co central metal, 3 NH2CH2CH2NH2’s joined via 6 coord bonds to the N atoms, octahedral shape, 2+ charge aroundentire complex (7 Marks)
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Q9 Purification calc/ Redox titration (16 marks)
Autocatalyst = a product of a reaction catalyses that reaction
i.e.Mn2+
(2 marks)
H2SO4 was the acid to tick (1 mark)
Reaction was slow as requires two negative ions to collide, which repulse each other, so high Ea. (I also just dropped in, little/no Mn2+formed so uncatalysed)
(3 marks)
Calculation:Ratio of C2O4^2- to MnO4^1 was 2:5
There were 26.4cm3 of 0.02 moldm3 MnO4- = 0.000528 mols MnO4 therefore therewas 0.00132 moles C2O4^2- in 25cm3 therefore 0.0132 moles in 250cm3,
There was 3 C2O4^2- per one mole of the salt therefore there were 4.4x10-3moles salt
Mr K3(Fe(c2o4)3).3h20 = 491.1, Mass = moles x mr = 4.4x10-3 x 491.1 = 2.16g
Original mass = 2.29 therefore 2.16/2.22 * 100 = 94.4% purity( 7 Marks) – Common wrong answer of 35% is likely to be 5/7marks
Colourimetry
Make solutions with known concs of MnO4- ions Plot calibration curve of concs to lightabsorpted/transmitted
Measure absorption/transmission of light in unknown soln,
extrapolate MnO4- conc (3 Marks)
*** Ongoing discussion as to whether to divide by two. General consensus is NO. It has been suggested that AQA won't give credit for dividing by two. Exam marking is generally standardized anyway, so it may not be a binary right or wrong way to do this question given that quite a few people have interpreted it in a different way
Preicted grade boundaries from isaisababy:
Full UMS: 93
A*: 86
A: 78
B: 70
C: 59?