AQA CHEM5 unofficial MS

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
(3 marks)

Al2O3
Ionic
Reacts with acids and bases
3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
Al2O3 + 6HCl -> 2AlCl3 + 3H2O
(4 marks)

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)

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)

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 (2 marks)V2O4 + 0.5 O2 -> V2O5
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)

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 thereactants? Products more ordered/less disordered (1 mark)DeltaG = -135 kJ.mol-1Delta G is less than/= 0 therefore feasible (3 marks)Reactants pure to prevent catalyst poisioning (1 mark)Q6 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 = CO2Na2CO3 / 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 + HCLBlue(2 marks)

Q7 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)



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?

nice one. u going for A*?
i think ill just be happy to sit on my AS UMS and get an A. A-levels have tired me out now

nice one. u going for A*?
i think ill just be happy to sit on my AS UMS and get an A. A-levels have tired me out now

I was hoping for an A*, although I'm not so sure after that horrible Chem4. And yeah aha I agree, I have just 2 more left and then freedom!

Is it not Zn2+|Zn||...
And the electron configuration of Co -1s2 2s2 2p6 3s2 3p6 3d9 ???

4s orbital gets filled before 3d

Does 3d not get filled first in all transition metals?

Is it not Zn2+|Zn||...
And the electron configuration of Co -1s2 2s2 2p6 3s2 3p6 3d9 ???

Oxidation on the left hand side right? So I think it was Zn|Zn2+||Co2+|Co

Honestly not sure about H2SO4 mate its not the standard answer

Hmm I also am unsure about the mno4- thing. I think that sounds a bit convoluted for what they asked. Fingers crossed for 2/3 marks if you have written the method and got the principle of a calibration curve and extrapolating

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)
____________________________________________________
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)
____________________________________________________
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)
____________________________________________________
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)
____________________________________________________
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)
____________________________________________________
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)
____________________________________________________
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)
____________________________________________________

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)
____________________________________________________


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?

For the first question for P i put simple molecular as I wasn't thinking straight instead of covalent will I still get the mark?

Will the one about solubility be CE= 0 if we said decrease but did the correct explanation?

2/3

2/3

Are you sure ?:0