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AQA CHEM5 A2 Chemistry - 19th June 2013 watch
- 20-06-2013 01:29
- 20-06-2013 01:30
Grade boundary predictions? Lol why is everyone ignoring my comments x
Posted from TSR Mobile
- 20-06-2013 01:31
- 20-06-2013 02:00
what did ppl get for Phosphorus with soduim hydroxide
- 20-06-2013 02:01
So slightly lower much than Jan 13
- 20-06-2013 02:08
The grade boundaries won't change too much. I remember everyone thinking june 12 was horrific last year and they pretty much stayed the same (and tons of people complained to AQA.
- 20-06-2013 02:11
ISNT 7a WRONG PPL
- 20-06-2013 03:34
Posted from TSR Mobile
- 20-06-2013 08:21
do you think for question 7 they will accept
[fe(h20)6]2+ + 2NaOH --> Fe(h20)4(OH)2 + 2H20 + 2Na+
I think you were just supposed to put ionic oh-
(Original post by esl_94)
- 20-06-2013 08:26
Sorry for the delay - I think these are Questions 6 and 7 Attachment 228381Attachment 228382Attachment 228383Attachment 228384
Again if I have missed any pages just let me know I seem to have 18 photos on my phone and have put 17 up here but I can't work out which I have missed!
Cant you just put up the whole paper??
- 20-06-2013 08:27
What's this question with substance Z (Co(NH3)2 cl2) ?? I don't remember it at all
- 20-06-2013 08:29
- 20-06-2013 09:29
Isnt question 2(b) 26 kj mol???
The enthalpy of hydration for the potassium ion is -322 and the dissociation for potassium bromide is +670, calculate the enthalpy of solution of potassium bromide:
enthalpy of solution = lattice dissociation + sum of hydrations
+26 = +670 + (-322 x 2)
am I wrong??
(Original post by nimbusquaffle)
- 20-06-2013 09:35
2a) Cl- ion is a smaller ion than Br- so has greater electronegativity. water is polar so contains a ~+ H+ ion. therefore Cl-is more attracted to H2O than Br- so more energy is released when hydration happens
c) increase in entropy and 1mole ---> 2moles. more disordered. ΔH<tΔS. therefore ΔG<0 nd feasible.
d) have no idea!
3a)i at 0K all substances are completely ordered. moles are constant with no vibrations. absolute entropy.
ii as temperature increases entropy increases, molecules vibrate more violently becoming more disordered
iii in line with L2
iv phase change causes a sudden increase in entropy. boiling has a greater increase in entropy than melting so L2 is greater than L1.
b)i ΔG=ΔH-TΔS can be rearranged to ΔG=-ΔST + ΔH. ΔS is m (gradient) ΔH is the constant c. ΔG is y and T is x. therefore follows the same equation. ΔS is negative due to negative gradient.
ii as T increases TΔS, because ΔH is constant. as T increases ΔH - TΔS decreases, ΔG decreases.
iii becomes feasible
c)i -44.5 Jk-1mol-1
ii 242 kJmol-1
hopefully help, correction ma be needed!
3 part 3, its unfeasible below 500K as G is positive.
- 20-06-2013 09:41
Guys do you think they would accept [CoCl3(NH3)] as in all fairness, knowing the coordination number of such a random complex is not on the spec?
(Original post by AspiringGenius)
- 20-06-2013 10:03
I found ti quite hard. I know I messed up the qmcDT equation and some of the cobalt chemistry equalling about 10 marks. Although if grade boundaries are low and I did well enough in Chem4 and the EMPA, An A* is achievable (I hope).
- 20-06-2013 10:06
(Original post by Croydon'Stepper)
- 20-06-2013 10:06
2a, you don't talk about electro negativity as it's not a covalent bond.
2b I did the correct method, got -7kJmol-1, but I think I ****ed up the actually calculations
Other than that, got the same as you!
- 20-06-2013 10:16
i forgot to turn to the back page
- 20-06-2013 10:18
"This is nearly complete unofficial markscheme:
Q1: a) what is the perfect ionic model (1)
Ions are perfect spheres, only electrostatic attraction/no covalent character
b) define 'standard enthalpy of lattice dissociation' (2)
The standard enthalpy change to convert one mole of a solid ionic lattice into its gaseous, constituent ions
c) Give 2 factors that affect the lattice enthalpy of dissociation (2)
Size of the ion/ionic radius
Charge on the ion
d) calculate the lattice enthalpy of dissociation for AgBr (3)
+905 kJ mol-1
e) How would you expect your value to 1d) to differ with that calculated from the PIModel. Explain (2)
as there is additional covalent character/bonding
tot = 10
4a) What is the bonding in MgO. How could you prove that MgO has this bonding (3)
Heat until molten
Conducts electricity/gives a current when being electrolysed
b) Why is SiO2 insoluble in water (3)
many, strong covalent bonds
lots of energy needed to be supplied to break bonds/ enthalpy of solution is very large
c) What has a higher melting point - SiO2 or P4O10? Explain (3)
P4O10 is a simple covalent molecule
Weak VdWs (and dipole-dipole) IM forces
Less energy needed to be supplied to break VdWs forces
d) Write an equation to show that MgO is basic (2)
MgO + 2HCl => MgCl2 + H2O
e) Write an equation for the reaction of P4O10 with MgO (1)
P4O10 + 6MgO => 2Mg3(PO4)2
tot = 12
5a) why can KCl not be used as a salt bridge (1)
It is not inert/it reacts with the cell reagents/Cl- oxidized to Cl2 by Cu2+
b) why do the electrons flow from right to left (2)
LHS has a higher conc. of Cu2+ / LHS is positive electrode
Equilibrium Cu2+ + 2e- => Cu lies more to right at the left hand electrode (or converse)
c) Why does the current fall to zero after some time (1)
Cu2+ concentration is the same in both half cells
d) calculate the electrode potential (1)
e) Write the equation for the cell when it is being recharged (2)
LiMnO2 => Li + MnO2
f) Give one reason why this cell produces CO2 (1)
CO2 released generating the electricity needed to recharge the cell
tot = 8
6a) calculate the frequency of visible light absorbed and state the units (2)
4.28x10^14 s^-1 or Hz
b) Why does this compound appear blue (2)
d-orbital electrons absorb visible light energy and are promoted/excited to a higher energy level;
we observe the blue colour that is not absorbed;
c) Red/blue complex. What is the deltaE value of a red solution in comparison with a blue one? (2)
absorbs blue light which is of a higher frequency;
d) State 3 factors that affect the frequency of visible light absorbed (3)
tot = 9
7) original colours, final observation, equation for each reaction
a) [Fe(H2O)6]2+ and [Fe(H2O)6]3+ with sodium carbonate (5 marks)
[Fe(H2O)6]2+ = pale green solution
observation = green ppt.
[Fe(H2O)6]3+ = violet/yellow/brown solution
observation = brown ppt and effervescence
[Fe(H2O)6]2+ + CO32- => FeCO3 + 3CO2 + 3H2O
2[Fe(H2O)6]3+ + 3CO32- => 2[Fe(OH)3(H2O)3] + 3CO2 + 3H2O
b) [Cu(H2O)6]2+ and [Co(H2O)6]2+ with HCL (4 marks)
[Cu(H2O)6]2+ = pale blue soln.
observation = green/yellow soln.
[Co(H2O)]2+ = pink soln.
observation = blue solution
[Cu(H2O)6]2+ + 4Cl- => [CuCl4]2- + 6H2O
[Co(H2O)6]2+ + 4Cl- => [CoCl4]2- + 6H2O
c) [Cr(H2O)6]3+ and [Fe(H2O)6]2+ with excess OH- (4 marks)
[Cr(H2O)6]3+ = green or ruby soln.
observation = green soln.
[Fe(H2O)6]2+ = pale green soln.
observation = green ppt.
[Cr(H2O)6]3+ + 6OH- => [Cr(OH)6]3- + 6H2O
[Fe(H2O)6] + 2OH- => [Fe(OH)2(H2O)4] + 2H2O
d) [Al(H2O)6]3+ and [Ag(H2O)2]+ with excess NH3 (4 marks)
[Al(H2O)6]3+ = colourless soln.
observation = white precipitate
[Ag(H2O)2]+ = colourless soln.
observation = colourless soln/no visible change
[Al(H2O)6]3+ + 3NH3 => [Al(OH)3(H2O)3] + 3NH4+
[Ag(H2O)2]+ + 2NH3 => [Al(NH3)2]+ + 2H2O
tot = 17
8a) ethanal can be oxidised to ethanoic acid by O2. Explain why Co2+ ions can increase the rate of this oxidation. Write 2 equations showing how Co2+ does this. (4 marks)
provides alternative mechanism/route with a lower Ea;
as has variable oxidation states;
2Co3+ + CH3CHO + H2O -> 2Co2+ + CH3COOH + 2H+
4Co2+ + O2 + 4H+ -> 4Co3+ + 2H2O
b) [Co(H2O)6] reacts with an excess of ethane-1,2-diamine. Write an equation for this reaction and explain why ethane-1,2-diamine is the more stable complex (3 marks)
more moles of products than reactants
Increase in disorder
Entropy change large and positive
c) Draw the complex ion. (3 marks)
arrows showing coordinate bonds from N atom to Co;
d) Cobalt ratio question (5 marks)
don’t know :P
tot = 15
[Fe(H2O)6]2+ + CO32- => FeCO3 + 3CO2 + 3H2O is wrong its:
[Fe(H2O)6]2+ + CO32- => FeCO3 + 6H2O
Can somebody finish this off"