[AlexBamsey]

(Original post by Emissionspectra)
Ye, it get destroyed.

Oh i really liked that too I don't remember much of it but I remember liking it..

[djb2]

(Original post by AlexBamsey)
I thought the first question regarding solubility was weird. I left it and just scribbled down some rubbish at the very end, couldn't think what to do.

The solubility was easy i thought? it was the only thing on the paper i could do. times both the solubility values by 10 so it's per litre and then take the difference between them. At least that's what I put...
most of the paper i blagged. bit of a non-starter as i got an e last year.

[AlexBamsey]

(Original post by djb2)
The solubility was easy i thought? it was the only thing on the paper i could do. times both the solubility values by 10 so it's per litre and then take the difference between them. At least that's what I put...
most of the paper i blagged. bit of a non-starter as i got an e last year.

Ah i was trying to remember how many dm^3 are in a litre lol. So i just left it out, only 2 marks I guess.

[Emissionspectra]

I think i got question 1 by the sounds of it, rest of the paper was pretty good I think .

[AlexBamsey]

I wish the person with the paper would post it on here.

[Emissionspectra]

Hope you put Green and Royal Blue for the colour of the complexes

[AlexBamsey]

Yellow-green and royal blue, as it is in the revision guide. :P

[AlexBamsey]

I was shocked they didn't put anything synoptic in this paper.

[GSM.]

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(Original post by AlexBamsey)
I wish the person with the paper would post it on here.

I've got one, I'll try scanning it in now x

[AlexBamsey]

(Original post by GSM.)

---

I've got one, I'll try scanning it in now x

Wooo I love you

[GSM.]

CH5 June 2012.pptx

(Original post by AlexBamsey)
Wooo I love you

Haha :L Here it is x

[AlexBamsey]

(Original post by GSM.)
CH5 June 2012.pptx

Haha :L Here it is x

tyvm

[rsp]

(Original post by AlexBamsey)
rsp can you please scan it so we can get an unofficial mark scheme up?

Really sorry my scanner has packed in , it will only scan a quarter of the page its a HP all in one printer if anyone has any suggestions to fix it then I will be more than willing

[rsp]

(Original post by rsp)
Really sorry my scanner has packed in , it will only scan a quarter of the page its a HP all in one printer if anyone has any suggestions to fix it then I will be more than willing

Ooo just realise someone else has scanned it so all ok

[Holby_fanatic]

That exam was painful.

[AlexBamsey]

Markscheme:

1)
a)35.4g

b)
i) 0.1 mol of potassium persulfate.
2:1 molar ratio so 0.05 moles of oxygen.

1 mol gives 29dm^3
1/0.05 = 20
29/20 = 1.45dm^3

ii) You could use a gas syringe to measure the volume of gas produced with respect to time.

c)
i)Pt electrode is placed in the solution to provide electrons for the reduction

ii) 2.01 - 0.54 = 1.47V. This positive EMF means the reaction is feasible, it means the electrons will move from the left to right, as the S208 2- is a more powerful oxidising agent than the iodine half cell.

d)
i) As you double the concentration of both of them, the rate increases by a factor of 4, and so as the iodine is order one, the persulfate must be order 1 also, making a total of order 2 of the reaction.

ii) Rate = k[I-][S208 2-]

k = Rate/[I-][S208 2-]
= 8.64x10-6 / (0.04 x 0.01)
= 0.0216
Units are mol^-1 dm^3 s^-1

iii) Step 1 must be the rate determining step. There must be one mole of S208 2- and one mole of iodide ions, and step 1 coincides with this.

c)
i) When sodium iodide is added to concentrated sulfuric acid initially you see misty fumes of HI and this results in the formation of SO2 which is a gas, so bubbles will be seen. Then a yellow solid of S is made, and then bubbles of H2S.

ii) As the iodide has a lower electrode potential value, this means it is a stronger reducing agent than the chloride ions. This means that it is powerful enough to reduce the sulfur in the H2SO4 all the way to H2S whereas with chloride ions, they can only reduce the sulfur to HSO4- due to being a weaker reducing agent.


2)
a)
-288 - 417 = -705kJmol^-1

b)
i)

ii)Adding alkaline would make the equilibrium shift to the left, thus reducing the amount of yellow iodine.

d)
i) Cl2 + 2NaOH -> NaCl + NaClO + H20

ii) It is used in bleaches, or is a mild antiseptic.

3)
a)
i) 19.2/1000 x 0.01 = 1.92 x10^-4

ii) It's a 1:1 ratio so 1.92x10^-4 moles of copper ions.

iii) Mass of copper = moles x mr = 1.92 x10^-4 x 63.5 = 0.0122g

Concentration in gdm^-3 = mass/volume = 0.0122 / 50/1000 = 0.244 gdm^-3

iv) The mass of the entire sample was 11.56g and the mass of copper was just found to be 0.0122g.
0.244/11.56 * 100 = 2.11%

b)
A transition metal is defined as a metal which it's d orbitals are partially filled in it's atom or ion. In copper it's +2 oxidiation state, it has a partially filled D orbital (3d9). In zinc even it's +2 oxidation state, it has a full d-orbital (3d10) which means it is not a transition metal.

c)
Shape and colour of [CuCl4]2- Tetrahedral and yellow-green.

Shape and colour of [Cu(NH3)4(H2O)2]2+ Octahedral and royal blue.

d) Both oxides are fairly stable as they both have negative enthalpies, meaning their products have less energy than reacants, so are less reactive and more stable. PbO has an even lower enthalpy than CuO so is even more stable than CuO.

e)
i) They have variable oxidation states which allows them to oxidise a species to make it more reactive.
They can also reduce a species for the same reasons above.

ii) It can reduce the amount of energy required in a reaction which could make it more efficient and reduce the energy needed.

4)
a) In CO the carbon has an oxidation state of +2, and in CO2 it has an oxidation state of +4. It is being oxidised, and so is acting as a reducing agent, while the iodine in I205 is being reduced.

b) The +2 oxidation state increases down group 4, and the +4 oxidiation states decreases down group 4.

c)
i) Both Al3+ and Pb2+ will react with NaOH to form white precipitates of Al(OH)3 and Pb(OH)2 respectively. However you can then add dropwise, NaOH until it is in excess, and both of these precipitates will re-dissolve which means the aluminium and lead are both amphoteric.

Al(OH)3 = aluminium hydroxide
Pb(OH)2 = lead hydroxide
NaOH = Sodium hydroxide

ii) A yellow precipitate of PbI2 is produced.

Pb2+ + 2I- -> PbI2

d)
i) In AlCl3 the aluminium only has 6 electrons in it's octet, which is not full. This means it is electron deficient. If two molecules of AlCl3 come together, a lone pair from a chlorine from each molecule will form a dative bond with the aluminium of the other molecule, which gives the aluminium 8 electrons in it's octet, i.e. making it full and more stable.

Drawing: Just draw it..

ii) AlCl3 can be used to produce a Cl+ ion in the chlorination of benzene.

iii)
I) You go from one mole of gas to two moles of gas meaning there is more freedom of movement for the molecules, thus increasing the entropy.
II) Let G = 0

T = DeltaH/DeltaS
= 60/0.088
= 681.81 K
= 682K

d) As it is a weak acid, the [H+] = [Al(H20)5(OH)]2+
Thus..

Kc = [H+]^2 / [Al(H20)6]3+]
[H+] = square root of (Kc x [Al(H20)6]3+)
= square root of (1.26 x 10^-5 x 0.1)
-log(ans) = 2.95


5)
a)
i) Kp = P(SO3) x P(NO) / P(NO2) x P(SO2)

It has no units.

ii) The part shaded is dynamic equilibrium and is thus incorrectly labelled

From the graph also, if you take 0.6 for the pressures of the reactants and 0.4 for the pressures of the products, and work out kP, you get 0.4 and not 2.5 as given in the question, so the values must be wrong.

iii) The forward reaction is exothermic. Increasing the temperature would cause the equilibrium to resist this change and follow the endothermic route, or the reverse reaction in this case. This would result in a smaller value of Kp as more reactants would be produced.

b)
i)The pH of the graph starts at 1 due to the nitric acid being a strong acid. At around 12.5 cm^3 oof ammonia added the pH slowly begins to increase. At 20cm^3 of the ammonia added, the equivalence point is reached as the ammonia has neutralised the nitric acid. After 45cm^3 of ammonia being added, the pH is around 11.6 and this is due to ammonia being a weak acid.

ii) It is a 1:1 ratio because after 20cm^3 of the ammonia being added, which is the same concentration as the nitric acid, the equivalence point is reached.

iii)
I)
II) It's around 6.6

iv) It would remain blue. The pH of the vertical region is 6.6 which is above 4.7 and the table stipulates that this would be a blue colour.

c)
Moles of ammonium nitrate in 200g solution is mass/mr = 40/80 = 0.5.

1 mol of ammonium nitrate dissolved to make 1000g solution produces 6.2oC drop in temperature..

There are 0.5 moles in 200g
So in 1000g there are (0.5x5) = 2.5 moles of ammonium nitrate

2.5 * 6.2 = 15.5oC temperature drop.



DONE! Woohoo lol. Sorry if i missed any out, it means I didn't know. Also please let me know if anything is wrong, and also don't take this as 100% right, as it is unofficial afterall.

[dave101]

2a) No idea how you got this value, you seem to have gone the correct way along some arrows, and then the wrong way along others without changing values, i believe the answer to be - 705, consistent with the idea lattice enthalpy of formations is always highly negative
2aiii) adding acid would mean more H+ ions and thus shift equilibrium to the right, this is wrong, add alkali, the OH- ions will shift the equilibrium to the left in order to create more H+ ions to neutralise the acid.
3iii) You have worked out the mass of copper in the solution that was titrated, your concentration in grams is correct, but this is the value you must carry forward to the next question. The total was 11.56 g, and you have 0.244g per dm, therefore percentage = (0.244/11.56)*100=2.11 percent, you have carried the wrong value forward
3e) They have variable oxidation states will probably be worth one mark, what you then go on to describe is kind of the same thing, the other mark will probably be for stating they have vacant/ partially filled d orbitals which allows that to take place, you may get full marks for that anyway
5aii) Right idea, wrong wording, the whole graph is in equilibrium, the part shaded is dynamic equilibrium and is thus incorrectly labelled


Other answers: 1a) 35.4g

[AlexBamsey]

(Original post by dave101)
2a) No idea how you got this value, you seem to have gone the correct way along some arrows, and then the wrong way along others without changing values, i believe the answer to be - 705, consistent with the idea lattice enthalpy of formations is always highly negative
2aiii) adding acid would mean more H+ ions and thus shift equilibrium to the right, this is wrong, add alkali, the OH- ions will shift the equilibrium to the left in order to create more H+ ions to neutralise the acid.
3iii) You have worked out the mass of copper in the solution that was titrated, your concentration in grams is correct, but this is the value you must carry forward to the next question. The total was 11.56 g, and you have 0.244g per dm, therefore percentage = (0.244/11.56)*100=2.11 percent, you have carried the wrong value forward
3e) They have variable oxidation states will probably be worth one mark, what you then go on to describe is kind of the same thing, the other mark will probably be for stating they have vacant/ partially filled d orbitals which allows that to take place, you may get full marks for that anyway
5aii) Right idea, wrong wording, the whole graph is in equilibrium, the part shaded is dynamic equilibrium and is thus incorrectly labelled


Other answers: 1a) 35.4g

Thank you! On 3iii why would you use the concentration not the mass? Just wondering

Have editted the others.

[AlexBamsey]

(Original post by dave101)
2a) No idea how you got this value, you seem to have gone the correct way along some arrows, and then the wrong way along others without changing values, i believe the answer to be - 705, consistent with the idea lattice enthalpy of formations is always highly negative
2aiii) adding acid would mean more H+ ions and thus shift equilibrium to the right, this is wrong, add alkali, the OH- ions will shift the equilibrium to the left in order to create more H+ ions to neutralise the acid.
3iii) You have worked out the mass of copper in the solution that was titrated, your concentration in grams is correct, but this is the value you must carry forward to the next question. The total was 11.56 g, and you have 0.244g per dm, therefore percentage = (0.244/11.56)*100=2.11 percent, you have carried the wrong value forward
3e) They have variable oxidation states will probably be worth one mark, what you then go on to describe is kind of the same thing, the other mark will probably be for stating they have vacant/ partially filled d orbitals which allows that to take place, you may get full marks for that anyway
5aii) Right idea, wrong wording, the whole graph is in equilibrium, the part shaded is dynamic equilibrium and is thus incorrectly labelled


Other answers: 1a) 35.4g

Also, were the other answers correct as far as you can see? If so I've had around 60 marks, yay

[Tulian]

What was the ratio question towards the end ? I think I got 4/9 , which I had no clue about doing though !