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# Aqa chem 4/ chem 5 june 2016 thread watch

1. (Original post by Suits101)
The question asked you to calculate the STANDARD enthalpy and entropy change, which was asked on Jan 13 I believe and you had to divide by two.
Oh, yes, your right, it did.
2. (Original post by jjsnyder)
Just a bit of an answer dump, missing loads if someone gives me question topics I'll fill if in.
Q1.
Covalent
P
P4O10 + 6H2O -> H3PO4
Ionic
Na
Na2O + H2O -> 2NaOH
Al2O3
Reacts with an acid and a base
3H2O + Al2O3 + 2NaOH -> 2NaAl(OH)4
Al2O3 + 6HCl -> 2AlCl3 + 3H2O

Q2. Standard Born Haber stuff
Lattice dissociation = +2328? Something around this value

Q7.
Enthalpy change -196
Entropy ~-160
G was around -185
Reaction is feasible as G is less than or equal to zero

Q8. First equation was just reacting it with 6 water molecules
P was Cr(H2O)3(OH)3
Reagent was anything with OH- ions

Q was CO2
Reagent was anything with CO3 2- ions

Reaction 4 product was [Cr(OH)6]3-
Excess reagent containing OH-

Not sure what the Reagent was for the last part, final thing was a blue solution.
Q9.
Calculation was 94.4%
Colorimetry talk about MnO4- ions being coloured use the intensity of the colour to measure the concentration etc.

Unassigned:
Hydrogen-Oxygen fuel cell Q:
Overall equation 2H2 + O2 -> 2H2O or multiples

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Q8. First equation was just reacting it with 6 water moleculesP was Cr(H2O)3(OH)3Reagent was anything with OH- ions

NH3 can be used as a reagent here too!

Reagent for cr3+ to cr2+ is Zn and H+

Hydrogen fuel cell half eqn was

O2 + 4e- + 2H2O ---> 4OH- postive electrode

H2 + 2OH- ---> 2H2O + 2e- negative electrode

Overall: o2 + 2H2 ---> 2H2O

No effect if pressure increases as you are only increasing the pressure of oxygen so the same number of electron are released so same number of redox reactions??? (Not sure though)

Graph was a straight horizontal line

No effect on emf for increase in surface area of pt

Environmental advantage was h2o doesnt contribute to global warming as much as co2 and/or acid rain
3. Someone make a proper unofficial markscheme?
4. (Original post by EverydayHell)
Oh I've seen something in the 1st question about Na the equation asked for the element reacting with water not the oxide of the element I think? What did everyone else write? I did 2Na + H2O ---> Na2O + H2.
The other equation asked for the equation with oxide which is why I did that.
You are right however i did it 2Na + 2H2O -> 2NaOH + H2
5. Example 4Using the values in the above table of standard enthalpies of formation, calculate the ΔHreactiono for the formation of NO2(g).SOLUTIONNO2(g) is formed from the combination of NO(g) and O2(g) in the following reaction:
2NO(g)+O2(g)⇋2NO2(g)
To find the ΔHreactiono, use the formula for the standard enthalpy change of formation:
ΔHoreaction=∑ΔHof(products) ˆ’âˆ‘Î”Hof(Reactants)
The relevant standard enthalpy of formation values from Table 1 are:
• O2(g): 0 kJ/mol
• NO(g): 90.25 kJ/mol
• NO2(g): 33.18 kJ/mol
Plugging these values into the formula above gives the following:
Î”Horeaction=(2mol)(33.18kJ/mol)âˆ’[(2mol)(90.25 kJ/mol)+(1mol)(0kJ/mol)]Î”Horeaction=âˆ’114.1kJ
To stop people losing about 7 marks in total bad luck for the clever clogs
6. (Original post by RME11)
No you didn't? The question didn't ask about Au+ ions the question asked about the reaction between gold solid and water because they don't react at all under standard conditions- that was the nature of the question. If it asked about what would happen if Au+ ions were placed in water then you'd be right, but it didn't so you're wrong... lol.

The question didn't say anything about standard enthalpy of formation of SO3, it asked for the enthapy change of the reaction given which was the formation of 2 moles of SO3 from 2 moles of SO2 and one mole of O2.
We said the exact same thing with E (Au+/Au) > E (O2/H2O) = 1 mark

I said Au+ would react with water instead of gold hence gold does not react with with Au...

It said standard enthalpy change.
7. (Original post by chzm)
what did everyone get for the tick box acid that should be used? i put H2SO4 and for the reducing agent for chromium??

i reckon 88/89 for an A* (maybe even 90)
82/83 for an A
70 something for a B
I actually think 87 is A*, 81 is A and it caps at around 95

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8. (Original post by hi-zen-berg)
So delta G is negative and rxn is feasible
Structure is attached photo with 2+ charge around entire complex in brackets

Attachment 554912
Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
9. (Original post by Suits101)
Excellent I agree

For delta G being negative you can have <= 0 and < 0, either one is fine.

The arrows should be from N to Co (co-ordinate bond)

H2SO4

Boundaries will not be that high, people still found it hard and many stupid mistakes were made.
I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
10. (Original post by FireBLue97)
Example 4Using the values in the above table of standard enthalpies of formation, calculate the Î”Hreactiono for the formation of NO2(g).SOLUTIONNO2(g) is formed from the combination of NO(g) and O2(g) in the following reaction:
2NO(g)+O2(g)â‡‹2NO2(g)
To find the Î”Hreactiono, use the formula for the standard enthalpy change of formation:
Î”Horeaction=âˆ‘Î”Hof(products)â ˆ’âˆ‘Î”Hof(Reactants)
The relevant standard enthalpy of formation values from Table 1 are:
• O2(g): 0 kJ/mol
• NO(g): 90.25 kJ/mol
• NO2(g): 33.18 kJ/mol
Plugging these values into the formula above gives the following:
Î”Horeaction=(2mol)(33.18kJ/mol)âˆ’[(2mol)(90.25 kJ/mol)+(1mol)(0kJ/mol)]Î”Horeaction=âˆ’114.1kJ
To stop people losing about 7 marks in total bad luck for the clever clogs
The question said STANDARD enthalpy change, not enthalpy change as this question says.

Regardless, one mark for each question would be dropped when calculating values, which means -2.

Gibbs free-energy change would allow ecf meaning no marks dropped.
11. (Original post by isaisababy)
Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
yh ur 100% right mate, I just googled the image and attached it.

Anyone remembering specific marks now? MS pls someone
12. (Original post by hi-zen-berg)
Thanks a lot;
For last part of Chromium Q, Zn + Hcl were reagents

I'll just add on some stuff Beginning of section B:

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 (5 marks)

[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 is attached photo with 2+ charge around entire complex in brackets

Attachment 554912

Would it be alright to use [Ar] for the electron configurations ???
13. Which acid was it for the redox titration?
14. (Original post by thomaarrss)
i put this as well
Yea same its right because the further to the right that the equilibrium lies the more positive the voltage (emf) of the cell so because equlibrium shift left emf would decrease
15. (Original post by User2334541)
I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
Hi,

No that's wrong.

Co is [Ar] 4s2 3d7
Co2+ is [Ar] 3d7 (you lose 4s electrons first - google it)

Sorry.
16. (Original post by koolgurl14)
You are right however i did it 2Na + 2H2O -> 2NaOH + H2
Thanks you might be right there the question threw me off a bit because it the said the oxide dissolved in water, but only the mark scheme will tell.
17. (Original post by Suits101)
The question said STANDARD enthalpy change, not enthalpy change as this question says.

Regardless, one mark for each question would be dropped when calculating values, which means -2.

Gibbs free-energy change would allow ecf meaning no marks dropped.
If I could rep this man x10 then I would
18. (Original post by User2334541)
I don't agree with Co2+ ion electronic configuration, its supposed to be 4s2 3d5 and because its a transition element, that does have variable oxidation state due to its incomplete 3d sub-shell.
It's 3d7 you lose 4s electrons first

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19. (Original post by isaisababy)
Shouldnt the arrows be going towards to Co since the Nitrogens are donating the electron pairs?
was looking for this comment i drew the arrows the other way round. I drew the lone pairs as well so they should know what i mean
20. (Original post by K2Cr2O7)
Would it be alright to use [Ar] for the electron configurations ???
It didn't say "full" electron configuration so you're probably okay.

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