OCR Chemistry A paper 1 (very) unofficial mark-scheme 14th June 2017

12/3/18 Update: I, as per forum rules, will not be distributing the papers I acquired after my exams. However, if people sat them for a mock or otherwise attained them, and would like some worked examples I'd be more than happy to help if you send me a PM, granted I can find the papers when i return home from Uni at the end of this term (27/3/18).

I've had a second look at the paper. here's a few of my answers (not in order):

AlH4: 3 covalent bonds and one dative bond
NH4+ - Tetrahedral, 109.5
NH2- - non-linear, 104.5
NH3>PH3 because H bonds. AsH3>PH3 as more E's so more dipole-dipole interactions so stronger London forces.
Group 2 metals increase in reactivity as shielding increases, atomic radius increases, nuclear charge increase outweighed by inc. in radius and shielding so outer electrons less attracted by nucleus so easier to remove hence more reactive.

Ba(NO3)2(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaNO3(aq) | Entropy decreases as solid formed which is more ordered
1/2 I2(s) -> I(g) | Entropy increases as gas formed which is less ordered/more disordered
Acid/base pairs: CH3COOH(B2) + H2FCCOOH(A1) -> H2FCCOO-(B1) + CH3COOH2+(A2)
Kp = p(CO)^4
Kc = [SO3]^2/([SO2]^2[O2]) dm^3mol^-1, 2.45 mol so3

pH = 4.50002 (this is right, 4.86 IS wrong.) pH is constant on dilution as [H+] = Ka ([HA]/[A-]) and conc's cancel out the volume.
-110.5Kj/mol for Hf of CO
690cm^3
first order, rate at t=0: 1.84*10^-3, k=7.7-8*10^-4 units: s^-1 or per second (range of answers as from graph) halflife was about 900ish. graph had weird jumps so ppl might have messed up.
962 Kelvin, 467KJ for at 25 degrees which > 0 hence not feasible.
Disproportionation is when one species is both oxidised and reduced: Cl reduced from +5 to -1. Cl oxidised from +5 to +7.
Potassium Chlorate(VII) [maybe they'll give potassium perchlorate but that isn't the systematic name afaik.]
[Cu(H2O)6]2+ : with HCl = [CuCl4]2-, with dropwise NH3 = Cu(oh)2 with excess NH3 = [Cu(NH3)4(H2O)]2+ and with KI was I2(brown solution) and CuI(white ppt)
named reactions for above: Ligand substitution and Redox.
Free Energy question: y = mx + c : G = -S(T) + H, explain grad = -S, y int is c = H
Le chatelier's principle: Forward reaction was endothermic so high temperature favors it, forward reaction was going to more moles of gas so a low pressure favors forward reaction. Actual conditions may vary as cost of maintaining high temperature and low pressure is high so a catalyst is used in favor of 'ideal' conditions.
0.753 for something.
Zn was best reducing agent, MnO4- was best oxidising

https://gyazo.com/b77d7fda10b60699b9873d28a6417ff8
All solutions 1mol/dm^3, 298K

last question: DISCLAIMER: 'en' is NH2CH2CH2NH2 just wrote en for speed
A = [Ni(en)3]cl2*2H2O
B = [Ni(en)3]cl2
C = [Ni(en)3]2+ - correct optical isomers drawn
D = en = H2NCH2CH2NH2

if you can remind me of any questions I might remember my answer. I know what I put for Multiple choice but not what letter they were.
PS: these are not canon - they are MY answers.

MY grade boundary predictions:
89 - A*
80 - A
72 - B
64 - C
56 - E

Topics to revise for paper 3:
Born Haber
Titration with I2/(S2O3)2- and MnO4-
Arrhenius Equation/temperature effect on Kc
Cation and Anion tests (Cr3+, Fe2+, Fe3+, Cu2+, Mn2+, NH4+ | CO32-, SO4^2-, Cl-, Br-. I-, order for anions is important!
Relative reactivity of Halogens (Cl>Br>I)
Enthalpy
Boltzmann
Calculating Kp from initial amounts + kp units

(edited 6 years ago)

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Reply 1

eydalef

do you remember the marks for each question?

Reply 2

Toffo132

Original post by eydalef

do you remember the marks for each question?

Give me a question and i'll tell you what I think it was, not gonna be 100% sure though

Reply 3

numan1234

Chromate or chlorate?

Reply 4

username2485161

was it potassium chlorate (V) ? Would have only been a mark though if I messed this up

Reply 5

Toffo132

Original post by numan1234

Chromate or chlorate?

Chlorate. Thanks!

Reply 6

Toffo132

Original post by maoism123

was it potassium chlorate (V) ? Would have only been a mark though if I messed this up

(VII). They sneakily asked for the systematic name of KClO4!

Reply 7

username2485161

Original post by Toffo132

(VII). They sneakily asked for the systematic name of KClO4!

Ahhh **** I thought it was KClO3, ah well should only be one mark right? I can't remember the question with 0.753 at all can you remember the numbers involved?

Reply 8

Toffo132

Original post by maoism123

Ahhh **** I thought it was KClO3, ah well should only be one mark right? I can't remember the question with 0.753 at all can you remember the numbers involved?

It was just one mark yes. I think it was concentration or moles of CH3COONa? something to do with that question.

Reply 9

Gaby123

Original post by maoism123

was it potassium chlorate (V) ? Would have only been a mark though if I messed this up

I put that too!

Reply 10

karunvsn

for the last one I wrote NH2C2H4NH2 in all my answers instead of NH2CH2CH2NH2, would I get a mark deducted for this?

Reply 11

DoubleDoors

Original post by Toffo132

400ish for at 25 degrees which > 0 hence not feasible.

I got 467K

Reply 12

Toffo132

Original post by karunvsn

for the last one I wrote NH2C2H4NH2 in all my answers instead of NH2CH2CH2NH2, would I get a mark deducted for this?

I'm not sure. it explicitly said show the structure, but I don't think you're chemically wrong in writing it like that. I personally would give you the BOD and mart it fully correct, but I'm not a marker nor a teacher.

Reply 13

DoubleDoors

Enthalpy of formation for (think it was CO?)

-111 kJ mol^-1

(given to 3sf because enthalpy change was only to this number of sf)

Reply 14

Claireharg

Original post by Toffo132

It was just one mark yes. I think it was concentration or moles of CH3COONa? something to do with that question.

I don't remember this question either. Do you remember other questions on this page? I feel like I've missed it out

Reply 15

Toffo132

Original post by DoubleDoors

I got 467K

For the temp at which feasibility changes it was:
G=0 so rearranging gives: T = H/S
T = (~674.3*1000)/(701) = 961.9=962K
[THESE NUMBERS ARE ROUGHLY WHAT WAS GIVEN NOT EXACTLY!]

Reply 16

DoubleDoors

Original post by Toffo132

For the temp at which feasibility changes it was:
G=0 so rearranging gives: T = H/S
T = (~674.3*1000)/(701) = 961.9=962K
[THESE NUMBERS ARE ROUGHLY WHAT WAS GIVEN NOT EXACTLY!]

I also got that :smile: