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

Last q ran out of time and wrote the molecular formula for A and B w . 2H2O after the molecular formula for A and Same for B without the additional . 2H2O
Got C and D fine
how many marks dyou think that suffices...
how dyou think the marks are allocated for that question

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

For the conjugate acid-base pair I've put down CH3COOH as the acid? Is it wrong?
And for that graph I said that the line shows that as the temperature increases, the feasibility decreases because the value for delta G is >1 with a positive value. This shows that as the temperature increases, the reaction is not feasible, no matter the changes in entropy and enthalpy. How many marks will I get out of 4?
I've also put down colorimeter for the question asked about another method to calculate volume of gas?

I suspect it will be like 80 for an A*.
People made too many errors for the A* boundary to be 89.

What are your predictions for A,B,C etc. I know in the linear qualification the grade boundaries for individual papers are less significant but I would still like to know

I don't understand what you mean by "less significant"

UMS marks have been removed so the mark that will determine the overall grade is the overall subject-level mark.

well I see the grade boundaries as lower than in previous years

something like

80% A*
75% A
68% B
58% C

etc

but there is no controlled assessment to up people's grade so I struggle to find the grade boundaries at 89 and 80 for an A* and A respectively

I would agree. B could be 65-68 I think.

According to Wikipedia, you still have to put square brackets around a neutral complex

I think the fact that people make mistakes from AS chemistry and also the fact that people didn't have enough time to revise for the Year 2 chemistry and the host of other factors will pull it down.

I'm still in shock how the admin thinks it's 89 for an A* because that exam was very very strange in my opinion for inorganic chem mind you I do think I did very well but still 89 for an A* would be a record and the topics such as the gibbs free graph, the 6 markers based on quality of writing and analysis, and acid errors only points to lower marks.

Very few people I know could understand that graph or the AlH4- bonding etc, and I mean very few. that's 5 marks already out the door and add that with the 6 marker from the Nickel analysis, the 2 marks from the rates equation, acid chemistry being awkward to people for around 3 marks, kP being written using solids for 1 mark and lack of people maybe justifying the equilibrium question as per year totals around 80 for an A* and 75 for an A, and yes like you said 65 for a B.

See I put 963K too because I got 962.02 so minimum temp must be above this. In previous papers I have sometimes seen two answers acceptable, hopefully thats the case this time

does anyone have the multiple choice answers ?