Equilibria, Energetics and Elements (F325) - June 2011 Exam.

Oh sorry, I meant the book you posted the pic of earlier. Is that available on pdf? And does that book also have the stretch and challenge questions?

Are you using pencil for your diagrams? Or should we draw our diagrams in pen? Did you know that examiners don't award marks for diagrams, especially the mechanisms unless it looks exactly like how it is in the textbook...a few friends lost marks because some of their diagrams did'nt get detected when the script was scanned. I got an A for my A2 Practical, so I'm certainly able to achieve the A*, although it is F324 that is going to make or break my A* grade.

Would highly advise you to attempt all the Past Questions, Trends and Patterns, Transition Elements and Unifying Concepts. The only difference between those old questions and the new is the deeper understanding, thinking and applying ones knowledge. The old papers were merely a matter of doing as many past questions and exam techniques. The new spec questions at A2 especially so are ones which are intentionally set to allow far less exam techniques, but more on application of knowledge. Calculations will be relatively straightforward marks, the Examiners Report cites how Calculations were generally very well done.

Right. So anyone fancy explaining how you get the answer to this redox Q?

Using Oxidation states construct a redox equation to :
Aqueous thiosulphate ions, S2O3 2-, are oxidised to sulphate(VI) ions, SO4 2-, by chlorine gas, which reduced to chloride ions.

The answer: S2O3 2- + 4CL2 + 5H2O -----------> 2SO4 2- + 8Cl- + 10H+

The 2 half equations might help if possible with explaining. I understand the electron balancing but not in this question :/

ok 1st S2o3- + cl2 --------> SO42- +CL- write the species that are in the reaction
2nd indentify oxidation of chlorine so:
0-----------------------> -1 on the LHS there 2 cholrine mocules so the balance it times cl by 2, so you have 2CL-, you also multiply the oxidation numbers so: 2*0-----------------------> -2 (cos -1*2 u get me)
diff in oxo numba is: A -2
oxo numba in s2o32- is +2------- +6 but u have to balance tne number of s, multiply s on rhs by2 and oxo numbas so:
+4--------+12 difference in oxo numba is
+8 balance this so you times so that you have -8 difference for chlorine A SO YOU need 8CLl- so times cl2 by 4 and 2cl- by 4 so effectively you get 8cli, ie 8 electrons which balances the electron difference for S ( from 4 to 12- difference of 8 electrons)
so for cholrine it will be 4CL2---------> 8CL- difference of 8 electrons ( 1 cl2 needs to eletrons to form 2cholride ions).

now your equation will be s2o32- + 4cl2-----> 2so42- +4cl2 = c (u notice the sulfur n chlorine are balanced. Now your merely add the H+, or O, H2o to balance out the oxygen's and hydrogen's

from C on lhs u have 3 oxgens on rhs side u have 8 oxygens , by convention u cannot add o on its own, so if you add 5H2O you balance the oxgens and you left with 10h+

so your overall equation will be
S2032- +5H2O + 4CL2------> 2SO42- 8CL- + 10H+
HALF EQUATIONS
they will always give u an unbalanced half equations TO WORK OUT OVERALL.
to start off with unbalanced half equations
4cl2 + 2e- ---------> 2cl-
S2O32- +5H20 + e- --------> SO42- +1oh+
les start by solving cl ok 4cl2 plus 8 elctrons and u form 8 ions-, each gainin an e-
4cl2 +8- -----> 8cl-
with sulphur you have
+2 ------------------> +6 bu there is 2 on LHS of S and 1 on rhs so times SO42- by 2 as well as both of the oxidation numbers now you have
s203- + 5h20 ---------> 2so42- + 10 h+ + 8e- in terms of oxo . you have
+4 --------------------> + 12 it has gain eight electrons that is balanced with chlorine so just cancel out elctrons in bold to get overall equation.

could anyone please explain me how to do 4bii on january 2011 paper? It is only worth one mark but I don't understand how they get the equation

I kinda worked backwards on that one:

So we know one half equation is:
o2 + 4h+ + 4e- --> 2h2o

and the overall is ch3oh + 1 1/2 o2 --> co2 + 2h2o (complete combustion)

so, when the first equation is combined with our second, we should get complete combustion with some cancelling out. So first things first, its definitely going to be

ch3oh ---> co2 (stuff that are missing)

We need one and half oxygen in overall equation, so multiply the original half equation by 1.5 to get:

1 1/2 o2 + 6h+ + 6e- --> 3h2o

6h+ + 6e- are not in the overall equation, so add them to the other side of the second half equation so they cancel out and we get:

ch3oh ---> 6h+ + 6e- + co2

Now if we add the two half equations, there is an extra h2o on the right hand side of the overall equation, so we need to add a h2o to the left of the equation we're working out so they cancel

ch3oh + h2o ---> 6h+ + 6e- + co2

Hope thats ok...

Thank you

Yes, I find it to be far more effective than the Heinemann one. Simply so because there are answers to them. They could set any kind of questions, but the Extension material and Stretch and Challenge questions are supposedly similar to the expected questions in the exam...At the conference the examiners strongly recommended we buy their textbooks :P lol Although it must be said some questions in the AS book featured in the AS exams

http://www.mediafire.com/?1wd9uec8nc7sraf Enjoy Stretch and Challenge questions really are 'challenging'

which Unit do you think is harder? Unit 4 or 5? Even though I got 88/90 for Unit 4, I first thought that Unit 5 is quite straightforward and much easier because I always get 95+%s in the old past papers. But you are scaring me now, with saying that most of the questions are challenging?!?? How challenging are they going to be?? Because I done Unit 4 in Jan and found it ok.

And is Unit 5 a synoptic Unit? Does that mean we can get asked anything? e.g. substitution of bromine with chlorine (unit 1 stuff)???

Did you do all the old past papers for Chains and Spectroscopy and Methods of Analysis & Detection? 88/90 is quite an impressive result, and is indeed an A*, as you need to average 270+ UMS you only need a good A in F325 and you've got the A* Unit 4 is harder in the sense you either know it or don't know it. Unit 5, calculations at least you can gain some marks and definitions. Unit 5 is a third Synoptic, and hence we're expected to know the AS content as well as the F324 content. So pretty much anything in all the units as well as F325 could appear in the exam. Equilibria and Energetics theory from AS Unit 2 is expected in Unit 5, as is the spdf electronic configurations and the Redox theory and moles calculations etc Hess's Law content from AS is assumed for Born Haber...ie When revising for F325, you really must review your AS notes and your F324 content throughly. My tutor kept stressing the importance of relating the AS content to the A2 content, as some questions will be Synoptic...although it cannot be assumed from the lack of Organic Chemistry in F325 papers so far that there will not be in this Summer's F325. Kinetics, Organic Reaction Mechanisms, Equilibria, Ionic equations were in AS, but they're also in F325...lots of Calculations...Acid Base titrations/curves..Entropy is a relatively new topic, and is'nt in the old spec...it relates to the Energetics content in F322. Isomerism occurs in Organic compounds but also in some Inorganic compounds...so F325 may well appear easier than F324, but it is actually more demanding as a third of it is Synoptic.

I was able to cope with Unit 4 and It was synoptic so I think i'll be able to cope with Unit 5 as well.
By the way how much do you need for an A*??

ok 1st S2o3- + cl2 --------> SO42- +CL- write the species that are in the reaction
2nd indentify oxidation of chlorine so:
0-----------------------> -1 on the LHS there 2 cholrine mocules so the balance it times cl by 2, so you have 2CL-, you also multiply the oxidation numbers so: 2*0-----------------------> -2 (cos -1*2 u get me)
diff in oxo numba is: A -2
oxo numba in s2o32- is +2------- +6 but u have to balance tne number of s, multiply s on rhs by2 and oxo numbas so:
+4--------+12 difference in oxo numba is
+8 balance this so you times so that you have -8 difference for chlorine A SO YOU need 8CLl- so times cl2 by 4 and 2cl- by 4 so effectively you get 8cli, ie 8 electrons which balances the electron difference for S ( from 4 to 12- difference of 8 electrons)
so for cholrine it will be 4CL2---------> 8CL- difference of 8 electrons ( 1 cl2 needs to eletrons to form 2cholride ions).

now your equation will be s2o32- + 4cl2-----> 2so42- +4cl2 = c (u notice the sulfur n chlorine are balanced. Now your merely add the H+, or O, H2o to balance out the oxygen's and hydrogen's

from C on lhs u have 3 oxgens on rhs side u have 8 oxygens , by convention u cannot add o on its own, so if you add 5H2O you balance the oxgens and you left with 10h+

so your overall equation will be
S2032- +5H2O + 4CL2------> 2SO42- 8CL- + 10H+
HALF EQUATIONS
they will always give u an unbalanced half equations TO WORK OUT OVERALL.
to start off with unbalanced half equations
4cl2 + 2e- ---------> 2cl-
S2O32- +5H20 + e- --------> SO42- +1oh+
les start by solving cl ok 4cl2 plus 8 elctrons and u form 8 ions-, each gainin an e-
4cl2 +8- -----> 8cl-
with sulphur you have
+2 ------------------> +6 bu there is 2 on LHS of S and 1 on rhs so times SO42- by 2 as well as both of the oxidation numbers now you have
s203- + 5h20 ---------> 2so42- + 10 h+ + 8e- in terms of oxo . you have
+4 --------------------> + 12 it has gain eight electrons that is balanced with chlorine so just cancel out elctrons in bold to get overall equation.

It's 270/300 at A2.. what have you got out of 40 in the practical? I can scale that up for you and tell you exactly what you need in F325 to get it

Hiya, does the practical mark counts towards the A* grade ? Oh yeah, it must if you need 270. um, can you work out what id need for an A and an A* grade. I got 81/90 and 39/40. No worries if not

oh last yr, i got 271/300 i think

Yeah but I am pretty sure that they won't go into detail with the stuff from previous units?? Because Unit 5 builds on the knowledge you've gained from Unit 1/2 and 4. I mean they can't just ask you about NMR spectroscopy, coz it hasn't even been mentioned in the Unit 5 specification. I am really good with the calculations because I do maths, so that's not a problem. I also always do extended calculations, for example when trying to find Ka rather than making [HA-]-[H+]=[HA-] I don't assume that the [HA-] stays undissociated and so for the calculation will need to use a quadratic equation.

But yeah F325 is hard on its own, so I don't think examiners will make it even harder by putting loads of stuff from previous Units > I mean I have looked at the two available past papers, haven't done them yet but when looking briefly through it I didn't recognize anything that requires in depth knowledge of other topics not included in Unit 5. For example there are no specific questions on organic chemistry.

Edit -> I just checked the spec and it says this:

Synoptic assessment involves the explicit drawing together of knowledge, understanding and skills
learned in different parts of the Advanced GCE course. The emphasis of synoptic assessment is to
encourage the development of the understanding of the subject as a discipline. All A2 units,
whether internally or externally assessed, contain synoptic assessment.
Synoptic assessment requires candidates to make and use connections within and between
different areas of chemistry at AS and A2, for example, by:
1. applying knowledge and understanding of more than one area to a particular situation or
context;
2.using knowledge and understanding of principles and concepts in planning experimental and
investigative work and in the analysis and evaluation of data;
3. bringing together scientific knowledge and understanding from different areas of the subject
and applying them.
All A2 units, F324–F326, contain some synoptic assessment.



I was able to cope with Unit 4 and It was synoptic so I think i'll be able to cope with Unit 5 as well.
By the way how much do you need for an A*??

What are the Redox Equations people learning prior to the exam, other than these ones:

Fe^2+ ---> Fe^3+ + e-

MnO4^- + 8H+ + 5e- ---> Mn^2+ + H2O

I2 + 2S2O3^2- ---> 2I- + S4O6

H2O2 ---> O2 + 2H+ + 2e-

Cr2O7^2- + 14H+ + 6e- ---> 2Cr3+ + 7H2O

Does anyone know why lattice enthalpy is more important that enthalpy of hydration for the enthalpy change of solution?