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

1. (Original post by voices1)
The doc is that the word document contain bundles of questions?
Most of them are easy and way2 similar
Well you cant really revise otherwise! All the past papers are gonna be like that!

(Original post by voices1)
So for exothermic reaction enthalpy outweighs entropy so reaction occurs free energy is less than zero. Correct?
you said -ve +ve Always -ve Feasible
will temperature need to be negative so (-ve)-(-temp*postive entropy) for a negatvie free energy, correct?

I'm confident with all calculation and written work regarding this but apart from the calculations most of it is memorizing that pure understanding with regards to free energy.

e.g for an endothermic reaction, entropy must be high enuf to outwiegh enthalpy and temp must be high.... can apply to particular reactions but have no clue what that means
Yeah that is correct! If H is negative and S is positive, the temperature doesn't matter because the reaction will always be feasible:

?H | ?S | ?G | Feasibility
-ve |+ve | Always -ve | Feasible
+ve | -ve | Always +ve | Never Feasible
-ve | -ve | -ve at low temp | Feasible at low temp
+ve | +ve | -ve at high temp | Feasible at high temp

Does this make more sense??
2. Guys, what is the oxidation number of cr in cr2(so4)3
I'm getting +1
o=-24 sulpur= +18 there are 3cr2 ie 6 cr that leaves each cr as +1
but from jan 2011 paper answer is +2
Confused here.....
3. (Original post by voices1)
Guys, what is the oxidation number of cr in cr2(so4)3
I'm getting +1
o=-24 sulpur= +18 there are 3cr2 ie 6 cr that leaves each cr as +1
but from jan 2011 paper answer is +2
Confused here.....

I keep getting +3 as Chromiums oxidation state.

Cr2(SO4)3

You know SO4 has a charge of -2. Multiply -2 by 3= -6.
Because it has no net overall charge, the oxidation number of chromium should equal= +6. Divide this by 2 gives you the charge of each chromium to be +3.

Dunno why the answer is +2.
4. (Original post by INeedToRevise)
I keep getting +3 as Chromiums oxidation state.

Cr2(SO4)3

You know SO4 has a charge of -2. Multiply -2 by 3= -6.
Because it has no net overall charge, the oxidation number of chromium should equal= +6. Divide this by 2 gives you the charge of each chromium to be +3.

Dunno why the answer is +2.
I made a mistake in my original post. the answer is +3

i was wondering dont you multiply the number out the bracket.. 3 with 2 molecules with cr2 as well- making it 3Cr2- ie 6 cr.....
so cr is+1.
if you apply the method that you use then for example

2mg(no3)2----> 2mgo +2no2 the equation shud be balanced???
confusing 2 way street here, do u get my point
5. Sorry if this has already been asked, but do we have to know the acid into base titration graphs, as well as the base into acid graphs? Only the base into acid graphs are shown in the revision book and textbook.
6. (Original post by voices1)
I made a mistake in my original post. the answer is +3

i was wondering dont you multiply the number out the bracket.. 3 with 2 molecules with cr2 as well- making it 3Cr2- ie 6 cr.....
so cr is+1.
if you apply the method that you use then for example

2mg(no3)2----> 2mgo +2no2 the equation shud be balanced???
confusing 2 way street here, do u get my point
You multiply everything in the brackets by the number outside the brackets. So for the Cr2(SO4)3, the SO4 should only be multiplied by 3.
7. (Original post by INeedToRevise)
You multiply everything in the brackets by the number outside the brackets. So for the Cr2(SO4)3, the SO4 should only be multiplied by 3.

Yep, that equation is balanced.
You have 2 Mg on each side.
2 N on each side.
And 4 O on each side.
Hang on, how is that equation balanced? you have to multiply reactants by the big numbers in front of them as well. So there are 4 N on the left etc.
8. can some give me help on the rate determining step, I know that it is the slowest step, and the rate equation contains the molecules in the rate determining step, but Im preety lost after that. Can someone give me questions based on the rate determining step. PLEASE HELPPPPPpp
9. (Original post by student777)
Hang on, how is that equation balanced? you have to multiply reactants by the big numbers in front of them as well. So there are 4 N on the left etc.
Whoops. Yeah your right Missed that 2.
10. Hi,
Im doing lots of old papers, 'trends and patterns' and 'unifying concepts' and all those and im confused as to what we need to know.. :/
do we need to know about things dissolving in water and the colour changes? giant, simple structures and types of bonding (synoptic i guess.. ) ?decomposition? absorption spectrums?
cheers
11. (Original post by tallysingh)
can some give me help on the rate determining step, I know that it is the slowest step, and the rate equation contains the molecules in the rate determining step, but Im preety lost after that. Can someone give me questions based on the rate determining step. PLEASE HELPPPPPpp
also do you guys think they could ask a rate detrermining step with an actual mechanism with curly arrows etc
12. (Original post by tallysingh)
also do you guys think they could ask a rate detrermining step with an actual mechanism with curly arrows etc
Hey Tally, just on your 2nd question, no I don't think that's likely;
we could be asked to suggest an intermediate, but it's probably undergrad. level to start drawing curly arrow mechanisms for them!

Now onto your first question; try not to make things more complicated than they are which can be easy for this topic.

Okay lets go,
Overall: NO2 + CO -> NO + CO2

Say our rate equation is Rate=[NO2]2
This equation can only be determined by experiments! Not from the overall equation!

we know that the rate-determining step is: the slowest step in the mechanism of a multi-step reaction

therefore by looking at our rate equation we know that the slowest step in our reaction is:

NO2 + NO2

now this is the part most susceptible to over-complication, my general rule of thumb is to just clump the two molecules together:
NO2 + NO2 -> N2O4

we now know that we have to get from N2O4 to NO and CO2, by using CO:

N2O4 + CO -> NO + CO2

but this doesn't balance with the overall equation, so we know to add NO2 to the other side of the equation to balance giving the following mechanism

NO2+ NO2 -> N204
N204 + CO -> NO + CO2 + NO2*(

* Cancels with the extra NO2 added in the first step
and this cancels down to form the overall equation:
NO2 + CO -> CO2 + NO

I hope this has helped!
13. (Original post by haydyb123)
Hey Tally, just on your 2nd question, no I don't think that's likely;
we could be asked to suggest an intermediate, but it's probably undergrad. level to start drawing curly arrow mechanisms for them!

Now onto your first question; try not to make things more complicated than they are which can be easy for this topic.

Okay lets go,
Overall: NO2 + CO -> NO + CO2

Say our rate equation is Rate=[NO2]2
This equation can only be determined by experiments! Not from the overall equation!

we know that the rate-determining step is: the slowest step in the mechanism of a multi-step reaction

therefore by looking at our rate equation we know that the slowest step in our reaction is:

NO2 + NO2

now this is the part most susceptible to over-complication, my general rule of thumb is to just clump the two molecules together:
NO2 + NO2 -> N2O4

we now know that we have to get from N2O4 to NO and CO2, by using CO:

N2O4 + CO -> NO + CO2

but this doesn't balance with the overall equation, so we know to add NO2 to the other side of the equation to balance giving the following mechanism

NO2+ NO2 -> N204
N204 + CO -> NO + CO2 + NO2*(

* Cancels with the extra NO2 added in the first step
and this cancels down to form the overall equation:
NO2 + CO -> CO2 + NO

I hope this has helped!
thanks mate, really appreciated.
14. please can someone tell me about the hydrogen oxygen fuel cell
ive read about it from three different books but im still a bit unsure as to what i need to know eg what equations

anyone?
also, do u reckon itll be a good paper this time?
15. helppp
i think theres a mistake in the HEINEMAN BOOK
PAGE 117 question 3!
i got 2000 and units as dm3 mol-1 s-1

but answer is different?
can someone do the q and tell me what they get please! THANXXX
16. (Original post by 786girl)
helppp
i think theres a mistake in the HEINEMAN BOOK
PAGE 117 question 3!
i got 2000 and units as dm3 mol-1 s-1

but answer is different?
can someone do the q and tell me what they get please! THANXXX
i get 2000 dm3 mol-1 s-1 too. i remember doing that question and thinking 'huh? how did i get that wrong? ' but yeah, book must be wrong.

k=rate/[O3][C2H4]
17. (Original post by wilsea05)
i get 2000 dm3 mol-1 s-1 too. i remember doing that question and thinking 'huh? how did i get that wrong? ' but yeah, book must be wrong.

k=rate/[O3][C2H4]
(Original post by 786girl)
helppp
i think theres a mistake in the HEINEMAN BOOK
PAGE 117 question 3!
i got 2000 and units as dm3 mol-1 s-1

but answer is different?
can someone do the q and tell me what they get please! THANXXX
Yeah the book is wrong. I asked my teacher - he done the question with me again, we got 2000.
18. (Original post by 786girl)
please can someone tell me about the hydrogen oxygen fuel cell
ive read about it from three different books but im still a bit unsure as to what i need to know eg what equations

anyone?
also, do u reckon itll be a good paper this time?
1)know the half-cell equations of oxygen and hydrogen in alkali solutions.
2)Know that the electrolyte is OH- and that it never runs out, because in a fuel cell you continously supply the system with a "fuel" in this case hydrogen.
3) know that it is impossible to use oxygen-hydrogen fuel as a source of energy in FCVs because scientists are still trying to work out ways how to safely store the hydrogen, because adsorbers and absorbers are not efficient atm + hydrogen gas cannot be safely compressed into a liquid yet so that's why we use hydrogen-rich fuels rather than hydrogen on its own. Scientists are trying to find ways of using hydrogen more in the future but do not know whether it is an efficient way of generating energy (i.e. Hydrogen economy)
19. Hey, Someone who is good at redox, please i beg you explain it to me. I know the basic but never know anything apart from OILRIG. please help
20. Hello people

Could someone please explain to me how to do 6d on jan11 peper. I understand what points we have to mention but I don't understand why the enthalpy change of solution of KF is less negative than of RBF. if K+ ion has a smaller radius, whould it not have more negative enthalpy change of solution as its lattice enthalpy would be more negative...

I would also appreciate some help on 4bii of the same paper.

Thank you

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