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

Someone please help me with rate determining step && entropy, feasibility, free energy etc!!

MAJOR REP FOR LIFE!

That's very true my friend, this is the exam I really need to destroy actually, I just haven't done as well I should have in Chemistry exams. I need to get roughly 126-130 UMS to get an A overall, wasn't helped by my coursework. But still I've only got myself to blame, no excuses. I'm going to be embarrassed if I end-up going to study a masters in Chemistry, with a B in Chemistry.

How about you?

Ahh I'm sure you'll be fine and have revised sufficiently. I need 104 which I feel is still quite high compared to other people!

Reply 442

haydyb123

9

Original post by blush.ox

I'd like them plz if you dont mind :flutter:

xD

[Cu(H2O)6]2+ Shape: Octahedral
[Cu(H2O)4(NH3)2]2+ Shape: Octahedral

Now this is a cool one, a bit of Ligand Sub:
[Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2-
Blue to Yellow, but you will observe green as the reaction is at equilibrium and is the mixture of the two colours
and last but by no means least

[Co(H2O)6]2+ + 4Cl- -> [CoCl4]2-
Pink to blue
:smile:

Reply 443

haydyb123

9

Original post by sportycricketer

Ahh I'm sure you'll be fine and have revised sufficiently. I need 104 which I feel is still quite high compared to other people!

The thing I'll say to that is that you clearly haven't accumulated marks for no reason, there's no reason why you shouldn't continue and rake in a similar amount of marks on this paper so, the key is just not to be complacent and it will work out fine!

Reply 444

haydyb123

9

Original post by *QueenBeee

Someone please help me with rate determining step && entropy, feasibility, free energy etc!!

MAJOR REP FOR LIFE!

If you have a specific question I'll help you out, as those topics are pretty large to go over completely! :smile:

Reply 445

sportycricketer

Original post by haydyb123

The thing I'll say to that is that you clearly haven't accumulated marks for no reason, there's no reason why you shouldn't continue and rake in a similar amount of marks on this paper so, the key is just not to be complacent and it will work out fine!

yeah agree with that! Never be complacent with anything at all cos that lead to failure. It's kinda better of someone needs high marks cos at least they'll work hard for it!

Reply 446

INeedToRevise

8

This exam is my second exam, but by far the one I'm most worried about. Its weird how this one is earlier than F324.

How is everyone spending these last 10 days?

Reply 447

blush.ox

15

Original post by haydyb123

[Cu(H2O)6]2+ Shape: Octahedral
[Cu(H2O)4(NH3)2]2+ Shape: Octahedral

Now this is a cool one, a bit of Ligand Sub:
[Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2-
Blue to Yellow, but you will observe green as the reaction is at equilibrium and is the mixture of the two colours
and last but by no means least

[Co(H2O)6]2+ + 4Cl- -> [CoCl4]2-
Pink to blue
:smile:

thankyou soo berry much, easier to take in like this i think :lol:

and yes i stayed off in order to 'revise' but TSR is actually helpful like right now :redface:

Reply 448

TheAwkwardBalloon0

Original post by arvin_infinity

There is a question in jan 2007 paper..
says what colour change occurs?
in cr207 2- + h20 <--> 2 cro4 2- + 2h+

Am not sure if its in the textbook

dichromate is orange and CrO42- is yellow :smile:

Reply 449

arvin_infinity

3

Original post by TheAwkwardBalloon0

dichromate is orange and CrO42- is yellow :smile:

Thanks for reply but I knew the answer ..just was worried if I am missing something in the textbook

Reply 450

arvin_infinity

3

Original post by haydyb123

Cu2+(aq) Cu(OH)2
Co2+ (aq) Co(OH)2 **
Fe2+ (aq) Fe(OH)2 *
Fe3+ (aq) Fe(OH)3

*Oxidised to Fe3+ on standing, and changes colour accordingly.
**Oxidised to Co3+ on standing, changes colour to Beige

I can do complexes as well if you would like?

+rep

Just found this equation in past p
A mean question!!!!!!!!!!
Just to confirm its not in the textbook so how the f... am I supposed to answer it :eek:

Any tips for these type of Q!?
Ca3(PO4)2 + 3H2SO4 ?? 2H3PO4 + 3CaSO4

Reply 451

J DOT A

Original post by arvin_infinity

+rep

Just found this equation in past p
A mean question!!!!!!!!!!
Just to confirm its not in the textbook so how the f... am I supposed to answer it :eek:

Any tips for these type of Q!?
Ca3(PO4)2 + 3H2SO4 ?? 2H3PO4 + 3CaSO4

What is the full question? Is it an ionic or redox?

Reply 452

haydyb123

9

Original post by arvin_infinity

+rep

Just found this equation in past p
A mean question!!!!!!!!!!
Just to confirm its not in the textbook so how the f... am I supposed to answer it :eek:

Any tips for these type of Q!?
Ca3(PO4)2 + 3H2SO4 ?? 2H3PO4 + 3CaSO4

My tips are remember what you have learned, and just apply your knowledge of acid and base reactions... though this is a complete guess, not sure if this is in our spec:

so you have 3 Ca2+ which presumably reacts with the 3 So42- to form 3CaSO4
which means that PO4 must have a charge of of -3 (-6 for both) which surprise surprise, would fit perfectly with the 6 H+s we have to form 2H3(PO4)
Just realised the answer was above...

But yes, remember the key principles and just apply them, don't be put off because they're unfamiliar. :smile:

Reply 453

nestle

has anyone got the Jan 2011 F325 paper??

cos there is only two past papers on the OCR website and i hate doing the legacy ones :/

thanks

Reply 454

haydyb123

9

Original post by nestle

has anyone got the Jan 2011 F325 paper??

cos there is only two past papers on the OCR website and i hate doing the legacy ones :/

thanks

Here you go :smile:

f325 january 2011 qp.pdf207.5KB

F325 January 2011 MS.pdf319.5KB

Reply 455

stellanocte

Hi, has anyone done the specimen paper? I don't get q. 1a)iii. - why is rate of formation O2 quarter? Is it to do with mols in the equation - do we assume that it's methanal rates??

Reply 456

haydyb123

9

Original post by stellanocte

Hi, has anyone done the specimen paper? I don't get q. 1a)iii. - why is rate of formation O2 quarter? Is it to do with mols in the equation - do we assume that it's methanal rates??

It's because whatever your rate of Methanal formation it is forming 2 moles of CH3OH, and O2 formation is a quarter of this, so yes as you said it's to do with the mols in the equation. It also says in the question that the rates are for methanol formation :smile:

Reply 457

stellanocte

Original post by haydyb123

It's because whatever your rate of Methanal formation it is forming 2 moles of CH3OH, and O2 formation is a quarter of this, so yes as you said it's to do with the mols in the equation. It also says in the question that the rates are for methanol formation :smile: