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

1. (Original post by Rocks:))
Surely you need to divide your 0.399 by 25/1000 because you've figured out the grams in 25cm3 but you need in dm3 so I get 15.9g dm3. which seems big. Hmm maybe I'm wrong, someone correct me.
there's no need as that is the right answer

the way I did it was to find the concentration of H2O2 in 250 which was 0.0469 mol dm-3

n=mass/molar mass, so mass = nMr

so 0.0469x34 = 1.5946g dm-3 in the 250cm3 solution

the concentration in 250 is going to be 1/10 of the concentration in the original 25cm3

so x10 = 15.9g dm3
2. (Original post by viksta1000)
Fe has oxidation state 6+ in FeO4 (2-) and oxidation state 3+ in Fe3+, so has changed by -3

O has oxidation state 2- in FeO4(2-) and oxidation state 0 in O2, so the change is +3, however, 2 oxygen atoms have been used to form O2, so the overall change is +4

to balance the lowest common factor of -3 and +4 is 12, so you multiplu FeO4(2-) by 4 to get:

4FeO4(2-) + n[H+] --> 4Fe(3+) + H2O + 02

now look at the overall molecular charges, as they have to balance on both sides:

we have (4x-2) + n(1) --> (4x3+) + 0 + 0

which is -8 + n1+ --> 12+
we need to get 12+ on both sides, so we need to add 20H+ ions to the left to get:

4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 02

simple balancing from here, 20H+ on right, so you need 10H2O:

4FeO4(2-) + 20[H+] --> 4Fe(3+) + 10H2O + 02

16 oxygens on left, 10 on the right, so we need 6 more, so times O2 by 3

4FeO4(2-) + 20[H+] --> 4Fe(3+) + H2O + 302

is the change in the oxidation no. of O from FeO42- to O2 should be +2 ?
3. do we need to know percentage purity?
4. (Original post by viksta1000)
there's no need as that is the right answer

the way I did it was to find the concentration of H2O2 in 250 which was 0.0469 mol dm-3

n=mass/molar mass, so mass = nMr

so 0.0469x34 = 1.5946g dm-3 in the 250cm3 solution

the concentration in 250 is going to be 1/10 of the concentration in the original 25cm3

so x10 = 15.9g dm3
Thank you for putting my mind at ease, I'm hopeless at following other peoples calculations but your way does seem a little more sophisticated than mine
5. (Original post by warrior_22)
is the change in the oxidation no. of O from FeO42- to O2 should be +2 ?
yes, for each O atom it is 2, but 2 atoms have been used from the FeO4 to make O2

so 2 atoms have changed by +2 = 2x+2 = +4 overall
6. Could someone help me with Q4,b,ii

2011 Jan.

Thanks
Attached Images
7. F325Jan11.pdf (206.7 KB, 53 views)
8. (Original post by viksta1000)
fe has oxidation state 6+ in feo4 (2-) and oxidation state 3+ in fe3+, so has changed by -3

o has oxidation state 2- in feo4(2-) and oxidation state 0 in o2, so the change is +3, however, 2 oxygen atoms have been used to form o2, so the overall change is +4

to balance the lowest common factor of -3 and +4 is 12, so you multiplu feo4(2-) by 4 to get:

4feo4(2-) + n[h+] --&gt; 4fe(3+) + h2o + 02

now look at the overall molecular charges, as they have to balance on both sides:

We have (4x-2) + n(1) --&gt; (4x3+) + 0 + 0

which is -8 + n1+ --&gt; 12+
we need to get 12+ on both sides, so we need to add 20h+ ions to the left to get:

4feo4(2-) + 20[h+] --&gt; 4fe(3+) + h2o + 02

simple balancing from here, 20h+ on right, so you need 10h2o:

4feo4(2-) + 20[h+] --&gt; 4fe(3+) + 10h2o + 02

16 oxygens on left, 10 on the right, so we need 6 more, so times o2 by 3

4feo4(2-) + 20[h+] --&gt; 4fe(3+) + h2o + 302

what is this!?!!?
Since when do we need to know this!?
What are you doing to me!?
9. (Original post by viksta1000)
there's no need as that is the right answer

the way I did it was to find the concentration of H2O2 in 250 which was 0.0469 mol dm-3

n=mass/molar mass, so mass = nMr

so 0.0469x34 = 1.5946g dm-3 in the 250cm3 solution

the concentration in 250 is going to be 1/10 of the concentration in the original 25cm3

so x10 = 15.9g dm3
but isn't the 250cm3 solution diluted? they are asking for undiluted
just a thought. correct me if i'm wrong
10. (Original post by jlcf)
Hi, here some more recently LEGACY QP and MS, has similar questions to the F325 specification. However there may have some question which is not relevant to our specification.

2816-01 - Unifying Concepts in Chemistry questions - Kc, Ka, Equilibrium, rate- determining step, Orders, Buffers
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions - Born-Haber Cycles, Standard enthalpy, Transiton metal questions as well.

You can also find these question paper on the OCR website.
These questions are good! have you got anymore? do you reckon these questions will be similar to what we get tommorow?
11. (Original post by CoffeeStinks)
Could someone help me with Q4,b,ii

2011 Jan.

Thanks
I get CH3OH + 1/2O2 -------> C02 + 4H+ +4e-

Is this right?
12. (Original post by hellosarah)
what is this!?!!?
Since when do we need to know this!?
What are you doing to me!?
Thank you, they make me look absolutely DUMB
13. (Original post by viksta1000)
Fe has oxidation state 6+ in FeO4 (2-) and oxidation state 3+ in Fe3+, so has changed by -3

O has oxidation state 2- in FeO4(2-) and oxidation state 0 in O2, so the change is +3, however, 2 oxygen atoms have been used to form O2, so the overall change is +4

to balance the lowest common factor of -3 and +4 is 12, so you multiplu FeO4(2-) by 4 to get:

4FeO4(2-) + n[H+] --&gt; 4Fe(3+) + H2O + 02

now look at the overall molecular charges, as they have to balance on both sides:

we have (4x-2) + n(1) --&gt; (4x3+) + 0 + 0

which is -8 + n1+ --&gt; 12+
we need to get 12+ on both sides, so we need to add 20H+ ions to the left to get:

4FeO4(2-) + 20[H+] --&gt; 4Fe(3+) + H2O + 02

simple balancing from here, 20H+ on right, so you need 10H2O:

4FeO4(2-) + 20[H+] --&gt; 4Fe(3+) + 10H2O + 02

16 oxygens on left, 10 on the right, so we need 6 more, so times O2 by 3

4FeO4(2-) + 20[H+] --&gt; 4Fe(3+) + H2O + 302

Hey thanks for this, I aksed the same question here 3 days ago i think, but no one really explained.
Thank you +rep
14. can some one please go over Question 6e in the June 2010 exam...

I got NH4CrO4 but its (NH4)2Cr2O7?
15. (Original post by Flux_Pav)
but isn't the 250cm3 solution diluted? they are asking for undiluted
just a thought. correct me if i'm wrong

if you work out the concentration of H2O2 in the 250, you know the diluted concentration of H2O2

the amount of H2O2 in the dilute sample is 25cm3 in 250cm3, which is the same as 1/10th of the original sample, so if you x10 you get the amount in the original sample
16. mole calculations, bonding and energetics from AS is synoptic so its a good idea to review those topics as well
17. (Original post by Rocks:))
I get CH3OH + 1/2O2 -------> C02 + 4H+ +4e-

Is this right?
I got something like that too but the MS says CH3OH + H2O -------->6H+ +6e- +CO2

I just can't seem to figure out why.
18. (Original post by SmartFool)
These questions are good! have you got anymore? do you reckon these questions will be similar to what we get tommorow?
http://www.ocr.org.uk/qualifications...nts/index.html

Here more in the LEGACY past paper sections. Once you download it, remember to use these files:

2816-01 - Unifying Concepts in Chemistry questions
2815-06 - Transition metal questions
2815-01 - Trends and Patterns questions

The questions are probably similar but they may throw in some curveball like they did in Jan 2011 paper. All you can do is prepare and hope something paper would be easy.
19. (Original post by CoffeeStinks)
I got something like that too but the MS says CH3OH + H2O -------->6H+ +6e- +CO2

I just can't seem to figure out why.
Because the examiners are wrong I have no idea. I give up, four exams in three days is killing me.
20. (Original post by viksta1000)

if you work out the concentration of H2O2 in the 250, you know the diluted concentration of H2O2

the amount of H2O2 in the dilute sample is 25cm3 in 250cm3, which is the same as 1/10th of the original sample, so if you x10 you get the amount in the original sample
cheers
say if the question asked for the diluted h202 concentration. what would you do differently? this time would you do :

1.1725x10^-2 x 1000
___________________

250 (insted of 25)

21. (Original post by CoffeeStinks)
I got something like that too but the MS says CH3OH + H2O -------->6H+ +6e- +CO2

I just can't seem to figure out why.
You know how you add two half equations to make an overall equation? To deduce the methanol half equation, you should kind of make a maths sum of it. So write the overall equation and beneath it, write the oxygen half equation and "subtract" it - see if you can work from there (btw, you need to make sure the electrons are balanced, multiply the oxygen half equation by 1.5 to before you subtract)

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