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# ocr a f325 revision thread watch

1. http://www.ocr.org.uk/Images/81033-u...s-specimen.pdf
3dii
The mark schemes says:
H negative because high temperature favours the endothermic direction
I got H negative but didn't get the reason, what do they mean by "because high temperature favours the endothermic direction"
2. Can someone please run through the Carbonic acid-hydrogencarbonate buffer system?
3. (Original post by janjua)
2.c no2 reacts with co emitted from car exhausts in the following rxn:

No2 + co -----> no + co2

rate equation for this reaction is rate={no2}2

suggest 2 step rxn mechanism for this rxn that is consistent with the kinetic data and overall equation.

P.s june 2006 paper at http://pastpapers.org/a2/chemistry/u...ne_2816_01.pdf

thnx
no2---> no + o
o + co ----> co2
4. (Original post by Holz888)
I was never taught this :/
I guess my best example is a esterification - the H2O is always in the Kc expression. But yeah, that does make sense
Hmm, I'll have a look to see which answer is correct because I've always been told that water won't have a concentration especially when being the solvent present in a reaction :-/
5. (Original post by v :))
Attachment 225863

How do I do c i? I get so confused when working out the second half equation! Can anyone suggest a method to approach these questions with? Thanks x

Posted from TSR Mobile
Minus the half equation known from the full equation.
For example, in this question you get the full equation as:
MH+NiO(OH)---->M+Ni(OH)2
and the half equation as:
NiO(OH)+H2O+e----->Ni(OH)2+OH-
So the other half equation must be:
MH+NiO(OH)---->M+Ni(OH)2 - NiO(OH)+H2O+e----->Ni(OH)2+OH-
=MH+Ni(OH) - NiO(OH) - H2O - e- ----> M+Ni(OH)2 - Ni(OH)2 - OH-
=MH - H2O - e-----> M - OH-
=MH+OH----->H2O + e- + M
6. (Original post by v :))
Attachment 225863

How do I do c i? I get so confused when working out the second half equation! Can anyone suggest a method to approach these questions with? Thanks x

Posted from TSR Mobile
Balance

http://imgur.com/Wcm2Mjd
7. Anyone got any question involving water of crystallization?
8. (Original post by Janjua)
Doesn't the rate determining step supposed to lead to the original equation in this case, the rxn btw NO2 + CO ----> NO + CO2 ?

Sorry for bothering you just get stuck at rate determining step.
It's OK, it's not bothering me
Well, the entire mechanism must show the overall equation. In mine, the first step was the rate determining step.
So we started with 2NO2 (it looks like two molecules need to bang into each other for the reaction to proceed, rather than one just falling apart). I decided to do
2NO2 --> 2NO + O2
then a fast (NOT rate determining) step could be O2 +2CO --> 2CO2

Now, if you take the entire mechanism and condense it, you get:
2NO2 + O2 + 2CO --> O2 + 2CO2
O2s on both sides cancel
2NO2 + 2CO --> 2CO2
Simplify
NO2 + CO --> CO2
9. January 2013 paper question 5 A, the Kc question.

The mark scheme for some reason multiplied the moles by 5 to figure out the concentration. In the question it did not say it was 5 dm. it said 200 cm volume and 500k temp. How did they get the 5?
10. (Original post by _sparks)
Okay quick question!
January 2011, q7b)ii)

They got the formula: 2H2O + 2I- + 2NO2- = 2NO + I2 + 4OH-

But using the method given in the book I get:
H20 + 2I- + NO2- = I2 + 2OH- + NO

Both equations are balanced(or are they?), but the mark scheme does not even talk about the one I got. The strange thing is that theyre not even multiples of each other!

Your minuses don't balance - you've 3 - on the LHS and 2- on the RHS.
What I always do is spend time just making sure everything is balanced - atoms, charges , electrons etc. When I did that question I got your answer at first, and I only realised that it was wrong by double checking. I hope that helps
11. DOES NO ONE KNOW THE ANSWER TO THIS?

question 7 january 2011 about DOC

According to the mark scheme we do it like this:

we get no of mol of 2S2O3 2-

2S2O3 2- --> I2 this means divide by 2

then

I2 --> 2Mn(OH)3 this means multiply by 2

then

2Mn(OH)3 --> 4Mn(OH)3
4Mn(OH)3 --> O2

^
now i thought no of mol of 2Mn(OH)3 should be multiplied by two and then the no of mol of 4Mn(OH)3 should be divided by four. BUT instead you have to say :

no of mol of 2Mn(OH)3 = no of mol of 4Mn(OH)3
and then divide by four.

why?
12. (Original post by afrodan98)
January 2013 paper question 5 A, the Kc question.

The mark scheme for some reason multiplied the moles by 5 to figure out the concentration. In the question it did not say it was 5 dm. it said 200 cm volume and 500k temp. How did they get the 5?
c = n/v

volume is 200/1000

multiplying by 5 and dividing by 200/1000 gives you the same answer
13. (Original post by afrodan98)
January 2013 paper question 5 A, the Kc question.

The mark scheme for some reason multiplied the moles by 5 to figure out the concentration. In the question it did not say it was 5 dm. it said 200 cm volume and 500k temp. How did they get the 5?
So to get conc, it needs to be out of 1dm3.

To go from 200cm3 to 1dm3, you need to divide your moles by 200 (to get 1cm3) and multiply by 1000 (to get 1000cm3 = 1dm3). This is the same as multiplying by 5
14. (Original post by needtosucceed=))
c = n/v

volume is 200/1000

multiplying by 5 and dividing by 200/1000 gives you the same answer
Oh right thanks.
15. (Original post by nightsky/)
Your minuses don't balance - you've 3 - on the LHS and 2- on the RHS.
What I always do is spend time just making sure everything is balanced - atoms, charges , electrons etc. When I did that question I got your answer at first, and I only realised that it was wrong by double checking. I hope that helps
Thanks a lot mate!
16. (Original post by Gold Phoenix)
Any last minute predictions on whats going to come up?
Most likely to be some sort of chemistry, but don't hold me to it.
17. DOES ANYONE KNOW THE ANSWER TO THIS?

question 7 january 2011 about DOC

According to the mark scheme we do it like this:

we get no of mol of 2S2O3 2-

2S2O3 2- --> I2 this means divide by 2

then

I2 --> 2Mn(OH)3 this means multiply by 2

then

2Mn(OH)3 --> 4Mn(OH)3
4Mn(OH)3 --> O2

^
now i thought no of mol of 2Mn(OH)3 should be multiplied by two and then the no of mol of 4Mn(OH)3 should be divided by four. BUT instead you have to say :

no of mol of 2Mn(OH)3 = no of mol of 4Mn(OH)3
and then divide by four.

why?
18. (Original post by fayled)
Most likely to be some sort of chemistry, but don't hold me to it.
Crap! I thought we were doing physics :|
19. (Original post by _sparks)
Thanks a lot mate!
No problem, good luck
20. (Original post by A-New-Start)
DOES NO ONE KNOW THE ANSWER TO THIS?

question 7 january 2011 about DOC

According to the mark scheme we do it like this:

we get no of mol of 2S2O3 2-

2S2O3 2- --> I2 this means divide by 2

then

I2 --> 2Mn(OH)3 this means multiply by 2

then

2Mn(OH)3 --> 4Mn(OH)3
4Mn(OH)3 --> O2

^
now i thought no of mol of 2Mn(OH)3 should be multiplied by two and then the no of mol of 4Mn(OH)3 should be divided by four. BUT instead you have to say :

no of mol of 2Mn(OH)3 = no of mol of 4Mn(OH)3
and then divide by four.

why?
So you have the moles of Mn(OH)3 as 0.0000246?

If you do, the first equation given tells you that the ratio of Mn(OH)3:O2 is:
4:1
0.0000246:0.0000246/4=0.000006150

Is that the bit you were having problems with? I'm not too sure which part you were stuck on....

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