Equilibria, Energetics and Elements (F325) - January 2012 Exam.

http://www.thestudentroom.co.uk/showthread.php?t=1252151

Have a look at that thread. There's 3 parts to the syllabus - Unifying concepts, Trends and patterns and Transition elements, if that's what you meant :redface:

thanks for that link really helped good luck with your exams.

Reply 181

Tateco

14

Not looking forward to the inevitable question I won't be able to do despite how much revision i do :mad:

Reply 182

010494

Is this exam at January 2011 or February the 1st 2012??? :/

Reply 183

Fatima0065

5

Original post by 010494

Is this exam at January 2011 or February the 1st 2012??? :/

February 1st 2012 :smile:

and its for 2 hours lol!!! good luck :smile:

Reply 184

Fatima0065

5

Original post by Tateco

Not looking forward to the inevitable question I won't be able to do despite how much revision i do :mad:

I know. Ive done so much revison but I'm guessing there will that one daunting question that i'll look at at think WTH!!! lol :biggrin:

Reply 185

TourettesUnicorn

Guys how do you do questions like this:

The overall reaction in a fuel cell is the complete combustion of methanol. Methanol is supplied at one electrode; oxygen at the other. Oxygen reacts at the negative electrode - O2 + 4H+ + 4e- ----> 2H2O

Deduce the half equation at the positive electrode.

The question if from January 2011. Thanks :smile:

Reply 186

Fatima0065

5

Original post by With an Accent

You would have to mention that bonding pairs are repelled more stronger by lone pairs than other bonding pairs.

Here's my answer, can you tell me if its right:
An ammonia molecule has 3 bonding pairs and 1 lone pair.
An ammonia ligand has no lone pairs and 4 bonding pairs
Lone pairs repel more than bonding pairs, which is why the ligand has 109.5 degree angle rather than a 107 degree
Is that right??? please feel free to add anything!!!
:biggrin:

Thankyou

Reply 187

sumsum123

2

Original post by TourettesUnicorn

Guys how do you do questions like this:

The overall reaction in a fuel cell is the complete combustion of methanol. Methanol is supplied at one electrode; oxygen at the other. Oxygen reacts at the negative electrode - O2 + 4H+ + 4e- ----> 2H2O

Deduce the half equation at the positive electrode.

The question if from January 2011. Thanks :smile:

Hey there, :smile:

They tell you that the fuel cell is about the combustion of methanol, in part one you write out the combustion of methonal, this is the final equation. So, they tell you that oxygen reacts at the negative electrode and you can see from the final combustion equation of methanol that methanol goes from

CH3OH ------> C02

Then it is a case of balancing. You start with water on the left hand side to balance the oxygens.

CH3OH + H20 -----> CO2 +

And then balance the hydrogen ions and the electrons on the right hand side.

CH3OH + H20 ----> CO2 + 6H+ + 6e-

Reply 188

cobra2k10

1

guys, i need help with question (4e) on the june 2011 paper. Why do you have to choose lactic acid?

Reply 189

kurdishboy94

1

Original post by cobra2k10

guys, i need help with question (4e) on the june 2011 paper. Why do you have to choose lactic acid?

because in the book on page 153 of the ocr book it states ph=pka and so this means the closest acid to the pka of the table is lactic acid. i was the only student in the class to see this everyone else made it complicated lool :biggrin:

hope that helps.

Reply 190

Tateco

14

Original post by cobra2k10

guys, i need help with question (4e) on the june 2011 paper. Why do you have to choose lactic acid?

Because its the closest pH to the desired buffer pH

Reply 191

lfcjamie

4

On that June 11 exam page 5 2) c ii) can someone do a guide for it. I can normally do Kc questions with ease but this one has just confused me. Don't get why EQ amount of O2 is 0.4 for example. So if someone would put up a step my step guide for me I'd very much appreciate it.

Reply 192

cobra2k10

1

Original post by kurdishboy94

because in the book on page 153 of the ocr book it states ph=pka and so this means the closest acid to the pka of the table is lactic acid. i was the only student in the class to see this everyone else made it complicated lool :biggrin:

hope that helps.

Original post by Tateco

Because its the closest pH to the desired buffer pH

thanks... Are there any predictions of what is likely to come up based on the previous 3 papers??

Reply 193

sumsum123

2

Original post by lfcjamie

On that June 11 exam page 5 2) c ii) can someone do a guide for it. I can normally do Kc questions with ease but this one has just confused me. Don't get why EQ amount of O2 is 0.4 for example. So if someone would put up a step my step guide for me I'd very much appreciate it.

Are you talking about question 5) part c) ii)

Reply 194

lfcjamie

4

Original post by sumsum123

Are you talking about question 5) part c) ii)

No sorry question 2 part C ii) which is on page 5 of the exam

Reply 195

sumsum123

2

Original post by lfcjamie

No sorry question 2 part C ii) which is on page 5 of the exam

Ahh kool

, I was staring at the wrong paper ... haha!
Okay, so this is the way I always do it by making a little table.

2NO + O2 ---><--- 2NO2
start (concentrations) 0.4 0.35 0
end (concentrations) 0.1 0.20 0.30

So, now gonna explain my table. So you have 0.8 moles of NO and 0.7 moles of 02. I am first going to convert them to concentrations by dividing them by 2 dm3. They tell me that 75% of the NO has reacted , so I find 75% of 0.4 moldm3 of NO and take away that value from 0.4. This gives me 0.1 . Btw , 75% of it is 0.3.

Now this is where molar ratio's come in. It is 2:1 ,, NO:02. So only I take away only half 0.3 (0.15) from 0.35 to give me 0.20 moldm3 of oxygen. And seeing as I had nothing of NO2 to begin with I add 0.3 because the molar ratio of NO:NO2 is 2:2. So now you convert your concentrations back to moles to show the moles of everything at equilibrium. That is where 0.4 moles of 02 comes from, you multiply your 0.2 moldm3 by 2dm3 to give you 0.4 moles. And now you pop your concentration figures into your Kc equation to give you 45 dm3mol-1.

Tell me if you don't get anything, I tend to babble alot :tongue:

Reply 196

Syma

Hi, can someone help me with the jan 2011 paper, question 7a(i):
http://www.ocr.org.uk/download/pp_11_jan/ocr_61491_pp_11_jan_gce_f325.pdf
I can do the first part where you do n = c*v so to get the moles of Na2S2O3 you do 0.001 * (24.6/1000) and moles is 2.46*10^-5 mol.
Then according to the mark scheme, to get the moles of O2 you do (2.46*10^-5)/4
Why do you divide by 4? :s

Reply 197

lfcjamie

4

Original post by sumsum123

Ahh kool

, I was staring at the wrong paper ... haha!
Okay, so this is the way I always do it by making a little table.

2NO + O2 ---><--- 2NO2
start (concentrations) 0.4 0.35 0
end (concentrations) 0.1 0.20 0.30

So, now gonna explain my table. So you have 0.8 moles of NO and 0.7 moles of 02. I am first going to convert them to concentrations by dividing them by 2 dm3. They tell me that 75% of the NO has reacted , so I find 75% of 0.4 moldm3 of NO and take away that value from 0.4. This gives me 0.1 . Btw , 75% of it is 0.3.

Now this is where molar ratio's come in. It is 2:1 ,, NO:02. So only I take away only half 0.3 (0.15) from 0.35 to give me 0.20 moldm3 of oxygen. And seeing as I had nothing of NO2 to begin with I add 0.3 because the molar ratio of NO:NO2 is 2:2. So now you convert your concentrations back to moles to show the moles of everything at equilibrium. That is where 0.4 moles of 02 comes from, you multiply your 0.2 moldm3 by 2dm3 to give you 0.4 moles. And now you pop your concentration figures into your Kc equation to give you 45 dm3mol-1.

Tell me if you don't get anything, I tend to babble alot :tongue:

Got it thanks a lot!

Reply 198

With an Accent

Original post by Fatima0065

Here's my answer, can you tell me if its right:
An ammonia molecule has 3 bonding pairs and 1 lone pair.
An ammonia ligand has no lone pairs and 4 bonding pairs
Lone pairs repel more than bonding pairs, which is why the ligand has 109.5 degree angle rather than a 107 degree
Is that right??? please feel free to add anything!!!
:biggrin:

Thankyou

I think that's fine :smile:

Reply 199

coOkiiee

Hey, can someone help me with June 2011 Q7dii on how to get the two half equations? Thanks