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Edexcel (IAL) Unit 2 Chemistry June 10th

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Reply 20
Hey, I'm really struggling with Hess's cycle especially of a reaction.
I've been doing this question for ages and i still can't get my head around it could anyone help ?

Question
Lead forms several solid oxides the most common ones are PbO , PbO2 and Pb3O4.
Part 1 said write an equation for the formation of PbO
then Part 2 said, 3PbO+1/2O2= Pb3O4
enthalpies of formation are:
PbO=-219
Pb3O4=-735

I worked out the standard enthalpy change of the reaction and it came to 78kJmol-1

but this is is the part I'm confused by ...

when Pb3O4 is heated it decomposes to PbO and PbO2
enthalpy change of the reaction is 20kJmol-1
so now i need to find the enthalpy change of formation of PbO2 from this data
https://e91d34a60c48d35030cf61ddda06cffc30feeddf.googledrive.com/host/0B1ZiqBksUHNYaklpczRpVjg4ZjA/June%202010%20QP%20-%20Unit%202%20Edexcel%20Chemistry.pdf

Q19d) iv. - i dont get why we've mutliplied by 10? like i get that the concentration is multiplied by 10 bc we're going from 25cm^3 to 250cm^3 but isnt the original vinegar solution 25cm^3?
Reply 22
Original post by Trudy9
Hey, I'm really struggling with Hess's cycle especially of a reaction.
I've been doing this question for ages and i still can't get my head around it could anyone help ?

Question
Lead forms several solid oxides the most common ones are PbO , PbO2 and Pb3O4.
Part 1 said write an equation for the formation of PbO
then Part 2 said, 3PbO+1/2O2= Pb3O4
enthalpies of formation are:
PbO=-219
Pb3O4=-735

I worked out the standard enthalpy change of the reaction and it came to 78kJmol-1

but this is is the part I'm confused by ...

when Pb3O4 is heated it decomposes to PbO and PbO2
enthalpy change of the reaction is 20kJmol-1
so now i need to find the enthalpy change of formation of PbO2 from this data


May I ask are you doing the Edexcel IAL qualification? Because if that's the case, then you really don't need to worry about Unit 1 topics for the Unit 2 exam! Now for your question, I did actually get stuck on that on a previous past paper, however I watched this video explaining the formulation of Hess cycle's really well and if I find it I shall definitely send you the link. However I do suggest that you focus on Unit 2 revision for the time being that is if you are doing the Edexcel IAL. :smile:
Reply 23
Is everyone ready for tomorrow?
Reply 25
Original post by biochempsy
Is everyone ready for tomorrow?


I think so.. but edexcel will always find a way to make the paper tricky.. What about you?
I think so too, unit 2 is a hard exam and the fact that there are hardly any patterns between exam papers makes it all the more difficult
Original post by Trudy9
Hey, I'm really struggling with Hess's cycle especially of a reaction.
I've been doing this question for ages and i still can't get my head around it could anyone help ?

Question
Lead forms several solid oxides the most common ones are PbO , PbO2 and Pb3O4.
Part 1 said write an equation for the formation of PbO
then Part 2 said, 3PbO+1/2O2= Pb3O4
enthalpies of formation are:
PbO=-219
Pb3O4=-735

I worked out the standard enthalpy change of the reaction and it came to 78kJmol-1

but this is is the part I'm confused by ...

when Pb3O4 is heated it decomposes to PbO and PbO2
enthalpy change of the reaction is 20kJmol-1
so now i need to find the enthalpy change of formation of PbO2 from this data

I'm struggling with the same question! especially part b where we gotta find the enthalpy change for the formation of Pb3O4, i got 78 kJ/mol when the markscheme says -78 kJ/mol!
And i think heating potassium oxide in oxygen is considered combustion so why is the enthalpy change positive when i calculate it from PbO: 3*-219 KJ/MOL and Pb3O4: -734
Original post by Trudy9
Hey, I'm really struggling with Hess's cycle especially of a reaction.
I've been doing this question for ages and i still can't get my head around it could anyone help ?

Question
Lead forms several solid oxides the most common ones are PbO , PbO2 and Pb3O4.
Part 1 said write an equation for the formation of PbO
then Part 2 said, 3PbO+1/2O2= Pb3O4
enthalpies of formation are:
PbO=-219
Pb3O4=-735

I worked out the standard enthalpy change of the reaction and it came to 78kJmol-1

but this is is the part I'm confused by ...

when Pb3O4 is heated it decomposes to PbO and PbO2
enthalpy change of the reaction is 20kJmol-1
so now i need to find the enthalpy change of formation of PbO2 from this data

I think i figured out how to calculate the standard enthalpy change of formation of PbO2,

basically, as you know, enthalpy change = sum of bonds broken - sum of bonds made

so, the bond broken is -735 of Pb3O4 and the bonds made are 2*-219 of 2PbO
and they have already given you the enthalpy change of the reaction.


so -735 -(-438 (which is -219*2) + PbO2) = 20
-735+ 438 is -297 now we are left with: -297 + PbO2 = 20
so E(PbO2) = -317


I recalculated this many times and this is the only correct answer i got, however the mark scheme said the right answer to be is -277 which means they added 20 to -297 which doesn't make sense on why they'd do that!
Reply 29
Where did the 20 come from?
May I know where the 20 comes from? The qs didn’t mention 20 from the beginning till the end.
is it -317 or 317
Search on pdfcookie or look it up on google As chemistry edexcel CGP free pdf
Original post by Trudy9
Hey, I'm really struggling with Hess's cycle especially of a reaction.
I've been doing this question for ages and i still can't get my head around it could anyone help ?

Question
Lead forms several solid oxides the most common ones are PbO , PbO2 and Pb3O4.
Part 1 said write an equation for the formation of PbO
then Part 2 said, 3PbO+1/2O2= Pb3O4
enthalpies of formation are:
PbO=-219
Pb3O4=-735

I worked out the standard enthalpy change of the reaction and it came to 78kJmol-1

but this is is the part I'm confused by ...

when Pb3O4 is heated it decomposes to PbO and PbO2
enthalpy change of the reaction is 20kJmol-1
so now i need to find the enthalpy change of formation of PbO2 from this data


I have to say that I am struggling too... And then you are mentioning that the standard enthalpy change of decomposition reaction of Pb3O4 into PbO and PbO2 is 20 kJ.mol-1. I was wondering where you got this information from because in my book, no mention of it, which makes the whole question even harder... How did you get this value of 20 kJ.mol-1?
By the way, the standard enthalpy change of reaction for 3PbO+1/2O2= Pb3O4 is -78 kJ.mol-1 (negative)
If you have this value of 20 kJ.mol-1, then you can apply Hess's cycle considering all the standard enthalpy changes of formation:
Pb3O4 -> 2 PbO + PbO2 Delta H decomposition (Pb3O4)= 20 kJ.mol-1 (1)
Pb3O4 -> 3 Pb + 2 O2 - Delta H formation (Pb3O4) = 735 kJ.mol-1 (2) positive sign this time
2 Pb + O2 -> 2 PbO 2 x Delta H formation (PbO) = 2 x (-219) kJ.mol-1 (3)
So Delta H formation (PbO2)= (1) - (2) - (3) = 20 - 735 - 2x(-219) = 277 kJ.mol-1
But again, my main question is how you got the value of 20 kJ.mol-1

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