AQA - Unit 5 - Energetics, Redox and Inorganic Chemistry

Reply 220

508836

3

How many marks out of 100 should I be looking at if I want to get an A* in this paper?

Reply 221

jimmy303

1

Original post by OhNo_23

In the specification it says that we need to be able to define enthalpy of atomisation of an element and of a compound. I have never come across one for a compound :/ Does anyone have any ideas of how this would be defined or have an example of an equation where this is the case?
Thanks in advance :smile:

Enthalpy of atomisation of an element: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.

Eg. Na(s) --> Na(g)

Enthalpy of atomisation of a compound: the enthalpy change when one mole of a compound in its standard state is converted into its constituent atoms in gaseous form.

E.g. CH4(g) --> C(g) + 4H(g)

The enthalpy of atomisation of a compound is equal to the bond energies of the compound. E.g. in the above reaction atomisation energy is equal to 4(C-H) bonds.

Reply 222

asb15

has anyone got the jan 11 paper for chemistry unit 5 and the mark scheme? it would be very appreciated, thanks :smile:

Reply 223

OhNo_23

5

Original post by jimmy303

Enthalpy of atomisation of an element: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state.

Eg. Na(s) --> Na(g)

Enthalpy of atomisation of a compound: the enthalpy change when one mole of a compound in its standard state is converted into its constituent atoms in gaseous form.

E.g. CH4(g) --> C(g) + 4H(g)

The enthalpy of atomisation of a compound is equal to the bond energies of the compound. E.g. in the above reaction atomisation energy is equal to 4(C-H) bonds.

Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?

Reply 224

ncollier4

Original post by OhNo_23

Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?

Ermm.......you just need to know the definition and be able to apply and use it for Born-Haber cycles :smile:

Think only when you apply it you only need to know its role in the Born-Haber cycle so you can work out the lattice enthalpy etc!

Reply 225

jimmy303

1

Original post by OhNo_23

Brilliant, thanks for your help! So is it essentially the same thing as mean bond enthalpy, because I notice that is not listed in the spec?

Actually it isn't. It will be similar but a value using mean bond enthalpies will be less accurate.

It is the same principle as comparing calculations using mean bond enthalpies with calculations using the born haber cycle.

With mean bond enthalpies the value is the average energy needed to break a covalent bond over a range of different compounds.

The atomisation of a compound on the other hand will only be looking at the energy needed to break the covalent bonds in that one compound.

Does that make sense?

Reply 226

OhNo_23

5

Original post by jimmy303

Actually it isn't. It will be similar but a value using mean bond enthalpies will be less accurate.

It is the same principle as comparing calculations using mean bond enthalpies with calculations using the born haber cycle.

With mean bond enthalpies the value is the average energy needed to break a covalent bond over a range of different compounds.

The atomisation of a compound on the other hand will only be looking at the energy needed to break the covalent bonds in that one compound.

Does that make sense?

That makes perfect sense, thanks so much! :smile:

Reply 227

jimmy303

1

Original post by T-Toe

How many marks out of 100 should I be looking at if I want to get an A* in this paper?

For the last three papers an A* (i.e. 90% ums points) has consistently been around 82-84 marks out of 100.

Reply 228

Seasick Steve

12

Original post by laura123

Do we need to know all the little details about electrochemical cells? Pleeeeeease say no! :frown:

I don't think we need to know about specific non- / rechargeable cells apart from from I think the hydrogen fuel cell.

Reply 229

starburst92

2

Original post by asb15

has anyone got the jan 11 paper for chemistry unit 5 and the mark scheme? it would be very appreciated, thanks :smile:

http://www.thestudentroom.co.uk/showthread.php?t=1611471

Reply 230

BethBeth

Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? :s-smilie:

I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand :frown:

Any help would be much appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-CHEM5-W-QP-JAN10.PDF

Reply 231

starburst92

2

Original post by BethBeth

Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? :s-smilie:

I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand :frown:

Any help would be much appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-CHEM5-W-QP-JAN10.PDF

You times by 10^6 as right now you're only using a 25cm^3 sample. You need to make this 25000dm^3. times by 1000 to make it dm^3 and then another 1000 to make it 25000. :smile:

Reply 232

Cameronn

Original post by BethBeth

Hi, can anyone help me with question 8c) on January 2010 paper?? I understand up to calculation moles of H3PO4, but don't know where to go next? :s-smilie:

I looked at the mark scheme and it said the next mark was for times it by 10^6... I don't understand :frown:

Any help would be much appreciated.

http://store.aqa.org.uk/qual/gce/pdf/AQA-CHEM5-W-QP-JAN10.PDF

Hey, yeah this is a nasty mathy question for 5 marks.

Firstly H3PO4 + 3NaOH --> Na3PO4 + 3H20
- 1:3 ratio of H3PO4:NaoH

n(mole) of NaOH added = (21.2/1000)*0.500 = 0.0106 mol
=> therefore n(mole) H3PO4 in 25cm^3 sample = 0.0106/3 = 3.533333*10^-3 mol

container is 25000dm^3
25cm^3 = 0.025dm^3
=> 25000/0.025 = 1000000 25cm^3 in 25000dm^3 - hence where the factor of 1000000 came from

therefore n(mole) H3PO4 in 25dm^3 = (3.53333*10^-3) * 1000000 = 3533.33333 mol

P4O10 + 6H2O ---> 4H3PO4
=> therefore n(mole) P4O10 = 3533.33333/4 = 883.3333mol

n = mass/mr => mass = n*mr
=> mass = (883.333333)*((16*10)+(31*4)) = 250866.6667g
= 250.8666667kg = 251kg :smile:

Reply 233

BethBeth

Original post by starburst92

You times by 10^6 as right now you're only using a 25cm^3 sample. You need to make this 25000dm^3. times by 1000 to make it dm^3 and then another 1000 to make it 25000. :smile:

Original post by Cameronn

Hey, yeah this is a nasty mathy question for 5 marks.

Firstly H3PO4 + 3NaOH --> Na3PO4 + 3H20
- 1:3 ratio of H3PO4:NaoH

n(mole) of NaOH added = (21.2/1000)*0.500 = 0.0106 mol
=> therefore n(mole) H3PO4 in 25cm^3 sample = 0.0106/3 = 3.533333*10^-3 mol

container is 25000dm^3
25cm^3 = 0.025dm^3
=> 25000/0.025 = 1000000 25cm^3 in 25000dm^3 - hence where the factor of 1000000 came from

therefore n(mole) H3PO4 in 25dm^3 = (3.53333*10^-3) * 1000000 = 3533.33333 mol

P4O10 + 6H2O ---> 4H3PO4
=> therefore n(mole) P4O10 = 3533.33333/4 = 883.3333mol

n = mass/mr => mass = n*mr
=> mass = (883.333333)*((16*10)+(31*4)) = 250866.6667g
= 250.8666667kg = 251kg :smile:

Thank you! Really appreciate this, think I'll be able to do this if it turns up on friday now :smile:

Hope your revision is going well :biggrin:

x

(edited 12 years ago)

Reply 234

azaking

Original post by ncollier4

Thank youuuuuuuuuuuuuuuuuuuuuuu :biggrin:

But where does the catalyst come in? :confused:

According to the nelson thornes book it would seem you dont need to know it. After the two equations all it states is the catalyst that's used then starts waffling on about other common catalyst . :smile:

Reply 235

508836

3

Hey fellow chemists, do you think it's worth doing these exercises (example below)?

http://www.a-levelchemistry.co.uk/AQA%20Chemistry/AQA%20A2%20Chemistry/Unit%205/5.1%20Thermodynamics/5.1%20Thermodynamics%20home.htm

Or just sticking with past papers? I'm pretty content with revision so far so I'm not sure :/

Reply 236

l4ith

10

I think this is my least favourite unit, it's pretty hard tbh

Reply 237

Raimu

I need about 60/120 for an A in Chemistry overall. As a result I haven't really revised for this paper as much as I should have, hopefully I can pull through.

Reply 238

BethBeth

Original post by Raimu

I need about 60/120 for an A in Chemistry overall. As a result I haven't really revised for this paper as much as I should have, hopefully I can pull through.