OCR Chemistry A F322 Chains, Energy and Resources Wed 23 May 2012

I'm curious about people so I'm wondering if anyone here could answers these:

What other A levels do you take along with Chemistry?
What did you get in Chemistry at GCSE and what are you aiming to get at A level?
Are you going to carry on with chemistry next year?
What course are you looking to do at uni?

I'll answer for myself.
1) Biology, Maths, Psychology
2) C (yep, I know, don't look at me in shame :frown:

)
3) Yes

4) Chemistry or Optometry

I'd love if other peeps could take a few secs to answer!

1) Biology, Maths, Religious Studies
2) A and A* in core and additional science
3) Yeah :smile:

4) Medicine

Reply 1181

Pride

21

Original post by thescientist17

But the enthalpy of the products is greater as it says it's 872 or whatever it was which is bigger than 697?

I got 771 actually... I don't know...

anyway, remember there are two H-Cl bonds to be formed, not 1.

Reply 1182

1a2b

Original post by LegendX

cus, u know the catalyst helps to bring reactions to an equilibrium quicker without actually changing the position of the equilibirum, hence, one of the rates must decrease for the other one to increase for it to reach equilibrium....

comprende?

i do thanks!

Reply 1183

kingam

Original post by narli

wowwww....i suppose this exam isnt THAT hard
good luck!

Lol we will find out tomorrow and thanks , good luck to you too I'm
Sure you will do great :smile:

Reply 1184

Pride

21

Original post by ReTurd

I have a question
when doing enthalpy change of reaction, which reactant/product do you use the moles of for enthaply change?
E.g. For enthaply of combustion you use the moles of reactant being burnt. and for enthalpy of formation you use the moles of the product you want formed.
So which do you use for enthalpy of reaction?

well, you're involving all the moles in the equation, just as you would if you were working it out using average bond enthalpies instead.

So you multiply the particular parts of the equation, depending on the number of moles in the balanced equation.

does that make much sense? If you've got an example question...

Reply 1185

thescientist17

9

Just watched a video and very confused. He said the energy lost by the chemical system was -200. Enthalpy reactants 2600 enthalpy produts 2800. I thought exothermic was when enthalpy of products was smaller than enthalpy of reactants? Could someone please explain enthalpy to me :frown:

This was posted from The Student Room's iPhone/iPad App

Reply 1186

Synchrome

Original post by ReTurd

I have a question
when doing enthalpy change of reaction, which reactant/product do you use the moles of for enthaply change?
E.g. For enthaply of combustion you use the moles of reactant being burnt. and for enthalpy of formation you use the moles of the product you want formed.
So which do you use for enthalpy of reaction?

you look for the enthalpy change thats indicated in the question it can be comubstion or formation.

Reply 1187

poyyo

6

bleh i'm suddenly feeling nervous now :/

Reply 1188

DingDong!

7

Original post by D4rth

Oh it's one of you... Loreto ftw

This was posted from The Student Room's iPhone/iPad App

Just bitter because we beat you in the league tables. But it's fine, we're very modest of our achievements. :cool:

Reply 1189

sr2123

Hey guys, another question:

How would you determine quickly, the number of structural isomers of : C8H18 and C8H17O

Reply 1190

King Boo

14

Original post by med123

here is a question

chain length inceases, value of enthalphy change of combustion becomes more negative. Use your understanding of bond breaking and bond making to explain this trend?

As chain length increases, so does the amount of and in turn strength of the van der waals forces. (presuming we are talking about a hydrocarbon). If bond making is an exothermic process, then as more bonds need to be broken, which is an endothermic process, it means the enthalpy change of combustion becomes more negative, as more enthalpy is found due to the more presence of stronger bonds?

ISH? :s-smilie:

Reply 1191

Tania2k9

can someone give me the link to the jan 2012 f322 paper please? :wink:

Reply 1192

LegendX

15

hi, my question:

whats the answer for question 1b iv?
Originally i had assumed that, because in the stem of the question it stated that myrcene had a triple bond but my answer was not like theres.

http://www.ocr.org.uk/download/pp_10_jun/ocr_57549_pp_10_jun_gce_f322.pdf

chem question1.png16.5KB

Reply 1193

thescientist17

9

Q4 jan 2012 how is it exothermic if the enthalpy of products is (184 + 697) and enthalpy of reactants is 697? Shouldn't enthalpy of products be smaller? Someone please help me I'm really confused! :confused:

This was posted from The Student Room's iPhone/iPad App

Reply 1194

poyyo

6

Original post by sr2123

Hey guys, another question:

How would you determine quickly, the number of structural isomers of : C8H18 and C8H17O

i'd also like to know this

Reply 1195

myyrh

9

Original post by thescientist17

Q4 jan 2012 how is it exothermic if the enthalpy of products is (184 + 697) and enthalpy of reactants is 697? Shouldn't enthalpy of products be smaller? Someone please help me I'm really confused! :confused:

This was posted from The Student Room's iPhone/iPad App

Delta H= Bonds broken - Bonds made
So for bonds broken its 436+243=679
bonds made is 2x (we are trying to find out the H-Cl bond which is made twice)
679-2x=-184
rearrange to
679+184=2x
863=2x
x=431.5

Reply 1196

sr2123

Original post by LegendX

hi, my question:

whats the answer for question 1b iv?
Originally i had assumed that, because in the stem of the question it stated that myrcene had a triple bond but my answer was not like theres.

http://www.ocr.org.uk/download/pp_10_jun/ocr_57549_pp_10_jun_gce_f322.pdf

Yes that it what the original compound looks like. However, they are now asking for an isomer of what you drew.

Reply 1197

sr2123

Guys please:

How would you determine quickly, the number of structural isomers of : C8H18 and C8H17O

Reply 1198

thescientist17

9

Original post by myyrh

Delta H= Bonds broken - Bonds made
So for bonds broken its 436+243=679
bonds made is 2x (we are trying to find out the H-Cl bond which is made twice)
679-2x=-184
rearrange to
679+184=2x
863=2x
x=431.5

Thanks, I think I'm getting confused with delta H = delta H products - delta H reactants.. What do you use this equation for? And if the enthalpy is 863 why is it exothermic as this is greater than 679? Or is this different because it's a different equation? I don't really understand the difference!

Reply 1199

myyrh

9

Original post by thescientist17

Thanks, I think I'm getting confused with delta H = delta H products - delta H reactants.. What do you use this equation for? And if the enthalpy is 863 why is it exothermic as this is greater than 679? Or is this different because it's a different equation? I don't really understand the difference!

The question is regarding bond enthalpies, not enthalpy change of formation/combustion. When dealing with bond enthalpies always use the formula (bonds broken(endothermic process) -bonds made(exothermic process). The overall enthalpy change is exothermic (-184) because the bonds made> bonds broken. Think of it in numbers, 679 is less than 863 so overall it's exothermic.
What you seem to be thinking about is regarding enthalpy change of formation/ combustion. This is when you would use the relevant equations.

-If information is given in enthalpy change of formation (the little f) use the following
Products-reactants

-If information is given in enthalpy change of combustion (the little c) use the following
reactants-products

OCR Chemistry A F322 Chains, Energy and Resources Wed 23 May 2012

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