# Hess cycles.

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#1
Could someone please explain the differences between the two Hess cycles and when to use each one.
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#2
Are the two standard exam questions:
- calculate the enthalpy change of the reaction
- calculate the enthalpy change of formation
any help is appreciated.
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6 years ago
#3
(Original post by OL350)
Are the two standard exam questions:
- calculate the enthalpy change of the reaction
- calculate the enthalpy change of formation
any help is appreciated.
Hmm. I'm not terribly familar with two different versions of Hess cycles (I studied AQA).

I think another question AQA liked was to explain that they're theoretical and can differ in practice due to covalent bonding and not just ionic bonding.
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6 years ago
#4
There aren't two kinds of Hess cycles? I don't think you have understood it correctly. Lets say for example we have a reaction where A+B goes to C. In order to find the enthalpy of this reaction we will use the cycle. If we knew how to go directly from A and B to C then we would not need to apply the Hess Cycle, however we don't. However we do know that A and B break down into D and E, and the reaction of D and E also makes C. We know the enthalpy of formation for D and E, we also know the enthalpy of formation for C from D and E. Therefore, we find the enthalpy of A+B-->C by using the enthalpy of A+B--->D+E with the addition of enthalpy from D+E--->C.

This may be a confusing reply, but I hope it helps
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#5
(Original post by ElChapo)
There aren't two kinds of Hess cycles? I don't think you have understood it correctly. Lets say for example we have a reaction where A+B goes to C. In order to find the enthalpy of this reaction we will use the cycle. If we knew how to go directly from A and B to C then we would not need to apply the Hess Cycle, however we don't. However we do know that A and B break down into D and E, and the reaction of D and E also makes C. We know the enthalpy of formation for D and E, we also know the enthalpy of formation for C from D and E. Therefore, we find the enthalpy of A+B-->C by using the enthalpy of A+B--->D+E with the addition of enthalpy from D+E--->C.

This may be a confusing reply, but I hope it helps
Hmm. It's the direction of the arrows I'm struggling on and when to use what I each situation.
- Calculating the enthalpy change of combustion (both arrows point up)?
- Calculating the enthalpy change of formation (both arrows point down)?
Is that correct?
I seem to think there is another one...'calculate the enthalpy change of the reaction?'
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6 years ago
#6
(Original post by OL350)
Hmm. It's the direction of the arrows I'm struggling on and when to use what I each situation.
- Calculating the enthalpy change of combustion (both arrows point up)?
- Calculating the enthalpy change of formation (both arrows point down)?
Is that correct?
I seem to think there is another one...'calculate the enthalpy change of the reaction?'
Okay yes, the direction can be confusing to some from what I remember. I like to think of it similar to vectors in maths, If your enthalpy change going from the reactants (A and B) to the intermediates (D and E) is in terms of an enthalpy of combustion then the arrow will go down, however if it was an enthalpy of formation it would point up as you form A and B from D and E so the direction shows you the route. In calculations if you follow the arrow direction, you should make any enthalpy where you go against the arrow direction negative, and if you go with the direction of the arrow it is positive. Going from D and E to C will be a formation so your arrow will point downwards if you were to use another enthalpy of combustion, as C combusts to form D and E. It all depends on the data you get in a question, it will either be formation enthalpies or combustion enthalpies.

Calculating the enthalpy change of the reaction is the end result.
Its a bit hard to say without seeing example questions, I suggest you try some past papers, and see how that goes. Good luck.
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#7
(Original post by ElChapo)
Okay yes, the direction can be confusing to some from what I remember. I like to think of it similar to vectors in maths, If your enthalpy change going from the reactants (A and B) to the intermediates (D and E) is in terms of an enthalpy of combustion then the arrow will go down, however if it was an enthalpy of formation it would point up as you form A and B from D and E so the direction shows you the route. In calculations if you follow the arrow direction, you should make any enthalpy where you go against the arrow direction negative, and if you go with the direction of the arrow it is positive. Going from D and E to C will be a formation so your arrow will point downwards if you were to use another enthalpy of combustion, as C combusts to form D and E. It all depends on the data you get in a question, it will either be formation enthalpies or combustion enthalpies.

Calculating the enthalpy change of the reaction is the end result.
Its a bit hard to say without seeing example questions, I suggest you try some past papers, and see how that goes. Good luck.
To simplify this would I be correct in saying:
- if formation figures are given then both arrows point up
if combustion figures are given then both arrows point down
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6 years ago
#8
(Original post by OL350)
To simplify this would I be correct in saying:
- if formation figures are given then both arrows point up
if combustion figures are given then both arrows point down
Yes I believe thats correct.
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6 years ago
#9
Both arrows downwards for Combustion (bottom should have co2 and water), both arrows upwards for formation (bottom should have elements, yus

When calculating, make the arrow that has been flipped' delta H value negative.
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#10
(Original post by AnnekaChan173)
Both arrows downwards for Combustion (bottom should have co2 and water), both arrows upwards for formation (bottom should have elements, yus

When calculating, make the arrow that has been flipped' delta H value negative.

going by what you've said the top answer must be correct? How would I go about answering the second one though?
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#11
(Original post by ElChapo)
Yes I believe thats correct.
Any help on the above question^?
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6 years ago
#12
(Original post by OL350)
Any help on the above question^?
Work your way backwards in the cycle 0
#13
(Original post by ElChapo)
Work your way backwards in the cycle Am I right in saying the answer is (-718) - (-10) = +708?
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6 years ago
#14
(Original post by OL350)
Am I right in saying the answer is (-718) - (-10) = +708?
I don't know, check the mark scheme, I think you're riight though
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#15
(Original post by ElChapo)
I don't know, check the mark scheme?
I don't have one ynfortunately...they were questions made up by my teacher 0
6 years ago
#16
(Original post by OL350)
I don't have one ynfortunately...they were questions made up by my teacher Yeah I think you're right
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#17
(Original post by ElChapo)
Yeah I think you're right
Apparently it's -274...any ideas how?
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6 years ago
#18
(Original post by OL350)
x

Apparently it's -274...any ideas how?
You also form 2 moles of PbO so you have to take the enthalpy change associated with those 2 moles into account.
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#19
(Original post by ElChapo)
You also form 2 moles of PbO so you have to take the enthalpy change associated with those 2 moles into account.
how does this look to you?
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6 years ago
#20
(Original post by OL350)
how does this look to you?
x
I am not too sure sorry, it looks like you have got the answer, so as long as you can make sense of it then its good. This is all a distant memory to me, I'm now in my second year of Chemical Engineering 0
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