Hess cycles.

Are the two standard exam questions:
- calculate the enthalpy change of the reaction
- calculate the enthalpy change of formation
any help is appreciated.

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?'

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

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.

Yes I believe thats correct.

Any help on the above question^?

Work your way backwards in the cycle

Am I right in saying the answer is (-718) - (-10) = +708?

I don't know, check the mark scheme?

I don't have one ynfortunately...they were questions made up by my teacher

Yeah I think you're right

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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.

how does this look to you?
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