Lularose83
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Calculate the enthalpy change for thermal decomposition of lithium carbonate (li2c03) given enthalpy of formation for li2c03- -1216 kj mol-1, li02- -598 kj mol-1 and c02- -394 kj mol-1. I got 224 kj mol-1, but I don't think it's right.

Calculate the enthalpy change for the reaction; c(graphite)>> c(diamond)
Enthalpy of combustion of carbon- graphite is -393.5 kj mol-1 and carbon-diamond is -395.4 kj mol-1.
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Lularose83
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Bump
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Pigster
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Can you show us your try, please?
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Lularose83
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For the first one I subtracted the product values (lithium oxide and co2) from the reactant enthalpy value (lithium carbonate) and got 224 kj mol-1.

I haven't tried the second one yet because I don't know what to do!
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Pigster
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Have you drawn a Hess' cycle?

Which way do your arrows points? Up or down?
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Lularose83
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(Original post by Pigster)
Have you drawn a Hess' cycle?

Which way do your arrows points? Up or down?
They point down! I think I drew the cycle correctly, I just wanted to see the answer that someone else got
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Pigster
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The arrows represent the arrows in a normal balanced equation. You're suggesting that enthalpy of formation represents the conversion of compounds into their elements. Does that match up with the definition (you REALLY HAVE to know VERY WELL - they will ask it every other paper)?
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Lularose83
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(Original post by Pigster)
The arrows represent the arrows in a normal balanced equation. You're suggesting that enthalpy of formation represents the conversion of compounds into their elements. Does that match up with the definition (you REALLY HAVE to know VERY WELL - they will ask it every other paper)?
Yes, they match. I think the main question that I wanted to ask was whether you subtract enthalpy of products from enthalpy of reactants or the other way round?
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Pigster
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You're given enthalpy of formation values, the arrows must point from the elements (assuming you've put them underneath the equation) up to the compounds.

I teach it thus:

Starting with Li2CO3 you can either decompose it to CO2 and Li2O OR you can do the reverse of enthalpy of formation and form the elements.

Since 2Li + C + 3/2O2 -> Li2CO3 has DH = -1216, the reverse must therefore have DH = - (-1216)

Now you've got a bunch of separate elements one can re-combine them to form our products, i.e. do DH of formation of Li2O and CO2. In this case we're going in the same direction of the arrows (pointing upwards) so we'll add the DH values.

The algebra becomes: DH(decomposition) = - (-1216) + (-598) + (-394) = +214

Alternatively, you can do products - reactants, but it could be reactants - products, depending on whether you're using enthalpy of formation or combustion data (I can't remember which way round they go!).
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Lularose83
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(Original post by Pigster)
You're given enthalpy of formation values, the arrows must point from the elements (assuming you've put them underneath the equation) up to the compounds.

I teach it thus:

Starting with Li2CO3 you can either decompose it to CO2 and Li2O OR you can do the reverse of enthalpy of formation and form the elements.

Since 2Li + C + 3/2O2 -> Li2CO3 has DH = -1216, the reverse must therefore have DH = - (-1216)

Now you've got a bunch of separate elements one can re-combine them to form our products, i.e. do DH of formation of Li2O and CO2. In this case we're going in the same direction of the arrows (pointing upwards) so we'll add the DH values.

The algebra becomes: DH(decomposition) = - (-1216) + (-598) + (-394) = +214

Alternatively, you can do products - reactants, but it could be reactants - products, depending on whether you're using enthalpy of formation or combustion data (I can't remember which way round they go!).
That makes sense, thank you very much!
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Pigster
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What do you get for the other Q?
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