# Enthalpy QWatch

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#1
How would you calculate a value for the enthalpy of formation of a compound when given deltaH values?
0
14 years ago
#2
it depends on what deltaH values you are given.
0
#3
Sorry, I meant deltaH values of standard enthalpy change.

This question, for example:

Use the following data to calculate a value for the enthalpy of formation of methane

C + O2 ---> CO2 deltaH = -394

H2 + 1/2O2 ---> H2O deltaH = -242

CH4 + 2O2 ---> CO2 + 2H2O deltaH = -802
0
14 years ago
#4
(Original post by BarelyLegal)
Sorry, I meant deltaH values of standard enthalpy change.

This question, for example:

Use the following data to calculate a value for the enthalpy of formation of methane

C + O2 ---> CO2 deltaH = -394

H2 + 1/2O2 ---> H2O deltaH = -242

CH4 + 2O2 ---> CO2 + 2H2O deltaH = -802

I think it's basic Hess' law stuff: energy change for a reaction is independent of route taken.

you want to find out the value for C + 2O2 + H4 --> CH4 + 2O2
to do this you need to use the values you already have, by doing the reactions in this order:
1) C + O2 --> CO2 (-394)
2) 2H2 + O2 --> 2H2O (-242 * 2 because mole numbers have to be doubled to balance it all out)
3) CO2 + 2H2O --> CH4 + 2O2 (+802 because the reaction is going opposite way as for the number you are given)

enthalpy formation of methane = -394 -484 +802

unless of course i've misunderstood and got it completely wrong... which happens a lot, sadly ;p
0
14 years ago
#5
(Original post by BarelyLegal)
How would you calculate a value for the enthalpy of formation of a compound when given deltaH values?

http://www.thestudentroom.co.uk/t87099.html

It's all there
0
#6
(Original post by fairieboi)
I think it's basic Hess' law stuff: energy change for a reaction is independent of route taken.

you want to find out the value for C + 2O2 + H4 --> CH4 + 2O2
to do this you need to use the values you already have, by doing the reactions in this order:
1) C + O2 --> CO2 (-394)
2) 2H2 + O2 --> 2H2O (-242 * 2 because mole numbers have to be doubled to balance it all out)
3) CO2 + 2H2O --> CH4 + 2O2 (+802 because the reaction is going opposite way as for the number you are given)

enthalpy formation of methane = -394 -484 +802

unless of course i've misunderstood and got it completely wrong... which happens a lot, sadly ;p
Yah - thought it might be something like that ... cheers
0
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