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Hess's law cycle

I'm often confused with how to draw a Hess's law cycle and how you work out the alternate route for the reaction and then calculate the enthaply change of something. Can someone explain and please teach me by using the following example.

C3H8 has the standard enthaply change of combustion of -2220 kJmol-1
C has the standard enthaply change of combustion of -394 kJmol-1
H2 has the standard enthaply change of combustion of -286 kJmol-1

The equation that represents the standard enthaply change of formation of propane is shown below.

3C(s) + 4H2(g) -->C3H8(g)

Use Hess's law and the data about the enthaply change of combustion to calculate a value of the enthaply change of formation of propane.

Thanks.
Reply 1
Ok... enthalpy change = enthalpy change of combustion (reactants) - enthalpy change of combustion (products)

I just memorised that formulae, its the opposite for formation

therefore enthalpy change of formation of propane is:

(-394 + -286) - (-2220)

= -680 + 2220
= 1540 Kj per mol (and so is endothermic)

hope that helps! it took me ages to get that 2! :biggrin: xxx
Reply 2
Thanks. There are formulae for the enthaply change?

What would the formula be if it were to claculate enthaply change of formation and reaction? Thanks.
There isn't specific formulae but how i remeber it is when you construct it, the addition of the arrows going in the same direction = the value of the arrow going in the opposite direction. If you get what i mean
Reply 4
Always remeber than if the arrows are going in the oppsite direction the value changes and you're sorted.
Reply 5
:ditto: I know this is very very obvious, but lots of ppl make the same mistake: - (-) gives + !!!!!
Reply 6
I think you'll find i did put a plus thankyou :biggrin:

And the other formula is just switching the the reactants and products around.

Have you got the aqa modula books?
The formulas are in likle squicky boxes on the hess's law pages! :biggrin:

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