Hess's Law urgent

I tend to use the formulas when doing any questions related to energetics e.g. enthalpy of formation = products - reactants
rather than use the cycle. I have this question however which I've no idea how to approach:
Calculate the standard enthalpy change of formation of liquid Hydrogen Peroxide, H2O2, from the following information:
2H2(g) + O2(g) ---> 2H2O(I)
where the standard enthanlpy change of reaction = -572kJ mol^-1
2H2O2(I) ---> 2H2O(I) + O2(g)
where the standard enthanlpy change of reaction = --196kJ mol^-1

Can someone please guide me...thanks in advance...

Reply 1

TSR Jessica

Sorry you've not had any responses about this. :frown:

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Reply 2

Katarinanokat

Standard enthalpy change of formation would have the formula:

H₂ + O₂ --> H₂O₂

You know this information:

2H₂(g) + O₂(g) ---> 2H₂O(I) -572kJ mol^-1 (-1144kJ, as there are 2 mols)

2H₂O₂(I) ---> 2H₂O(I) + O₂(g) -196kJ mol^-1 (-392 kJ, as there are 2 mols)

So your enthalpy cycle will have 2 steps:
2H₂(g) + O₂(g) ---> 2H₂O(I)
2H₂O(I) + O₂(g) ---> 2H₂O₂(I)

Notice how the water cancels out, and on the left sides of the equations you have you only use 2 mols of hydrogen gas and 2 mols of oxygen gas (as the water cancels out)

Therefore we can begin to add the equations. The first step is the same as given to us so we have:
2H₂(g) + O₂(g) ---> 2H₂O(I) -1144kJ

But the next step is flipped that what is given to us, so we must flip the sign of the enthalpy change:
2H₂O₂(I) ---> 2H₂O(I) + O₂(g) 392 kJ

Adding our enthalpy values (using hess' law, which states that regardless of the pathway the reaction has the same enthalpy change)
392 - 1144= -752 kJ

but note we we have 2 mols of hydrogen peroxide formed, which is not the standard enthalpy of formation (as that is 1 mol) so we must half our value:
-376 kJmol^-1

However, I must note that this only works if the book wrote per mole, not for that entire reaction.

Does that work?