Hess's Law - Can anyone answer this?

[Amy J]

Hess’s Law states that “the standard enthalpy of a reaction can be calculated from the sum of the (delta r H) of reactions into which it can be decomposed”. Use Hess’s Law to calculate in a single step, (delta r H) for the reaction,

4NO(g)+ 6NO2(g)+12NH3(g)+O2(g) => 11.N2(g) +18H20(g)

given that

4NO(g)+4NH3(g)+O2(g) => 4N2(g)+6H2O(g), (delta r H)(1) = -1627.6 kJ mol-1

and

6NO2(g)+8NH3(g) => 7N2(g)+12H2O(g),
(delta r H)(2) = -2732.0 kJ mol-1

I'm struggling with this, when I understand hess's law. It's diffcult to write (delta r H) in symbols here, but to be exact it written as delta/subscript r/H/ degrees celcius symbol - if anyone could even tell what it stand for it would really help. I'd also really appreciate some help on the question itself as well thanks.

[j07]

You need to form a balanced Hess's cycle using these three given equations. Hess's cycle states that either route has the same energy change, so you can either take the route of ?H(reaction) or that of ?H(1) + ?H(2) as both give the same total products. Therefore:

?H(reaction) = ?H(1) + ?H(2)

?H(reaction) = (-1627.6) + (-2732.0) = -4359.6 kJ

Remember to take into account the negative signs which are important to state, along with the units. From the result this reaction is exothermic.

[drjaycat]

A nice easy example there....
you are adding the products and reactants from both these equations together to get your pverall reaction equation.
So you just do the same to the enthalpy values. In this case,

delta H overall = -1627.6 + -2732.0 kJ
= -4359.6 kJ

[minip2087]

been a while since first year chem....but i believe delta h of reaction is equal to -4359.6 kJ/mol. If you add the two reactions with known delta h you get the target reaction. Similarly the enthalpy can be added as it is a state function.