Enthalpy of formation using simulatenous equations? how?

Hey,

I know they're are two ways of doing this question the non-simulatenous equation way (which I can do), but the other way it says rearranging the below equations and solving the simulatenous equation. If someone could show me how to derive the equation it would be great

Given the following enthalpy changes for the indicated reactions, calculate the standard enthalpy change of formation of HNO_3(l)?

H₂0_2(l) + 2NO_2(g) => 2HNO_3(l) Delta H = -226.8 kJ mol^-1
N_2(g) + 2O_2(g) => 2NO_2(g) Delta H = 66.4 kJ mol^-1
H_2(g) (g) + O₂ (g) => H₂O_2(l) Delta H = -187.8 kJ mol^-1

H₂0_2(l) + 2NO_2(g) + N_2(g) + 2O_2(g) + H_2(g) (g) + O_2(g) => 2HNO_3(l) + 2NO_2(g) + H₂O_2(l)

The reagents in red and blue cancel (since in effect the values of the bond enthalpies cancel since we get the same bonds before and after) to give:
N_2(g) + 2O_2(g) + H_2(g) (g) + O_2(g) => 2HNO_3(l)

So the enthalpy change for the above reaction is:
= -226.8 kJ mol^-1+ 66.4 kJ mol^-1 -187.8 kJ mol^-1
= -348.2 kJ mol^-1

But the enthalpy of formation of HNO_3(l) is defined as per a mole of substance.

The reaction for the enthalpy of formation would be:
½N_2(g) + O_2(g) + ½H_2(g) (g) + ½O_2(g) => HNO_3(l)

½N_2(g) + 3/2O_2(g) + ½H_2(g) (g) => HNO_3(l)

This is half the enthalpy change of the calculated value.
= ½(-348.2) kJ mol^-1
= -174.1 kJ mol^-1

Using simultaneous equations
½H₂0_2(l) + NO_2(g) => HNO_3(l) (1)

N_2(g) + 2O_2(g) => 2NO_2(g) (2)

H_2(g) (g) + O₂ (g) => H₂O_2(l) (3)

Equations (3) and (2) into (1)
You get the same result as above.

E.g. (2) into (1)
½H₂0_2(l) + ½(N_2(g) + 2O_2(g)) => HNO_3(l) (1)