Question

Given the following thermochemical equations,
NO(g) + (1/2) Cl2(g) --> NOCl(g) -37.78 kJ (per mol NOCl)
NO(g) + (1/2) O2(g) --> NO2(g) -56.53 kJ (per mol NO2)
2 NO2(g) --> N2O4(g) -58.03 kJ (per mol N2O4)
what is the standard enthalpy change for the reaction below?
N2O4(g) + Cl2(g) --> 2 NOCl(g) + O2(g)
Give your answer in kJ (per mol N2O4), accurate to two decimal places.

Answer 1

I did this in my head and I BELIEVE (but you need to try it to see)
eqn 3 reversed. Add
to 2x eqn 1
and add to the reverse of 2 x equn 2.

Add the dH values. When you multiply a rxn by a coefficient do the same for dH. When you reverse and equation change the sign on dH.