### 4D.23 7th Ed.

Posted:

**Tue Jan 29, 2019 7:55 pm**The question asks to calculate the standard enthalpy formation of dinitrogen pentoxide from the info:

2NO+O2–>2NO2 delta H= -114.1kJ

4NO2+O2->2N2O5 delta H= -110.2 kJ

I’m confused on how to do this. The solutions show to add the first reaction with half of the second, but why are you allowed to multiply the second equation by 1/2. Then after it adds the two up it says the enthalpy of the reaction equals the enthalpy of formation of N2O5 minus twice the enthalpy formation of NO, how is that so? Can you step by step explain this problem, thank you!

2NO+O2–>2NO2 delta H= -114.1kJ

4NO2+O2->2N2O5 delta H= -110.2 kJ

I’m confused on how to do this. The solutions show to add the first reaction with half of the second, but why are you allowed to multiply the second equation by 1/2. Then after it adds the two up it says the enthalpy of the reaction equals the enthalpy of formation of N2O5 minus twice the enthalpy formation of NO, how is that so? Can you step by step explain this problem, thank you!