# enthalpy chem problem?

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Now consider the following set of reactions:

N2 + 2O2→N2O4 ,ΔH=−8 kJ/mol

N2 + O2→2NO ,ΔH=180 kJ/mol

What is the enthalpy for the following reaction?

overall: N2O4→2NO + O2

Well, Hess’ law states that there are different ways to reach a certain reaction. When you flip the equation, the enthalpy changes sign. When you double the amount of stuff you’re reacting, you double the enthalpy value. That said, we are looking to rearrange these two equations above to get N2O4 –> 2NO + O2

First we’d flip the first one to get

N2O4–>N2+2O2 which has an enthalpy of 8 kJ/mol

and we’d leave N2+O2–>2NO with an enthalpy of 180 kJ/mol

If you do it correctly, things on opposite sides will cancel each other out.

I’ll put them here again to make it easier for you to visualize it

N2O4–>N2+2O2 ΔH 8 kJ/mol

N2+O2–>2NO ΔH 180 kJ/mol

N2’s on both sides cancel, and we remove one O2 from the 2O2’s we have up there, then combine the equations.

N2O4–>2NO+O2 ΔH188 kJ/mol

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