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Given the following equations:
2 H2 (g) + O2 (g)--> 2 H2O (l) deltaH = -571.6 kJ

N2 (g) + O2 (g)-->2 NO (g) deltaH = +180.5 kJ

N2 (g) + 3 H2 (g) --->2 NH3 (g) deltaH = -92.22 kJ

Determine the enthalpy change (deltaH) for the following reaction:
2 NO (g) + 5 H2 (g)--> 2 NH3 (g) + 2 H2O (l)
A) 483.3 kJ/mol
B) -483.3 kJ/mol
C) -844.3 kJ/mol
D) 844.3 kJ/mol
E) -241.7 kJ/mol

My professor didn't explain this well enough at all. and I tried to use the book, but the example problem has a different set up. I know something has to be reversed. Is it the final equation? And I know the sign becomes the opposite when it is reversed? But step by step if anyone could explain what I need to do to find the answer that would be great. I am lost.

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  1. Add equation 1, reverse equation 2, and add equation 3. Add the delta Hs and reverse the sign of delta H if you reverse the reaction.

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