I don't know. Your second equation makes no sense. I think you must have meant 5/4 O2 and not 5/2. Assuming you meant 5/4, there are two steps to perform.
1. The equation is reversed; therefore, dH will tbe the negative of the forward reaction.
2. You've gone from 4 NO to NO and the new equation is 1/4 of of the old so take 1/4 of the new dH.
Given that 4NH3+5O2-4NO+6H2O ΔH= -906kJ.
What is the Enthalpy change for NO +3/2H2O - NH3 +5/2 O2
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1 answer