Question

Consider the reaction of nitrogen dioxide with molecular hydrogen: 2 NO2(g) + 7 H2(g) + 4H2O(l) + 2 NH3(g) Calculate AH°rxn for this reaction given by manipulation of reactions (i) and (ii): 2 NH3(g) → N2(g) + 3 H2(g) AH°rxn = +92.0 kJ (i) N2(g) +4 H2O(l) ---> 2 NO2(g) + 4 H2(g) AHºrxn = +3.40E2 kJ (ii)​

Answer 1

Reverse both reactions to get the final reaction proposed. When you reverse a reaction, you just flip the sign on the enthalpy of reaction. Then add the two.

-92.0 kJ - 340. kJ = -432 kJ