# Solving Enthalpy Changes for Chemical Reactions

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Calculate the enthalpy change for the following reactions

P4O6 + 2O2 ? P4O10

given the following enthalpies of reaction.

P4 + 3O2 ? P4O6 ?H = -1640.1 kJ equation 1

P4 + 5O2 ? P4O10 ?H = -2940.1 kJ equation 2

b) NO2(g) + (7/2) H2(g) ---> 2H2O(l) + NH3(g)

2NH3(g) ---> N2(g) + 3H2(g) ?HÂ° = +92 kJ equation 1

(1/2) N2(g) + 2H2O(l) ---> NO2(g) + 2H2(g) ?HÂ° = +170 kJ equation 2

Please provide detailed explanations.

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