Answer to Question #78250 in General Chemistry for Chloe Lozano

Question

What is the ΔHo of the equation 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (g)? Given: ΔHfo NH3 = –45.9 kJ/mol, ΔHfo NO = 90.3 kJ/mol, ΔHfo H2O = –242 kJ/mol.

A. ΔHo = 90.7 kJ
B. ΔHo = –90.7 kJ
C. ΔHo = 907 kJ
D. ΔHo = –907 kJ
E. None of the Above

Answer 1

ΔH0 = (4*ΔHf0(NO) + 6*ΔHf0(H2O)) - (4*ΔHf0(NH3) + 5*ΔHf0(O2))

ΔH0 = (4*(90.3) + 6*(-242)) - (4*(-45.9) + 5*(0)) = 361.2 – 1452 + 183.6 = -907.2 kJ

D. ΔH0 = -907 kJ