Ascientist measures the standard enthalpy change for the following reaction to be -953.2 kj: 4nh3(g) 5 o2(g)4no(g) 6 h2o(g) based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of h2o(g) is kj/mol.

ΔHºf H₂O = -249.6 kJ/mol

Explanation:

4NH3(g) +5 O2(g) ⇒ 4NO(g) + 6 H2O(g)

The given ΔºHrxn is equal according to Hess law to:

Since we can find in a reference book the standard enthalpies of formation for NO, NH₃ and having the ΔHºrxn, the problem reduces to an equation with one unknown. (Remember the standard enthalpies for elements, in this case oxygen, is zero)

ΔHºf NO = 90.2 kJ/mol

ΔHºf  NH₃ = -45.94 kJ/mol

therefore,

-953.2 kJ = 4 (90.2 kJ) + 6 ΔHºf H₂O  - (4(-45.94) kJ + 0 kJ ) = 360.8 kJ - 6 ΔHºf H₂O + 183.8 kJ

-953. kJ =   544.6 kJ + 6 ΔHºf H₂O

-1497.6 kJ /6=  ΔHºf H₂O

-249.6 kJ = ΔHºf H₂O