Question

A scientist 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.

Answer 1

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:

ΔºHrxn = 4 ( ΔHºf NO ) + 6 ( ΔHºf H₂O) - ( ΔHºf NH₃ + 5 ΔHºf O₂

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