A scientist measures the standard enthalpy change for the following reaction to be -899.6 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 NH3(g) is kJ/mol.
we have:
deltaHo rxn = -899.6 KJ/mol
Hof(O2(g)) = 0.0 KJ/mol
Hof(NO(g)) = 90.25 KJ/mol
Hof(H2O(g)) = -241.818 KJ/mol
we have the Balanced chemical equation as:
4 NH3(g) + 5 O2(g) ---> 4 NO(g) + 6 H2O(g)
deltaHo rxn = 4*Hof(NO(g)) + 6*Hof(H2O(g)) - 4*Hof( NH3(g)) - 5*Hof(O2(g))
-899.6 = 4*(90.25) + 6*(-241.818) - 4*Hof(NH3(g)) - 5*(0.0)
Hof(NH3(g)) = -47.6 KJ/mol
Answer: -47.6 KJ/mol
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