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A scientist measures the standard enthalpy change for the following reaction to be -899.6 kJ :...

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.

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Answer #1

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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