Question

One step in the synthesis of nitric acid is the conversion of ammonia to nitric oxide. 4NH₃(g) + 5O₂(g) ? 4NO(g) + 6H₂O(g) Calculate ΔHᵣxn for this reaction. ΔH_f [NH₃(g)] = -45.9 kJ/mol; ΔH_f [NO(g)] = 90.3 kJ/mol; ΔH_f [H₂O(g)] = -241.8 kJ/mol.

a) -906.0 kJ
b) -197.4 kJ
c) -105.6 kJ
d) 197.4 kJ
e) 906.0 kJ

Answer 1

The enthalpy change for the conversion of ammonia to nitric oxide is -197.4 kJ.

Step-by-step explanation:

The enthalpy change ($\Delta H_{rxn}$) for the conversion of ammonia to nitric oxide can be calculated using the given values of the standard enthalpies of formation. The equation for the reaction is:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

Using Hess's law, we can find the enthalpy change:

$\Delta H_{rxn}$ = (4 * $\Delta H_f [NO(g)]$) + (6 * $\Delta H_f [H₂O(g)]$) - (4 * $\Delta H_f [NH₃(g)]$)

Substituting the given values:

$\Delta H_{rxn}$ = (4 * 90.3 kJ/mol) + (6 * (-241.8 kJ/mol)) - (4 * (-45.9 kJ/mol)) = -197.4 kJ