An important step in the synthesis of nitric acid is the conversion of ammonia to nitric oxide. 4NH₃(g) + 5O₂(g) $\to$ 4NO(g) + 6H₂O(g)
Calculate $\Delta$ H°_rxn for this reaction.
$\Delta$ H°_f [NH₃(g) ] = -45.9 kJ/mol; $\Delta$ H°_f [NO(g) ] = 90.3 kJ/mol; $\Delta$ H°_f [H₂O(g) ] = -241.8 kJ/mol

Question

An important step in the synthesis of nitric acid is the conversion of ammonia to nitric oxide. 4NH₃(g) + 5O₂(g)

\to

4NO(g) + 6H₂O(g)
Calculate

\Delta

H°_rxn for this reaction.

\Delta

H°_f [NH₃(g) ] = -45.9 kJ/mol;

\Delta

H°_f [NO(g) ] = 90.3 kJ/mol;

\Delta

H°_f [H₂O(g) ] = -241.8 kJ/mol

Quizplus · Accepted Answer

Using Hess's Law, we can calculate the enthalpy change for the reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

ΔH°rxn = (4 × ΔH°f[NO(g)]) + (6 × ΔH°f[H2O(g)]) 
          - (4 × ΔH°f[NH3(g)]) - (5 × ΔH°f[O2(g)])

As given in the question,
ΔH°f[NH3(g)] = -45.9 kJ/mol
ΔH°f[NO(g)] = 90.3 kJ/mol
ΔH°f[H2O(g)] = -241.8 kJ/mol
ΔH°f[O2(g)] = 0 kJ/mol

Substituting the values, we get:

ΔH°rxn = (4 × 90.3) + (6 × (-241.8)) - (4 × (-45.9)) - (5 × 0)
        = 361.2 - 1450.8 + 183.6
        = -906 kJ/mol

Therefore, the answer is A: -906.0 kJ.

An Important Step in the Synthesis of Nitric Acid Is →\to→