Question

Calculate ΔH∘ in kilojoules for the reaction of ammonia NH₃ (ΔH∘f=−46.1kJ/mol) with O₂ to yield nitric oxide (NO) (ΔH∘f=91.3 kJ/mol) and H₂O(g) (ΔH∘f=−241.8kJ/mol), a step in the Ostwald process for the commercial production of nitric acid.

4NH₃(g)+5O₂(g)→4NO(g)+6H₂O(g)

Answer 1

The standard enthalpy change (ΔH°) for the reaction of ammonia with oxygen to yield nitric oxide and water is calculated using standard enthalpies of formation and is found to be -901.2 kilojoules.

Step-by-step explanation:

To calculate the standard enthalpy change (ΔH°) for the reaction of ammonia with oxygen to yield nitric oxide and water (a step in the Ostwald process), we use the standard enthalpies of formation (ΔH°f) for the compounds involved. The balanced chemical equation for the reaction is:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)

To find the ΔH° for the reaction, we apply Hess's law and use the formula:

ΔH° = [∑ (ΔH°f products) × (moles of products)] - [∑ (ΔH°f reactants) × (moles of reactants)]

Plugging in the values:

ΔH° = [(4 mol × 91.3 kJ/mol) + (6 mol × -241.8 kJ/mol)] - [(4 mol × -46.1 kJ/mol) + (5 mol × 0 kJ/mol)]

ΔH° = [(365.2 kJ) + (-1450.8 kJ)] - [(-184.4 kJ) + 0 kJ]

ΔH° = -1085.6 kJ + 184.4 kJ

ΔH° = -901.2 kJ

The standard enthalpy change for the reaction is -901.2 kilojoules.