Question

Ammonia, nh3 (delta.hf = –46.2 kj), reacts with oxygen to produce water (delta.hf = –241.8 kj) and nitric oxide, no (delta.hf = 91.3 kj), in the following reaction: 4 upper n upper h subscript 3 (g) plus 5 upper o subscript 2 (g) right arrow 6 upper h subscript 2 upper o (g) plus 4 upper n upper o (g). what is the enthalpy change for this reaction? use delta h r x n equals the sum of delta h f of all the products minus the sum of delta h f of all the reactants.. –900.8 kj –104.6 kj 104.6 kj 900.8 kj

Answer 1

The enthalpy change (ΔH) for the reaction given the data from the question is –900.8 KJ

Data obtained from the question

• 4NH₃ + 5O₂ —> 6H₂O + 4NO
• Enthalpy of ammonia, NH₃ = –46.2 KJ
• Enthalpy of Oxygen = 0 KJ
• Enthalpy of water, H₂O = –241.8 KJ
• Enthalpy of nitric oxide, NO = 91.3 KJ
• Enthalpy change (ΔH) =?

How to determine the enthalpy change

ΔHrxn = ∑ΔH(products) - ∑ΔH(reactants)

ΔHrxn = ∑[H(H₂O) + H(NO)] - ∑[H(NH₃) + H(O₂)]

ΔHrxn = [(6 × –241.8) + (4 × 91.3)] – [(4 × –46.2) + (5×0)]

ΔHrxn = –1085.6 + 184.8

ΔHrxn = –900.8 KJ

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