Question

ammonia, nh₃(δhᵢ = -45.9 kj), reacts with oxygen to produce water (δhᵢ = -241.8 kj) and nitric oxide, no (δhᵢ = 91.3 kj), in the following reaction:
4nh₃(g) + 5o₂(g) → 6h₂o(g) + 4no(g)
what is the enthalpy change for this reaction?
use δhᵣₓₙ = σ(δhᵢ,ₚᵣₒₙₙₑₜₛ) - σ(δhᵢ,ᵣₑₐᶜₜₐₙₜₛ).
○ -902 kj
○ -104.6 kj
○ 104.6 kj
○ 900.8 kj

Explanation:

Step1: Identify Products and Reactants

Products: \(6H_2O(g)\) and \(4NO(g)\)
Reactants: \(4NH_3(g)\) and \(5O_2(g)\) (Note: \(\Delta H_f\) for \(O_2(g)\) is 0 as it's an element in its standard state)

Step2: Calculate \(\sum(\Delta H_{f,products})\)

For \(H_2O(g)\): \(\Delta H_f = -241.8\ kJ\), moles = 6
For \(NO(g)\): \(\Delta H_f = 91.3\ kJ\), moles = 4
\(\sum(\Delta H_{f,products}) = (6 \times -241.8) + (4 \times 91.3)\)
\(= -1450.8 + 365.2 = -1085.6\ kJ\)

Step3: Calculate \(\sum(\Delta H_{f,reactants})\)

For \(NH_3(g)\): \(\Delta H_f = -45.9\ kJ\), moles = 4
For \(O_2(g)\): \(\Delta H_f = 0\), moles = 5
\(\sum(\Delta H_{f,reactants}) = (4 \times -45.9) + (5 \times 0)\)
\(= -183.6 + 0 = -183.6\ kJ\)

Step4: Calculate \(\Delta H_{rxn}\)

Using \(\Delta H_{rxn} = \sum(\Delta H_{f,products}) - \sum(\Delta H_{f,reactants})\)
\(\Delta H_{rxn} = -1085.6 - (-183.6) = -1085.6 + 183.6 = -902\ kJ\)

Answer:

-902 kJ