Determine the enthalpy:
4nh3(g) + 7o2(g) −→ 4no2(g) + 6h2o(ℓ)

∆h◦ f species ( kj mol)
nh3(g) −46.11
no2(g) +33.2
h2o(ℓ) −285.83

1. −1899 kj/mol rxn
2. +2034.4 kj/mol rxn
3. −298.7 kj/mol rxn
4. −1766 kj/mol rxn
5. −1397.6 kj/mol rxn

The enthalpy change of the reaction −1397.6 kJ/mol.

We can determine the enthalpy change of the reaction from the enthalpy of formation of each specie as follows;

∆H(reaction) = ∑∆H◦f(products) - ∑∆H◦f(reactants)

Given the following from the question;

NH3(g) = −46.11 kJ /mol

NO2(g) = +33.2 kJ /mol

H2O(ℓ) = −285.83 kJ /mol

We can now substitute values;

∆H(reaction) =   ∑(4 × 33.2) + (6 × (−285.83)) - ∑(4 × (−46.11)) + (7 × 0))  kJ /mol

The ∆H◦ f  of oxygen is zero because it is a pure substance.

∆H(reaction) = (132.8 - 1714.98) - (-184.44 + 0) kJ /mol

∆H(reaction) =−1397.6 kJ/mol

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Choice 5.: -1397.6 kJ/mol rxn.

Explanation:

Here's how to find the enthalpy change of a reaction from the standard enthalpy of formation of each species.

for the most stable allotrope of each elements under their standard state. For example,

As a result, for . Note the gaseous state symbol. This value is implied such that it's not provided in the question.

For this reaction:

.

.

Apply the equation:

.

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