Given
`N_(2)(g)+3H_(2)(g)rarr2NH_(3)(g),Delta_(r)H^(Ө)= -92.4 kJ mol^(-1)`
What is the standard enthalpy of formation of `NH_(3)` gas?

Explanation

To find the standard enthalpy of formation of ammonia gas (NH₃), we start with the given reaction:

\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]

with the standard enthalpy change of the reaction:

\[ \Delta_rH^\circ = -92.4 \, \text{kJ mol}^{-1} \]

### Step 1: Understand the Definition
The standard enthalpy of formation (\( \Delta_fH^\circ \)) of a compound is defined as the enthalpy change when one mole of the compound is formed from its elements in their standard states.

### Step 2: Identify the Reaction for Formation
In this case, the formation of ammonia from its elements (nitrogen and hydrogen) is represented by the reaction provided. However, the reaction produces 2 moles of NH₃.

### Step 3: Adjust the Reaction for One Mole
To find the standard enthalpy of formation for one mole of NH₃, we need to adjust the stoichiometry of the reaction. We can do this by dividing the entire reaction by 2:

\[ \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g) \]

### Step 4: Adjust the Enthalpy Change
Since we divided the reaction by 2, we also need to divide the enthalpy change by 2:

\[ \Delta_fH^\circ = \frac{\Delta_rH^\circ}{2} \]

### Step 5: Calculate the Enthalpy of Formation
Substituting the value of \( \Delta_rH^\circ \):

\[ \Delta_fH^\circ = \frac{-92.4 \, \text{kJ}}{2} = -46.2 \, \text{kJ mol}^{-1} \]

### Final Answer
Thus, the standard enthalpy of formation of ammonia gas (NH₃) is:

\[ \Delta_fH^\circ = -46.2 \, \text{kJ mol}^{-1} \]

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