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?

Answer

To find the standard enthalpy of formation of ammonia gas (NH₃), we can follow these steps:

### Step 1: Understand the Reaction
The given reaction is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
The enthalpy change for this reaction (\( \Delta_rH^\circ \)) is given as -92.4 kJ.

### Step 2: Definition of Standard Enthalpy of Formation
The standard enthalpy of formation (\( \Delta_fH^\circ \)) is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

### Step 3: Relate the Reaction to Formation
In the given reaction, 2 moles of NH₃ are produced. To find the standard enthalpy of formation for one mole of NH₃, we need to divide the total enthalpy change by the number of moles of NH₃ produced.

### Step 4: Calculate the Enthalpy Change for One Mole
Since the reaction produces 2 moles of NH₃, we can calculate the standard enthalpy of formation for one mole of NH₃ as follows:
\[ \Delta_fH^\circ(NH_3) = \frac{\Delta_rH^\circ}{\text{number of moles of } NH_3} \]
\[ \Delta_fH^\circ(NH_3) = \frac{-92.4 \text{ kJ}}{2} \]
\[ \Delta_fH^\circ(NH_3) = -46.2 \text{ kJ/mol} \]

### Step 5: Conclusion
Thus, the standard enthalpy of formation of ammonia gas (NH₃) is:
\[ \Delta_fH^\circ(NH_3) = -46.2 \text{ kJ/mol} \]