Given `N_(2)(g)+3H(g)=2NH_(3)(g),DeltaH^(@)=22` kcal. The standard enthalpy of formation of `NH_(3)` gas is

Options

A. `-11Kcal//mol`
B. `11Kcal//mol`
C. `-22Kcal//mol`
D. `22Kcal//mol`

Explanation

To find the standard enthalpy of formation of \( NH_3 \) gas from the given reaction and its enthalpy change, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \]
The enthalpy change (\( \Delta H \)) for this reaction is given as \( 22 \) kcal.

2. **Understand the Concept of Enthalpy of Formation**:
The standard enthalpy of formation (\( \Delta H_f \)) of a compound is the change in enthalpy when one mole of the compound is formed from its elements in their standard states.

3. **Set Up the Enthalpy Equation**:
The enthalpy change for the reaction can be expressed as:
\[ \Delta H = \Delta H_f (\text{products}) - \Delta H_f (\text{reactants}) \]
For our reaction:
\[ \Delta H = \Delta H_f (2NH_3) - [\Delta H_f (N_2) + 3 \Delta H_f (H_2)] \]

4. **Substituting Known Values**:
- The standard enthalpy of formation for elements in their standard states (like \( N_2 \) and \( H_2 \)) is \( 0 \).
- Therefore, we can simplify:
\[ \Delta H = \Delta H_f (2NH_3) - [0 + 3 \times 0] = \Delta H_f (2NH_3) \]

5. **Relate to the Given Enthalpy Change**:
Since \( \Delta H \) for the reaction is \( 22 \) kcal, we have:
\[ 22 \text{ kcal} = \Delta H_f (2NH_3) \]

6. **Calculate the Enthalpy of Formation for One Mole**:
Since the enthalpy of formation calculated is for \( 2 \) moles of \( NH_3 \), we divide by \( 2 \) to find the enthalpy of formation for \( 1 \) mole:
\[ \Delta H_f (NH_3) = \frac{22 \text{ kcal}}{2} = 11 \text{ kcal} \]

### Final Answer:
The standard enthalpy of formation of \( NH_3 \) gas is \( 11 \) kcal per mole.

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