Given the following data
`H_2(g) rarr2H(g),DeltaH=104.2kcal `
`Cl_2 (g) rarr 2Cl(g), DeltaH = 58 kcal `
`HCl = H(g) +Cl(g) , DeltaH = 103.2 kcal`
The standard enthalpy of formation of HCl(g) is

Options

A. `-143.2 kcal `
B. `-22.4 kcal `
C. `-22.1 kacl`
D. `58 kcal`

Answer

To find the standard enthalpy of formation of HCl(g), we can use the provided thermochemical data and Hess's law. The standard enthalpy of formation is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

### Step-by-Step Solution:

1. **Identify the Reaction for Formation of HCl**:
The formation of HCl from its elements in their standard states can be represented as:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g) \]

2. **Write Down the Given Data**:
- For the dissociation of hydrogen gas:
\[ H_2(g) \rightarrow 2H(g), \quad \Delta H = 104.2 \text{ kcal} \]
- For the dissociation of chlorine gas:
\[ Cl_2(g) \rightarrow 2Cl(g), \quad \Delta H = 58 \text{ kcal} \]
- For the formation of HCl from its atoms:
\[ H(g) + Cl(g) \rightarrow HCl(g), \quad \Delta H = 103.2 \text{ kcal} \]

3. **Adjust the Reactions for Formation**:
Since we need to form HCl from half a mole of H2 and half a mole of Cl2, we need to adjust the enthalpy values accordingly:
- For \( \frac{1}{2} H_2(g) \rightarrow H(g) \):
\[ \Delta H = \frac{104.2}{2} = 52.1 \text{ kcal} \]
- For \( \frac{1}{2} Cl_2(g) \rightarrow Cl(g) \):
\[ \Delta H = \frac{58}{2} = 29 \text{ kcal} \]

4. **Reverse the Formation of HCl**:
The reaction for the formation of HCl can be reversed to find the enthalpy change:
\[ HCl(g) \rightarrow H(g) + Cl(g), \quad \Delta H = -103.2 \text{ kcal} \]

5. **Apply Hess's Law**:
Now we can sum the enthalpy changes:
\[ \Delta H_{formation} = \Delta H(H_2) + \Delta H(Cl_2) + \Delta H(HCl) \]
Substituting the values:
\[ \Delta H_{formation} = 52.1 \text{ kcal} + 29 \text{ kcal} - 103.2 \text{ kcal} \]
\[ \Delta H_{formation} = 81.1 \text{ kcal} - 103.2 \text{ kcal} = -22.1 \text{ kcal} \]

6. **Conclusion**:
The standard enthalpy of formation of HCl(g) is:
\[ \Delta H_{f} = -22.1 \text{ kcal} \]

### Final Answer:
The standard enthalpy of formation of HCl(g) is \(-22.1 \text{ kcal}\).

---