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) from its elements, we can use the given reactions and their enthalpy changes. The standard enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.

### Step-by-Step Solution:

1. **Identify the Target Reaction**:
We want to find the enthalpy change for the reaction:
\[ \frac{1}{2} H_2(g) + \frac{1}{2} Cl_2(g) \rightarrow HCl(g) \]

2. **Write Down the Given Reactions**:
- Reaction 1:
\[ H_2(g) \rightarrow 2H(g), \quad \Delta H = 104.2 \text{ kcal} \]
- Reaction 2:
\[ Cl_2(g) \rightarrow 2Cl(g), \quad \Delta H = 58 \text{ kcal} \]
- Reaction 3:
\[ H(g) + Cl(g) \rightarrow HCl(g), \quad \Delta H = 103.2 \text{ kcal} \]

3. **Adjust the Reactions**:
- For Reaction 1, we need only half of the reaction:
\[ \frac{1}{2} H_2(g) \rightarrow H(g), \quad \Delta H = \frac{104.2}{2} = 52.1 \text{ kcal} \]
- For Reaction 2, we also need half:
\[ \frac{1}{2} Cl_2(g) \rightarrow Cl(g), \quad \Delta H = \frac{58}{2} = 29 \text{ kcal} \]
- For Reaction 3, we need to reverse it to get HCl on the product side:
\[ HCl(g) \rightarrow H(g) + Cl(g), \quad \Delta H = -103.2 \text{ kcal} \]

4. **Combine the Adjusted Reactions**:
Now we can combine the adjusted reactions:
\[ \frac{1}{2} H_2(g) \rightarrow H(g) \quad (\Delta H = 52.1 \text{ kcal}) \]
\[ \frac{1}{2} Cl_2(g) \rightarrow Cl(g) \quad (\Delta H = 29 \text{ kcal}) \]
\[ HCl(g) \rightarrow H(g) + Cl(g) \quad (\Delta H = -103.2 \text{ kcal}) \]

5. **Calculate the Total Enthalpy Change**:
Adding the enthalpy changes:
\[ \Delta H_{formation} = 52.1 + 29 - 103.2 \]
\[ \Delta H_{formation} = 81.1 - 103.2 = -22.1 \text{ kcal} \]

### Final Answer:
The standard enthalpy of formation of HCl(g) is:
\[ \Delta H_f = -22.1 \text{ kcal} \]