The bond dissociation energy of gaseous `H_(2),Cl_(2)` and `HCl` are `104,58` and `103 kcal mol^(-1)` respecitvely. Calculate the enthalpy of formation for `HCl` gas.

Options

A. (a) `-22` kcal
B. (b) `+22` kcal
C. (c) `+184` kcal
D. (d) `-184` kcal

Explanation

To calculate the enthalpy of formation for HCl gas, we can follow these steps:

### Step 1: Write the balanced chemical equation for the formation of HCl gas.
The formation of HCl gas from its elements can be represented as:
\[ \frac{1}{2} \text{H}_2(g) + \frac{1}{2} \text{Cl}_2(g) \rightarrow \text{HCl}(g) \]

### Step 2: Identify the bond dissociation energies (BDE) for the reactants and products.
From the question, we have the following bond dissociation energies:
- BDE of H2 = 104 kcal/mol
- BDE of Cl2 = 58 kcal/mol
- BDE of HCl = 103 kcal/mol

### Step 3: Apply the formula for the enthalpy change (ΔH) of the reaction.
The enthalpy change for the reaction can be calculated using the formula:
\[ \Delta H = \text{BDE of reactants} - \text{BDE of products} \]

### Step 4: Calculate the BDE for the reactants.
Since we are using half a mole of H2 and half a mole of Cl2, we calculate:
- BDE of \(\frac{1}{2} \text{H}_2\) = \(\frac{1}{2} \times 104 \text{ kcal/mol} = 52 \text{ kcal}\)
- BDE of \(\frac{1}{2} \text{Cl}_2\) = \(\frac{1}{2} \times 58 \text{ kcal/mol} = 29 \text{ kcal}\)

Adding these gives:
\[ \text{Total BDE of reactants} = 52 \text{ kcal} + 29 \text{ kcal} = 81 \text{ kcal} \]

### Step 5: Calculate the total BDE for the products.
Since we are forming one mole of HCl, we have:
- BDE of HCl = 103 kcal/mol

### Step 6: Substitute the values into the ΔH formula.
Now we can substitute the values into the formula:
\[ \Delta H = \text{Total BDE of reactants} - \text{BDE of products} \]
\[ \Delta H = 81 \text{ kcal} - 103 \text{ kcal} \]
\[ \Delta H = -22 \text{ kcal} \]

### Conclusion:
The enthalpy of formation for HCl gas is:
\[ \Delta H = -22 \text{ kcal/mol} \]