Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` is

Options

A. `93 KJ "mol"^(-1)`
B. `-245 KJ "mol"^(-1)`
C. `-93 KJ "mol"^(-1)`
D. `245 KJ "mol"^(-1)`

Explanation

To find the enthalpy of formation of HCl, we will use the bond dissociation enthalpies given for \( H_2 \), \( Cl_2 \), and \( HCl \). We will apply Hess's law to relate the enthalpy changes in the reaction.

### Step-by-Step Solution:

1. **Write the reaction for the formation of HCl:**
\[ H_2(g) + Cl_2(g) \rightarrow 2 HCl(g) \]

2. **Identify the bond dissociation enthalpies:**
- \( \Delta H \) for \( H_2 \) dissociation: \( 434 \, \text{kJ/mol} \)
- \( \Delta H \) for \( Cl_2 \) dissociation: \( 242 \, \text{kJ/mol} \)
- \( \Delta H \) for \( HCl \) formation: \( 431 \, \text{kJ/mol} \)

3. **Dissociate the reactants:**
- Dissociation of \( H_2 \):
\[ H_2(g) \rightarrow 2H(g) \quad (\Delta H_1 = 434 \, \text{kJ/mol}) \]
- Dissociation of \( Cl_2 \):
\[ Cl_2(g) \rightarrow 2Cl(g) \quad (\Delta H_2 = 242 \, \text{kJ/mol}) \]

4. **Combine the atoms to form HCl:**
- Formation of \( 2HCl \):
\[ 2H(g) + 2Cl(g) \rightarrow 2HCl(g) \quad (\Delta H_3 = -2 \times 431 \, \text{kJ/mol} = -862 \, \text{kJ/mol}) \]

5. **Calculate the total enthalpy change using Hess's law:**
\[ \Delta H_{formation} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H_{formation} = 434 + 242 - 862 \]
\[ \Delta H_{formation} = 676 - 862 = -186 \, \text{kJ/mol} \]

6. **Since the reaction produces 2 moles of HCl, divide by 2 to find the enthalpy of formation for 1 mole:**
\[ \Delta H_{formation} \, (1 \, \text{mol}) = \frac{-186}{2} = -93 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of formation of \( HCl \) is \( -93 \, \text{kJ/mol} \).