Calculate the bond enthalpy of H-CI given that the bond enthalpies of `H_(2)` and `CI_(2)` are 435.4 and 242.8 kJ ` "mol"^(-1)` respectively and enthalpy of formation of `HCI(g) is `-92.2 kJ mol`

Explanation

To calculate the bond enthalpy of H-Cl, we can use the given bond enthalpies of H₂ and Cl₂, along with the enthalpy of formation of HCl. We will follow these steps:

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

### Step 2: Use the Enthalpy of Formation
The enthalpy of formation (\( \Delta H_f \)) of HCl is given as -92.2 kJ/mol. This means that the formation of 2 moles of HCl releases 92.2 kJ of energy.

### Step 3: Calculate the Bond Enthalpy of H₂ and Cl₂
The bond enthalpy of H₂ is given as 435.4 kJ/mol. Since we are breaking 1 mole of H₂ to form 2 moles of HCl, the bond enthalpy for 1 mole of H is:
\[ \text{Bond Enthalpy of H} = \frac{435.4 \text{ kJ/mol}}{2} = 217.7 \text{ kJ/mol} \]

The bond enthalpy of Cl₂ is given as 242.8 kJ/mol. Similarly, for 1 mole of Cl, the bond enthalpy is:
\[ \text{Bond Enthalpy of Cl} = \frac{242.8 \text{ kJ/mol}}{2} = 121.4 \text{ kJ/mol} \]

### Step 4: Set Up the Equation
Using Hess's law, we can set up the equation for the enthalpy change of the reaction:
\[ \Delta H = \text{Bond Enthalpy of H} + \text{Bond Enthalpy of Cl} - \Delta H_f(\text{HCl}) \]
Substituting the values we have:
\[ \Delta H = 217.7 \text{ kJ/mol} + 121.4 \text{ kJ/mol} - (-92.2 \text{ kJ/mol}) \]

### Step 5: Calculate the Bond Enthalpy of H-Cl
Now, substituting the values into the equation:
\[ \Delta H = 217.7 + 121.4 + 92.2 \]
\[ \Delta H = 431.3 \text{ kJ/mol} \]

Thus, the bond enthalpy of H-Cl is:
\[ \text{Bond Enthalpy of H-Cl} = 431.3 \text{ kJ/mol} \]

### Summary of the Solution
The bond enthalpy of H-Cl is calculated to be 431.3 kJ/mol using the bond enthalpies of H₂ and Cl₂ and the enthalpy of formation of HCl.