Given the bond energies of `H - H` and `Cl - Cl` are `430 kJ mol^(-1)` and `240 kJ mol^(-1)`, respectively, and `Delta_(f) H^(@)` for `HCl` is `- 90 kJ mol^(-1)`. Bond enthalpy of `HCl` is

Options

A. 245 kJ `mol^(-1)`
B. 290 kJ `mol^(-1)`
C. 380 kJ `mol^(-1)`
D. 425 kJ `mol^(-1)`

Answer

To calculate the bond enthalpy of HCl using the given bond energies and the enthalpy of formation, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identify the bond energies:**
- The bond energy of H-H (hydrogen) is given as \( 430 \, \text{kJ/mol} \).
- The bond energy of Cl-Cl (chlorine) is given as \( 240 \, \text{kJ/mol} \).
- The enthalpy of formation \( \Delta_f H^\circ \) for HCl is given as \( -90 \, \text{kJ/mol} \).

3. **Use the bond enthalpy formula:**
The enthalpy change for the reaction can be expressed as:
\[ \Delta H = \text{Bond enthalpy of reactants} - \text{Bond enthalpy of products} \]
For the formation of 2 moles of HCl, we can write:
\[ \Delta H_f = \text{Bond enthalpy of H-H} + \text{Bond enthalpy of Cl-Cl} - \text{Bond enthalpy of HCl} \]

4. **Substitute the known values:**
Since we are forming 2 moles of HCl, we need to consider half the bond energies for the calculation:
\[ -90 = \frac{1}{2}(430) + \frac{1}{2}(240) - \text{Bond enthalpy of HCl} \]

5. **Calculate the bond energies:**
- Calculate the average bond energies:
\[ \text{Bond enthalpy of H-H} = 430 \, \text{kJ/mol} \quad \text{(for 1 mole)} \]
\[ \text{Bond enthalpy of Cl-Cl} = 240 \, \text{kJ/mol} \quad \text{(for 1 mole)} \]
Thus, the total bond energy for the reactants is:
\[ \frac{1}{2}(430) + \frac{1}{2}(240) = 215 + 120 = 335 \, \text{kJ/mol} \]

6. **Rearranging the equation:**
Now we can rearrange the equation to find the bond enthalpy of HCl:
\[ -90 = 335 - \text{Bond enthalpy of HCl} \]
\[ \text{Bond enthalpy of HCl} = 335 + 90 = 425 \, \text{kJ/mol} \]

### Final Answer:
The bond enthalpy of HCl is \( 425 \, \text{kJ/mol} \).