The `H-H` bond energy is `430KJmol^(-1)` and `Cl-Cl` bond energy is `240KJmol^(-1)` . `DeltaH` for `HCl` is `-90KJ` . The `H-Cl` bond energy is about

Options

A. `180 kJ mol^(-1)`
B. `360 kJ mol^(-1)`
C. `213 kJ mol^(-1)`
D. `425 kJ mol^(-1)`

Explanation

To calculate the H-Cl bond energy using the given bond energies and the enthalpy change for the formation of HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- H-H bond energy = 430 kJ/mol
- Cl-Cl bond energy = 240 kJ/mol
- ΔH for HCl = -90 kJ

2. **Write the Reaction for the Formation of HCl:**
The formation of HCl from its elements can be represented as:
\[ \frac{1}{2} H_2 + \frac{1}{2} Cl_2 \rightarrow HCl \]

3. **Express the Enthalpy Change in Terms of Bond Energies:**
The enthalpy change (ΔH) for the reaction can be expressed in terms of the bond energies:
\[ \Delta H = \text{Bond energy of reactants} - \text{Bond energy of products} \]
For our reaction:
\[ \Delta H = \left(\frac{1}{2} \times \text{Bond energy of H-H} + \frac{1}{2} \times \text{Bond energy of Cl-Cl}\right) - \text{Bond energy of H-Cl} \]

4. **Substitute the Known Values:**
Substitute the bond energies and ΔH into the equation:
\[ -90 = \left(\frac{1}{2} \times 430 + \frac{1}{2} \times 240\right) - \text{Bond energy of H-Cl} \]

5. **Calculate the Left Side:**
Calculate the bond energies of the reactants:
\[ -90 = \left(215 + 120\right) - \text{Bond energy of H-Cl} \]
\[ -90 = 335 - \text{Bond energy of H-Cl} \]

6. **Rearrange to Solve for Bond Energy of H-Cl:**
Rearranging the equation gives:
\[ \text{Bond energy of H-Cl} = 335 + 90 \]
\[ \text{Bond energy of H-Cl} = 425 \text{ kJ/mol} \]

7. **Conclusion:**
The H-Cl bond energy is approximately **425 kJ/mol**.