Bond energies of `H-H and Cl-Cl` are `"430 kJ mol"^(-1)` and `"242 kJ mol"^(-1)` respectively. `DeltaH_(f)` for HCl is `"91 kJ mol"^(-1)`.
What will be the bond energy of `H-Cl`?

Options

A. 672 kJ
B. 182 kJ
C. 245 kJ
D. 88 kJ

Explanation

To find the bond energy of H-Cl, we can use the given bond energies of H-H and Cl-Cl, along with the formation enthalpy of HCl. Here’s a step-by-step solution:

### Step 1: Write the relevant equations
We know the bond dissociation energies for the bonds involved:
- For H-H:
\[ \text{H}_2 \rightarrow 2 \text{H} \quad \Delta H = +430 \, \text{kJ/mol} \]
- For Cl-Cl:
\[ \text{Cl}_2 \rightarrow 2 \text{Cl} \quad \Delta H = +242 \, \text{kJ/mol} \]

### Step 2: Write the formation reaction for HCl
The formation of HCl from its elements can be represented as:
\[ \text{H} + \text{Cl} \rightarrow \text{HCl} \quad \Delta H_f = -91 \, \text{kJ/mol} \]

### Step 3: Relate bond energies to the enthalpy of formation
The bond energy of H-Cl can be found using the following relationship:
\[ \Delta H_f = \text{Bond energy of H-H} + \text{Bond energy of Cl-Cl} - \text{Bond energy of H-Cl} \]

### Step 4: Substitute known values
We know:
- Bond energy of H-H = \( \frac{430}{2} = 215 \, \text{kJ/mol} \) (since we are considering half the bond energy for one mole of H)
- Bond energy of Cl-Cl = \( \frac{242}{2} = 121 \, \text{kJ/mol} \) (similarly for Cl)

Substituting these values into the equation:
\[ -91 = 215 + 121 - \text{Bond energy of H-Cl} \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \text{Bond energy of H-Cl} = 215 + 121 + 91 \]

### Step 6: Calculate the bond energy of H-Cl
Now, calculate:
\[ \text{Bond energy of H-Cl} = 215 + 121 + 91 = 427 \, \text{kJ/mol} \]

### Final Answer
The bond energy of H-Cl is \( 427 \, \text{kJ/mol} \).

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