The bond enthalpies of `H-H` and `Cl-Cl` are `430` and `242 kJ mol^(-1)`, respectively. If `DeltaH_(f) (HCl)` is `-91 kJmol^(-1)` the bond enthalpy of `HCl` would be

Options

A. `-214 kJ mol^(-1)`
B. `-427 kJ mol^(-1)`
C. `214 kJ mol^(-1)`
D. `427 kJ mol^(-1)`

Explanation

To find the bond enthalpy of HCl, we can use the following steps:

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

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