Bond dissociation enthalpy of `H_(2)` , `Cl_(2)` and `HCl` are `434, 242` and `431KJmol^(-1)` respectively. Enthalpy of formation of `HCl` is

Options

A. `-245KJmol^(-1)`
B. `-63KJmol^(-1)`
C. `245KJmol^(-1)`
D. `-93KJmol^(-1)`

Explanation

To find the enthalpy of formation of HCl using the given bond dissociation enthalpies, we can follow these steps:

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

### Step 2: Identify the bond dissociation enthalpies
We are given the following bond dissociation enthalpies:
- \( \Delta H_{diss}(\text{H}_2) = 434 \, \text{kJ/mol} \)
- \( \Delta H_{diss}(\text{Cl}_2) = 242 \, \text{kJ/mol} \)
- \( \Delta H_{diss}(\text{HCl}) = 431 \, \text{kJ/mol} \)

### Step 3: Calculate the total energy required to break the bonds in the reactants
For the reaction, we need to break one mole of H-H bonds and half a mole of Cl-Cl bonds:
\[ \text{Energy required} = \frac{1}{2} \Delta H_{diss}(\text{H}_2) + \frac{1}{2} \Delta H_{diss}(\text{Cl}_2) \]
Substituting the values:
\[ = \frac{1}{2} \times 434 + \frac{1}{2} \times 242 \]
Calculating:
\[ = 217 + 121 = 338 \, \text{kJ} \]

### Step 4: Calculate the energy released during the formation of HCl
The energy released when forming 1 mole of HCl is simply the bond dissociation enthalpy of HCl:
\[ \text{Energy released} = \Delta H_{diss}(\text{HCl}) = 431 \, \text{kJ} \]

### Step 5: Apply Hess's Law to find the enthalpy of formation
Using Hess's Law, we can find the enthalpy change for the formation of HCl:
\[ \Delta H_f = \text{Energy required} - \text{Energy released} \]
Substituting the values:
\[ \Delta H_f = 338 - 431 = -93 \, \text{kJ/mol} \]

### Conclusion
The enthalpy of formation of HCl is:
\[ \Delta H_f = -93 \, \text{kJ/mol} \]

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