Calculate change in enthalpy for the reaction at `27^(@)C`
`H_(2)(g)+Cl_(2)(g)rarr2H-Cl(g)`
by using the bond energy and energy data
Bond energies of `H-H,Cl-Cl` and `H-Cl` bonds are `435 kJ mol^(_1), 240 kJ mol^(-1)` and `430 kJmol^(-1)` respectively.

Answer

To calculate the change in enthalpy (ΔH) for the reaction:

\[ H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \]

we will use the bond energies provided for the bonds involved in the reaction.

### Step 1: Identify the bonds broken and formed

In the reaction, we have:
- **Bonds broken**:
- 1 H-H bond in \( H_2 \)
- 1 Cl-Cl bond in \( Cl_2 \)

- **Bonds formed**:
- 2 H-Cl bonds in \( 2HCl \)

### Step 2: Write the bond energy values

The bond energies given are:
- H-H bond energy = 435 kJ/mol
- Cl-Cl bond energy = 240 kJ/mol
- H-Cl bond energy = 430 kJ/mol

### Step 3: Calculate the total bond energy of reactants

The total bond energy of the reactants is the sum of the bond energies of the bonds broken:

\[ \text{Total bond energy of reactants} = \text{Bond energy of H-H} + \text{Bond energy of Cl-Cl} \]
\[ = 435 \text{ kJ/mol} + 240 \text{ kJ/mol} = 675 \text{ kJ/mol} \]

### Step 4: Calculate the total bond energy of products

The total bond energy of the products is the sum of the bond energies of the bonds formed:

\[ \text{Total bond energy of products} = 2 \times \text{Bond energy of H-Cl} \]
\[ = 2 \times 430 \text{ kJ/mol} = 860 \text{ kJ/mol} \]

### Step 5: Calculate the change in enthalpy (ΔH)

Using the formula for change in enthalpy:

\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
\[ \Delta H = 675 \text{ kJ/mol} - 860 \text{ kJ/mol} = -185 \text{ kJ/mol} \]

### Final Answer

The change in enthalpy for the reaction is:

\[ \Delta H = -185 \text{ kJ/mol} \]

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