Calculate `P-Cl` bond enthalpy
`P(s)+(3)/(2)Cl_(2)(g)rarr PCl_(3)(g)` Given `: Delta_(f)H(PCl_(3),g)=306kJ //mol`
`DeltaH_("atomization")(P,s)=314kJ//mol`,
`Delta_(r)H(Cl,g)=1231kJ//mol`

Options

A. 123.66 kJ/mol
B. 371 kJ/mol
C. 19 kJ/mol
D. None of these

Explanation

To calculate the P-Cl bond enthalpy for the reaction:

\[ P(s) + \frac{3}{2}Cl_2(g) \rightarrow PCl_3(g) \]

we will use the given data:

- \(\Delta_f H(PCl_3, g) = 306 \, \text{kJ/mol}\)
- \(\Delta H_{\text{atomization}}(P, s) = 314 \, \text{kJ/mol}\)
- \(\Delta H_{\text{atomization}}(Cl_2, g) = 1231 \, \text{kJ/mol}\)

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data**:
The reaction involves solid phosphorus and gaseous chlorine forming gaseous phosphorus trichloride. We have the formation enthalpy for \(PCl_3\), the atomization enthalpy for phosphorus, and the atomization enthalpy for chlorine.

2. **Write the Equation for Bond Enthalpy**:
The bond enthalpy can be calculated using the following equation:
\[ \Delta H_{\text{reaction}} = \text{(Bond enthalpy of products)} - \text{(Bond enthalpy of reactants)} \]
Rearranging gives:
\[ \text{Bond enthalpy of P-Cl} = \Delta H_{\text{reaction}} + \text{(Bond enthalpy of reactants)} \]

3. **Calculate the Enthalpy of the Reaction**:
Since the formation enthalpy of \(PCl_3\) is given as \(306 \, \text{kJ/mol}\), we can say:
\[ \Delta H_{\text{reaction}} = 306 \, \text{kJ/mol} \]

4. **Calculate the Total Enthalpy of Reactants**:
- For phosphorus:
\[ \Delta H_{\text{atomization}}(P) = 314 \, \text{kJ/mol} \]
- For chlorine, we have \(\frac{3}{2}Cl_2\). The total enthalpy for chlorine is:
\[ \Delta H_{\text{atomization}}(Cl_2) = \frac{3}{2} \times 2 \times 1231 \, \text{kJ/mol} = 3 \times 1231 \, \text{kJ/mol} = 3693 \, \text{kJ/mol} \]

5. **Combine the Enthalpies of Reactants**:
The total enthalpy of the reactants is:
\[ \text{Total Enthalpy of Reactants} = 314 + 3693 = 4007 \, \text{kJ/mol} \]

6. **Set Up the Equation for Bond Enthalpy**:
Since \(PCl_3\) has three P-Cl bonds, we can express the bond enthalpy as:
\[ 3 \times \text{Bond enthalpy of P-Cl} = \Delta H_{\text{reaction}} + \text{Total Enthalpy of Reactants} \]
Substituting the values:
\[ 3 \times \text{Bond enthalpy of P-Cl} = 306 + 4007 \]
\[ 3 \times \text{Bond enthalpy of P-Cl} = 4313 \, \text{kJ/mol} \]

7. **Calculate the Bond Enthalpy**:
Dividing by 3 gives:
\[ \text{Bond enthalpy of P-Cl} = \frac{4313}{3} = 1437.67 \, \text{kJ/mol} \]

### Final Answer:
The bond enthalpy of the P-Cl bond is approximately \(1437.67 \, \text{kJ/mol}\).