For the reaction
`PCI _(3) (g) + CI_(2)(g) rightarrow PCI_(5)(g), Delta H = -x kJ`
If the `Delta H_(f)^(@)"PCI"_(3) is -y kJ, ` what is `Delta H_(f)^(@)PCI_(5)` ?

Options

A. `(x - y) kJ`
B. `(y -x)kJ`
C. `-(x + y) kJ`
D. `(x + y ) kJ`

Answer

To solve the problem, we need to find the standard heat of formation (\( \Delta H_f^\circ \)) of \( PCl_5(g) \) given the reaction:

\[ PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g) \]

with the enthalpy change of the reaction \( \Delta H = -x \, \text{kJ} \) and the standard heat of formation of \( PCl_3(g) \) as \( \Delta H_f^\circ (PCl_3) = -y \, \text{kJ} \).

### Step-by-Step Solution:

1. **Write the Reaction and Known Values:**
- The reaction is:
\[ PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g) \]
- Given:
- \( \Delta H = -x \, \text{kJ} \)
- \( \Delta H_f^\circ (PCl_3) = -y \, \text{kJ} \)
- \( \Delta H_f^\circ (Cl_2) = 0 \, \text{kJ} \) (since it is in its standard state)

2. **Use the Enthalpy Change Equation:**
- According to Hess's law, the enthalpy change for the reaction can be expressed as:
\[ \Delta H = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]
- For our reaction:
\[ \Delta H = \Delta H_f^\circ (PCl_5) - \left( \Delta H_f^\circ (PCl_3) + \Delta H_f^\circ (Cl_2) \right) \]

3. **Substitute Known Values:**
- Substitute the known values into the equation:
\[ -x = \Delta H_f^\circ (PCl_5) - \left( -y + 0 \right) \]
- This simplifies to:
\[ -x = \Delta H_f^\circ (PCl_5) + y \]

4. **Rearrange to Solve for \( \Delta H_f^\circ (PCl_5) \):**
- Rearranging the equation gives:
\[ \Delta H_f^\circ (PCl_5) = -x - y \]

5. **Express in Final Form:**
- We can factor out the negative sign:
\[ \Delta H_f^\circ (PCl_5) = - (x + y) \]

### Final Answer:
Thus, the standard heat of formation of \( PCl_5(g) \) is:
\[ \Delta H_f^\circ (PCl_5) = - (x + y) \, \text{kJ} \]