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 find the enthalpy of formation (\( \Delta H_f^\circ \)) for \( PCl_5(g) \), we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction.

### Step-by-Step Solution:

1. **Write the reaction and the given information:**
\[ PCl_3(g) + Cl_2(g) \rightarrow PCl_5(g), \quad \Delta H = -x \text{ kJ} \]
\[ \Delta H_f^\circ(PCl_3) = -y \text{ kJ} \]
Note that the enthalpy of formation for \( Cl_2(g) \) is zero because it is in its standard state.

2. **Define the enthalpy of formation for \( PCl_5(g) \):**
Let \( \Delta H_f^\circ(PCl_5) = z \text{ kJ} \).

3. **Apply Hess's law:**
According to Hess's law, the change in enthalpy for the reaction can be expressed as:
\[ \Delta H = \Delta H_f^\circ(PCl_5) - \left( \Delta H_f^\circ(PCl_3) + \Delta H_f^\circ(Cl_2) \right) \]
Since \( \Delta H_f^\circ(Cl_2) = 0 \), we can simplify this to:
\[ \Delta H = z - (-y) \]
This simplifies to:
\[ \Delta H = z + y \]

4. **Set up the equation based on the given enthalpy change:**
We know from the problem statement that:
\[ -x = z + y \]

5. **Rearrange the equation to solve for \( z \):**
\[ z = -x - y \]

6. **Express \( z \) in terms of \( -x \) and \( y \):**
\[ z = -x + y \text{ kJ} \]

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