For the reaction : `PCl_(5) (g) rarrPCl_(3) (g) +Cl_(2)(g)`:

Options

A. `DeltaH = DeltaU`
B. `DeltaH gt DeltaU`
C. `Delta lt DeltaU`
D. None of the above

Explanation

To solve the question regarding the reaction \( PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \), we need to analyze the relationship between the change in internal energy (\( \Delta U \)) and the change in enthalpy (\( \Delta H \)) for this reaction, which is exothermic and carried out at constant pressure.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction we are considering is:
\[ PCl_5 (g) \rightleftharpoons PCl_3 (g) + Cl_2 (g) \]

2. **Determine the Change in Moles of Gas (\( \Delta n_g \))**:
To find \( \Delta n_g \), we calculate the difference between the moles of products and the moles of reactants.
- Moles of products: \( PCl_3 (g) + Cl_2 (g) = 2 \) moles
- Moles of reactants: \( PCl_5 (g) = 1 \) mole
\[ \Delta n_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

3. **Use the Relationship Between \( \Delta H \) and \( \Delta U \)**:
The relationship between the change in enthalpy and the change in internal energy at constant pressure is given by:
\[ \Delta H = \Delta U + \Delta n_g RT \]
Here, \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin.

4. **Substitute \( \Delta n_g \) into the Equation**:
Since we have calculated \( \Delta n_g = 1 \), we can substitute this into the equation:
\[ \Delta H = \Delta U + (1)RT \]

5. **Analyze the Sign of \( \Delta H \)**:
Given that the reaction is exothermic, \( \Delta H \) is negative. Therefore, we can express this as:
\[ \Delta H < 0 \]
This implies that:
\[ \Delta U + RT > 0 \quad \text{(since \( \Delta H = \Delta U + RT \))} \]
Thus, we can conclude that:
\[ \Delta U < \Delta H \]

6. **Final Conclusion**:
Since \( \Delta H \) is negative and \( \Delta U \) is less than \( \Delta H \), we can summarize:
\[ \Delta U < \Delta H \]