For the gas phase reactions `PCl_(5)(g) hArr PCl_(3)(g) + Cl_(2)(g)`
Which of the following conditions are correct

Options

A. `Delta H lt 0 and Delta s lt 0`
B. `Delta H gt 0 and Delta S lt 0`
C. `Delta H= 0 and Delta S lt 0`
D. `Delta H gt 0 and Delta S gt 0`

Explanation

To analyze the gas phase reaction \( PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \) and determine which conditions are correct, we will follow these steps:

### Step 1: Write the Reaction
The reaction can be expressed as:
\[ PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g) \]

### Step 2: Determine the Change in Moles (\(\Delta n\))
To find \(\Delta n\), we calculate the difference between the moles of products and reactants:
- Moles of products: \(1 \, (PCl_3) + 1 \, (Cl_2) = 2\)
- Moles of reactants: \(1 \, (PCl_5) = 1\)

Thus,
\[ \Delta n = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

### Step 3: Calculate \(\Delta H\)
Using the relationship between enthalpy change (\(\Delta H\)), internal energy change (\(\Delta E\)), and the change in number of moles:
\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]
Since \(\Delta n = 1\), we can say:
\[ \Delta H = \Delta E + R \cdot T \]
Given that \(\Delta H\) is greater than 0, we conclude that the reaction is endothermic.

### Step 4: Determine the Change in Entropy (\(\Delta S\))
Entropy change can be calculated as:
\[ \Delta S = S_{products} - S_{reactants} \]
Since we have more moles of gas in the products than in the reactants, we can conclude that:
\[ \Delta S > 0 \]
This indicates an increase in disorder.

### Step 5: Analyze Gibbs Free Energy (\(\Delta G\))
The relationship between Gibbs free energy, enthalpy, and entropy is given by:
\[ \Delta G = \Delta H - T \Delta S \]
For a spontaneous reaction, \(\Delta G < 0\). Since \(\Delta H > 0\) and \(\Delta S > 0\), we need to analyze the temperature:
- If \(T\) is sufficiently high, the term \(T \Delta S\) can outweigh \(\Delta H\), making \(\Delta G\) negative, indicating spontaneity.

### Conclusion
Based on the analysis:
1. \(\Delta H > 0\) (endothermic reaction).
2. \(\Delta S > 0\) (increase in entropy).
3. The reaction can be spontaneous at high temperatures.

### Correct Conditions
The correct conditions regarding the reaction are:
- \(\Delta H > 0\)
- \(\Delta S > 0\)
- The reaction is spontaneous at high temperatures.