The equilibrium:
`P_(4)(g)+6Cl_(2)(g) hArr 4PCl_(3)(g)`
is attained by mixing equal moles of `P_(4)` and `Cl_(2)` in an evacuated vessel. Then at equilibrium:

Options

A. `[Cl_(2)]gt[PCl_(3)]`
B. `[CL_(2)]gt[P_(4)]`
C. `[P_(4)]gt[Cl_(2)]`
D. `[PCl_(3)]gt[P_(4)]`

Answer

To solve the equilibrium problem given by the reaction:

\[ P_4(g) + 6Cl_2(g) \rightleftharpoons 4PCl_3(g) \]

we will follow these steps:

### Step 1: Initial Moles
We start by mixing equal moles of \( P_4 \) and \( Cl_2 \). Let's assume we have 1 mole of each:

- Initial moles of \( P_4 = 1 \)
- Initial moles of \( Cl_2 = 1 \)
- Initial moles of \( PCl_3 = 0 \)

### Step 2: Change in Moles at Equilibrium
Let \( \alpha \) be the extent of the reaction that has occurred at equilibrium. According to the stoichiometry of the reaction:

- For every 1 mole of \( P_4 \) that reacts, 6 moles of \( Cl_2 \) react and 4 moles of \( PCl_3 \) are formed.

At equilibrium, the moles will be:

- Moles of \( P_4 = 1 - \alpha \)
- Moles of \( Cl_2 = 1 - 6\alpha \)
- Moles of \( PCl_3 = 4\alpha \)

### Step 3: Equilibrium Concentrations
To find the equilibrium concentrations, we need to express the moles in terms of volume \( V \) (assuming the volume is the same for all components):

- Concentration of \( P_4 = \frac{1 - \alpha}{V} \)
- Concentration of \( Cl_2 = \frac{1 - 6\alpha}{V} \)
- Concentration of \( PCl_3 = \frac{4\alpha}{V} \)

### Step 4: Comparing Concentrations
We need to determine which of the concentrations is greater at equilibrium.

1. **Comparing \( P_4 \) and \( Cl_2 \)**:
- Concentration of \( P_4 > \) Concentration of \( Cl_2 \)
- This translates to:
\[ 1 - \alpha > 1 - 6\alpha \]
Simplifying gives:
\[ 6\alpha > \alpha \quad \Rightarrow \quad 5\alpha > 0 \]
This is always true for \( \alpha > 0 \).

2. **Comparing \( Cl_2 \) and \( PCl_3 \)**:
- Concentration of \( Cl_2 > \) Concentration of \( PCl_3 \)
- This translates to:
\[ 1 - 6\alpha > 4\alpha \]
Simplifying gives:
\[ 1 > 10\alpha \quad \Rightarrow \quad \alpha < 0.1 \]
This is not universally true as \( \alpha \) can be greater than 0.1.

3. **Comparing \( PCl_3 \) and \( P_4 \)**:
- Concentration of \( PCl_3 > \) Concentration of \( P_4 \)
- This translates to:
\[ 4\alpha > 1 - \alpha \]
Simplifying gives:
\[ 5\alpha > 1 \quad \Rightarrow \quad \alpha > 0.2 \]
This is also not universally true as \( \alpha \) can be less than 0.2.

### Conclusion
From the comparisons, we conclude that:

- The concentration of \( P_4 \) is greater than \( Cl_2 \).
- The concentrations of \( Cl_2 \) and \( PCl_3 \) cannot be definitively compared without knowing \( \alpha \).
- The concentrations of \( PCl_3 \) and \( P_4 \) cannot be definitively compared without knowing \( \alpha \).

Thus, the only definitive conclusion is that:

**The concentration of \( P_4 \) is greater than the concentration of \( Cl_2 \) at equilibrium.**