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)]`

Explanation

To solve the equilibrium problem given by the reaction:

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

we will analyze the situation step by step.

### Step 1: Initial Moles
Initially, we have equal moles of \( P_4 \) and \( Cl_2 \). Let's assume we start with 1 mole of \( P_4 \) and 1 mole of \( Cl_2 \).

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

### Step 2: Change in Moles
Let \( x \) be the amount of \( P_4 \) that reacts at equilibrium. According to the stoichiometry of the reaction:

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

Thus, at equilibrium:
- Moles of \( P_4 \) remaining = \( 1 - x \)
- Moles of \( Cl_2 \) remaining = \( 1 - 6x \)
- Moles of \( PCl_3 \) formed = \( 4x \)

### Step 3: Equilibrium Concentrations
Assuming the volume of the vessel is \( V \), the equilibrium concentrations can be expressed as:

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

### Step 4: Comparing Concentrations
Now we need to compare the concentrations of \( P_4 \), \( Cl_2 \), and \( PCl_3 \) at equilibrium.

1. **Comparing \( P_4 \) and \( Cl_2 \)**:
- Concentration of \( P_4 \) = \( \frac{1 - x}{V} \)
- Concentration of \( Cl_2 \) = \( \frac{1 - 6x}{V} \)

Since \( 1 - 6x \) will decrease more rapidly than \( 1 - x \) as \( x \) increases, it follows that:
- If \( x \) is small, \( 1 - x \) will be greater than \( 1 - 6x \).
- Therefore, \( [P_4] > [Cl_2] \).

2. **Comparing \( Cl_2 \) and \( PCl_3 \)**:
- Concentration of \( Cl_2 \) = \( \frac{1 - 6x}{V} \)
- Concentration of \( PCl_3 \) = \( \frac{4x}{V} \)

As \( x \) increases, \( 4x \) will increase while \( 1 - 6x \) will decrease. Eventually, \( [Cl_2] \) will be less than \( [PCl_3] \) when enough \( PCl_3 \) is formed.

3. **Comparing \( P_4 \) and \( PCl_3 \)**:
- Concentration of \( P_4 \) = \( \frac{1 - x}{V} \)
- Concentration of \( PCl_3 \) = \( \frac{4x}{V} \)

Since \( 4x \) increases while \( 1 - x \) decreases, at equilibrium, \( [PCl_3] \) will be greater than \( [P_4] \) when \( x \) is sufficiently large.

### Conclusion
From the analysis, we find that at equilibrium:
- The concentration of \( P_4 \) is greater than that of \( Cl_2 \).
- The concentration of \( Cl_2 \) is less than that of \( PCl_3 \).

Thus, the correct answer is that the concentration of \( P_4 \) is greater than the concentration of \( Cl_2 \).

### Final Answer
**The concentration of \( P_4 \) is greater than the concentration of \( Cl_2 \).**

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