For the following equilibrium : `Si(s)+2Cl_(2)(g) hArr SiCl_(4)(g)`, if P is total pressure then degree of dissociation (x) is related to P as (Presume x to very small)

Options

A. `x prop P`
B. `x prop (1)/(P)`
C. `x prop sqrt(P)`
D. `x prop sqrt((1)/(P))`

Answer

To solve the problem regarding the equilibrium reaction \( \text{Si}(s) + 2\text{Cl}_2(g) \rightleftharpoons \text{SiCl}_4(g) \) and the relationship between the degree of dissociation \( x \) and total pressure \( P \), we can follow these steps:

### Step 1: Define the Initial Conditions
At the start of the reaction (T = 0), we assume:
- Moles of Si = 1 (solid, does not affect pressure)
- Moles of Cl2 = 2 (since it is a gas, it will contribute to pressure)

### Step 2: Define the Change in Moles
Let \( x \) be the degree of dissociation of \( \text{Si} \):
- Moles of \( \text{Si} \) at equilibrium = \( 1 - x \) (but since Si is solid, it does not contribute to pressure)
- Moles of \( \text{Cl}_2 \) at equilibrium = \( 2 - 2x \) (since 2 moles of \( \text{Cl}_2 \) are consumed)
- Moles of \( \text{SiCl}_4 \) formed = \( x \)

### Step 3: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium can be expressed as:
- Total moles = Moles of \( \text{Cl}_2 \) + Moles of \( \text{SiCl}_4 \)
- Total moles = \( (2 - 2x) + x = 2 - x \)

### Step 4: Relate Total Pressure to Moles
Using the ideal gas law, the total pressure \( P \) can be expressed in terms of the total number of moles:
- Since the reaction involves gases, we can write:
\[ P = \frac{(nRT)}{V} \]
Where \( n \) is the total number of moles, \( R \) is the gas constant, \( T \) is temperature, and \( V \) is volume. For simplicity, we can assume \( R \) and \( T \) are constant, and we can express pressure directly in terms of moles.

### Step 5: Substitute for Total Pressure
Assuming the volume is constant, we can relate the total pressure to the total moles:
- Let \( P \) be the total pressure at equilibrium:
\[ P = k(2 - x) \]
Where \( k \) is a constant that incorporates \( R \) and \( T \).

### Step 6: Approximation for Small \( x \)
Since \( x \) is very small, we can approximate:
- \( P \approx k \cdot 2 \) (since \( x \) is negligible)

### Step 7: Solve for Degree of Dissociation \( x \)
From the equation \( P = k(2 - x) \), we can rearrange to find \( x \):
- Rearranging gives:
\[ x = 2 - \frac{P}{k} \]

### Step 8: Establish the Relationship
If we assume \( k \) is a constant, we can express \( x \) in terms of \( P \):
- Thus, we can say:
\[ x \propto P \]
This indicates that the degree of dissociation \( x \) is directly proportional to the total pressure \( P \).

### Final Result
The degree of dissociation \( x \) is related to the total pressure \( P \) as:
\[ x \propto P \]