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 find the relationship between the degree of dissociation (x) and the total pressure (P) for the equilibrium reaction:

\[ \text{Si(s)} + 2\text{Cl}_2(g) \rightleftharpoons \text{SiCl}_4(g) \]

we will follow these steps:

### Step 1: Write the Initial and Equilibrium Conditions
- Initially, we have 1 mole of Si (solid) and 2 moles of Cl₂ (gas). The initial number of moles of SiCl₄ is 0.
- At equilibrium, let the degree of dissociation of Cl₂ be x. Therefore:
- Moles of Cl₂ that dissociate = 2x
- Moles of SiCl₄ formed = x
- The remaining moles of Cl₂ at equilibrium = \(2 - 2x\)

### Step 2: Calculate Total Moles at Equilibrium
The total number of moles at equilibrium can be calculated as:
\[ \text{Total moles} = \text{Moles of Si} + \text{Remaining moles of Cl₂} + \text{Moles of SiCl₄} \]
\[ = 1 + (2 - 2x) + x = 3 - x \]

### Step 3: Calculate the Partial Pressures
Using the total pressure (P), we can express the partial pressures of the gases:
- Partial pressure of SiCl₄:
\[ P_{\text{SiCl}_4} = \left(\frac{x}{3 - x}\right) P \]
- Partial pressure of Cl₂:
\[ P_{\text{Cl}_2} = \left(\frac{2 - 2x}{3 - x}\right) P \]

### Step 4: Write the Expression for Kp
The equilibrium constant \(K_p\) for the reaction is given by:
\[ K_p = \frac{P_{\text{SiCl}_4}}{(P_{\text{Cl}_2})^2} \]
Substituting the expressions for the partial pressures:
\[ K_p = \frac{\left(\frac{x}{3 - x}\right) P}{\left(\frac{2 - 2x}{3 - x}\right) P}^2 \]
\[ = \frac{x P}{\left(2 - 2x\right)^2} \cdot \frac{1}{(3 - x)^2} \]

### Step 5: Simplify Under the Assumption x is Very Small
Since x is very small, we can approximate:
- \(3 - x \approx 3\)
- \(2 - 2x \approx 2\)

Thus, we can simplify \(K_p\):
\[ K_p \approx \frac{x P}{(2)^2} \cdot \frac{1}{(3)^2} = \frac{x P}{4 \cdot 9} = \frac{x P}{36} \]

### Step 6: Rearranging to Find the Relationship Between x and P
From the above equation, we can express x in terms of P:
\[ x = \frac{36 K_p}{P} \]

### Conclusion
From this relationship, we can see that as the total pressure (P) increases, the degree of dissociation (x) decreases, indicating that x is inversely proportional to P.