For a gaseous reaction
`aA(g)+bB(g)hArrcC(g)+dD(g)`
equilibrium constants `K_(c),K_(p)` and `K_(x)` are represented by the following reation
`K_(c)=([C]^(c)[D]^(d))/([A]^(a)[B]^(b)),K_(p)=(Pc^(c).P_(D)^(d))/P_(A)^(a)` and `Kx=(x_(C)^(c).x_(D)^(d))/(x_(A)^(a).x_(B)^(b)`
where `[A]` represents molar concentrationof `A,p_(A)` represents partial pressure of A and P represents total pressure, `x_(A)` represents mole fraction of A
For the reaction `SO_(2)Cl_(2)(g)hArrSO_(2)(g)+Cl_(2)(g),K_(p)gtK_(x)` is obtained at :

Options

A. `0.5` atm
B. `0.8` atm
C. `1` atm
D. `2`atm

Answer

To solve the problem, we need to analyze the given reaction and the relationships between the equilibrium constants \( K_c \), \( K_p \), and \( K_x \).

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ \text{SO}_2\text{Cl}_2(g) \rightleftharpoons \text{SO}_2(g) + \text{Cl}_2(g) \]

2. **Determine the Change in Moles (\( \Delta N_g \))**:
- Reactants: 1 mole of \( \text{SO}_2\text{Cl}_2 \)
- Products: 1 mole of \( \text{SO}_2 \) + 1 mole of \( \text{Cl}_2 \) = 2 moles
- Therefore,
\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 1 = 1 \]

3. **Relate \( K_p \) and \( K_x \)**:
The relationship between \( K_p \) and \( K_x \) is given by:
\[ K_p = K_x \cdot P^{\Delta N_g} \]
where \( P \) is the total pressure.

4. **Substituting \( \Delta N_g \)**:
Since \( \Delta N_g = 1 \):
\[ K_p = K_x \cdot P^1 = K_x \cdot P \]

5. **Condition for \( K_p > K_x \)**:
For \( K_p \) to be greater than \( K_x \):
\[ K_x \cdot P > K_x \]
This simplifies to:
\[ P > 1 \]

6. **Determine the Pressure**:
Thus, \( K_p > K_x \) is obtained when the total pressure \( P \) is greater than 1 atmosphere.

7. **Evaluate the Options**:
The options provided are:
- 0.5 atm (less than 1)
- 0.8 atm (less than 1)
- 1 atm (equal to 1)
- 2 atm (greater than 1)

The only option that satisfies \( P > 1 \) is **2 atm**.

### Final Answer:
The correct answer is **2 atm**.

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