In the equilibrium `SO_2Cl_2(g) hArr SO_2(g) +Cl_2(g)`
at 2000 k and 10 atm pressure , % `CI_2 =% SO_2=40` (by volume) then

Options

A. `K_c=0.1 mol "It"^(-1)`
B. `(n(SO_2CI_2))/(n(SO_2))=1/4` at equilibrium
C. `n(SO_2CI_2)=n(SO_2)=n(CI_2)`
D. `K_p=8` atm

Answer

To solve the equilibrium problem for the reaction \( SO_2Cl_2(g) \rightleftharpoons SO_2(g) + Cl_2(g) \) at 2000 K and 10 atm pressure, where the percentage of \( Cl_2 \) and \( SO_2 \) is 40% by volume, we can follow these steps:

### Step 1: Understand the given data
- Total pressure \( P_{total} = 10 \, \text{atm} \)
- Percentage of \( Cl_2 = 40\% \)
- Percentage of \( SO_2 = 40\% \)

### Step 2: Calculate the partial pressures
Since the percentages of \( Cl_2 \) and \( SO_2 \) are both 40%, we can find their partial pressures:
- Volume of \( Cl_2 = 40\% \) of total volume
- Volume of \( SO_2 = 40\% \) of total volume
- Volume of \( SO_2Cl_2 \) can be calculated as the remaining volume.

Let’s assume the total volume is 100 mL (this is arbitrary and simplifies calculations):
- Volume of \( Cl_2 = 40 \, \text{mL} \)
- Volume of \( SO_2 = 40 \, \text{mL} \)
- Volume of \( SO_2Cl_2 = 100 \, \text{mL} - 40 \, \text{mL} - 40 \, \text{mL} = 20 \, \text{mL} \)

### Step 3: Calculate the partial pressures using the ideal gas law
Using the relationship between volume and pressure (at constant temperature):
- \( P_{Cl_2} = \frac{40}{100} \times P_{total} = 0.4 \times 10 \, \text{atm} = 4 \, \text{atm} \)
- \( P_{SO_2} = \frac{40}{100} \times P_{total} = 0.4 \times 10 \, \text{atm} = 4 \, \text{atm} \)
- \( P_{SO_2Cl_2} = \frac{20}{100} \times P_{total} = 0.2 \times 10 \, \text{atm} = 2 \, \text{atm} \)

### Step 4: Write the expression for the equilibrium constant \( K_p \)
The equilibrium constant \( K_p \) for the reaction is given by:
\[ K_p = \frac{P_{SO_2} \cdot P_{Cl_2}}{P_{SO_2Cl_2}} \]

### Step 5: Substitute the values into the equation
Substituting the partial pressures we calculated:
\[ K_p = \frac{(4 \, \text{atm}) \cdot (4 \, \text{atm})}{2 \, \text{atm}} = \frac{16}{2} = 8 \, \text{atm} \]

### Conclusion
The equilibrium constant \( K_p \) for the reaction at the given conditions is \( 8 \, \text{atm} \).

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