In a closed system : `A(s)hArr 2B(g)+3C`, if the partial pressure of C is doubled, then partial pressure of B will be

Options

A. Twice the orignal pressure
B. Halfof its original pressure
C. `1/(2sqrt2)` times the original pressure
D. `2sqrt2` times its original pressure

Answer

To solve the problem, we need to analyze the equilibrium reaction and how changing the partial pressure of one component affects the others.

### Step-by-Step Solution:

1. **Write the equilibrium expression**:
The reaction is given as:
\[ A(s) \rightleftharpoons 2B(g) + 3C(g) \]
The equilibrium constant \( K_p \) for this reaction can be expressed as:
\[ K_p = \frac{(P_B)^2 (P_C)^3}{1} \]
Since \( A \) is a solid, it does not appear in the equilibrium expression.

2. **Define the initial conditions**:
Let the initial partial pressures be:
- \( P_B = P_B \) (initial partial pressure of B)
- \( P_C = P_C \) (initial partial pressure of C)

3. **Change in partial pressure of C**:
According to the problem, the partial pressure of C is doubled:
\[ P_C' = 2P_C \]

4. **Write the new equilibrium expression**:
After the change, the new equilibrium constant \( K_p' \) can be expressed as:
\[ K_p' = \frac{(P_B')^2 (P_C')^3}{1} = (P_B')^2 (2P_C)^3 = 8(P_B')^2 (P_C)^3 \]

5. **Equate the two expressions for \( K_p \)**:
Since \( K_p \) remains constant, we have:
\[ K_p = (P_B)^2 (P_C)^3 = 8(P_B')^2 (P_C)^3 \]

6. **Cancel \( (P_C)^3 \) from both sides**:
This gives us:
\[ (P_B)^2 = 8(P_B')^2 \]

7. **Solve for \( P_B' \)**:
Rearranging the equation:
\[ (P_B')^2 = \frac{(P_B)^2}{8} \]
Taking the square root of both sides:
\[ P_B' = \frac{P_B}{\sqrt{8}} = \frac{P_B}{2\sqrt{2}} \]

8. **Final result**:
Thus, the new partial pressure of B when the partial pressure of C is doubled is:
\[ P_B' = \frac{1}{2\sqrt{2}} P_B \]