In the reaction ,
`A(s) +B(g) + "heat" hArr 2C(s) + 2D(g) `
at equilibrium, pressure of B is doubled to re-establish the equilibrium. The factor, by which conc. Of D is changed, is

Options

A. `sqrt2`
B. `2`
C. `3`
D. `sqrt3`

Answer

To solve the problem, we need to analyze the equilibrium reaction given and how the change in pressure of B affects the concentration of D.

### Step-by-Step Solution:

1. **Write the Equilibrium Expression**:
The reaction is given as:
\[ A(s) + B(g) + \text{heat} \rightleftharpoons 2C(s) + 2D(g) \]
The equilibrium constant \( K_p \) for the reaction can be expressed in terms of the partial pressures of the gases involved:
\[ K_p = \frac{(P_D)^2}{(P_B)} \]
where \( P_D \) is the partial pressure of D and \( P_B \) is the partial pressure of B.

2. **Initial Conditions**:
Let the initial partial pressure of B be \( P_B \) and the initial partial pressure of D be \( P_D \). Thus, we have:
\[ K_p = \frac{(P_D)^2}{P_B} \]

3. **Change in Pressure of B**:
According to the problem, the pressure of B is doubled:
\[ P_B' = 2P_B \]

4. **New Equilibrium Expression**:
When the pressure of B is doubled, we need to find the new concentration of D at equilibrium. The new equilibrium expression becomes:
\[ K_p = \frac{(P_D')^2}{P_B'} \]
Substituting \( P_B' \):
\[ K_p = \frac{(P_D')^2}{2P_B} \]

5. **Setting the Equilibrium Expressions Equal**:
Since \( K_p \) remains constant, we can set the two expressions for \( K_p \) equal to each other:
\[ \frac{(P_D)^2}{P_B} = \frac{(P_D')^2}{2P_B} \]

6. **Cancel \( P_B \) from Both Sides**:
Canceling \( P_B \) from both sides gives:
\[ (P_D)^2 = \frac{(P_D')^2}{2} \]

7. **Rearranging the Equation**:
Rearranging the equation to solve for \( P_D' \):
\[ (P_D')^2 = 2(P_D)^2 \]

8. **Taking the Square Root**:
Taking the square root of both sides:
\[ P_D' = P_D \sqrt{2} \]

9. **Conclusion**:
The concentration of D changes by a factor of \( \sqrt{2} \) when the pressure of B is doubled.

### Final Answer:
The factor by which the concentration of D is changed is \( \sqrt{2} \).

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