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. `sqrt(2)`
B. 2
C. 3
D. `sqrt(3)`

Explanation

To solve the problem, we need to analyze the equilibrium reaction given:

\[ A(s) + B(g) + \text{heat} \rightleftharpoons 2C(s) + 2D(g) \]

### Step 1: Write the expression for the equilibrium constant (K)

The equilibrium constant (K) for the reaction can be expressed in terms of the partial pressures of the gaseous components:

\[ K = \frac{(P_D)^2}{(P_B)} \]

Where:
- \( P_D \) is the partial pressure of D
- \( P_B \) is the partial pressure of B

### Step 2: Analyze the change in pressure of B

According to the problem, the pressure of B is doubled. Let’s denote the initial pressure of B as \( P_B \). After doubling, the new pressure of B becomes:

\[ P_B' = 2P_B \]

### Step 3: Substitute the new pressure into the equilibrium expression

Substituting \( P_B' \) into the equilibrium expression gives us:

\[ K = \frac{(P_D')^2}{(2P_B)} \]

Where \( P_D' \) is the new partial pressure of D after the change.

### Step 4: Set the two expressions for K equal to each other

Since K remains constant at equilibrium, we can set the two expressions for K equal to each other:

\[ \frac{(P_D)^2}{(P_B)} = \frac{(P_D')^2}{(2P_B)} \]

### Step 5: Simplify the equation

By cross-multiplying, we can simplify the equation:

\[ (P_D)^2 \cdot (2P_B) = (P_D')^2 \cdot (P_B) \]

Dividing both sides by \( P_B \) (assuming \( P_B \neq 0 \)) gives:

\[ 2(P_D)^2 = (P_D')^2 \]

### Step 6: Solve for \( P_D' \)

Taking the square root of both sides, we find:

\[ P_D' = \sqrt{2} \cdot P_D \]

### 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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