In the equilibrium,`AB(s) rarr A(g) + B(g)`, if the equilibrium concentration of A is doubled, the equilibrium concentration of B would become

Options

A. Half
B. Twice
C. `1//4th`
D. `1//8th`

Explanation

To solve the problem, we start with the equilibrium reaction:

\[ AB(s) \rightleftharpoons A(g) + B(g) \]

1. **Define the Equilibrium Constant (K)**:
The equilibrium constant \( K \) for the reaction can be expressed in terms of the concentrations of the gaseous products A and B. Since AB is a solid, its concentration does not appear in the expression for \( K \):
\[ K = [A][B] \]

2. **Initial Concentrations**:
Let the initial equilibrium concentrations of A and B be represented as:
\[ [A] = a \quad \text{and} \quad [B] = b \]

3. **Final Concentrations**:
According to the problem, the equilibrium concentration of A is doubled:
\[ [A]_{final} = 2a \]
We need to find the new equilibrium concentration of B, which we will denote as \( [B]_{final} \).

4. **Set Up the Equilibrium Constant Expression**:
Since the equilibrium constant does not change with the change in concentrations (as long as temperature remains constant), we can set up the equation:
\[ K = [A]_{initial}[B]_{initial} = [A]_{final}[B]_{final} \]
Substituting the known values:
\[ ab = (2a)[B]_{final} \]

5. **Solve for \( [B]_{final} \)**:
Rearranging the equation to solve for \( [B]_{final} \):
\[ [B]_{final} = \frac{ab}{2a} \]
Simplifying this gives:
\[ [B]_{final} = \frac{b}{2} \]

6. **Conclusion**:
Therefore, if the equilibrium concentration of A is doubled, the equilibrium concentration of B would become half of its initial concentration:
\[ [B]_{final} = \frac{b}{2} \]