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 need to analyze the equilibrium reaction given:

**Equilibrium Reaction:**
\[ AB(s) \rightleftharpoons A(g) + B(g) \]

**Step 1: Write the expression for the equilibrium constant (Kc).**
The equilibrium constant \( K_c \) for the reaction can be expressed as:
\[ K_c = \frac{[A][B]}{[AB]} \]
Since \( AB \) is a solid, its concentration does not appear in the expression.

**Step 2: Define the initial concentrations.**
Let the initial equilibrium concentrations be:
- \([A] = a\)
- \([B] = b\)

Thus, the equilibrium constant can be expressed as:
\[ K_c = [A][B] = ab \]

**Step 3: Consider the change in concentration of A.**
According to the problem, the equilibrium concentration of A is doubled:
\[ [A] = 2a \]

**Step 4: Relate the change in concentration of B to the change in concentration of A.**
Since the equilibrium constant \( K_c \) remains constant, we can write:
\[ K_c = [A][B] \]
Substituting the new concentration of A:
\[ K_c = (2a)[B] \]

**Step 5: Set the two expressions for Kc equal to each other.**
From the initial state, we have:
\[ K_c = ab \]
From the new state, we have:
\[ K_c = (2a)[B] \]

Setting these equal gives:
\[ ab = (2a)[B] \]

**Step 6: Solve for the new concentration of B.**
Dividing both sides by \( 2a \) (assuming \( a \neq 0 \)):
\[ b = \frac{ab}{2a} \]
\[ b = \frac{b}{2} \]

Thus, the new equilibrium concentration of B becomes:
\[ [B] = \frac{b}{2} \]

**Final Answer:**
If the equilibrium concentration of A is doubled, the equilibrium concentration of B would become half of its original concentration.

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