In a reversible reaction `Aunderset(K_2)overset(K_1)hArrB` the initial concentration of A and B are a and b in moles per litre and the equilibrium concentrations are (a-x) and (b+x) respectively, Express x in terms of `K_1, K_2, a and b`.

Options

A. `(K_1a-K_2b)/(K_1+K_2)`
B. `(K_1a-K_2b)/(K_1-K_2)`
C. `(K_1a-K_2b)/(K_1K_2)`
D. `(K_1a+K_2b)/(K_1+K_2)`

Answer

To solve the problem, we need to express \( x \) in terms of \( K_1, K_2, a, \) and \( b \) for the reversible reaction \( A \underset{K_2}{\overset{K_1}{\rightleftharpoons}} B \).

### Step-by-step Solution:

1. **Define Initial Concentrations**:
- The initial concentration of \( A \) is \( a \) moles per liter.
- The initial concentration of \( B \) is \( b \) moles per liter.

2. **Define Equilibrium Concentrations**:
- At equilibrium, the concentration of \( A \) will be \( a - x \).
- At equilibrium, the concentration of \( B \) will be \( b + x \).

3. **Write the Expression for the Equilibrium Constant**:
- The equilibrium constant \( K \) for the reaction can be expressed as:
\[ K = \frac{[B]_{eq}}{[A]_{eq}} = \frac{b + x}{a - x} \]

4. **Relate the Equilibrium Constant to Forward and Backward Rate Constants**:
- The equilibrium constant can also be expressed in terms of the rate constants:
\[ K = \frac{K_1}{K_2} \]

5. **Set the Two Expressions for \( K \) Equal**:
- Equating the two expressions for \( K \):
\[ \frac{b + x}{a - x} = \frac{K_1}{K_2} \]

6. **Cross Multiply to Solve for \( x \)**:
- Cross multiplying gives:
\[ K_2(b + x) = K_1(a - x) \]
- Expanding both sides:
\[ K_2b + K_2x = K_1a - K_1x \]

7. **Rearranging the Equation**:
- Combine like terms:
\[ K_2x + K_1x = K_1a - K_2b \]
- This simplifies to:
\[ (K_1 + K_2)x = K_1a - K_2b \]

8. **Solve for \( x \)**:
- Finally, divide both sides by \( K_1 + K_2 \):
\[ x = \frac{K_1a - K_2b}{K_1 + K_2} \]

### Final Expression:
Thus, the expression for \( x \) in terms of \( K_1, K_2, a, \) and \( b \) is:
\[ x = \frac{K_1a - K_2b}{K_1 + K_2} \]