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. (A) `(K_(1)a-K_(2)b)/(K_(1)+K_(2))`
B. (B) `(K_(1)a-K_(2)b)/(K_(1)-(K_(2))`
C. (C) `(K_(1)a-K_(2)b)/(K_(1)K_(2))`
D. (D) `(K_(1)a-K_(2)b)/(K_(1)+K_(2))`

Explanation

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. **Write the Reaction and Initial Conditions**:
The reaction is given as:
\[ A \underset{K_2}{\overset{K_1}{\rightleftharpoons}} B \]
The initial concentrations are:
- Concentration of \( A \) = \( a \) moles/L
- Concentration of \( B \) = \( b \) moles/L

2. **Define Equilibrium Concentrations**:
At equilibrium, the concentrations of \( A \) and \( B \) will change as follows:
- Concentration of \( A \) = \( a - x \)
- Concentration of \( B \) = \( b + x \)

3. **Write the Expression for Equilibrium Constant**:
The equilibrium constant \( K_c \) for the reaction can be expressed in terms of the equilibrium concentrations:
\[ K_c = \frac{[B]}{[A]} = \frac{b + x}{a - x} \]

4. **Relate \( K_c \) to \( K_1 \) and \( K_2 \)**:
The relationship between the forward and backward rate constants is given by:
\[ K_c = \frac{K_1}{K_2} \]
Therefore, we can set up the equation:
\[ \frac{K_1}{K_2} = \frac{b + x}{a - x} \]

5. **Cross-Multiply to Eliminate the Fraction**:
Cross-multiplying gives:
\[ K_1(a - x) = K_2(b + x) \]

6. **Expand and Rearrange the Equation**:
Expanding both sides:
\[ K_1a - K_1x = K_2b + K_2x \]
Rearranging to group \( x \) terms on one side:
\[ K_1a - K_2b = K_2x + K_1x \]
This simplifies to:
\[ K_1a - K_2b = (K_1 + K_2)x \]

7. **Solve for \( x \)**:
Now, isolate \( x \):
\[ x = \frac{K_1a - K_2b}{K_1 + K_2} \]

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