For a reaction scheme `Aoverset(k_(1))toBoverset(k_(2))toD` , If the rate of formation if `B` is set to be Zero then the concentration of `B` is given by :

Options

A. `(K_(1)-K_(2))[A]`
B. `K_(1)K_(2)[A]`
C. `(K_(1)+K_(2))[A]`
D. `(K_(1)/K_(2))[A]`

Answer

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

1. **Identify the reactions**:
- The first reaction is \( A \overset{k_1}{\rightarrow} B \)
- The second reaction is \( B \overset{k_2}{\rightarrow} D \)

2. **Write the rate of formation of B**:
The rate of formation of B can be expressed as:
\[ \text{Rate of formation of B} = k_1 [A] - k_2 [B] \]
where:
- \( k_1 [A] \) is the rate at which B is formed from A.
- \( k_2 [B] \) is the rate at which B is consumed to form D.

3. **Set the rate of formation of B to zero**:
According to the problem, the rate of formation of B is set to zero:
\[ k_1 [A] - k_2 [B] = 0 \]

4. **Rearranging the equation**:
Rearranging the equation gives:
\[ k_1 [A] = k_2 [B] \]

5. **Solving for [B]**:
To find the concentration of B, we can isolate [B]:
\[ [B] = \frac{k_1}{k_2} [A] \]

Thus, the concentration of B when the rate of formation is set to zero is:
\[ [B] = \frac{k_1}{k_2} [A] \]

### Final Answer:
The concentration of B is given by:
\[ [B] = \frac{k_1}{k_2} [A] \]