For `2A overset(k_(1)) rarr B overset(k_(2)) rarr3C. K_(1) =2xx10^(-4) sec^(-1) and k_(2) =3xx10^(-4)`/mol-sec. {d[B]/dt} equal to :

Options

A. `k_(1) [A] -k_(2)[B]`
B. `k_(1) [A]^(2)-k_(2) [B]`
C. `k_(1)[A]-k_(2)[B]^(2)`
D. `k_(1)[A]^(2)-k_(2) [C]^(3)`

Answer

To solve the problem, we need to find the rate of change of concentration of species B, denoted as \(\frac{d[B]}{dt}\), for the given reactions:

1. \(2A \overset{k_1}{\rightarrow} B\)
2. \(B \overset{k_2}{\rightarrow} 3C\)

Where:
- \(k_1 = 2 \times 10^{-4} \, \text{s}^{-1}\) (first-order reaction)
- \(k_2 = 3 \times 10^{-4} \, \text{mol}^{-1}\text{s}^{-1}\) (second-order reaction)

### Step-by-Step Solution:

1. **Identify the Rate of Formation of B**:
The formation of B from A is given by the rate equation:
\[ \text{Rate of formation of B} = k_1 [A]^2 \]
Since the stoichiometry of the reaction is \(2A \rightarrow B\), we use \([A]^2\) because two moles of A are consumed to produce one mole of B.

2. **Identify the Rate of Consumption of B**:
The consumption of B to form C is given by the rate equation:
\[ \text{Rate of consumption of B} = k_2 [B]^2 \]
Here, B is consumed in a second-order reaction, hence the \([B]^2\) term.

3. **Write the Net Rate of Change of B**:
The net rate of change of concentration of B is the difference between the rate of formation and the rate of consumption:
\[ \frac{d[B]}{dt} = \text{Rate of formation of B} - \text{Rate of consumption of B} \]
Substituting the expressions from steps 1 and 2:
\[ \frac{d[B]}{dt} = k_1 [A]^2 - k_2 [B]^2 \]

4. **Final Expression**:
Thus, the final expression for the rate of change of concentration of B is:
\[ \frac{d[B]}{dt} = k_1 [A]^2 - k_2 [B]^2 \]