`3A to 2B`, rate of reaction `(+d [B])/( dt)` is equal to:

Options

A. `- (3)/(2) (d[A])/(dt)`
B. `- (2)/(3) (d[A])/( dt)`
C. `- (1)/(3) (d[A])/( dt)`
D. `+2 (d [A])/(dt)`

Explanation

To solve the question regarding the rate of reaction for the equation \( 3A \rightarrow 2B \), we can derive the relationship between the rate of disappearance of reactant A and the rate of appearance of product B.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction is given as:
\[ 3A \rightarrow 2B \]

2. **Define the rates of change**:
The rate of reaction can be defined in terms of the change in concentration of reactants and products over time. For reactant A, the rate of disappearance is given by:
\[ -\frac{d[A]}{dt} \]
For product B, the rate of appearance is given by:
\[ \frac{d[B]}{dt} \]

3. **Relate the rates to stoichiometry**:
The stoichiometric coefficients from the balanced equation indicate the relationship between the rates of disappearance of A and the appearance of B. According to the coefficients:
- For every 3 moles of A that disappear, 2 moles of B are produced.

4. **Set up the relationship**:
Using the stoichiometric coefficients, we can express the rates as:
\[ -\frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt} \]

5. **Rearranging the equation**:
To find the rate of formation of B in terms of the rate of disappearance of A, we can rearrange the equation:
\[ \frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

6. **Final expression**:
Thus, the rate of reaction in terms of the rate of change of concentration of B is:
\[ \frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

### Conclusion:
The rate of reaction \( \frac{d[B]}{dt} \) is equal to \( -\frac{2}{3} \frac{d[A]}{dt} \).