In the reaction `3Ararr2B`, rate of reaction `+(d[B])/(dt)` is equal to :

Options

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

Explanation

To solve the question regarding the rate of reaction for the equation \(3A \rightarrow 2B\), we can follow these steps:

### Step 1: Write the general expression for the rate of reaction
For any chemical reaction, the rate can be expressed in terms of the change in concentration of reactants and products. The general form is:
\[ \text{Rate} = -\frac{1}{\nu_A} \frac{d[A]}{dt} = \frac{1}{\nu_B} \frac{d[B]}{dt} \]
where \(\nu_A\) and \(\nu_B\) are the stoichiometric coefficients of A and B, respectively.

### Step 2: Identify the stoichiometric coefficients
In the reaction \(3A \rightarrow 2B\):
- The stoichiometric coefficient for A (\(\nu_A\)) is 3.
- The stoichiometric coefficient for B (\(\nu_B\)) is 2.

### Step 3: Write the rate of disappearance of A
The rate of disappearance of A can be expressed as:
\[ -\frac{d[A]}{dt} = \frac{1}{3} \text{Rate} \]

### Step 4: Write the rate of appearance of B
The rate of appearance of B can be expressed as:
\[ \frac{d[B]}{dt} = \frac{1}{2} \text{Rate} \]

### Step 5: Relate the rates of A and B
From the expressions above, we can relate the rates:
\[ \frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

### Step 6: Find the expression for \(+\frac{d[B]}{dt}\)
To express the rate of formation of B (which is positive), we can rearrange the equation:
\[ +\frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

### Final Answer
Thus, the rate of reaction \(+\frac{d[B]}{dt}\) is equal to:
\[ +\frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]