If `3A rarr 2B`, then the rate of reaction of `+ (dB)/(d t)` 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 problem, we need to determine the rate of formation of product B in the reaction \(3A \rightarrow 2B\).

### Step-by-Step Solution:

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

2. **Understand the stoichiometry of the reaction**:
From the balanced equation, we can see that 3 moles of A produce 2 moles of B.

3. **Define the rate of reaction**:
The rate of reaction can be expressed in terms of the change in concentration of reactants and products over time. For reactant A, the rate can be expressed as:
\[ \text{Rate} = -\frac{1}{3} \frac{d[A]}{dt} \]
For product B, the rate can be expressed as:
\[ \text{Rate} = +\frac{1}{2} \frac{d[B]}{dt} \]

4. **Set the rates equal**:
Since the rate of reaction is the same regardless of whether we are considering the reactants or the products, we can set the two expressions for the rate equal to each other:
\[ -\frac{1}{3} \frac{d[A]}{dt} = \frac{1}{2} \frac{d[B]}{dt} \]

5. **Solve for \(\frac{d[B]}{dt}\)**:
To find \(\frac{d[B]}{dt}\), we can rearrange the equation:
\[ \frac{d[B]}{dt} = -\frac{1}{3} \frac{d[A]}{dt} \cdot \frac{2}{1} \]
Simplifying this gives:
\[ \frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

6. **Final expression**:
Therefore, the rate of formation of B is:
\[ \frac{d[B]}{dt} = -\frac{2}{3} \frac{d[A]}{dt} \]

### Conclusion:
The rate of reaction of \(+\frac{dB}{dt}\) is equal to \(-\frac{2}{3} \frac{dA}{dt}\).

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