For the reaction : `3A_((g))rarr 2B_((g))`, the rate of formation of 'B' at 298 K is represented as `ln((d[B])/(dt))=-4.606+2ln[A]`. The order of reaction is

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine the order of the reaction given by the equation \(3A_{(g)} \rightarrow 2B_{(g)}\) and the rate expression \(\ln\left(\frac{d[B]}{dt}\right) = -4.606 + 2\ln[A]\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Rate Expression**:
The rate of formation of \(B\) is given as:
\[ \ln\left(\frac{d[B]}{dt}\right) = -4.606 + 2\ln[A] \]

2. **Rearranging the Rate Expression**:
We can rewrite the equation in exponential form to express the rate in terms of \(A\):
\[ \frac{d[B]}{dt} = e^{-4.606} \cdot [A]^2 \]
Here, \(e^{-4.606}\) is a constant (let's denote it as \(k\)), so we can express the rate as:
\[ \frac{d[B]}{dt} = k[A]^2 \]

3. **Relating the Rate of Formation of B to A**:
From the stoichiometry of the reaction, the rate of formation of \(B\) is related to the consumption of \(A\):
\[ \frac{1}{2} \frac{d[B]}{dt} = -\frac{1}{3} \frac{d[A]}{dt} \]
This implies:
\[ \frac{d[B]}{dt} = -\frac{3}{2} \frac{d[A]}{dt} \]

4. **Setting Up the Rate Expression**:
Now, we can equate the two expressions for the rate:
\[ -\frac{3}{2} \frac{d[A]}{dt} = k[A]^2 \]

5. **Finding the Order of Reaction**:
The rate expression can be written as:
\[ \frac{d[A]}{dt} = -\frac{2}{3} k[A]^2 \]
This indicates that the rate of reaction depends on the concentration of \(A\) raised to the power of 2. Therefore, the order of the reaction with respect to \(A\) is 2.

6. **Conclusion**:
The overall order of the reaction is the sum of the powers of the concentration terms in the rate law. Since we have:
\[ n = 2 \]
The order of the reaction is:
\[ \text{Order of reaction} = 2 \]

### Final Answer:
The order of the reaction is **2**.