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 \(3A \rightarrow 2B\) and the rate of formation of \(B\) expressed as:

\[ \ln\left(\frac{d[B]}{dt}\right) = -4.606 + 2\ln[A] \]

we can follow these steps:

### Step 1: Understand the Rate Expression
The rate of formation of \(B\) can be expressed in terms of the change in concentration of \(B\) over time, divided by its stoichiometric coefficient. For the reaction, we have:

\[ \text{Rate of formation of } B = \frac{1}{2} \frac{d[B]}{dt} \]

### Step 2: Relate the Rate to Concentration of \(A\)
According to the rate law, the rate can also be expressed as:

\[ \text{Rate} = k[A]^n \]

where \(k\) is the rate constant and \(n\) is the order of the reaction with respect to \(A\).

### Step 3: Equate the Two Expressions
From the above two expressions, we can equate them:

\[ \frac{1}{2} \frac{d[B]}{dt} = k[A]^n \]

### Step 4: Rearranging the Equation
Rearranging gives:

\[ \frac{d[B]}{dt} = 2k[A]^n \]

### Step 5: Taking the Natural Logarithm
Taking the natural logarithm of both sides:

\[ \ln\left(\frac{d[B]}{dt}\right) = \ln(2k) + n\ln[A] \]

### Step 6: Compare with Given Equation
Now, we compare this with the given equation:

\[ \ln\left(\frac{d[B]}{dt}\right) = -4.606 + 2\ln[A] \]

From this comparison, we can see that:

- The term \(n\ln[A]\) corresponds to \(2\ln[A]\).
- Therefore, we can conclude that \(n = 2\).

### Step 7: Conclusion
Since the order of the reaction \(n\) is equal to \(2\), we conclude that the order of the reaction is:

\[ \text{Order of reaction} = 2 \]