If `n_(A)` and `n_(B)` are the number of moles at any instant in the reaction : `2A(g) to 3B g` (carried out in a vessel of VL, the rate of the reaction at that instant is given by

Options

A. `-(1)/(2)(dn_(A))/(dt)=(1)/(3)(dn_(B))/(dt)`
B. `-(1)/(V)(dn_(A))/(dt)=(1)/(V)(dn_(B))/(dt)`
C. `-(1)/(2V)(dn_(A))/(dt)=(1)/(3V)(dn_(B))/(dt)`
D. `-(1)/(V)(n_(A))/(t)=(1)/(V)(n_(B))/(t)`

Explanation

To find the rate of the reaction for the given equation \(2A(g) \rightarrow 3B(g)\), we will follow these steps:

### Step 1: Write the rate expression
The rate of a reaction can be expressed in terms of the change in concentration of reactants and products. For the reaction \(2A \rightarrow 3B\), we can write the rate as:

\[ \text{Rate} = -\frac{1}{2} \frac{d[n_A]}{dt} = \frac{1}{3} \frac{d[n_B]}{dt} \]

### Step 2: Define concentration
Concentration (\(C\)) is defined as the number of moles (\(n\)) divided by the volume (\(V\)). Therefore, we can express the concentrations of \(A\) and \(B\) as:

\[ C_A = \frac{n_A}{V} \quad \text{and} \quad C_B = \frac{n_B}{V} \]

### Step 3: Substitute concentrations into the rate expression
Substituting the expressions for concentration into the rate equation, we get:

\[ \text{Rate} = -\frac{1}{2} \frac{1}{V} \frac{d[n_A]}{dt} = \frac{1}{3} \frac{1}{V} \frac{d[n_B]}{dt} \]

### Step 4: Rearranging the equation
To express the rate in a more standard form, we can rearrange the above equations. Multiplying through by \(V\) gives:

\[ \text{Rate} = -\frac{1}{2V} \frac{d[n_A]}{dt} = \frac{1}{3V} \frac{d[n_B]}{dt} \]

### Step 5: Final expression for the rate
Thus, the final expression for the rate of the reaction is:

\[ \text{Rate} = -\frac{1}{2V} \frac{d[n_A]}{dt} = \frac{1}{3V} \frac{d[n_B]}{dt} \]