Rate of a chemical reaction 2A (g) `to` B(g) is defined as :
`'r_B'=1/V(dn_B)/(dt)`
Where `n_B`=number of moles of B formed and `C_B` =Concentration of B then which of the following relation correct .

Options

A. `r_B=(dC_B)/(dt)+(C_B)/V(dV)/(dt)` (If volume V is not constant )
B. `r_B=(dC_B)/(dt)`(If volume V is constant)
C. `r_B=(dC_B)/(dt)-(C_B)/V(dV)/(dt)` (If volume V is not constant )
D. `r_B=-(dC_B)/(dt)+(C_B)/V(dV)/(dt)` (If volume V is not constant )

Explanation

To solve the problem, we need to analyze the rate of the chemical reaction given by the equation:

\[ R_B = \frac{1}{V} \frac{dN_B}{dt} \]

where:
- \( R_B \) is the rate of formation of B,
- \( V \) is the volume of the reaction,
- \( N_B \) is the number of moles of B formed.

We also know that the concentration of B (\( C_B \)) is related to the number of moles and volume by the equation:

\[ C_B = \frac{N_B}{V} \]

### Step 1: Express the rate in terms of concentration
Using the definition of concentration, we can express the rate \( R_B \) in terms of concentration:

\[ R_B = \frac{1}{V} \frac{dN_B}{dt} \]

Substituting \( N_B = C_B \cdot V \):

\[ R_B = \frac{1}{V} \frac{d(C_B \cdot V)}{dt} \]

### Step 2: Apply the product rule
Using the product rule of differentiation:

\[ \frac{d(C_B \cdot V)}{dt} = C_B \frac{dV}{dt} + V \frac{dC_B}{dt} \]

Substituting this back into the expression for \( R_B \):

\[ R_B = \frac{1}{V} \left( C_B \frac{dV}{dt} + V \frac{dC_B}{dt} \right) \]

### Step 3: Simplify the expression
Now, we can simplify the expression:

\[ R_B = \frac{C_B}{V} \frac{dV}{dt} + \frac{dC_B}{dt} \]

### Step 4: Analyze the case when volume is constant
If the volume \( V \) is constant, then \( \frac{dV}{dt} = 0 \). Therefore, the equation simplifies to:

\[ R_B = \frac{dC_B}{dt} \]

### Step 5: Conclusion
Thus, we find that when the volume is constant, the rate of formation of B is equal to the rate of change of concentration of B:

\[ R_B = \frac{dC_B}{dt} \]

### Final Answer
The correct relation is that when the volume is constant, the rate of formation of B is equal to the rate of change of concentration of B.