For a reaction `B+2D to 3T`, it is given that `- (dC_a)/(dt)= kC_a C_D^2`. The expression for `-(dC_D)/(dt)` will be:

Options

A. `2KC_BC_D^2`
B. `1/2 k C_B C_D^2`
C. `4KC_BC_D^2`
D. `1/4 k C_BC_D^2`

Answer

To solve the problem, we need to derive the expression for the rate of change of concentration of D, given the reaction and the rate law.

### Step-by-Step Solution:

1. **Write the Balanced Reaction:**
The given reaction is:
\[ B + 2D \rightarrow 3T \]

2. **Identify the Rate Law:**
The rate of the reaction is given by:
\[ -\frac{dC_B}{dt} = k C_B C_D^2 \]
where \( C_B \) and \( C_D \) are the concentrations of B and D, respectively.

3. **Relate the Rates of Change:**
According to stoichiometry, the rates of change of the concentrations are related as follows:
\[ -\frac{dC_B}{dt} = \frac{1}{1} \cdot -\frac{dC_B}{dt} = \frac{1}{2} \cdot -\frac{dC_D}{dt} \]
This means:
\[ -\frac{dC_B}{dt} = \frac{1}{2} \left(-\frac{dC_D}{dt}\right) \]

4. **Substitute the Rate Law into the Stoichiometric Relationship:**
From the rate law, we can substitute \( -\frac{dC_B}{dt} \):
\[ \frac{1}{2} \left(-\frac{dC_D}{dt}\right) = k C_B C_D^2 \]

5. **Solve for \(-\frac{dC_D}{dt}\):**
Rearranging the equation gives:
\[ -\frac{dC_D}{dt} = 2 k C_B C_D^2 \]

6. **Final Expression:**
Thus, the expression for the rate of change of concentration of D is:
\[ -\frac{dC_D}{dt} = 2 k C_B C_D^2 \]