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 \(-\frac{dC_D}{dt}\) based on the given rate equation for the reaction \(B + 2D \rightarrow 3T\).

### Step-by-Step Solution:

1. **Understand the Reaction**:
The reaction is given as \(B + 2D \rightarrow 3T\). Here, \(B\) and \(D\) are the reactants, and \(T\) is the product.

2. **Rate of Reaction**:
The rate of the reaction is given as:
\[ -\frac{dC_B}{dt} = kC_B C_D^2 \]
This means that the rate of disappearance of \(B\) is proportional to the concentration of \(B\) and the square of the concentration of \(D\).

3. **Using Stoichiometry**:
According to the stoichiometry of the reaction:
- For every 1 mole of \(B\) that reacts, 2 moles of \(D\) are consumed.
- Therefore, the rate of disappearance of \(D\) can be expressed as:
\[ -\frac{dC_D}{dt} = 2 \left(-\frac{dC_B}{dt}\right) \]

4. **Substituting the Rate of \(B\)**:
We can substitute the expression for \(-\frac{dC_B}{dt}\) into the equation for \(-\frac{dC_D}{dt}\):
\[ -\frac{dC_D}{dt} = 2 \left(kC_B C_D^2\right) \]

5. **Final Expression**:
Thus, we can write:
\[ -\frac{dC_D}{dt} = 2kC_B C_D^2 \]

This is the required expression for the rate of change of concentration of \(D\).

### Final Answer:
\[ -\frac{dC_D}{dt} = 2kC_B C_D^2 \]