In the reaction , `A+2Bto6C+2D`, if the initial rate `-(d[A])/(dt)` at t= 0 is `2.6xx10^(-2)Msec^(-1)`, what will be the value of `-(d[B])/(dt)` at t=0?

Options

A. `8.5xx10^(-2)Msec^(-1)`
B. `2.5xx10^(-2)Msec^(-1)`
C. `5.2xx10^(-2)Msec^(-1)`
D. `7.5xx10^(-2)Msec^(-1)`

Answer

To solve the problem, we need to analyze the given reaction and the rate of change of the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction**:
The reaction given is:
\[ A + 2B \rightarrow 6C + 2D \]

2. **Identify the Rate Expressions**:
The rate of the reaction can be expressed in terms of the change in concentration of the reactants and products:
\[ -\frac{d[A]}{dt} = \frac{1}{2} \left(-\frac{d[B]}{dt}\right) = \frac{1}{6}\frac{d[C]}{dt} = \frac{1}{2}\frac{d[D]}{dt} \]
This means that for every 1 mole of A that reacts, 2 moles of B react, and 6 moles of C and 2 moles of D are produced.

3. **Given Information**:
We are given that the initial rate of change of concentration of A at \( t = 0 \) is:
\[ -\frac{d[A]}{dt} = 2.6 \times 10^{-2} \, \text{M/s} \]

4. **Relate the Rates**:
From the rate expressions, we can relate the rate of change of B to that of A:
\[ -\frac{d[A]}{dt} = \frac{1}{2} \left(-\frac{d[B]}{dt}\right) \]
Rearranging gives:
\[ -\frac{d[B]}{dt} = 2 \left(-\frac{d[A]}{dt}\right) \]

5. **Substitute the Known Value**:
Now, substituting the known value of \(-\frac{d[A]}{dt}\):
\[ -\frac{d[B]}{dt} = 2 \times (2.6 \times 10^{-2} \, \text{M/s}) = 5.2 \times 10^{-2} \, \text{M/s} \]

6. **Final Answer**:
Therefore, the value of \(-\frac{d[B]}{dt}\) at \( t = 0 \) is:
\[ -\frac{d[B]}{dt} = 5.2 \times 10^{-2} \, \text{M/s} \]