In the reaction, `A + 2B to 6C + 2D`, if the initial rate `-(d[A])/(dt)` at t = 0 is `2.6 xx 10^(-2) m sec^(-1)`, what will be the value of `-(d[B])/(dt)" at " t = 0`?

Options

A. `8.5 xx 10^(-2) msec^(-1) `
B. `2.5 xx 10^(-2)m sec^(-1)`
C. `5.2 xx 10^(-2)m sec^(-1)`
D. `7.5 xx 10^(-2) m sec^(-1)`

Explanation

To solve the problem, we need to relate the rates of change of the concentrations of the reactants and products in the given chemical reaction:

**Reaction:**
\[ A + 2B \rightarrow 6C + 2D \]

### Step-by-Step Solution:

1. **Identify the Rate of Reaction:**
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, the rate can be defined as:
\[ -\frac{d[A]}{dt} = \frac{1}{1} \cdot -\frac{d[A]}{dt} = \frac{1}{2} \cdot -\frac{d[B]}{dt} = \frac{1}{6} \cdot \frac{d[C]}{dt} = \frac{1}{2} \cdot \frac{d[D]}{dt} \]

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

3. **Relate the Rates:**
From the stoichiometry of the reaction, we can relate the rate of change of B to that of A:
\[ -\frac{d[A]}{dt} = \frac{1}{2} \cdot -\frac{d[B]}{dt} \]
Rearranging gives:
\[ -\frac{d[B]}{dt} = 2 \cdot -\frac{d[A]}{dt} \]

4. **Substituting the Known Value:**
Substitute the known value of \(-\frac{d[A]}{dt}\) into the equation:
\[ -\frac{d[B]}{dt} = 2 \cdot (2.6 \times 10^{-2}) = 5.2 \times 10^{-2} \, \text{m/s} \]

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

### Summary:
The value of \(-\frac{d[B]}{dt}\) at \( t = 0 \) is \( 5.2 \times 10^{-2} \, \text{m/s} \).

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