`A+ 2B rarr C+D." If "-(d[A])/(dt)=5xx10^(-4)" mol L"^(-1) s^(-1)," then "-(d[B])/(dt)` is :

Options

A. `2.5xx10^(-4)" mol L"^(-1) s^(-1)`
B. `5.0xx10^(-4)" mol L"^(-1) s^(-1)`
C. `2.5xx10^(-3)" mol L"^(-1) s^(-1)`
D. `1.0xx10^(-3)" mol L"^(-1) s^(-1)`

Explanation

To solve the problem, we need to determine the rate of disappearance of species B in the reaction:

\[ A + 2B \rightarrow C + D \]

Given that the rate of disappearance of A is:

\[ -\frac{d[A]}{dt} = 5 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

We can use the stoichiometry of the reaction to find the rate of disappearance of B.

### Step-by-step Solution:

1. **Identify the stoichiometric coefficients**:
In the reaction, the stoichiometric coefficients are:
- For A: 1
- For B: 2
- For C: 1
- For D: 1

2. **Write the rate expression**:
The rate of the reaction can be expressed in terms of the change in concentration of each reactant and product:
\[ \text{Rate} = -\frac{d[A]}{dt} = -\frac{1}{2} \frac{d[B]}{dt} \]
This means that for every 1 mole of A that disappears, 2 moles of B disappear.

3. **Relate the rates**:
From the stoichiometry, we can relate the rate of disappearance of A to the rate of disappearance of B:
\[ -\frac{d[B]}{dt} = 2 \left(-\frac{d[A]}{dt}\right) \]

4. **Substitute the given value**:
Now we substitute the given rate of disappearance of A into the equation:
\[ -\frac{d[B]}{dt} = 2 \times (5 \times 10^{-4}) = 10 \times 10^{-4} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Final answer**:
Therefore, the rate of disappearance of B is:
\[ -\frac{d[B]}{dt} = 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \]

### Summary:
The rate of disappearance of B is \( 1 \times 10^{-3} \, \text{mol L}^{-1} \text{s}^{-1} \).

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