The rate constant of the reaction
`2H_(2)O_(2)(aq) rarr 2H_(2)O(l) + O_(2)(g)` is `3 xx 10^(-3) min^(-1)`
At what concentration of `H_(2)O_(2)`, the rate of the reaction will be `2 xx 10^(-4) Ms^(-1)` ?

Options

A. `6.67 xx 10^(-3) (M)`
B. `2 (M)`
C. `4 (M)`
D. `0.08 (M)`

Explanation

To solve the problem, we need to determine the concentration of \( H_2O_2 \) at which the rate of the reaction is \( 2 \times 10^{-4} \, \text{M s}^{-1} \). Given that the rate constant \( k \) is \( 3 \times 10^{-3} \, \text{min}^{-1} \), we can use the rate law for a first-order reaction.

### Step-by-Step Solution:

1. **Identify the Rate Law**:
For a first-order reaction, the rate of the reaction can be expressed as:
\[ R = k [H_2O_2] \]
where \( R \) is the rate of the reaction, \( k \) is the rate constant, and \([H_2O_2]\) is the concentration of hydrogen peroxide.

2. **Substitute the Given Values**:
We know:
- Rate \( R = 2 \times 10^{-4} \, \text{M s}^{-1} \)
- Rate constant \( k = 3 \times 10^{-3} \, \text{min}^{-1} \)

3. **Convert Units**:
Since the rate constant \( k \) is given in minutes, we need to convert the rate from \( \text{M s}^{-1} \) to \( \text{M min}^{-1} \):
\[ R = 2 \times 10^{-4} \, \text{M s}^{-1} \times 60 \, \text{s/min} = 1.2 \times 10^{-2} \, \text{M min}^{-1} \]

4. **Set Up the Equation**:
Now we can substitute the values into the rate equation:
\[ 1.2 \times 10^{-2} = (3 \times 10^{-3}) [H_2O_2] \]

5. **Solve for \([H_2O_2]\)**:
Rearranging the equation to find \([H_2O_2]\):
\[ [H_2O_2] = \frac{1.2 \times 10^{-2}}{3 \times 10^{-3}} = 4 \, \text{M} \]

6. **Conclusion**:
The concentration of \( H_2O_2 \) at which the rate of the reaction will be \( 2 \times 10^{-4} \, \text{M s}^{-1} \) is \( 4 \, \text{M} \).

### Final Answer:
\[ [H_2O_2] = 4 \, \text{M} \]