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. `6.67xx10^(-3)` M
B. 2 M
C. 4 M
D. 0.08 M

Answer

To solve the problem, we need to determine the concentration of hydrogen peroxide (H₂O₂) when the rate of the reaction is given. We can use the rate equation for the reaction:

**Step 1: Write the rate equation.**
The rate of the reaction can be expressed as:
\[ \text{Rate} = k \cdot [H_2O_2]^n \]
For the reaction \( 2H_2O_2 \rightarrow 2H_2O + O_2 \), it is a second-order reaction with respect to H₂O₂, which means \( n = 1 \).

**Step 2: Substitute the known values into the rate equation.**
Given:
- Rate constant, \( k = 3 \times 10^{-3} \, \text{min}^{-1} \)
- Rate of the reaction, \( \text{Rate} = 2 \times 10^{-4} \, \text{M s}^{-1} \)

Since the rate constant is in minutes, we need to convert the rate from M/s to M/min:
\[ 2 \times 10^{-4} \, \text{M s}^{-1} \times 60 \, \text{s/min} = 1.2 \times 10^{-2} \, \text{M min}^{-1} \]

Now we can substitute these values into the rate equation:
\[ 1.2 \times 10^{-2} = (3 \times 10^{-3}) \cdot [H_2O_2] \]

**Step 3: Solve for the concentration of H₂O₂.**
Rearranging the equation to solve for \([H_2O_2]\):
\[ [H_2O_2] = \frac{1.2 \times 10^{-2}}{3 \times 10^{-3}} \]

Calculating this gives:
\[ [H_2O_2] = \frac{1.2 \times 10^{-2}}{3 \times 10^{-3}} = 4 \, \text{M} \]

**Final Answer:**
The concentration of \( H_2O_2 \) required for the reaction rate to be \( 2 \times 10^{-4} \, \text{M s}^{-1} \) is **4 M**.

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