Consider the following first order reaction.
` 2H_2 O_2 (aq) overset(I^(-) (aq))to 2H_2 O (l) +O_2 (g)`
If the rate constant of the reaction is `1.01 xx 10^(-2) min^(-)`
Calculate rate of reaction when `[H_2 O_2]=0.5 mol L^(-)`

Explanation

To solve the problem, we need to calculate the rate of the first-order reaction given the rate constant and the concentration of the reactant. Here are the steps to find the solution:

### Step 1: Identify the reaction order and rate law
The reaction given is a first-order reaction with respect to \( H_2O_2 \). The rate law for a first-order reaction can be expressed as:
\[ \text{Rate} (R) = k [H_2O_2] \]
where \( k \) is the rate constant and \([H_2O_2]\) is the concentration of hydrogen peroxide.

### Step 2: Substitute the known values
We are given:
- Rate constant \( k = 1.01 \times 10^{-2} \, \text{min}^{-1} \)
- Concentration of \( H_2O_2 = 0.5 \, \text{mol L}^{-1} \)

Now we can substitute these values into the rate equation:
\[ R = (1.01 \times 10^{-2} \, \text{min}^{-1}) \times (0.5 \, \text{mol L}^{-1}) \]

### Step 3: Calculate the rate
Now, we perform the multiplication:
\[ R = 1.01 \times 0.5 \times 10^{-2} \, \text{mol L}^{-1} \, \text{min}^{-1} \]
Calculating \( 1.01 \times 0.5 \):
\[ 1.01 \times 0.5 = 0.505 \]
Thus, we have:
\[ R = 0.505 \times 10^{-2} \, \text{mol L}^{-1} \, \text{min}^{-1} \]

### Step 4: Convert to standard scientific notation
To express \( 0.505 \times 10^{-2} \) in standard scientific notation:
\[ 0.505 \times 10^{-2} = 5.05 \times 10^{-3} \, \text{mol L}^{-1} \, \text{min}^{-1} \]

### Step 5: State the final answer
The rate of the reaction when \([H_2O_2] = 0.5 \, \text{mol L}^{-1}\) is:
\[ R = 5.05 \times 10^{-3} \, \text{mol L}^{-1} \, \text{min}^{-1} \]