Decompsition of `H_(2)O_(2)` follows a frist order reactions. In 50 min the concentrations of `H_(2)O_(2)` decreases from 0.5 to 0.125 M in one such decomposition . When the concentration of `H_(2)O_(2)` reaches 0.05 M, the rate of fromation of `O_(2)` will be

Options

A. `6.93xx10^(-4) "mol min"^(-1)`
B. `2.66L " min"^(-1)"at STP" `
C. `1.34xx10^(-2) "mol min"^(-1)`
D. `6.93xx10^(-2) "mol min"^(-1)`

Answer

To solve the problem regarding the decomposition of hydrogen peroxide (H₂O₂) and the rate of formation of oxygen (O₂), we will follow these steps:

### Step 1: Write the balanced equation for the decomposition of H₂O₂.
The decomposition of hydrogen peroxide can be represented as:
\[ 2 \text{H}_2\text{O}_2 \rightarrow 2 \text{H}_2\text{O} + \text{O}_2 \]

### Step 2: Determine the change in concentration of H₂O₂.
The concentration of H₂O₂ decreases from 0.5 M to 0.125 M in 50 minutes. We can calculate the change in concentration:
- Initial concentration, \([H_2O_2]_0 = 0.5 \, M\)
- Final concentration, \([H_2O_2]_t = 0.125 \, M\)

### Step 3: Calculate the number of half-lives.
The concentration of H₂O₂ decreases from 0.5 M to 0.125 M, which is a quarter of the original concentration. This indicates that two half-lives have occurred:
- First half-life: 0.5 M to 0.25 M
- Second half-life: 0.25 M to 0.125 M

Since this change occurs in 50 minutes, we can conclude:
- One half-life = \( \frac{50 \, \text{minutes}}{2} = 25 \, \text{minutes} \)

### Step 4: Calculate the rate constant (k) for the first-order reaction.
For a first-order reaction, the half-life (\(t_{1/2}\)) is related to the rate constant (k) by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
Substituting the half-life:
\[ 25 \, \text{minutes} = \frac{0.693}{k} \]
Rearranging gives:
\[ k = \frac{0.693}{25} \, \text{min}^{-1} \]
Calculating k:
\[ k \approx 0.02772 \, \text{min}^{-1} \]

### Step 5: Write the rate law for the reaction.
For a first-order reaction, the rate of reaction can be expressed as:
\[ \text{Rate} = k [H_2O_2] \]

### Step 6: Calculate the rate of formation of O₂ when \([H_2O_2] = 0.05 \, M\).
Substituting the value of k and the concentration of H₂O₂ into the rate law:
\[ \text{Rate} = k [H_2O_2] = 0.02772 \, \text{min}^{-1} \times 0.05 \, M \]
Calculating the rate:
\[ \text{Rate} = 0.001386 \, \text{M/min} \]

### Step 7: Relate the rate of formation of O₂ to the rate of disappearance of H₂O₂.
From the balanced equation, we see that 2 moles of H₂O₂ produce 1 mole of O₂. Therefore, the rate of formation of O₂ is half the rate of disappearance of H₂O₂:
\[ \text{Rate of formation of O}_2 = \frac{1}{2} \times \text{Rate of disappearance of H}_2O_2 \]
Thus:
\[ \text{Rate of formation of O}_2 = \frac{1}{2} \times 0.001386 \, \text{M/min} \]
Calculating this gives:
\[ \text{Rate of formation of O}_2 = 0.000693 \, \text{M/min} \]
or in scientific notation:
\[ \text{Rate of formation of O}_2 = 6.93 \times 10^{-4} \, \text{M/min} \]

### Final Answer:
The rate of formation of O₂ when the concentration of H₂O₂ reaches 0.05 M is:
\[ 6.93 \times 10^{-4} \, \text{M/min} \]