Half life of reaction `: H_(2)O_(2)(aq) rarr H_(2)O(l)+(1)/(2)O_(2)(g)`is independent of initial concentration of `H_(2)O_(2)` volume of `O_(2)` gas after 20 minute is `5L` at `1 atm` and `27^(@)C` and after completion of reaction is `50L`. The rate constant is `:`

Options

A. `(1)/(20)log 10 mi n^(-1)`
B. `(2.303)/(20)log 10 mi n^(-1)`
C. `(2.303)/(20)log. (50)/(45)mi n^(-1)`
D. `(2.303)/(20)log. (45)/(50)mi n^(-1)`

Explanation

To solve the problem, we need to determine the rate constant (k) for the given reaction, which is first order. The steps to find the rate constant are as follows:

### Step 1: Understand the Reaction
The reaction is:
\[ H_2O_2(aq) \rightarrow H_2O(l) + \frac{1}{2} O_2(g) \]

### Step 2: Identify the Given Information
- Volume of \( O_2 \) gas after 20 minutes (\( V_t \)) = 5 L
- Total volume of \( O_2 \) gas after completion of the reaction (\( V_{\infty} \)) = 50 L
- Time (\( t \)) = 20 minutes

### Step 3: Use the First Order Reaction Formula
For a first order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A_0]}{[A]} \right) \]
Where:
- \( [A_0] \) = initial concentration of \( H_2O_2 \)
- \( [A] \) = concentration of \( H_2O_2 \) at time \( t \)

### Step 4: Relate Concentrations to Volume
In this case, we can relate the concentrations to the volumes of \( O_2 \) collected:
- Initial concentration of \( H_2O_2 \) can be represented as \( [A_0] = V_{\infty} = 50 \, \text{L} \)
- Concentration of \( H_2O_2 \) at time \( t \) can be represented as \( [A] = V_{\infty} - V_t = 50 \, \text{L} - 5 \, \text{L} = 45 \, \text{L} \)

### Step 5: Substitute Values into the Formula
Now we can substitute the values into the rate constant formula:
\[ k = \frac{2.303}{20} \log \left( \frac{50}{45} \right) \]

### Step 6: Calculate the Logarithm
Calculate the logarithm:
\[ \log \left( \frac{50}{45} \right) = \log(1.1111) \approx 0.045757 \]

### Step 7: Calculate the Rate Constant
Now substitute the logarithm value back into the equation for \( k \):
\[ k = \frac{2.303}{20} \times 0.045757 \]
\[ k \approx \frac{2.303 \times 0.045757}{20} \]
\[ k \approx \frac{0.1055}{20} \]
\[ k \approx 0.005275 \, \text{min}^{-1} \]

### Final Answer
The rate constant \( k \) is approximately:
\[ k \approx 0.005275 \, \text{min}^{-1} \]

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