`H_(2)O_(2)to H_(2)O+O_(2)` the rate constant of this reaction is `2xx10^(-2)"min"^(-1)`(at certain temp.) & reaction is started with`0.5M` conc. of `H_(2)O_(2)`. After 23.03 min, the reaction was suddenly stopped by lowering the temp & the remaining `H_(2)O_(2)` was completely reacted with `NaOCl`. If no. of moles of `O_(2)` produced in second reaction is `1/a` find `a. (log1.5=0.2)`

Explanation

To solve the problem step by step, we will analyze the reaction and apply the first-order kinetics to find the value of \( a \).

### Step 1: Understand the Reaction
The reaction given is:
\[ H_2O_2 \rightarrow H_2O + O_2 \]
This reaction indicates that hydrogen peroxide decomposes into water and oxygen.

### Step 2: Identify Given Data
- The rate constant \( k = 2 \times 10^{-2} \, \text{min}^{-1} \)
- Initial concentration of \( H_2O_2 \) (denoted as \( a \)) = 0.5 M
- Time \( t = 23.03 \, \text{min} \)

### Step 3: Use the First-Order Rate Equation
For a first-order reaction, the integrated rate law is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{a}{a - x} \right) \]
Where:
- \( a \) = initial concentration
- \( x \) = amount of \( H_2O_2 \) that has reacted
- \( a - x \) = concentration of \( H_2O_2 \) remaining

### Step 4: Substitute Values into the Rate Equation
Substituting the known values into the equation:
\[ 2 \times 10^{-2} = \frac{2.303}{23.03} \log \left( \frac{0.5}{0.5 - x} \right) \]

### Step 5: Solve for \( a - x \)
Rearranging gives:
\[ 0.02 = \frac{2.303}{23.03} \log \left( \frac{0.5}{0.5 - x} \right) \]
Multiplying both sides by \( 23.03 \):
\[ 0.02 \times 23.03 = 2.303 \log \left( \frac{0.5}{0.5 - x} \right) \]
Calculating \( 0.02 \times 23.03 \):
\[ 0.46 = 2.303 \log \left( \frac{0.5}{0.5 - x} \right) \]
Dividing both sides by \( 2.303 \):
\[ \log \left( \frac{0.5}{0.5 - x} \right) = \frac{0.46}{2.303} \approx 0.2 \]

### Step 6: Exponentiate to Remove Logarithm
Using the property of logarithms:
\[ \frac{0.5}{0.5 - x} = 10^{0.2} = 1.58 \]

### Step 7: Solve for \( a - x \)
Cross-multiplying gives:
\[ 0.5 = 1.58(0.5 - x) \]
Expanding:
\[ 0.5 = 0.79 - 1.58x \]
Rearranging:
\[ 1.58x = 0.79 - 0.5 \]
\[ 1.58x = 0.29 \]
\[ x = \frac{0.29}{1.58} \approx 0.1835 \]

### Step 8: Calculate Remaining \( H_2O_2 \)
Now, substituting \( x \) back to find \( a - x \):
\[ a - x = 0.5 - 0.1835 \approx 0.3165 \]

### Step 9: Relate to Second Reaction
In the second reaction, the remaining \( H_2O_2 \) reacts completely with \( NaOCl \), producing an equal amount of \( O_2 \):
\[ \text{Moles of } O_2 = a - x \approx 0.3165 \]
Given that the number of moles of \( O_2 \) produced is \( \frac{1}{a} \):
\[ \frac{1}{a} = 0.3165 \implies a \approx \frac{1}{0.3165} \approx 3.15 \]

### Step 10: Final Value of \( a \)
Thus, rounding gives:
\[ a \approx 3 \]

### Conclusion
The value of \( a \) is approximately 3.