Decomposition of $H_2O_2$, follows a first order reaction. In 50 min, the concentration of $H_2O_2$, decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of $H_2O_2$, reaches 0.05 M, the rate of formation of $O_2$ will be (Given log 2=0.3010)

$6.93×10^{-4}$ mol/min

The correct answer is Option (3) → $6.93×10^{-4}$ mol/min

To find the rate of formation of \(O_2\) when the concentration of \(H_2O_2\) reaches \(0.05 \, M\), we can follow these steps:

Given that the decomposition of \(H_2O_2\) is a first-order reaction, we can use the first-order rate equation:

\(k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right)\)

Where:

\([A]_0\) = initial concentration = \(0.5 \, M\)

\([A]\) = final concentration = \(0.125 \, M\)

\(t\) = time = \(50 \, min\)

Substituting the values:

\(k = \frac{2.303}{50} \log \left( \frac{0.5}{0.125} \right)\)

Calculating \(\frac{0.5}{0.125} = 4\):

\(\log(4) = 2 \log(2) = 2 \times 0.3010 = 0.602\)

Now substituting into the equation:

\(k = \frac{2.303}{50} \times 0.602 = \frac{1.384206}{50} = 0.02768412 \, min^{-1}\)

For a first-order reaction, the rate (\(r\)) can be expressed as:

\(r = k [H_2O_2]\)

Substituting \(k\) and the concentration:

\(r = 0.02768412 \, min^{-1} \times 0.05 \, M\)

Calculating:

\(r = 0.001384206 \, mol/min\)

The decomposition of \(H_2O_2\) produces \(O_2\) according to the reaction:

\(2H_2O_2 \rightarrow 2H_2O + O_2\)

From the stoichiometry, \(2 \, mol\) of \(H_2O_2\) produces \(1 \, mol\) of \(O_2\), so:

\(\text{Rate of formation of } O_2 = \frac{1}{2} \times r\)

Substituting \(r\):

\(\text{Rate of } O_2 = \frac{1}{2} \times 0.001384206 \approx 0.000693 \, mol/min\)

The rate of formation of \(O_2\) is: \(6.93 \times 10^{-4} \, mol/min\)