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)
Answer & explanation
Correct answer: option 3
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: \(6.93 \times 10^{-4} \, mol/min\)