Question

Decomposition of $H_2O_2$ follows a first order reaction. In fifty minutes 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 :

• A. $6.93 \times 10^{-2} \,mol \, min^{-1}$
• B. $6.93 \times 10^{-4} \,mol \, min^{-1}$
• C. $2.66 \,L \, min^{-1} $ at $STP$
• D. $1.34 \times 10^{-2} mol \, min^{-1}$

Answer 1

The Correct Option is B

To determine the rate of formation of O_2 when the concentration of H_2O_2 reaches 0.05 \, M, we need to follow the steps given below:

1. The decomposition of H_2O_2 is a first-order reaction, which means the rate of reaction depends on the concentration of H_2O_2 raised to the first power.

2. The first-order reaction can be described by the equation:

   k = \frac{1}{t} \ln\frac{[A]_0}{[A]}

   Where k is the rate constant, [A]_0 is the initial concentration, [A] is the concentration at time t.

3. Given: [A]_0 = 0.5 \, M, [A] = 0.125 \, M, t = 50 \, \text{minutes}.

4. Let's calculate the rate constant k using the given concentrations:

   k = \frac{1}{50} \ln\frac{0.5}{0.125} = \frac{1}{50} \ln 4

   Using \ln 4 \approx 1.386,

   k = \frac{1.386}{50} = 0.02772 \, \text{min}^{-1}

5. The rate of decomposition of H_2O_2 when [H_2O_2] = 0.05 \, M is given by:

   -\frac{d[H_2O_2]}{dt} = k[H_2O_2]

   Substitute the values:

   -\frac{d[H_2O_2]}{dt} = 0.02772 \times 0.05 = 0.001386 \, \text{mol} \, \text{min}^{-1}

6. Decomposition of H_2O_2 is:

   2H_2O_2 \rightarrow 2H_2O + O_2

   One mole of O_2 is produced from two moles of H_2O_2, so the rate of formation of O_2 is half the rate of the decomposition of H_2O_2.

7. Therefore, the rate of formation of O_2 is:

   =\frac{1}{2} \times 0.001386 = 0.000693 \, \text{mol} \, \text{min}^{-1} = 6.93 \times 10^{-4} \, \text{mol} \, \text{min}^{-1}

Thus, the correct answer is 6.93 \times 10^{-4} \, \text{mol} \, \text{min}^{-1}.