Problem 16 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) follows a first order reaction. In fifty minutes the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) decreases from \(0.5\) to \(0.125 \mathrm{M}\) in one such decomposition. When the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) reaches \(0.05 \mathrm{M}\), the rate of formation of \(\mathrm{O}_{2}\) will be: [Main 2016] (a) \(2.66 \mathrm{~L} \mathrm{~min}^{-1}\) at STP (b) \(1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\) (c) \(6.96 \times 10^{-2} \mathrm{~mol} \min ^{-1}\) (d) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The correct option is (d) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\).

Step 1: Determine Rate Constant

Since the decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) is a first-order reaction, we can use the first-order kinetics formula given by \( \ln \left( \frac{[\mathrm{A}]_0}{[\mathrm{A}]} \right) = kt \). Here, \([\mathrm{A}]_0\) is the initial concentration (0.5 M), \([\mathrm{A}]\) is the final concentration (0.125 M), and \(t\) is time (50 minutes). Substitute these values and solve for \(k\).

Step 2: Calculate Rate Constant

Using the formula for first-order reactions, \( \ln \left( \frac{0.5}{0.125} \right) = k \times 50 \). This simplifies to \( \ln(4) = 50k \). Solve for \(k\): \( k = \frac{\ln(4)}{50} \approx 0.0277 \text{ min}^{-1}\).

Step 3: Determine Rate of Reaction

The rate of a first-order reaction is given by \( \text{Rate} = k [\mathrm{H}_2\mathrm{O}_2] \). When \([\mathrm{H}_2\mathrm{O}_2] = 0.05 \text{ M}\), we have \( \text{Rate} = 0.0277 \times 0.05 = 0.001385 \text{ M min}^{-1} \).

Step 4: Relate Rate of Formation of \(\mathrm{O}_2\)

The decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) can be represented as \( \mathrm{2H}_2\mathrm{O}_2 \rightarrow \mathrm{2H}_2\mathrm{O} + \mathrm{O}_2 \). For every 2 moles of \(\mathrm{H}_{2}\mathrm{O}_2\) decomposed, 1 mole of \(\mathrm{O}_2\) is formed. Thus, the rate of formation of \(\mathrm{O}_2\) is half the rate of consumption of \(\mathrm{H}_2\mathrm{O}_2\), yielding \( \text{Rate}_{\mathrm{O}_2} = \frac{0.001385 M \text{ min}^{-1}}{2} = 0.0006925 \text{ M min}^{-1} \).

Step 5: Convert to Moles per Minute

Finally, calculate the rate of \(\mathrm{O}_2\) formation in moles per minute. Since the rate was calculated as \(0.0006925 M \text{ min}^{-1}\), assume 1 L for molarity conversion: \(0.0006925 \text{ mol min}^{-1}\) is equivalent to \(6.93 \times 10^{-4} \text{ mol min}^{-1}\).

Key Concepts

Decomposition Kinetics

Understanding decomposition kinetics is crucial for predicting how reactions progress over time. In the case of hydrogen peroxide (\(\mathrm{H}_{2}\mathrm{O}_{2}\)), it decomposes through first-order kinetics. This means that the rate of decomposition depends on the concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\).

First-order reactions have a distinct characteristic: the rate of reaction decreases over time as the reactant concentration decreases. To understand this, think of first-order reactions using the equation:

• \( \ln \left( \frac{[\mathrm{A}]_0}{[\mathrm{A}]} \right) = kt \)

Here, \([\mathrm{A}]_0\) and \([\mathrm{A}]\) are the initial and current concentrations respectively, \(k\) is the rate constant, and \(t\) is time.

This formula highlights how the concentration of a reactant affects the reaction rate, making it vital for accurately predicting reaction behaviors.

Hydrogen Peroxide

Hydrogen peroxide is an essential compound with various applications. Its formula is \(\mathrm{H}_2\mathrm{O}_2\), and it is known for being a strong oxidizer. This characteristic makes it useful for bleaching, cleaning, and even as a mild antiseptic for wounds.

In this context, the decomposition of hydrogen peroxide is of interest, especially considering safety and environmental factors. The reaction leads to water and oxygen, depicted as:

• \( \mathrm{2H}_2\mathrm{O}_2 \rightarrow \mathrm{2H}_2\mathrm{O} + \mathrm{O}_2 \)

As \(\mathrm{H}_{2}\mathrm{O}_{2}\) decomposes, it releases oxygen, a process significant in various industrial and biological settings. Despite its utility, \(\mathrm{H}_{2}\mathrm{O}_{2}\) must be handled with care due to its reactive nature.

Rate Constant Calculation

To understand how fast a reaction occurs, we calculate the rate constant \(k\). For hydrogen peroxide decomposition, this is done using the first-order kinetics equation:

For example, from the problem provided, using the equation \( \ln \left( \frac{0.5}{0.125} \right) = kt \) over a period of 50 minutes, we simplify to:

• \(\ln(4) = 50k\)

This simplifies to solving for \(k\) as follows:

• \( k = \frac{\ln(4)}{50} \approx 0.0277 \text{ min}^{-1}\)

Here, \(k\) gives us an understanding of the speed of the reaction. A higher \(k\) would mean faster reaction rates, helping chemists and engineers optimize processes involving \(\mathrm{H}_{2}\mathrm{O}_{2}\).

The rate constant captures the intrinsic reaction rate irrespective of reactant concentrations.

Oxygen Formation Rate

The rate of oxygen formation provides insights into the efficiency of the decomposition process. The decomposition of \(\mathrm{H}_{2}\mathrm{O}_{2}\) ensures the steady release of \(\mathrm{O}_2\), important in both laboratories and industries.

In our context, once the concentration of \(\mathrm{H}_{2}\mathrm{O}_{2}\) is reduced to \(0.05 \text{ M}\), the rate of formation of \(\mathrm{O}_2\) is calculated using the rate equation.

The relationship is:

• \( \text{Rate}_{\mathrm{O}_2} = \frac{\text{Rate}_{\mathrm{H}_2\mathrm{O}_2}}{2} \)
• = \( \frac{0.001385 \text{ M min}^{-1}}{2} = 0.0006925 \text{ M min}^{-1} \)

Thus, the converted rate of oxygen formation becomes \(6.93 \times 10^{-4} \text{ mol min}^{-1}\).

This calculation is vital for quantifying oxygen release, ensuring the safe and effective use of \(\mathrm{H}_{2}\mathrm{O}_{2}\) in practical scenarios.