Problem 223 - A Complete Resource Book in Chemistry for JEE Main

Question

Deomposition 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: (a) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\) (b) \(2.66 \mathrm{~L} \mathrm{~min}^{-1}\) at \(\mathrm{STP}\) (c) \(1.34 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\) (d) \(6.93 \times 10^{-2} \mathrm{~mol} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The answer is (a) \(6.93 \times 10^{-4} \mathrm{~mol} \mathrm{~min}^{-1}\).

Step 1: Understand the Reaction Order

The given reaction is first order with respect to the concentration of \(\mathrm{H}_2\mathrm{O}_2\). This implies that the rate of the reaction is directly proportional to the concentration of \(\mathrm{H}_2\mathrm{O}_2\).

Step 2: First Order Reaction Rate Law Equation

The rate law for a first-order reaction is given by: \[\ln \left( \frac{[A]_0}{[A]} \right) = kt\]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), and \(k\) is the rate constant.

Step 3: Calculate the Rate Constant \(k\)

Substitute the given values into the equation:\[ \ln \left( \frac{0.5}{0.125} \right) = k \times 50\] Solve for \(k\):\[k = \frac{\ln(4)}{50} = \frac{1.386}{50} = 0.02772 \, \text{min}^{-1}\]

Step 4: Calculate the Concentration Rate Change

The rate of a first-order reaction can also be expressed as:\[\text{Rate} = k \times [\mathrm{H}_2\mathrm{O}_2]\]Substitute \(k = 0.02772 \, \text{min}^{-1}\) and \([\mathrm{H}_2\mathrm{O}_2] = 0.05 \, M\):\[\text{Rate} = 0.02772 \times 0.05 = 0.001386 \, M \, \text{min}^{-1}\]

Step 5: Determine the Rate of Formation of \(\mathrm{O}_2\)

The equation for decomposition of \(\mathrm{H}_2\mathrm{O}_2\) is:\[2\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2\]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 disappearance of \(\mathrm{H}_2\mathrm{O}_2\):\[ \text{Rate of } \mathrm{O}_2 = \frac{1}{2} \times 0.001386 = 0.000693 \, \text{mol min}^{-1} \]

Step 6: Match with Given Options

Comparing with the given options, the rate of formation of \(\mathrm{O}_2\) is \(6.93 \times 10^{-4} \, \text{mol} \, \text{min}^{-1}\), which corresponds to option (a).

Key Concepts

Understanding Reaction Rate

The reaction rate refers to the speed at which a chemical reaction proceeds. In our example, the decomposition of hydrogen peroxide (\( \mathrm{H}_2\mathrm{O}_2 \)), the reaction rate determines how quickly the concentration of \( \mathrm{H}_2\mathrm{O}_2 \) decreases over time. Reaction rates can be influenced by factors such as temperature, concentration, and the presence of a catalyst. For this first-order reaction, the rate is directly proportional to the concentration of \( \mathrm{H}_2\mathrm{O}_2 \). This means that as the concentration of \( \mathrm{H}_2\mathrm{O}_2 \) decreases, the rate of reaction also decreases, and vice versa.

• Higher concentration = faster reaction
• Lower concentration = slower reaction

Understanding the reaction rate helps predict how long a reaction will take to reach a certain stage or completion.

Importance of Rate Constant

The rate constant (\( k \)) is a crucial variable in the rate equation of a first-order reaction. It provides a link between the concentration of a reactant and the reaction rate. Importantly, the rate constant is independent of the concentration and depends mostly on factors like temperature and catalysts. In our decomposition reaction of hydrogen peroxide, the calculated rate constant was \( k = 0.02772 \, \text{min}^{-1} \). This means that at any given concentration, the decomposition follows this constant value to determine the rate of the reaction.
The rate constant helps to:

• Predict the speed of reaction at various stages
• Understand how external conditions influence the reaction

Consistently, knowing \( k \) allows chemists to anticipate the behavior of a reaction under different circumstances.

Exploring Decomposition Reactions

Decomposition reactions involve a single compound breaking down into two or more simpler substances. In the context of the decomposition of hydrogen peroxide, this process results in the formation of water and oxygen gas. Such reactions are significant in both natural processes and industrial applications. In biological systems, hydrogen peroxide plays a role as a signaling molecule and a defense mechanism. However, here we focus on its chemical breakdown:

• Equation: \( 2\mathrm{H}_2\mathrm{O}_2 \rightarrow 2\mathrm{H}_2\mathrm{O} + \mathrm{O}_2 \)
• Outcome: Formation of water and release of oxygen gas

Studying this reaction helps in understanding the dynamics of breakdown and product formation.

Role of Hydrogen Peroxide

Hydrogen peroxide is a simple yet significant molecule with a wide array of uses. Chemically represented by \( \mathrm{H}_2\mathrm{O}_2 \), it can act as both an oxidizing and reducing agent in reactions. Its decomposition into water and oxygen is a crucial reaction for both everyday applications and industrial uses.

• Common uses: Antiseptic, bleach, and a cleaner
• Significance in reactions: Its decomposition releases oxygen, making it useful in various chemical and biological processes

When exploring first-order reactions, hydrogen peroxide's breakdown provides a perfect case study to understand fundamental kinetics concepts.