Problem 74 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The half-life for the first-order reaction: \(\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})\) is \(30 \mathrm{~min}\) If the volume of \(\mathrm{O}_{2}(\mathrm{~g})\) collected is \(100 \mathrm{ml}\) after a long time, then the volume of \(\mathrm{O}_{2}(\mathrm{~g})\) collected (at the same pressure and temperature) after \(60 \mathrm{~min}\) from the start of reaction is (a) \(25 \mathrm{ml}\) (b) \(12.5 \mathrm{ml}\) (c) \(75 \mathrm{~m}\) ] (d) \(50 \mathrm{ml}\)

Step-by-Step Solution

Answer

25 ml

Step 1: Understanding the Half-Life Concept

The half-life of a first-order reaction is the time required for the concentration of the reactant to fall to half its initial value. Given a half-life of 30 minutes, after 30 minutes, the original concentration of the reactant would have decreased by half, and therefore the volume of oxygen produced would also be half of the volume produced after a long time.

Step 2: Determine the Total Reaction Time in Terms of Half-Lives

The problem states that the total reaction time is 60 minutes, which is exactly two half-lives (60 minutes / 30 minutes per half-life = 2 half-lives).

Step 3: Calculate the Volume of Oxygen After Each Half-Life

After the first half-life (30 minutes), half of the oxygen would be produced. Therefore, the volume would be 100 ml / 2 = 50 ml. After the second half-life (another 30 minutes, for a total of 60 minutes), the volume would be halved again, so it would be 50 ml / 2 = 25 ml.

Key Concepts

Chemical Kinetics

Chemical kinetics is the study of the speed at which chemical processes occur and the factors that affect these rates. It involves the analysis of how different variables such as temperature, pressure, concentration, and the presence of catalysts influence the rate of chemical reactions. For reactions such as the decomposition of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2), kinetics provides insight into the speed of oxygen production.In a first-order reaction like the one between hydrogen peroxide and water, the rate of reaction is directly proportional to the concentration of one reactant. This means as the concentration of hydrogen peroxide decreases, the rate of oxygen production slows down. Understanding this relationship is crucial for predicting how much oxygen will be produced over time and how the reaction velocity changes as the reactants are consumed.

Half-Life Calculation

Half-life is a term most commonly used in nuclear physics and chemistry, referring to the duration required for a substance to decrease to half its initial amount. In the context of chemical kinetics, the half-life of a first-order reaction does not depend on the initial concentration of reactants. It's a constant value specific to each reaction under constant conditions.To calculate the half-life for a first-order reaction, one can use the formula: \br\[ t_{1/2} = \frac{\ln(2)}{k} \]where \br\( t_{1/2} \) is the half-life and \br\( k \) is the rate constant of the reaction. In our textbook exercise, the half-life of the reaction is given as 30 minutes, and after each half-life, the volume of oxygen produced is reduced by half. If students remember this principle, it becomes easier to understand the progression of the reaction over time, as illustrated in the step-by-step solution provided with the original problem.

Oxygen Production Volume

In chemical reactions involving gases, the volume of gas produced can often be measured to determine the extent of a reaction. In the case of our exercise, hydrogen peroxide decomposes to produce water and oxygen gas, and the volume of oxygen is used as an indicator of reaction progress. For a complete reaction, the exercise states that 100 ml of oxygen is collected. The volume of a gas produced in a chemical reaction is directly related to the number of moles of gas generated. As the reaction progresses, the moles of hydrogen peroxide decrease, and so does the volume of oxygen produced. Since the reaction has a half-life of 30 minutes, after 60 minutes, which equates to two half-lives, the volume of oxygen produced would be a quarter of the original volume if no other changes occur in pressure and temperature. This is because the volume produced after the first half-life would be halved, and then halved again after the second half-life. Students need to remember that for first-order reactions, the volume of gas produced at constant pressure and temperature is a good proxy for reaction progress.