Problem 15 - Analytical Chemistry

Question

Hydrogen peroxide decomposes by a second-order reaction, $$ 2 \mathrm{H}_{2} \mathrm{O}_{2} \stackrel{\text { catalyst }}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ If \(35.0 \%\) of a \(0.1000 M\) solution decomposes in \(8.60 \mathrm{~min}\), how much time is required for the evolution of \(100 \mathrm{~mL} \mathrm{O}_{2}\) from \(100.0 \mathrm{~mL}\) of a \(0.1000 \mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at standard temperature and pressure?

Step-by-Step Solution

Answer

Calculate the time using second-order kinetics and the known rate constant.

Step 1: Understand the Decomposition Reaction

Hydrogen peroxide decomposes through a second-order reaction. The reaction involves two hydrogen peroxide molecules decomposing into two water molecules and one oxygen molecule, which is given by \[2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} + \mathrm{O}_{2}\].

Step 2: Determine Initial Concentration and Decomposition

Initially, we have a solution with a concentration of 0.1000 M. As per the problem, 35.0% of it decomposes. To find out how much still remains (not decomposed), calculate:\[35.0\% \text{ decomposed } \Rightarrow 0.1000 \times 0.350 = 0.0350 \, \text{M decomposed} \]The remaining concentration is:\[0.1000 - 0.0350 = 0.0650 \, \text{M}\]

Step 3: Calculate the Half-Life of the Reaction

For a second-order reaction, the half-life \(t_{1/2}\) is given by: \[t_{1/2} = \frac{1}{k[A]_0}\] where \(k\) is the rate constant and \([A]_0\) is the initial concentration. Use the known decomposition data (35% decomposed in 8.60 minutes) to determine \(k\) by solving for it. We have \([A]\) as 0.0650 M at time 8.60 minutes from an initial 0.1000 M.

Step 4: Calculate Rate Constant (k) using Integrated Rate Equation

The integrated rate law for a second-order reaction is:\[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]Plug in given information to solve for \(k\):\[ \frac{1}{0.0650} - \frac{1}{0.1000} = k(8.60)\]Solve for \(k\).

Step 5: Determine Moles of Oxygen Evolved

Convert the volume of \(O_2\) produced at STP to moles:\[100\, \text{mL} = 0.100\, \text{L}\]Using the ideal gas law at STP, 1 mole of gas occupies 22.4 L:\[\text{moles of } O_2 = \frac{0.100}{22.4} \approx 0.004464\, \text{mol} \].

Step 6: Relate Moles of Oxygen to Decomposed Hydrogen Peroxide

Since 1 mole of \(O_2\) corresponds to 2 moles of \(H_2O_2\) decomposed:\[\text{moles of } H_2O_2 = 2 \times 0.004464 = 0.008928\, \text{mol} \]

Step 7: Calculate Time for Complete Evolution of 100 mL Oxygen

Using the moles calculated and initial volume of \(H_2O_2\) solution:\[100.0 \text{ mL of } 0.1000 \, \text{M} \Rightarrow 0.0100 \text{ mol initial } H_2O_2 \]Using the previously calculated rate constant \(k\), and second-order kinetics:\[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]With \([A]_0 = 0.1000 \text{ M}\) and \([A] = 0.1000 - \frac{0.008928}{0.1} = 0.01072 \text{ M}\), solve for time \(t\).

Key Concepts

Decomposition reaction

Decomposition reactions are a type of chemical reaction where a single substance breaks down into two or more simple substances. In this exercise, we specifically look at the decomposition of hydrogen peroxide (\( \mathrm{H}_2 \mathrm{O}_2 \)).

• Hydrogen peroxide decomposes into water (\( \mathrm{H}_2 \mathrm{O} \)) and oxygen (\( \mathrm{O}_2 \)).
• The reaction involved in the decomposition is: \( 2 \mathrm{H}_2 \mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} + \mathrm{O}_2 \)
• This type of reaction typically requires a catalyst to occur efficiently, although it's not always mentioned.

In a decomposition reaction like this, the original compound, hydrogen peroxide, loses some of its stability and breaks apart. Understanding this reaction helps us calculate various kinetic parameters involved in the experiment, such as rate constants and reaction rates.

Rate constant

The rate constant (\( k \)) is a crucial parameter in chemical reactions because it indicates how fast a reaction proceeds. For second-order reactions, such as the decomposition of hydrogen peroxide, the rate law expression is given by:\[\text{rate} = k[A]^2 \]where \( [A] \) is the concentration of the substance.

• The half-life (\( t_{1/2} \)) for a second-order reaction is given by \( \frac{1}{k[A]_0} \)
• The integrated rate law is:\[ \frac{1}{[A]} - \frac{1}{[A]_0} = kt \]
• At the beginning of the reaction in the original problem, \( [A]_0 = 0.1000 \text{ M} \)

Using initial concentrations and elapsed time, we can solve for \( k \) when analyzing how much of the compound has decomposed after a specific period. The rate constant provides insight into the reaction's speed under set conditions.

Ideal gas law

The Ideal Gas Law is a fundamental equation in chemistry that relates the volume, pressure, temperature, and number of moles of a gas. It is expressed as \( PV = nRT \). Each symbol represents:

• \( P \) = Pressure
• \( V \) = Volume
• \( n \) = Number of moles
• \( R \) = Ideal gas constant
• \( T \) = Temperature

In the context of the decomposition of hydrogen peroxide, we use this law to determine the number of moles of oxygen gas (\( \mathrm{O}_2 \)) produced at standard temperature and pressure (STP). At STP, 1 mole of an ideal gas occupies 22.4 L, which simplifies computations involving volume and moles of gas. Therefore, when you have 100 mL (or 0.1 L) of oxygen, \( n = \frac{0.1}{22.4} \approx 0.004464 \text{ moles} \).

Chemical kinetics

Chemical kinetics studies the rates of chemical processes and the factors affecting them. The focus is on understanding how reactants turn into products.

• Kinetics helps us determine the speed at which a reaction proceeds.
• It involves various components like rate laws, reaction order, and rate constants.
• For a second-order reaction, the rate is proportional to the square of the concentration of one reactant.

In this particular exercise, we're applying principles of chemical kinetics to solve for time and calculate how much of a substance decomposes over a given period. By understanding kinetics, we can predict how long it takes for reactions to proceed under varying conditions. The step-by-step determination of the rate constant and subsequent calculations are practical applications of kinetic principles, allowing the calculation of unknown variables like the evolution of gases from a decomposing substance.