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Ch.14 - Chemical Kinetics
McMurry - Chemistry 8th Edition
McMurry8th EditionChemistryISBN: 9781292336145Not the one you use?Change textbook
All textbooksMcMurry 8th EditionCh.14 - Chemical KineticsProblem 142
Chapter 14, Problem 142

A 0.500 L reaction vessel equipped with a movable piston is filled completely with a 3.00% aqueous solution of hydrogen peroxide. The H2O2 decomposes to water and O2 gas in a first-order reaction that has a half-life of 10.7 h. As the reaction proceeds, the gas formed pushes the piston against a constant external atmospheric pressure of 738 mm Hg. Calculate the PV work done (in joules) after a reaction time of 4.02 h. (You may assume that the density of the solution is 1.00 g/mL and that the temperature of the system is maintained at 20 °C.)

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Calculate the initial amount of hydrogen peroxide (H2O2) in moles. First, find the mass of H2O2 in the solution by using the density and the percentage concentration. Density of the solution is 1.00 g/mL, so the mass of the solution in the vessel is 500 g (since 0.500 L = 500 mL). 3.00% of this mass is H2O2, which equals 15 g of H2O2. Convert this mass to moles using the molar mass of H2O2 (34.01 g/mol).
Determine the amount of H2O2 left after 4.02 hours using the first-order decay formula: \( [H2O2] = [H2O2]_0 \times e^{-kt} \), where \( [H2O2]_0 \) is the initial concentration, \( k \) is the rate constant derived from the half-life (\( k = \frac{\ln(2)}{t_{1/2}} \)), and \( t \) is the time elapsed.
Calculate the amount of O2 produced. The stoichiometry of the decomposition reaction \( 2H2O2 \rightarrow 2H2O + O2 \) shows that 1 mole of O2 is produced for every 2 moles of H2O2 that decompose. Use the difference in the initial and remaining moles of H2O2 to find the moles of O2 produced.
Use the ideal gas law to find the volume of O2 gas produced at the given conditions. The ideal gas law is \( PV = nRT \), where \( P \) is the pressure (convert 738 mm Hg to atm), \( V \) is the volume of the gas, \( n \) is the number of moles of gas, \( R \) is the ideal gas constant (0.0821 L atm/mol K), and \( T \) is the temperature in Kelvin (convert 20°C to Kelvin).
Calculate the PV work done using the formula \( W = -P\Delta V \), where \( W \) is the work done by the system, \( P \) is the external pressure, and \( \Delta V \) is the change in volume of the gas (which is the volume of O2 produced). Remember that work done by the system against external pressure is considered negative in thermodynamics.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Reactions

First-order reactions are chemical reactions where the rate is directly proportional to the concentration of one reactant. In this case, the decomposition of hydrogen peroxide follows first-order kinetics, meaning that as the concentration of H2O2 decreases, the rate of reaction also decreases. The half-life of a first-order reaction is constant, allowing for straightforward calculations of concentration over time.
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Gas Laws and PV Work

The ideal gas law relates the pressure, volume, and temperature of a gas, and is crucial for calculating work done by gases. In this scenario, the work done (PV work) is calculated using the formula W = PΔV, where P is the external pressure and ΔV is the change in volume due to gas production. Understanding how to apply this law is essential for determining the energy changes in the system.
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Density and Volume Calculations

Density is defined as mass per unit volume and is critical for converting between mass and volume in solutions. Given that the density of the hydrogen peroxide solution is 1.00 g/mL, this allows for the calculation of the mass of H2O2 present in the reaction vessel. Accurate volume calculations are necessary to determine how much gas is produced and how it affects the piston movement.
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Related Practice
Textbook Question
For the thermal decomposition of nitrous oxide, 2 N2O1g2S 2 N21g2 + O21g2, values of the parameters in the Arrhenius equation are A = 4.2 * 109 s-1 and Ea = 222 kJ>mol. If a stream of N2O is passed through a tube 25 mm in diameter and 20 cm long at a flow rate of 0.75 L/min at what temperature should the tube be maintained to have a partial pressure of 1.0 mm of O2 in the exit gas? Assume that the total pressure of the gas in the tube is 1.50 atm.
Textbook Question
You may have been told not to mix bleach and ammonia. The reason is that bleach (sodium hypochlorite) reacts with ammonia to produce toxic chloramines, such as NH2Cl. For example, in basic solution: OCl-1aq2 + NH31aq2S OH-1aq2 + NH2Cl1aq2 (a) The following initial rate data for this reaction were obtained in basic solution at 25 °C What is the rate law for the reaction? What is the numerical value of the rate constant k, including the correct units?
Textbook Question

The rate constant for the first-order decomposition of gaseous N2O5 to NO2 and O2 is 1.7 * 10-3 s-1 at 55 °C. (a) If 2.70 g of gaseous N2O5 is introduced into an evacuated 2.00 L container maintained at a constant temperature of 55 °C, what is the total pressure in the container after a reaction time of 13.0 minutes?

Textbook Question

The rate constant for the first-order decomposition of gaseous N2O5 to NO2 and O2 is 1.7 * 10-3 s-1 at 55 °C. (b) Use the data in Appendix B to calculate the initial rate at which the reaction mixture absorbs heat (in J/s). You may assume that the heat of the reaction is independent of temperature.