Domanda del libro di testo
The half-life for radioactive decay (a first-order process) of plutonium- 239 is 24,000 years. How many years does it take for one mole of this radioactive material to decay until just one atom remains?
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Ch.15 - Chemical Kinetics
Tro5th EditionChemistry: A Molecular ApproachISBN: 9780134874371
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving117
- Ch.2 - Atoms & Elements106
- Ch.3 - Molecules and Compounds128
- Ch.4 - Chemical Reactions and Chemical Quantities56
- Ch.5 - Introduction to Solutions and Aqueous Solutions84
- Ch.6 - Gases112
- Ch.7 - Thermochemistry96
- Ch.8 - The Quantum-Mechanical Model of the Atom60
- Ch.9 - Periodic Properties of the Elements86
- Ch.10 - Chemical Bonding I: The Lewis Model91
- Ch.11 - Chemical Bonding II: Molecular Shapes, VSEPR & MO Theory85
- Ch.12 - Liquids, Solids & Intermolecular Forces51
- Ch.13 - Solids & Modern Materials48
- Ch.14 - Solutions93
- Ch.15 - Chemical Kinetics114
- Ch.16 - Chemical Equilibrium68
- Ch.17 - Acids and Bases137
- Ch.18 - Aqueous Ionic Equilibrium154
- Ch.19 - Free Energy & Thermodynamics91
- Ch.20 - Electrochemistry105
- Ch.21 - Radioactivity & Nuclear Chemistry74
- Ch.22 - Organic Chemistry141
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Capitolo 15, Problema 109
A certain substance X decomposes. Fifty percent of X remains after 100 minutes. How much X remains after 200 minutes if the reaction order with respect to X is (c) second order?
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Identify the reaction order: The problem states that the reaction is second order with respect to X.
Use the second-order integrated rate law: \( \frac{1}{[X]} = kt + \frac{1}{[X]_0} \), where \([X]_0\) is the initial concentration, \([X]\) is the concentration at time \(t\), and \(k\) is the rate constant.
Determine the rate constant \(k\): Use the information that 50% of X remains after 100 minutes. This means \([X] = \frac{1}{2}[X]_0\) at \(t = 100\) minutes. Substitute these values into the second-order rate law to solve for \(k\).
Calculate \([X]\) at 200 minutes: Use the value of \(k\) obtained from the previous step and substitute \(t = 200\) minutes into the second-order rate law to find \([X]\) at this time.
Interpret the result: The calculated \([X]\) will tell you how much of the substance X remains after 200 minutes.
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Reaction Order
Reaction order refers to the power to which the concentration of a reactant is raised in the rate law of a chemical reaction. It indicates how the rate of reaction depends on the concentration of reactants. For example, a second-order reaction means that the rate is proportional to the square of the concentration of the reactant, which significantly influences how the concentration changes over time.
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Average Bond Order
Half-Life in Second-Order Reactions
The half-life of a substance in a second-order reaction is inversely proportional to the initial concentration of the reactant. Unlike first-order reactions, where the half-life is constant, the half-life for second-order reactions increases as the reaction proceeds. This means that as time goes on, it takes longer for the concentration of the reactant to decrease by half.
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01:44Second-Order Half-Life
Integrated Rate Law for Second-Order Reactions
The integrated rate law for a second-order reaction can be expressed as 1/[X] = kt + 1/[X]₀, where [X] is the concentration of the reactant at time t, k is the rate constant, and [X]₀ is the initial concentration. This equation allows us to calculate the concentration of the reactant at any given time, which is essential for determining how much of substance X remains after a specified duration.
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01:59Second-Order Reactions