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.14 - Chemical Kinetics
Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831
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TOC.tableOfContents
- Ch.1 - Matter, Measurement & Problem Solving136
- Ch.2 - Atoms & Elements112
- Ch.3 - Molecules, Compounds & Chemical Equations159
- Ch.4 - Chemical Quantities & Aqueous Reactions140
- Ch.5 - Gases118
- Ch.6 - Thermochemistry106
- Ch.7 - Quantum-Mechanical Model of the Atom62
- Ch.8 - Periodic Properties of the Elements103
- Ch.9 - Chemical Bonding I: The Lewis Model104
- Ch.10 - Chemical Bonding II: Molecular Shapes & Valence Bond Theory89
- Ch.11 - Liquids, Solids & Intermolecular Forces73
- Ch.12 - Solids and Modern Material59
- Ch.13 - Solutions110
- Ch.14 - Chemical Kinetics140
- Ch.15 - Chemical Equilibrium78
- Ch.16 - Acids and Bases155
- Ch.17 - Aqueous Ionic Equilibrium176
- Ch.18 - Free Energy and Thermodynamics108
- Ch.19 - Electrochemistry118
- Ch.20 - Radioactivity and Nuclear Chemistry82
- Ch.21 - Organic Chemistry150
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Capitolo 14, 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?
Guida verificata passo dopo passo1
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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00:36
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
Pratica correlata
Domanda del libro di testo
Consider the reaction: 2 NH3(aq) + OCl-(aq) → N2H4(aq) + H2O(l) + Cl- (aq) This three-step mechanism is proposed: NH3(aq) + OCl- (aq) Δk1k2 NH2Cl(aq) + OH- (aq) Fast NH2Cl(aq) + NH3(aq) →k3 N2H5+ (aq) + Cl- (aq) Slow N2H5+ (aq) + OH-(aq) →k4 N2H4(aq) + H2O(l) Fast a. Show that the mechanism sums to the overall reaction.
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