Ch.14 - Chemical Kinetics

Tro - Chemistry: A Molecular Approach 4th Edition

Tro4th EditionChemistry: A Molecular ApproachISBN: 9780134112831Non è quello che usi tu?Cambia libro di testo

Tro 4th Edition

Ch.14 - Chemical Kinetics

Problema 97b

Capitolo 14, Problema 97b

The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide is found to be first order with a rate constant of 0.128/s at 150 K. b. If the surface is initially completely covered with n-butane at 150 K, how long will it take for 25% of the molecules to desorb (leave the surface)? For 50% to desorb?

Guida verificata passo dopo passo

Identify that the desorption process is first-order, which means the rate of desorption depends on the concentration of n-butane on the surface.

Use the first-order rate equation: \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \), where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), \(k\) is the rate constant, and \(t\) is the time.

For 25% desorption, 75% of the n-butane remains on the surface. Set \([A] = 0.75[A]_0\) and solve for \(t\) using the equation: \( \ln \left( \frac{[A]_0}{0.75[A]_0} \right) = 0.128t \).

For 50% desorption, 50% of the n-butane remains on the surface. Set \([A] = 0.50[A]_0\) and solve for \(t\) using the equation: \( \ln \left( \frac{[A]_0}{0.50[A]_0} \right) = 0.128t \).

Calculate the time \(t\) for both scenarios using the natural logarithm and the given rate constant.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Durata del video:

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

First-Order Kinetics

First-order kinetics refers to a reaction rate that is directly proportional to the concentration of one reactant. In this case, the desorption of n-butane follows first-order kinetics, meaning that the rate of desorption decreases as the concentration of n-butane on the surface decreases. The mathematical representation is given by the equation ln([A]0/[A]t) = kt, where [A]0 is the initial concentration, [A]t is the concentration at time t, k is the rate constant, and t is time.

Video consigliato:

Percorso guidato

02:29

First-Order Reactions

Rate Constant

The rate constant (k) is a proportionality factor in the rate equation that provides insight into the speed of a reaction. For first-order reactions, the units of the rate constant are typically s⁻¹. In this scenario, the rate constant of 0.128/s indicates that the desorption process occurs relatively quickly at 150 K, allowing for calculations of how long it will take for specific percentages of n-butane to desorb from the surface.

Video consigliato:

Percorso guidato

00:45

Rate Constant Units

Exponential Decay

Exponential decay describes the process by which a quantity decreases at a rate proportional to its current value. In the context of desorption, the number of n-butane molecules remaining on the surface decreases exponentially over time. This relationship can be expressed mathematically, allowing us to determine the time required for a certain percentage of molecules to desorb, such as 25% or 50%, using the first-order kinetics equation.

Video consigliato:

Percorso guidato

03:01

Beta Decay

Pratica correlata

Domanda del libro di testo

The kinetics of this reaction were studied as a function of temperature. (The reaction is first order in each reactant and second order overall.)

C₂H₅Br(aq) + OH^- (aq) → C₂H₅OH(l) + Br^- (aq)

Temperature (°C) k (L,mol •s)

25 8.81⨉10^-5

35 0.000285

45 0.000854

55 0.00239

65 0.00633

a. Determine the activation energy and frequency factor for the reaction.

Domanda del libro di testo

The evaporation of a 120-nm film of n-pentane from a single crystal of aluminum oxide is zero order with a rate constant of 1.92⨉1013 molecules/cm²•s at 120 K. a. If the initial surface coverage is 8.9⨉1016 molecules/cm², how long will it take for one-half of the film to evaporate?

881

views

rank

Domanda del libro di testo

Consider the reaction in which HCl adds across the double bond of ethene: HCl + H₂C=CH₂ → H₃C-CH₂Cl The following mechanism, with the accompanying energy diagram, has been suggested for this reaction:

Step 1 HCl + H₂C=CH₂ → H₃C=CH₂⁺ + Cl^-

Step 2 H₃C=CH₂⁺ + Cl^- → H₃C-CH₂Cl

b. What is the expected order of the reaction based on the proposed mechanism?

585

views

Domanda del libro di testo

The desorption (leaving of the surface) of a single molecular layer of n-butane from a single crystal of aluminum oxide is found to be first order with a rate constant of 0.128/s at 150 K. a. What is the half-life of the desorption reaction?-

873

views