Ch.14 - Chemical Kinetics

Tro - Chemistry: A Molecular Approach 4th Edition

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Tro 4th Edition

Ch.14 - Chemical Kinetics

Problem 99a

Chapter 14, Problem 99a

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.

Verified step by step guidance

Identify the Arrhenius equation: \(k = A e^{-\frac{E_a}{RT}}\), where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol·K), and \(T\) is the temperature in Kelvin.

Convert the given temperatures from Celsius to Kelvin by adding 273.15 to each temperature value.

Take the natural logarithm of both sides of the Arrhenius equation to linearize it: \(\ln(k) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T}\).

Plot \(\ln(k)\) versus \(\frac{1}{T}\) to obtain a straight line, where the slope of the line is \(-\frac{E_a}{R}\) and the y-intercept is \(\ln(A)\).

Calculate the slope of the line to find \(E_a\) using the relation \(E_a = -\text{slope} \times R\), and determine \(A\) by taking the exponential of the y-intercept: \(A = e^{\text{y-intercept}}\).

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Key Concepts

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

Reaction Kinetics

Reaction kinetics is the study of the rates of chemical reactions and the factors that affect these rates. It involves understanding how concentration, temperature, and catalysts influence the speed of a reaction. In this case, the reaction is first order in each reactant, indicating that the rate depends linearly on the concentration of each reactant.

Arrhenius Equation

The Arrhenius equation relates the rate constant of a reaction to temperature and activation energy. It is expressed as k = A * e^(-Ea/RT), where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. This equation is crucial for determining how temperature affects reaction rates and for calculating activation energy.

Activation Energy

Activation energy (Ea) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to form products. A higher activation energy means that fewer molecules have sufficient energy to react at a given temperature, which typically results in a slower reaction rate. Understanding Ea is essential for analyzing the temperature dependence of reaction rates.

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Textbook Question

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?

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Textbook Question

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

b. Determine the rate constant at 15 °C.

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Textbook Question

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

c. If a reaction mixture is 0.155 M in C₂H₅Brand 0.250 M in OH^-, what is the initial rate of the reaction at 75 °C?

Textbook Question

Consider the two reactions:

O + N₂ → NO + N E_a = 315 kJ/mol

Cl + H₂ → HCl + H E_a = 23 kJ/mol

a. Why is the activation barrier for the first reaction so much higher than that for the second?

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Textbook Question

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?

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