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Chapter 12: Problem 107

The decomposition of iodoethane in the gas phase proceeds according to the following equation: $$ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{HI}(g) $$ At \(660 . \mathrm{K}, k=7.2 \times 10^{-4} \mathrm{~s}^{-1}\); at 720. \(\mathrm{K}, k=1.7 \times 10^{-2} \mathrm{~s}^{-1}\). What is the value of the rate constant for this first-order decomposition at \(325^{\circ} \mathrm{C} ?\) If the initial pressure of iodoethane is 894 torr at \(245^{\circ} \mathrm{C}\), what is the pressure of iodoethane after three half-lives?

Short Answer

Expert verified
The rate constant for the first-order decomposition of iodoethane at \(325^{\circ}C\) is approximately \(k_3 = 1.13 \times 10^{-3} s^{-1}\). After three half-lives, the pressure of iodoethane is approximately \(P_t = 111.75\) torr.

Step by step solution

01

Convert the given Celsius temperatures to Kelvin

To calculate the rate constant at the desired temperature, we need to convert the given Celsius temperatures to Kelvin. \[ T_1 = 660 K \] \[ T_2 = 720 K \] \[ T_3 = 325^{\circ}C + 273.15 = 598.15 K\]
02

Use Arrhenius equation to find activation energy

We can use the Arrhenius equation to relate rate constants and temperatures: \[ k = A e^{\frac{-E_{a}}{RT}} \] where k is the rate constant, A is the pre-exponential factor, E_{a} is the activation energy, R is the gas constant, and T is the temperature. We have values for k and T for two different temperatures: \[ k_1 = 7.2 \times 10^{-4} s^{-1}, T_1 = 660 K \] \[ k_2 = 1.7 \times 10^{-2} s^{-1}, T_2 = 720 K\] We can express the activation energy as follows: \[ \frac{k_1}{k_2} = \frac{A e^{\frac{-E_{a}}{R T_1}}}{A e^{\frac{-E_{a}}{R T_2}}} \Rightarrow \frac{E_a}{R} = \frac{T_1 T_2}{T_2 - T_1} \ln{\frac{k_2}{k_1}} \] Plug in the values and solve for E_{a}: \[ E_a = R \cdot \frac{660 K \cdot 720 K}{720 K - 660 K} \ln{\frac{1.7 \times 10^{-2} s^{-1}}{7.2 \times 10^{-4} s^{-1}}} \]
03

Calculate the rate constant at desired temperature

Now that we have found the activation energy, E_{a}, we can determine the rate constant at the desired temperature of \(T_3 = 598.15 K\) using the Arrhenius equation. We can rewrite the equation as follows: \[ k_3 = k_1 \cdot e^{\frac{E_{a}}{R} ( \frac{1} { T_1 } - \frac{1} { T_3 } )} \] Plug in the values and solve for k_3: \[ k_3 = 7.2 \times 10^{-4} s^{-1} \cdot e^{\frac{E_{a}}{R} ( \frac{1} { 660 K } - \frac{1} { 598.15 K } )}\]
04

Determine the pressure of iodoethane after three half-lives

For a first-order reaction, we can use the half-life formula to calculate the concentration (or pressure) after a certain number of half-lives: \[ P_t = P_0 \cdot \left(\frac{1}{2}\right)^n \] where \(P_t\) is the pressure at a given time, \(P_0\) is the initial pressure, and n is the number of half-lives. Given the initial pressure of iodoethane as \(P_0 = 894 torr\) and the number of half-lives as 3, we can calculate the pressure of iodoethane after three half-lives: \[ P_t = 894 torr \cdot \left(\frac{1}{2}\right)^3 \] Calculate the value of \(P_t\) to find the final pressure of iodoethane after three half-lives.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Arrhenius Equation
The Arrhenius equation is a fundamental formula that describes how the rate of a chemical reaction depends on temperature and activation energy. It is expressed as
\[ k = Ae^{\frac{-E_{a}}{RT}} \]
where:
  • \( k \) is the rate constant of the reaction,
  • \( A \) is the pre-exponential factor (also known as the frequency factor), which represents the number of times that reactants approach the activation barrier per unit time,
  • \( E_{a} \) is the activation energy required for the reaction to occur,
  • \( R \) is the universal gas constant (8.314 J/mol·K),
  • \( T \) is the temperature in Kelvin.

The equation indicates that as the temperature increases, the rate constant \( k \) also increases, meaning the reaction rate speeds up. This is because a higher temperature increases the number of molecules that have sufficient energy to overcome the activation energy barrier.
Activation Energy
Activation energy, denoted as \( E_{a} \), is the minimum energy required for reactants to transform into products during a chemical reaction. It represents the height of the energy barrier that must be overcome for the reaction to proceed. In the context of the problem, the activation energy for the decomposition of iodoethane can be found using the rate constants at two different temperatures. Using the Arrhenius equation, it's possible to solve for \( E_{a} \) and then apply this value to predict the reaction rate at any temperature. Lower activation energy means that the molecules need less energy to react, leading to a faster reaction, while a higher activation energy suggests a slower reaction rate.
First-Order Reaction
A first-order reaction is a type of chemical reaction where the rate is directly proportional to the concentration of one reactant. This means that as the reactant's concentration decreases by half, the rate of the reaction also halves. The rate law for a first-order reaction is given by:
\[ \text{Rate} = k[\text{Reactant}] \]
where \( k \) is the rate constant and the square brackets denote the concentration of the reactant. In our textbook problem, the decomposition of iodoethane is a first-order reaction, meaning the rate at which it decomposes depends on its concentration at a given time.
Half-Life
Half-life is the time required for the concentration of a reactant to decrease by half in a first-order reaction. It is an important concept because it provides an intuitive appreciation for the speed of a reaction. The half-life of a first-order reaction is constant and is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( t_{1/2} \) is the half-life and \( k \) is the rate constant. This means that regardless of the concentration of the reactant, it will always take the same amount of time for it to reduce by half. In the problem provided, knowing the initial pressure and the rate constant could help us find out how many half-lives have passed and thus determine the remaining pressure of iodoethane after a certain time.
Rate Constant
The rate constant, \( k \), is a crucial coefficient in the rate law of a chemical reaction that links the reaction rate to the reactant concentrations. It is determined by several factors, including temperature, activation energy, and the presence of a catalyst. In a first-order reaction, the rate constant can be used along with initial concentrations to determine the reaction's progress over time. As the Arrhenius equation shows, the rate constant is temperature-dependent; therefore, adjusting the temperature will alter the rate at which the reaction proceeds. In the problem, by calculating the rate constant at a different temperature, one can predict how fast the reaction will occur under new conditions.

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Most popular questions from this chapter

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

Consider the reaction $$ 3 \mathrm{~A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{~A}]^{2}[\mathrm{~B}][\mathrm{C}] $$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} M\) a. If after \(3.00 \mathrm{~min},[\mathrm{~A}]=3.26 \times 10^{-5} M\), calculate the value of \(k\). b. Calculate the half-life for this experiment. c. Calculate the concentration of \(\mathrm{B}\) and the concentration of \(\mathrm{A}\) after \(10.0 \mathrm{~min}\).

Consider the general reaction $$ \mathrm{aA}+\mathrm{bB} \longrightarrow \mathrm{cC} $$ and the following average rate data over some time period \(\Delta t\) : $$ \begin{aligned} -\frac{\Delta \mathrm{A}}{\Delta t} &=0.0080 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ -\frac{\Delta \mathrm{B}}{\Delta t} &=0.0120 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \\ \frac{\Delta \mathrm{C}}{\Delta t} &=0.0160 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s} \end{aligned} $$ Determine a set of possible coefficients to balance this general reaction.
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