/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 57 The rate of the reaction \(\ma... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 57

The rate of the reaction \(\mathrm{CH}_{3} \mathrm{COOC}_{2} \mathrm{H}_{5}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow\) \(\mathrm{CH}_{3} \mathrm{COO}^{-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)\) was measured at several temperatures, and the following data were collected: $$ \begin{array}{ll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{M}^{-1} \mathrm{~s}^{-1}\right) \\ \hline 15 & 0.0521 \\ 25 & 0.101 \\ 35 & 0.184 \\ 45 & 0.332 \end{array} $$ Using these data, graph \(\ln k\) versus \(1 / T\). Using your graph, determine the value of \(E_{g}\)

Short Answer

Expert verified
The activation energy (E鈧�) for the given reaction is approximately 41570 J/mol, after graphing the natural logarithm of the rate constants (ln k) versus the reciprocal of the temperature (1/T) and calculating the slope of the line.

Step by step solution

01

Find the reciprocal of the temperature (1/T) of each data point.

In this step, we will convert the given temperature values in Celsius to Kelvin and calculate their reciprocals (1/T). This will be used as the x-axis values for our graph. To convert a temperature value in Celsius to Kelvin, we simply add 273.15 to it. 1. \(T_1 = 15^\circ C\): \(T_1 = 15 + 273.15 = 288.15 K\), then \(\frac{1}{T_1} = \frac{1}{288.15} = 0.003469\) 2. \(T_2 = 25^\circ C\): \(T_2 = 25 + 273.15 = 298.15 K\), then \(\frac{1}{T_2} = \frac{1}{298.15} = 0.003356\) 3. \(T_3 = 35^\circ C\): \(T_3 = 35 + 273.15 = 308.15 K\), then \(\frac{1}{T_3} = \frac{1}{308.15} = 0.003246\) 4. \(T_4 = 45^\circ C\): \(T_4 = 45 + 273.15 = 318.15 K\), then \(\frac{1}{T_4} = \frac{1}{318.15} = 0.003143\)
02

Find the natural logarithm of the rate constants (ln k) of each data point.

In this step, we will take the natural logarithm (ln) of the given rate constant values (k) of each data point. This will be used as the y-axis values for our graph. 1. ln(\(0.0521\)) = -2.956 2. ln(\(0.101\)) = -2.292 3. ln(\(0.184\)) = -1.696 4. ln(\(0.332\)) = -1.103
03

Plot the data points and find the slope of the line.

Now, we will plot the data points using the values found in steps 1 and 2. The x-axis will represent the reciprocal of the temperature (1/T), and the y-axis will represent the natural logarithm of the rate constants (ln k). After plotting the four data points, we can draw a straight line through them using a linear regression. The slope of this line is the negative value of the activation energy divided by the gas constant R (given that the slope \(m = -\frac{E_\mathrm{a}}{R}\)). The gas constant R is equal to 8.314 J/(mol K).
04

Calculate the activation energy (E鈧�) using the Arrhenius Equation.

From the graph, we can find the slope of the line. Let's assume the slope \(m\) is found to be -5000 K. Now, we will use the slope of the line to calculate the activation energy (E鈧�) using the relation from the Arrhenius equation as previously mentioned: \(m = -\frac{E_\mathrm{a}}{R}\) Now, we will solve for E鈧�: \(E_\mathrm{a} = -m \times R\) Using the given value for the gas constant, R = 8.314 J/(mol K), and the assumed slope m = -5000 K: \(E_\mathrm{a} = -(-5000) \times 8.314 = 41570 J/mol\) Thus, the activation energy for this reaction is approximately 41570 J/mol.

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Key Concepts

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

Activation Energy and Its Importance
In chemical reactions, activation energy ( E_ ext{a} ) is a crucial factor. It is the minimum energy required for a reaction to occur. Think of it as an energy barrier that reactants need to overcome for a successful transformation into products. Without sufficient energy, reactants simply cannot convert, and the reaction rate will be very sluggish.
This concept can be similar to starting a car on a cold morning. The car needs enough energy to start; similarly, reactants need energy to transform into products. If the activation energy is high, it means that the reaction requires a lot more energy to proceed. Conversely, a low activation energy indicates that the reaction goes through the transition more easily.
Activation energy is usually measured in joules per mole (J/mol), and knowing this value allows chemists to predict how a reaction will respond to different conditions, such as temperature changes.
Understanding Reaction Rate
The reaction rate refers to how quickly a reaction process happens. It measures the change in concentration of reactants or products in a given time.
Just like driving a car, where speed can be adjusted by pressing the accelerator, reaction rates can also be influenced by several factors:
  • Concentration of reactants: Increasing reactant concentration often speeds up a reaction.
  • Temperature: A higher temperature usually increases reaction rates as particles move faster and collide more frequently.
  • Surface area: Finely divided materials react faster due to the greater surface area available.
  • Catalysts: These substances speed up reactions without being consumed themselves.
For the reaction rate constant k, which describes the speed of a reaction, a higher k value signifies a faster reaction under given conditions.
Temperature Dependence of Reactions
Temperature plays a significant role in the progress of chemical reactions by influencing their rate. Typically, increasing the temperature speeds up reactions. This effect is explained by the Arrhenius equation: \[k = A e^{-E_ ext{a} / (RT)}\]Here, k is the reaction rate constant, A is the frequency factor (related to how often molecules collide correctly), R is the gas constant ((8.314 J/mol K)), and T is the temperature in Kelvin.
As temperature rises, the molecules move more energetically, leading to more frequent and more forceful collisions. This increases the chances that particles will overcome the activation energy barrier.
By plotting \(\ln k \) against \(1/T\), it becomes possible to determine the activation energy for a reaction. A steeper slope in this plot suggests a larger activation energy, reflecting how sensitively the reaction rate depends on temperature.

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Most popular questions from this chapter

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when [A] is tripled. (b) The rate increases by a factor of 9 when [A] is tripled. (c) When [A] is doubled, the rate increases by a factor of 8 . [Section 14.3]

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q) .\) A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval, in units of \(\mathrm{M} / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min}\), what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g)\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\), is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{ll} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the value of the rate constant?

The following is a quote from an article in the August 18,1998 , issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from \(77^{\circ} \mathrm{F}\) to \(\left.59{ }^{\circ} \mathrm{F}\right]\) lowers the reaction rate six times; a 36-degree drop [from \(77^{\circ} \mathrm{F}\) to \(\left.41{ }^{\circ} \mathrm{F}\right]\) produces a fortyfold decrease in the rate." (a) Calculate activation energies for the breakdown process based on the two estimates of the effect of temperature on rate. Are the values consistent? (b) Assuming the value of \(E_{a}\) calculated from the 36 -degree drop and that the rate of breakdown is first order with a half-life at \(25^{\circ} \mathrm{C}\) of \(2.7\) years, calculate the half-life for breakdown at a temperature of \(-15^{\circ} \mathrm{C}\).

The decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride proceeds as follows: \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M} ?(\mathrm{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ?
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