/*! 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 65 The rate of the reaction $$ ... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 65

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

Short Answer

Expert verified
The short answer to the question is: 1. Convert the given temperatures to Kelvin and find their inverse: 0.00347 K鈦宦�, 0.00335 K鈦宦�, 0.00325 K鈦宦�, and 0.00314 K鈦宦�. 2. Calculate the natural logarithm of the rate constants: -3.255, -2.291, -1.693, and -1.101. 3. Plot the points (\(\frac{1}{T}\), ln{k}) on a graph and determine the slope (m) using linear regression. 4. Calculate the activation energy using the equation \(E_a = -R \cdot m\), where R is the gas constant (8.314 J/mol路K), and m is the slope obtained in the previous step.

Step by step solution

01

Recall the Arrhenius equation

The Arrhenius equation relates the rate constant (k), the activation energy (\(E_a\)), the temperature (T) in Kelvin, and the pre-exponential factor (A): \[k = A \cdot e^{-\frac{E_a}{RT}}\] In this equation, R is the gas constant, and its value is 8.314 J/mol路K.
02

Linearize the Arrhenius equation

To linearize the Arrhenius equation, we can take the natural logarithm of both sides: \[\ln{k} = \ln{A} - \frac{E_a}{RT}\] Now we have a linear equation of the form y = mx + b, where y = \(\ln{k}\), m = -\(\frac{E_a}{R}\), x = \(\frac{1}{T}\), and b = \(\ln{A}\).
03

Convert given temperatures to Kelvin and find the inverse

We need to convert all temperatures from Celsius to Kelvin and find the inverse (1/T): 15掳C = 288.15 K; \(\frac{1}{T_1}\) = 0.00347 K鈦宦� 25掳C = 298.15 K; \(\frac{1}{T_2}\) = 0.00335 K鈦宦� 35掳C = 308.15 K; \(\frac{1}{T_3}\) = 0.00325 K鈦宦� 45掳C = 318.15 K; \(\frac{1}{T_4}\) = 0.00314 K鈦宦�
04

Calculate the natural logarithm of the rate constants

Now, calculate the natural logarithm of the k values given in the table: \(\ln{k_1}\) = -3.255 \(\ln{k_2}\) = -2.291 \(\ln{k_3}\) = -1.693 \(\ln{k_4}\) = -1.101
05

Construct the graph and determine the slope

Plot the points (\(\frac{1}{T}\), ln{k}) on a graph: (0.00347, -3.255) (0.00335, -2.291) (0.00325, -1.693) (0.00314, -1.101) Using linear regression, determine the slope (m) of the graph.
06

Calculate the activation energy

The slope of the graph is equal to -\(\frac{E_a}{R}\). We can rearrange the equation as follows: \[E_a = -R \cdot m\] Substitute the value of R and the slope obtained in the previous step into this equation to calculate the activation energy \(E_a\) in J/mol.

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Key Concepts

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

Reaction Rate Constant
In the study of chemical kinetics, the reaction rate constant, represented as k, is a crucial parameter which provides insight into the speed at which a reaction occurs.

This constant is influenced by various factors, including the temperature, the presence of catalysts, and the physical state of the reactants. A higher value of k suggests a reaction that occurs more quickly. The Arrhenius equation provides a way to connect the temperature of the system with the rate constant, illustrating how k varies with temperature changes.

To put it simply, at higher temperatures, reactants have more energy, and so the likelihood of successful collisions between them increases, leading to an increased reaction rate constant.
Activation Energy
Another fundamental concept in chemical reactions is activation energy, designated as \(E_a\). This refers to the minimum amount of energy needed for reactants to transform into products during a chemical reaction.

Thinking of it as a barrier, only reactants with sufficient energy to overcome this hurdle can result in a successful reaction. The magnitude of \(E_a\) reflects the reaction's sensitivity to temperature changes. A lower activation energy means that more molecules possess the necessary energy even at lower temperatures, thereby speeding up the rate at which the reaction proceeds.

The Arrhenius equation's exponential term \(e^{-\frac{E_a}{RT}}\) shows the direct relationship between \(E_a\) and the reaction rate constant \(k\); as \(E_a\) decreases, the rate constant \(k\) increases, enhancing the reaction rate.
Arrhenius Plot
When studying reaction kinetics, the graphical representation known as an Arrhenius plot is an important tool used for analyzing the effects of temperature on the rate constant of a reaction.

By plotting the natural logarithm of the reaction rate constant \(\ln(k)\) against the inverse temperature \(1/T\), a straight line is usually formed, which is indicative of the linear relationship described by the Arrhenius equation after logarithmic transformation. The slope of this line corresponds to \(-\frac{E_a}{R}\) and allows us to determine the activation energy.

The y-intercept, on the other hand, gives the natural logarithm of the pre-exponential factor \(\ln(A)\), which represents the frequency or likelihood of reaction collisions occurring. These plots are not just theoretical; in practice, they provide a practical method for experimentally determining the activation energy and pre-exponential factor from temperature-dependent reaction rate data, as seen in the exercise provided.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

Indicate whether each statement is true or false. If it is false, rewrite it so that it is true. (a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy change for the reaction. (b) Exothermic reactions are faster than endothermic reactions. (c) If you double the temperature for a reaction, you cut the activation energy in half.

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of \(3.924 \AA\). (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \AA=1 \times 10^{-10} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm} \mathrm{Pt}\) sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one \(\mathrm{Pt}\) atom can be estimated from its atomic diameter of \(2.8 \AA\). (c) Using your results from (a) and (b), calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

What is the molecularity of each of the following elementary reactions? Write the rate law for each. (a) \(2 \mathrm{NO}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{2}(g)\)

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?
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