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If the rate constant for a reaction triples when the temperature rises from \(3.00 \times 10^{2} \mathrm{K}\) to \(3.10 \times 10^{2} \mathrm{K},\) what is the activation energy of the reaction?

Short Answer

Expert verified
The activation energy is approximately 84.9 kJ/mol.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation describes the temperature dependence of the rate constant and is given by \( k = Ae^{-\frac{E_a}{RT}} \), 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.
02

Setting Up the Given Information

We are given that the rate constant triples when the temperature increases from \(300\, \text{K} \) to \(310\, \text{K}\). Hence, \( k_2 = 3k_1 \), where \( k_1 \) is the rate constant at \( T_1 = 300 \text{ K} \) and \( k_2 \) is the rate constant at \( T_2 = 310 \text{ K} \).
03

Use the Arrhenius Equation Ratio

Using the Arrhenius equation for both temperatures, we can write \( \frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{RT_2}}}{Ae^{-\frac{E_a}{RT_1}}} = e^{\frac{E_a}{R}(\frac{1}{T_1} - \frac{1}{T_2})} \). Since \( \frac{k_2}{k_1} = 3 \), it becomes \( 3 = e^{\frac{E_a}{R}(\frac{1}{300} - \frac{1}{310})} \).
04

Solve for Activation Energy \( E_a \)

To find \( E_a \), take the natural logarithm of both sides: \( \ln(3) = \frac{E_a}{R} \left( \frac{1}{300} - \frac{1}{310} \right)\). Simplify this to solve for \( E_a \): \[ E_a = R \cdot \ln(3) \cdot \left( \frac{1}{300} - \frac{1}{310} \right)^{-1} \].
05

Calculate the Activation Energy

Substitute in the values: \( R = 8.314 \text{ J/mol} \cdot \text{K} \) and calculate the difference in reciprocal temperatures: \( \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} \approx 0.0001075 \).Therefore, \( E_a = 8.314 \cdot \ln(3) / 0.0001075 \approx 8.314 \cdot 1.0986 / 0.0001075 \approx 84857 \text{ J/mol} \) or \( 84.9 \text{ kJ/mol} \).

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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 concept in chemistry that explains how reaction rates depend on temperature. This equation is expressed as \( k = Ae^{-\frac{E_a}{RT}} \). Here:
  • \( k \) is the rate constant, indicating how fast a reaction proceeds.
  • \( A \) is the pre-exponential factor, representing the frequency of collisions.
  • \( E_a \) is the activation energy, the minimum energy required for a reaction to occur.
  • \( R \) is the gas constant, valued at approximately 8.314 J/mol·K.
  • \( T \) is the absolute temperature in Kelvin.
The equation shows that as the temperature increases, the rate constant usually increases, leading to faster reaction rates.
The exponential function \( e^{-\frac{E_a}{RT}} \) highlights how an increase in the temperature reduces the fraction \( \frac{E_a}{RT} \), thereby increasing \( k \), assuming all other factors remain constant.
Rate constant
The rate constant \( k \) is a crucial parameter in kinetics, providing specific information about the speed at which a chemical reaction occurs. It is a proportional constant in the rate law expression of a reaction. For example, in the rate equation \( ext{Rate} = k[A][B] \) for a bimolecular reaction, \( k \) determines the rate based on concentrations \( [A] \) and \( [B] \).

The value of the rate constant is influenced by several factors:
  • Temperature: As indicated by the Arrhenius equation, the rate constant typically increases with temperature.
  • Activation energy: A lower activation energy results in a higher rate constant, meaning the reaction can proceed more easily.
  • Reaction mechanism: Different mechanisms can have different rate constants.
Understanding these factors helps in controlling reaction speeds in various applications, such as industrial synthesis or biochemical reactions.
Temperature dependence
Temperature plays a crucial role in determining chemical reaction rates. According to the Arrhenius equation, the rate constant \( k \) is directly correlated with temperature. This is often referred to as the "temperature dependence" of reaction rates.

When temperature increases, more molecules have the requisite energy (equal to or greater than the activation energy \( E_a \)) to react. This results in:
  • Increased frequency of effective collisions.
  • Higher probability of overcoming the activation energy barrier.
  • An increase in the rate constant \( k \), speeding up the reaction.
This dependency highlights the importance of temperature control in chemical processes, ensuring optimal reaction rates for efficiency and safety. Even a small change in temperature can significantly affect the pace of a reaction.
Natural logarithm
The natural logarithm function (denoted \( \ln \)) is essential in the context of the Arrhenius equation, especially when deriving activation energies. This logarithm is based on the constant \( e \approx 2.718 \,\), which is fundamental in exponential growth and decay problems.

In practical applications, the natural logarithm helps in linearizing the Arrhenius equation. By taking the \( \ln \) of both sides of \( k = Ae^{-\frac{E_a}{RT}} \), the equation becomes:\[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \]This linear form allows comparison and graphing of experimental data. If we plot \( \ln k \) against \( \frac{1}{T} \), the slope of the resulting straight line gives \(-\frac{E_a}{R}\).

Utilizing the natural logarithm simplifies calculations, making it easier to determine the activation energy \( E_a \) from empirical data.

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Most popular questions from this chapter

Radioactive gold-198 is used in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 5.6-mg sample of the isotope, how much of this sample remains after 1.0 day?

Draw a reaction coordinate diagram for an exothermic reaction that occurs in a single step. Identify the activation energy and the net energy change for the reaction on this diagram. Draw a second diagram that represents the same reaction in the presence of a catalyst, assuming a single step reaction is involved here also. Identify the activation energy of this reaction and the energy change. Is the activation energy in the two drawings different? Does the energy evolved in the two reactions differ?

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We want to study the hydrolysis of the beautiful green, cobalt-based complex called trans-dichlorobis(ethylenediamine) cobalt(III) ion, (Check your book to see figure) In this hydrolysis reaction, the green complex ion trans\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en \(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) ). The reaction progress is followed by observing the color of the solution. The original solution is green, and the final solution is red, but at some intermediate stage when both the reactant and product are present, the solution is gray. Changes in color with time as \(\mathrm{Cl}^{-}\) ion is replaced by \(\mathrm{H}_{2} \mathrm{O}\) in a cobalt(III) complex. The shape in the middle of the beaker is a vortex that arises because the solutions are being stirred using a magnetic stirring bar in the bottom of the beaker. Reactions such as this have been studied extensively, and experiments suggest that the initial, slow step in the reaction is the breaking of the Co-Cl bond to give a five-coordinate intermediate. The intermediate is then attacked rapidly by water. Slow: $$\begin{aligned}\operatorname{trans}-\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\end{aligned}$$ Fast: $$\begin{aligned}\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) & \rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})\end{aligned}$$ (a) Based on the reaction mechanism, what is the predicted rate law? (b) As the reaction proceeds, the color changes from green to red with an intermediate stage where the color is gray. The gray color is reached at the same time, no matter what the concentration of the green starting material (at the same temperature). How does this show the reaction is first order in the green form? Explain. (c) The activation energy for a reaction can be found by plotting In \(k\) versus \(1 / T .\) However, here we do not need to measure \(k\) directly. Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\) the time needed to achieve the gray color is a measure of \(k .\) Use the data below to find the activation energy. $$\begin{array}{cc}\text { Temperature }^{\circ} \mathrm{C} & \text { (for the Same Initial Concentration) } \\\\\hline 56 & 156 \mathrm{s} \\\60 & 114 \mathrm{s} \\\65 & 88 \mathrm{s} \\\75 & 47 \mathrm{s} \\\\\hline\end{array}$$

When heated, cyclopropane is converted to propene (see Example 15.5 ). Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=\) \(1.02 \times 10^{-3} \mathrm{s}^{-1},\) respectively. Determine the activation energy, \(E_{\omega},\) from these data.

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