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The activation energy of a certain reaction is \(65.7 \mathrm{~kJ} / \mathrm{mol}\) How many times faster will the reaction occur at \(50^{\circ} \mathrm{C}\) than at \(0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The reaction will occur approximately 8.64 times faster at 50°C compared to 0°C.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we add 273.15 to each temperature: 0°C = 273.15 K, 50°C = 273.15 + 50 = 323.15 K.
02

Calculate \(\frac{k1}{k2}\)

We will be solving for the rate constants at both temperatures, then finding the ratio of these rate constants to determine how many times faster the reaction will be. Let \(k1\) be the rate constant at 0°C (273.15 K) and \(k2\) be the rate constant at 50°C (323.15 K). Lets find the ratio \(\frac{k1}{k2}\) using the Arrhenius equation. Dividing the Arrhenius equation for \(k1\) by the equation for \(k2\), we get: \[\frac{k1}{k2} = \frac{A\text{exp}\left(-\frac{E_a}{8.314 \times 273.15}\right)}{A\text{exp}\left(-\frac{E_a}{8.314 \times 323.15}\right)}\]
03

Simplify the equation

We can now simplify the equation to solve for the ratio between the rate constants. Observe that the pre-exponential factor, \(A\), cancels out: \[\frac{k1}{k2} = \frac{\text{exp}\left(-\frac{E_a}{8.314 \times 273.15}\right)}{\text{exp}\left(-\frac{E_a}{8.314 \times 323.15}\right)}\] Next, we can combine the exponential terms: \[\frac{k1}{k2} = \text{exp}\left(\frac{E_a}{8.314}\times\left(\frac{1}{273.15} -\frac{1}{323.15}\right)\right)\]
04

Substitute given values and calculate

Now, substitute the given value for the activation energy (65,700 J/mol) into the equation, and calculate the ratio: \[\frac{k1}{k2} = \text{exp}\left(\frac{65700}{8.314}\times\left(\frac{1}{273.15} -\frac{1}{323.15}\right)\right)\] \[\frac{k1}{k2} \approx 8.64\]
05

Final answer

The reaction will occur approximately 8.64 times faster at 50°C compared to 0°C.

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Key Concepts

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

Activation Energy
Activation energy is a crucial concept in chemical kinetics. It refers to the minimum amount of energy required for a chemical reaction to occur. Think of it like a hill that reactants must climb over to transform into products. Without sufficient energy, the reactants can't make it over the hill, and the reaction won't happen.

In our lesson, we found that the activation energy of the reaction is 65.7 kJ/mol. A higher activation energy means reactions proceed more slowly because fewer molecules have enough energy to get over the energy barrier. Conversely, a lower activation energy means the reaction can happen more quickly.

Understanding activation energy helps in controlling how fast reactions occur, which is essential in industrial processes and even in everyday life, like cooking.
Arrhenius Equation
The Arrhenius equation is a formula that shows the relationship between the temperature and the rate constant of a reaction. It provides a way to calculate how fast a reaction is based on temperature and activation energy. This equation is given by:
  • \[ k = A \times e^{-\frac{E_a}{RT}} \]
where:
  • \( k \) is the rate constant
  • \( A \) is the pre-exponential factor
  • \( E_a \) is the activation energy
  • \( R \) is the universal gas constant (8.314 J/mol K)
  • \( T \) is the temperature in Kelvin
With this equation, by knowing the activation energy and the temperatures involved, we can calculate the change in the rate constant.

In the original exercise, we used the Arrhenius equation to determine how many times faster the reaction proceeds by comparing the rate constants at two different temperatures. This approach helps illustrate how temperature affects chemical processes, making it a pivotal tool for both chemists and engineers.
Rate Constant
The rate constant, denoted as \( k \), is a significant measurement in the study of chemical kinetics. It is a proportionality constant in the rate equation of a reaction that provides the relationship between the concentration of reactants and the rate of the reaction.

The value of \( k \) is influenced by factors like temperature and activation energy. Higher values of \( k \) generally indicate a faster reaction rate.

In our example, we compared the rate constants \( k_1 \) and \( k_2 \) at two different temperatures using the Arrhenius equation. This allowed us to see how the rate of reaction changes with temperature changes.

Understanding the rate constant helps in predicting how changing conditions will affect the speed of a chemical process, offering invaluable insights for controlling reactions in both laboratory and industry settings.

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Most popular questions from this chapter

The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)-\rightarrow 2 \mathrm{NOCl}(g)\) obeys the rate law, rate \(=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}\right]\). The following mechanism has been proposed for this reaction: $$ \begin{aligned} \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}) & \longrightarrow \mathrm{NOCl}_{2}(g) \\ \mathrm{NOCl}_{2}(g)+\mathrm{NO}(g) & \cdots & 2 \mathrm{NOCl}(g) \end{aligned} $$ (a) What would the rate law be if the first step were rate determining? (b) Based on the observed rate law, what can we conclude about the relative rates of the two steps?

Molecular iodine, \(\mathrm{I}_{2}(g)\), dissociates into iodine atoms at \(625 \mathrm{~K}\) with a first-order rate constant of \(0.271 \mathrm{~s}^{-1}\). (a) What is the half-life for this reaction? (b) If you start with \(0.050 \mathrm{M} \mathrm{I}_{2}\) at this temperature, how much will remain after \(5.12 \mathrm{~s}\) assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(\mathrm{g})+\mathrm{HCl}(\mathrm{g})\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{Cl}(\mathrm{g}) \longrightarrow \mathrm{CH}_{3}(\mathrm{~g})+\mathrm{HCl}(\mathrm{g})\) Reaction 4: \(\mathrm{CH}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g)\) The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{kJ} / \mathrm{mol}) & E_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(\mathrm{~g})\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

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