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Chapter 14: Problem 87

The activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

Step by step solution

01

Convert temperatures to Kelvin

To calculate the reaction rates, we must first convert the given temperatures to Kelvin. The conversion formula is: \(T_{K} = T_{C} + 273.15\) For the two given temperatures, the Kelvin values are: \(T_{25^{\circ}\mathrm{C}} = 25 + 273.15 = 298.15\,K\) \(T_{125^{\circ}\mathrm{C}} = 125 + 273.15 = 398.15\,K\)
02

Calculate the reaction rate constants

Using the Arrhenius equation, we can calculate the reaction rate constants for both the catalyzed and uncatalyzed reactions. We have: \(k_{uncatalyzed} = Ae^{-E_a(uncatalyzed) / RT}\) \(k_{catalyzed} = Ae^{-E_a(catalyzed) / RT}\) Since the collision factor A remains the same in both equations, we can find the ratio of catalyzed reaction rate to uncatalyzed reaction rate as: \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-E_a(catalyzed) / RT}}{e^{-E_a(uncatalyzed) / RT}}\)
03

Calculate the rate increase factors

Now, we can plug in the given activation energies and temperature values to calculate the rate increase factors for both temperatures. (a) For \(T = 298.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 298.15)}}{e^{-(95 \times 10^3) / (8.314 \times 298.15)}} 鈮� 211.89\) (b) For \(T = 398.15 K\): \(\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-(55 \times 10^3) / (8.314 \times 398.15)}}{e^{-(95 \times 10^3) / (8.314 \times 398.15)}} 鈮� 69.40\) Therefore, the catalyst will increase the rate of reaction by approximately 211.89 times at \(25^{\circ} \mathrm{C}\) and 69.40 times at \(125^{\circ} \mathrm{C}\).

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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 the minimum amount of energy that reacting particles need to undergo a chemical reaction. It acts like a barrier, and only particles with energy equal to or greater than this barrier will successfully react. Let's imagine activation energy as a hill that reactants have to climb. If you add a catalyst, it lowers the height of the hill, making it easier for reactions to happen.
An uncatalyzed reaction has a higher activation energy compared to a catalyzed one. In our example, the activation energy decreases from 95 kJ/mol to 55 kJ/mol when a catalyst is introduced. This significant reduction implies that more particles will have sufficient energy to overcome this barrier and participate in the reaction, making the reaction faster.
Arrhenius Equation
The Arrhenius equation is a formula used to express how reaction rates are affected by temperature and activation energy. It is given by:
  • \( k = Ae^{-E_a/RT} \)
Here, \( k \) is the reaction rate constant, \( A \) is the pre-exponential factor or frequency factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
The equation shows that the reaction rate \( k \) increases with a decrease in activation energy (as we see with catalysis) or with an increase in temperature. Exponent \(-E_a/RT\) is a critical factor; as \( E_a \) decreases or \( T \) increases, the exponent becomes less negative, which increases the value of \( k \). This equation helps explain why catalysis is so effective in increasing reaction rates.
Reaction Rate Constants
The reaction rate constant, \( k \), is a crucial part of the Arrhenius equation. It determines the speed of a chemical reaction at a given temperature and activation energy. With catalysis, even if the collision factor \( A \) remains the same, the lower activation energy reduces the negative impact on the exponent of the Arrhenius equation, leading to a higher \( k \).
In the context of our example, inserting the activation energies and temperature values into the Arrhenius equation for both catalyzed and uncatalyzed reactions helps us compute the reaction rate constants. The ratio of these constants gives us a quantitative understanding of how much faster (or slower) a reaction occurs鈥攁 key insight for optimizing chemical processes.
Temperature Conversion
Temperature conversion, especially from Celsius to Kelvin, is fundamental in calculations involving the Arrhenius equation. This is because reaction rates depend on the absolute temperature, not simply the temperature difference.
The formula for converting Celsius to Kelvin is:
  • \( T_K = T_C + 273.15 \)
In our exercise, the temperatures 25掳C and 125掳C were converted to 298.15 K and 398.15 K, respectively. Kelvin is used because it ensures that temperature measurements start at absolute zero, providing an accurate scale for kinetic calculations as required by thermodynamic principles. Proper temperature conversion is key to correctly using the Arrhenius equation and assessing reaction kinetics.

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Most popular questions from this chapter

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{array}{c}{\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \text { (slow) }} \\ {\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{g})+\mathrm{I}^{-}(a q) \text { (fast) }}\end{array} $$ \(\begin{array}{l}{\text { (a) Write the chemical equation for the overall process. }} \\ {\text { (b) Identify the intermediate, if any, in the mechanism. }} \\ {\text { (c) Assuming that the first step of the mechanism is rate }} \\ {\text { determining, predict the rate law for the overall process. }}\end{array}\)

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) , what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{s}^{-1}\) at \(100^{\circ} \mathrm{C} .\) In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{s}^{-1}\) at \(21^{\circ} \mathrm{C}\) . (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{Cas} \mathrm{com}-\) pared to that at \(21^{\circ} \mathrm{C} ?(\mathbf{d})\) On the basis of parts \((\mathrm{c})\) and \((\mathrm{d}),\) what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?
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