/*! 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 87 The activation energy of an unca... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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
At 25掳C (298.15 K), the catalyst increases the rate of the reaction by a factor of about 713.95. At 125掳C (398.15 K), the catalyst increases the rate of the reaction by a factor of about 49.65.

Step by step solution

01

Understand the Arrhenius equation

The Arrhenius equation is used to calculate the rate constant (k) of a reaction, given the activation energy (Ea) and temperature (T): \[k = Ae^{\frac{-Ea}{RT}}\] where A is the pre-exponential factor (also known as the collision factor), R is the gas constant (8.314 J/(mol K)), and T is the temperature in Kelvin. Since we are only interested in the ratio of the rate constants for the uncatalyzed and catalyzed reactions, the pre-exponential factor will cancel out. Therefore, we only need to compare the exponents for the two reactions.
02

Convert temperatures to Kelvin

First, convert the given temperatures from Celsius to Kelvin: (a) \(T_1 = 25 + 273.15 = 298.15 K\) (b) \(T_2 = 125 + 273.15 = 398.15 K\)
03

Calculate the ratio of the rate constants

For both temperatures, calculate the ratio of the rate constants for the catalyzed reaction (Ea = 55 kJ/mol) to the uncatalyzed reaction (Ea = 95 kJ/mol): (a) At 298.15 K: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-55 \times 10^3}{8.314 \times 298.15}}}{e^{\frac{-95 \times 10^3}{8.314 \times 298.15}}} = e^{\frac{40 \times 10^3}{8.314 \times 298.15}} \approx 713.95\] (b) At 398.15 K: \[\frac{k_{cat}}{k_{uncat}} = \frac{e^{\frac{-55 \times 10^3}{8.314 \times 398.15}}}{e^{\frac{-95 \times 10^3}{8.314 \times 398.15}}} = e^{\frac{40 \times 10^3}{8.314 \times 398.15}} \approx 49.65\]
04

Interpret the results

The factor by which the catalyst increases the rate of the reaction at each temperature is: (a) At 25掳C (298.15 K): The catalyst increases the rate of the reaction by a factor of about 713.95 (b) At 125掳C (398.15 K): The catalyst increases the rate of the reaction by a factor of about 49.65

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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 \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C}\) ? (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

The enzyme carbonic anhydrase catalyzes the reaction \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{H}^{+}(a q)\). In water, without the enzyme, the reaction proceeds with a rate constant of \(0.039 \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(1.0 \times 10^{6} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

Many primary amines, \(\mathrm{RNH}_{2}\), where \(\mathrm{R}\) is a carbon- containing fragment such as \(\mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}_{2}\), and so on, undergo reactions where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary amine (just use a \(\mathrm{C}\) atom for " \(\mathrm{R}^{w}\) ). (b) What kind of reactant with a primary amine can produce a tetrahedral intermediate?

Consider the hypothetical reaction \(2 \mathrm{~A}+\mathrm{B} \longrightarrow 2 \mathrm{C}+\mathrm{D}\). The following two-step mechanism is proposed for the reaction: Step 1: \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{X}\) Step 2: A \(+\mathrm{X} \longrightarrow \mathrm{C}+\mathrm{D}\) \(\mathrm{X}\) is an unstable intermediate. (a) What is the predicted rate law expression if Step 1 is rate determining? (b) What is the predicted rate law expression if Step 2 is rate determining? (c) Your result for part (b) might be considered surprising for which of the following reasons: (i) The concentration of a product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither reasons (i) nor (ii).

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. Near room temperature, the rate constant equals \(7.0 \times 10^{-4} \mathrm{~s}^{-1}\). Calculate the half-life at this temperature. (b) At \(415^{\circ} \mathrm{C},\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}\) decomposes in the gas phase, \(\left(\mathrm{CH}_{2}\right)_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{4}(g)+\mathrm{CO}(g)\). If the reaction is first order with a half-life of \(56.3 \mathrm{~min}\) at this temperature, calculate the rate constant in \(\mathrm{s}^{-1}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.