Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 74

A first-order reaction has rate constants of $4.6 \times 10^{-2} \mathrm{s}^{-1}\( and \)8.1 \times 10^{-2} \mathrm{s}^{-1}\( at \)0^{\circ} \mathrm{C}\( and \)20 .^{\circ} \mathrm{C},$ respectively. What is the value of the activation energy?

Short Answer

Expert verified
The activation energy for this first-order reaction can be determined using the Arrhenius equation and the given rate constants at two different temperatures. After converting the temperatures to Kelvin and solving for the activation energy (Ea), we find that the activation energy is approximately \(41.28 \frac{\textrm{kJ}}{\textrm{mol}}\).

Step by step solution

01

Convert temperatures to Kelvin

In order for the Arrhenius equation to be used, the temperatures need to be in Kelvin (K). Celsius can be converted to Kelvin by simply adding 273.15: \(T_1 = 0^{\circ} C + 273.15 K = 273.15 K\) \(T_2 = 20^{\circ} C + 273.15 K = 293.15 K\) Now we have the two temperatures, \(T_1\) and \(T_2\), in Kelvin: 273.15 K and 293.15 K.
02

Write down the Arrhenius equation

The Arrhenius equation relates the temperature, rate constant, and activation energy as follows: \(k = A \cdot e^{-\frac{Ea}{RT}}\) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant (8.314 J/mol·K), and T is the temperature (in Kelvin). Since we have two sets of values for k and T, we can set up two equations using the Arrhenius equation: \(k_1 = A \cdot e^{-\frac{Ea}{R \cdot T_1}}\) \(k_2 = A \cdot e^{-\frac{Ea}{R \cdot T_2}}\)
03

Solve for the activation energy (Ea)

We can find Ea by dividing the two equations (k1 and k2) and eliminating the pre-exponential factor A: \(\frac{k_1}{k_2} = \frac{A \cdot e^{-\frac{Ea}{R \cdot T_1}}}{A \cdot e^{-\frac{Ea}{R \cdot T_2}}}\) The A terms will cancel out, and we can rearrange the equation to solve for Ea: \(Ea = R \cdot (\frac{T_1 \cdot T_2 \cdot ln(\frac{k_1}{k_2})}{T_2 - T_1})\) Now, we can plug in the values for k1, k2, T1, and T2: \(Ea = 8.314 \frac{\textrm{J}}{\textrm{mol} \cdot \textrm{K}} \cdot (\frac{273.15 \textrm{K} \cdot 293.15 \textrm{K} \cdot ln(\frac{4.6 \times 10^{-2} \mathrm{s}^{-1}}{8.1 \times 10^{-2} \mathrm{s}^{-1}})}{293.15 \textrm{K} - 273.15 \textrm{K}})\)
04

Compute the activation energy

\( Ea = 8.314 \frac{\textrm{J}}{\textrm{mol} \cdot \textrm{K}} \cdot (\frac{273.15 \textrm{K} \cdot 293.15 \textrm{K} \cdot ln(0.568)}{20 \textrm{K}})\) Now, calculate the activation energy: \(Ea = 41283.96 \frac{\textrm{J}}{\textrm{mol}}\) Convert the activation energy to kilojoules per mole: \(Ea = 41.28 \frac{\textrm{kJ}}{\textrm{mol}}\) The activation energy for this first-order reaction is approximately 41.28 kJ/mol.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

The thiosulfate ion \(\left(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\right)\) is oxidized by iodine as follows: $$ 2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+2 \mathrm{I}^{-}(a q) $$ In a certain experiment, \(7.05 \times 10^{-3}\) mol/L of $\mathrm{S}_{2} \mathrm{O}_{3}^{2-}$ is consumed in the first 11.0 seconds of the reaction. Calculate the rate of consumption of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) . Calculate the rate of production of iodide ion.

For enzyme-catalyzed reactions that follow the mechanism $$ \begin{aligned} \mathrm{E}+\mathrm{S} & \rightleftharpoons \mathrm{E} \cdot \mathrm{S} \\ \mathrm{E} \cdot \mathrm{S} & \rightleftharpoons \mathrm{E}+\mathrm{P} \end{aligned} $$ a graph of the rate as a function of [S], the concentration of the substrate, has the following appearance: Note that at higher substrate concentrations the rate no longer changes with [S]. Suggest a reason for this.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

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.

Sign-up for free