/*! 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 66 The temperature dependence of th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The temperature dependence of the rate constant for a reaction is tabulated as follows: $$ \begin{array}{lc} \hline \text { Temperature (K) } & k\left(M^{-1} \mathrm{~s}^{-1}\right) \\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{array} $$ Calculate \(E_{a}\) and \(A\).

Short Answer

Expert verified
After plotting the natural logarithm of the rate constants (ln(k)) against the inverse of the temperatures (1/T) and finding the linear regression, we can calculate the activation energy (Ea) and pre-exponential factor (A) using the equations: \(Ea = -mR\) and \(A = e^b\), where m is the slope and b is the y-intercept from the linear regression.

Step by step solution

01

Logarithmic form of Arrhenius equation

\[ ln(k) = ln(A) - \frac{Ea}{RT} \] #Step 2: Calculate slope and intercept of the linear regression# Now, we need to find the best linear regression for the given data to determine the activation energy (Ea) and the pre-exponential factor (A). To do this, we can plot the natural logarithm of the given rate constants (ln(k)) against the inverse of the temperatures (1/T), which should give us a linear result.
02

Linear regression equations

\[ ln(k) = b + m(1/T) \] \[ b = ln(A) \] \[ m = -\frac{Ea}{R} \] Where b is the y-intercept, m is the slope, and R is the gas constant. #Step 3: Calculate the slope and intercept from the data# Using the data given and some standard statistical methods, you can find the slope (m) and intercept (b) of the best-fit line. One can do this in either a spreadsheet program or using a calculator with regression capabilities. #Step 4: Calculate Ea and A from the slope and intercept#
03

Calculate Ea and A

Now that we have the slope (m) and intercept (b) from the best-fit line, we can use the equations we derived above to find the values of Ea and A: \[ Ea = -mR \] \[ A = e^b \] Once you calculate Ea and A using these equations, you will have solved the problem.

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 91Ó°ÊÓ!

Key Concepts

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

Chemical Kinetics
Chemical kinetics is a branch of chemistry that deals with the rates of chemical reactions, how they change under different conditions, and what steps happen during the transition from reactants to products. Understanding chemical kinetics is crucial for various applications, such as the design of chemical reactors, the preservation of food, and even the development of medications. The speed of a reaction, or the rate at which it occurs, is affected by several factors such as temperature, pressure, and concentration of reactants.

For students diving into the realm of chemical kinetics, it's essential to understand the rate constant, which is a measure of how quickly a reaction takes place. The rate constant is dictated by the nature of the reactants, the activation energy, and the reaction's conditions. The problem provided explores how temperature influences the rate constant, emphasizing the importance of temperature in reaction kinetics.
Activation Energy
Activation energy, often symbolized as \(E_a\), is the minimum amount of energy required to initiate a chemical reaction. It can be visualized as a barrier that reactants must overcome to transform into products. The higher the activation energy, the fewer the number of molecules that have enough kinetic energy to react upon collision.

To overcome the activation energy barrier, molecules often need to collide with proper orientation and sufficient force, which can be facilitated by raising the temperature or using a catalyst. In the context of the exercise, activation energy is a vital parameter that can be extracted from data using the Arrhenius equation. Understanding \(E_a\) allows chemists to predict the speed of reactions and manipulate conditions for desired reaction rates.
Rate Constant
The rate constant \(k\) is an essential factor in chemical kinetics that provides a quantitative measure of the speed of a chemical reaction. It relates the concentration of the reactants to the rate of the reaction, reflecting how quickly reactants are converted into products. The value of the rate constant changes with different conditions, particularly temperature, and is influenced by the activation energy of the reaction.

As seen in the exercise, students are asked to comprehend how the rate constant varies with temperature, using real data to find the rate constant values at different temperatures. The Arrhenius equation, which incorporates the rate constant, activation energy, and temperature, allows students to calculate these important variables and develop a deeper understanding of reaction kinetics.
Temperature Dependence
Temperature plays a critical role in chemical kinetics by influencing the rate of a reaction. According to the Arrhenius equation, as the temperature increases, the rate constant of a reaction also increases, generally leading to a faster reaction. This is because a higher temperature means that more molecules have the kinetic energy required to overcome the activation energy barrier, resulting in more effective collisions.

In the exercise given, this temperature dependence is exhibited through a set of data connecting various temperatures with corresponding rate constants. By analyzing this data with the Arrhenius equation, students can see first-hand how a seemingly abstract concept like temperature dependence has practical implications in the calculation and understanding of a reaction's rate constant and activation energy.

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

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 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2} ?\)

Consider a hypothetical reaction between \(A, B,\) and \(C\) that is first order in \(A,\) zero order in \(B,\) and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when \([\mathrm{A}]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when \([\mathrm{B}]\) is tripled and the other reactant concentrations are held constant? (d) How does the rate change when [C] is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled? (f) By what factor does the rate change when the concentrations of all three reactants are cut in half?

(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 (a)? (c) How do the half-lives of first-order and second-order reactions differ?

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant? (c) Does the rate constant for a reaction generally increase or decrease with an increase in reaction temperature?

A reaction \(A+B \longrightarrow C\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? Explain. (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B}\) ? What is the overall reaction order? (c) What are the units of the rate constant?

See all solutions

Recommended explanations on Chemistry Textbooks

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.