/*! 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 21 If for a reaction 1 and 2 , acti... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If for a reaction 1 and 2 , activation energies are \(\mathrm{E}_{\mathrm{A}}\) and \(E_{A_{2}}\) and rate constants are \(K_{1}\) and \(K_{2}\) at temperature \(\mathrm{T}\) and \(\mathrm{K}_{1}^{\prime}\) and \(\mathrm{K}_{2}^{\prime}\) at temperature \(\mathrm{T}\) ' where \(\mathrm{T}\) is higher than \(\mathrm{T}\) then which among the following is true if \(\mathrm{E}_{\mathrm{A}_{1}}>\mathrm{E}_{\mathrm{A}_{2}} ?\) (a) \(\frac{K_{1}^{\prime}}{K_{1}}>\frac{K_{2}^{\prime}}{K_{2}}\) (b) \(\frac{K_{1}^{\prime}}{K_{1}}<\frac{K_{2}^{\prime}}{K_{2}}\) (c) \(\frac{K_{1}^{\prime}}{K_{1}}=\frac{K_{2}^{\prime}}{K_{2}}\) (d) None

Short Answer

Expert verified
\( \frac{K_1'}{K_1} > \frac{K_2'}{K_2} \), so the answer is (a).

Step by step solution

01

Understand the Arrhenius equation

The Arrhenius equation defines the relationship between the rate constant \( K \) and the temperature \( T \) of a reaction. It is expressed as \( K = Ae^{-E_A/RT} \), where \( A \) is the pre-exponential factor, \( E_A \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the absolute temperature.
02

Apply the Arrhenius equation to both reactions

For reaction 1: \( K_1 = A_1 e^{-E_{A_1}/RT} \) and at the higher temperature \( T' \), \( K_1' = A_1 e^{-E_{A_1}/RT'} \). For reaction 2: \( K_2 = A_2 e^{-E_{A_2}/RT} \) and \( K_2' = A_2 e^{-E_{A_2}/RT'} \).
03

Find the ratio of rate constants at different temperatures

For reaction 1, the ratio of the rate constants is given by: \( \frac{K_1'}{K_1} = e^{(E_{A_1}/R)(1/T - 1/T')} \). For reaction 2, the corresponding expression is \( \frac{K_2'}{K_2} = e^{(E_{A_2}/R)(1/T - 1/T')} \).
04

Compare the effect of activation energy

Since \( E_{A_1} > E_{A_2} \) and both share the same temperature change \( 1/T - 1/T' \), the exponential factor for reaction 1 will increase more significantly than for reaction 2. Therefore, \( \frac{K_1'}{K_1} > \frac{K_2'}{K_2} \).
05

Select the correct option

Option (a) \( \frac{K_1'}{K_1} > \frac{K_2'}{K_2} \) is the correct choice, as a greater activation energy (\( E_{A_1} > E_{A_2} \)) leads to a larger increase in the ratio of rate constants when temperature is raised.

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.

Activation Energy
Activation energy is a crucial concept in understanding chemical reaction kinetics. It's the minimum amount of energy needed for reactants to transform into products. Think of it like a hurdle that molecules need enough energy to get over. Without reaching this energy level, a reaction simply won't proceed efficiently or quickly.

In the scientific notation, this is often represented as \(E_A\). It appears in the Arrhenius equation, which describes how temperature influences reaction rates. Higher activation energy implies that the molecules need more energy to get going, which typically means the reaction happens more slowly under the same conditions. In the context of our exercise, reaction 1 has a higher activation energy compared to reaction 2. As a result, when temperature changes, reaction 1 shows a more significant change in its rate constant than reaction 2.
Rate Constants
Rate constants are values that represent the speed at which a chemical reaction occurs. In the Arrhenius equation \( K = Ae^{-E_A/RT} \), the value of \( K \) is our rate constant. It is influenced by factors like the pre-exponential factor \( A \), activation energy \( E_A \), and the temperature \( T \) of the system.

The role of rate constants is vital since they allow chemists to predict how a reaction's rate will change under different conditions. Greater values of rate constants mean faster reactions. However, these vary with changes in temperature, as seen in the exercise — higher temperatures generally increase \( K \) since they lower the exponential term \( e^{-E_A/RT} \), making the reaction faster.
Temperature Dependence
The influence of temperature on reaction rates is a fundamental principle in chemical kinetics. According to the Arrhenius equation, temperature is inversely related to the exponential factor in the rate constant. The expression \( e^{-E_A/RT} \) implies that as temperature \( T \) increases, \( R \cdot T \) increases. This reduces the exponential term, and the rate constant \( K \) increases.

In layman's terms, warming up a reaction generally speeds it up. For instance, in the exercise, with \( T' > T \), the rate constants for both reactions, \( K_1' \) and \( K_2' \), generally increase compared to their values \( K_1 \) and \( K_2 \) at the lower temperature. This highlights how sensitively reactions can respond to temperature shifts, explaining why temperature control is crucial in industrial and laboratory reactions.
Chemical Reaction Kinetics
Chemical reaction kinetics involves studying the rates of chemical processes and understanding the factors that affect these rates. It's the science of speeding up or slowing down reactions, and it heavily features concepts like activation energy and rate constants.

Within this field, the Arrhenius equation is a foundational tool. It lets chemists quantify how reactants transform into products at different temperatures, providing insights into reaction mechanisms. The exercise touched on these principles by showing how differing activation energies between two reactions affect their rate constants when the temperature changes. By understanding kinetics, we can control reaction conditions to optimize yields, reduce side reactions, and enhance safety.

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

\(\mathrm{A} \rightarrow\) Product, \(\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}[\mathrm{A}]^{3}\) If a is the initial concentration and \((a-x)\) is the concentration of A after time \(t\), then rate constant is given by: (a) \(\mathrm{k}=\frac{1}{\mathrm{t}}\left[\frac{1}{(\mathrm{a}-\mathrm{x})}-\frac{1}{\mathrm{a}}\right]\) (b) \(\mathrm{k}=\frac{1}{2 \mathrm{t}}\left[\frac{1}{(\mathrm{a}-\mathrm{x})^{2}}-\frac{1}{\mathrm{a}^{2}}\right]\) (c) \(\mathrm{k}=\frac{2.303}{\mathrm{t}} \log \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)\) (d) \(\mathrm{k}=\frac{\mathrm{x}}{\mathrm{t}}\)

On introducing a catalyst at \(500 \mathrm{~K}\) the rate of a first order reaction increases by \(1.718\) times. The activation energy in presence of catalyst is \(4.15 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The slope of the plot of \(\ln \mathrm{k}\) (in \(\mathrm{sec}^{-1}\) ) \(\mathrm{vs} \frac{1}{\mathrm{~T}}\) (T in Kelvin) in absence of catalyst is \(\left(\mathrm{R}=8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\) (a) \(+1\) (b) \(-1\) (c) \(+1000\) (d) \(-1000\)

Consider the following elementary process: \(\frac{1}{2} \mathrm{H}_{2}+\frac{1}{2} \mathrm{I}_{2} \longrightarrow \mathrm{HI}\) The molecularity of the above reaction is (a) 1 (b) \(\frac{1}{2}\) (c) 2 (d) 4

Two substances \(\mathrm{A}\) and \(\mathrm{B}\) are present such that \(\left[\mathrm{A}_{0}\right]\) \(=4\left[\mathrm{~B}_{0}\right]\) and half-life of \(\mathrm{A}\) is 5 minute and that of \(\mathrm{B}\) is 15 minute. If they start decaying at the same time following first order kinetics how much time later will the concentration of both of them would be same? (a) 15 minute (b) 10 minute (c) 5 minute (d) 12 minute

The reaction, Sucrose \(\stackrel{\mathrm{H}^{+}}{\longrightarrow}\) Glucose \(+\) Fructose, takes place at certain temperature while the volume of solution is maintained at 1 litre. At time zero the initial rotation of the mixture is \(34^{\circ}\). After 30 minutes the total rotation of solution is \(19^{\circ}\) and after a very long time, the total rotation is \(-11^{\circ} \mathrm{C}\). Find the time when solution was optically inactive. (a) \(135 \mathrm{~min}\) (b) \(103.7 \mathrm{~min}\) (c) \(38.7 \mathrm{~min}\) (d) \(45 \mathrm{~min}\)

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.