/*! 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 93 A small sample of \(c i s\) -dic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small sample of \(c i s\) -dichloroethene in which one carbon atom is the radioactive isotope \({ }^{14} \mathrm{C}\) is added to an equilibrium mixture of the cis and trans isomers at a certain temperature. Eventually, \(40 \%\) of the radioactive molecules are found to be in the trans configuration at any given time. (a) Determine the value of \(K_{\mathrm{c}}\) for the cis \(\rightleftharpoons\) trans equilibrium. (b) What would have happened if a small sample of radioactive trans isomer had been added instead of the cis isomer?

Short Answer

Expert verified
(a) \(K_{c} = 0.667\). (b) 60% radioactive molecules would be in the cis configuration.

Step by step solution

01

Analyze the Given Information

We're given that initially, **cis-dichloroethene** with one carbon atom as radioactive isotope \\(^{14}\mathrm{C} \) is added to the mixture of cis and trans isomers. Over time, 40% of these radioactive molecules are found in the trans configuration.
02

Identify Initial and Equilibrium Conditions

Initially, let's say we have \(x\) moles of radioactive cis isomers. At equilibrium, \(40\%\) of these \(x\) moles are trans isomers. Therefore, \(0.4x\) moles are trans, and \(0.6x\) moles remain as cis isomers.
03

Express the Equilibrium Constant Formula

The equilibrium constant \(K_{c}\) for the reaction will be defined as the ratio of products' concentration to reactants' concentration. Therefore, \[K_{c} = \frac{\text{[trans]}}{\text{[cis]}} = \frac{0.4x}{0.6x}\] Notice that the \(x\) cancels out, simplifying to: \[K_{c} = \frac{0.4}{0.6}\]
04

Calculate the Equilibrium Constant Value

Simplify the fraction: \[K_{c} = \frac{0.4}{0.6} = \frac{2}{3} = 0.667\] Thus, the equilibrium constant value for the equilibrium between the cis and trans isomers is \(0.667\).
05

Evaluate the Scenario with Radioactive Trans Isomer

If a small sample of radioactive trans isomer were added instead of the cis isomer, by equilibrium, **60%** of the radioactive molecules would shift into the cis configuration. Thus, the percentage values for cis and trans configurations will simply be reversed compared to the original scenario.

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.

Cis-trans isomerization
Cis-trans isomerization is a fascinating type of stereoisomerism where two isomers, referred to as "cis" and "trans," differ based on the position of substituent groups around a double bond or a ring structure. In chemical structures with a double bond, rotation around the bond is restricted, leading to this type of isomerism.
The cis isomer has substituent groups on the same side of the double bond, whereas in the trans isomer, they are on opposite sides. This distinct difference in spatial arrangement confers unique physical and chemical properties to each isomer.
  • **Cis Isomer**: Substituents on the same side.
  • **Trans Isomer**: Substituents on opposite sides.
This conversion between cis and trans isomers can occur under certain conditions such as light exposure, heat, or through catalysts. In the exercise given, a radioactive carbon-labeled cis-dichloroethene isomer undergoes isomerization, affecting the equilibrium positions as part of a dynamic process.
Equilibrium constant (Kc)
The equilibrium constant, denoted as \(K_c\), represents the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction. It's a key concept in understanding chemical equilibrium, indicating the extent to which reactants are converted into products. High \(K_c\) values typically suggest a reaction favoring product formation, while lower values indicate reactant preference.
In the problem, the equilibrium between cis and trans isomers was analyzed. When 40% of radioactive cis molecules converted to trans form, the \(K_c\) was calculated using:\[K_c = \frac{\text{[trans]}}{\text{[cis]}}\]Given that 60% of the original cis molecules remain, and 40% converted to trans, the equilibrium constant became:
  • \(K_c = \frac{0.4}{0.6} = 0.667\)
Therefore, this indicates the trans form is slightly less favored under the given conditions. Understanding \(K_c\) helps predict and interpret the direction and completeness of reactions in equilibrium.
Radioactive isotopes
Radioactive isotopes, also known as radioisotopes, have unstable atomic nuclei which decay over time, emitting radiation. These isotopes are crucial in a variety of scientific and medical applications.
  • They are used as tracers in biochemical experiments to track chemical processes and paths.
  • In medicine, they help in diagnostic imaging and cancer treatments.
In the given problem, a radioactive carbon isotope \(^{14}C\) is used to study the dynamic equilibrium between cis and trans isomers of dichloroethene. The labeled (radioactive) molecules allow us to trace how isotopic atoms distribute between isomers at equilibrium.
This radioisotope approach helps in understanding not just chemical equilibria but also reaction mechanisms and molecular interactions, providing insights that stable isotopes could not easily offer. The emission of radiation during decay is detectable, thus making it possible to follow reactions and transformations at a molecular level.

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 equilibrium constants for dissolving silver sulfate and silver sulfide in water are \(1.7 \times 10^{-5}\) and \(6 \times 10^{-30}\), respectively. (a) Write the balanced dissociation reaction equation and the associated equilibrium constant expression for each process. (b) Which compound is more soluble? Explain your answer. (c) Which compound is less soluble? Explain your answer.

Phosphorus pentachloride decomposes at high temperatures. $$ \mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ An equilibrium mixture at some temperature consists of \(3.120 \mathrm{~g} \mathrm{PCl}_{5}, 3.845 \mathrm{~g} \mathrm{PCl}_{3},\) and \(1.787 \mathrm{~g} \mathrm{Cl}_{2}\) in a sealed 1.00-L flask. (a) If you add \(1.418 \mathrm{~g} \mathrm{Cl}_{2}\) without changing the volume, how will the equilibrium be affected? (b) Calculate the concentrations of all three substances when equilibrium is reestablished.

The hydrocarbon cyclohexane, \(\mathrm{C}_{6} \mathrm{H}_{12},\) can isomerize, changing into methylcyclopentane, a compound with the same molecular formula but a different molecular structure. $$ \mathrm{C}_{6} \mathrm{H}_{12}(\mathrm{~g}) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{9} \mathrm{CH}_{3}(\mathrm{~g}) $$ \(\begin{array}{l}\text { cyclohexane } & \text { methylcyclopentane }\end{array}\) The equilibrium constant \(K_{\mathrm{c}}\) has been estimated to be 0.12 at \(25^{\circ} \mathrm{C}\). If you place \(3.79 \mathrm{~g}\) cyclohexane in an empty \(2.80-\mathrm{L}\) flask and seal the flask, calculate the mass of cyclohexane that is present when equilibrium is established.

Mustard gas was used in chemical warfare in World War I. Mustard gas can be produced according to this reaction: $$ \mathrm{SCl}_{2}(\mathrm{~g})+2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g}) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(\mathrm{~g}) $$ An evacuated 5.00 -L flask at \(20.0^{\circ} \mathrm{C}\) is filled with \(0.258 \mathrm{~mol} \mathrm{SCl}_{2}\) and \(0.592 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{4}\) and sealed. After equilibrium is established, 0.0349 mol mustard gas is present. (a) Calculate the partial pressure of each gas at equilibrium. (b) Calculate \(K_{\mathrm{c}}\) at \(20.0^{\circ} \mathrm{C}\).

Samples of \(\mathrm{N}_{2} \mathrm{O}_{4}\) can be prepared in which both nitrogen atoms are the heavier isotope \({ }^{15} \mathrm{~N}\). Designating this isotope as \(\mathrm{N}^{*}\), we can write the formula of the molecules in such a sample as \(\mathrm{O}_{2} \mathrm{~N}^{*}-\mathrm{N}^{*} \mathrm{O}_{2}\) and the formula of typical \(\mathrm{N}_{2} \mathrm{O}_{4}\) as \(\mathrm{O}_{2} \mathrm{~N}-\mathrm{NO}_{2}\). When a tiny quantity of \(\mathrm{O}_{2} \mathrm{~N}^{*}-\mathrm{N}^{*} \mathrm{O}_{2}\) is introduced into an equilibrium mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\), the \({ }^{15} \mathrm{~N}\) immediately becomes distributed among both \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) molecules, and in the \(\mathrm{N}_{2} \mathrm{O}_{4}\) it is invariably in the form \(\mathrm{O}_{2} \mathrm{~N}^{*}-\mathrm{NO}_{2}\). Explain how this observation supports the idea that equilibrium is dynamic.

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.