Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Standard state Gibbs free energy change for isomerization reaction, cis-2-pentene \(\rightleftharpoons\) trans-2-pentene is \(-3.67 \mathrm{~kJ} / \mathrm{mol}\) at \(400 \mathrm{~K}\). if more trans- 2 -pentene is added to the reaction vessel, then: (a) More cis-2-pentene is formed (b) Additional trans-2-pentene is formed (c) Equilibrium remains unaffected (d) Equilibrium is shifted in the forward direction

Short Answer

Expert verified
(a) More cis-2-pentene is formed.

Step by step solution

01

Understand the Reaction

The reaction given is an isomerization reaction between cis-2-pentene and trans-2-pentene. The standard Gibbs free energy change, \( \Delta G^{\circ} \), is \(-3.67 \, \text{kJ/mol}\), which indicates that the reaction is spontaneous under standard conditions towards the formation of trans-2-pentene.
02

Analyze Le Châtelier's Principle

Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium will shift to counteract the change. Adding more trans-2-pentene to the system will initially increase its concentration.
03

Predict the Equilibrium Shift

According to Le Châtelier's principle, increasing the concentration of trans-2-pentene will shift the equilibrium to the left to reduce the concentration of the trans-2-pentene. This means more cis-2-pentene will be formed in response to the addition.
04

Conclusion

Based on the above analysis, the system will respond to the added trans-2-pentene by producing more cis-2-pentene, shifting the equilibrium towards the formation of cis-2-pentene.

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 Vaia!

Key Concepts

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

Isomerization Reaction
An isomerization reaction involves the conversion of one isomer into another. In this exercise, the focus is on the isomerization of cis-2-pentene to trans-2-pentene, two isomers of pentene. Isomers are molecules that have the same molecular formula but different structural arrangements. This reaction keeps the molecular makeup the same, but changes how the atoms are connected or oriented in space.
In these reactions, the transformation happens without breaking major chemical bonds, except for those dictating the spatial arrangement of the atoms. The isomerization from cis to trans (or vice versa) often affects physical properties like boiling point and stability. In our case, this process is spontaneous under standard conditions, tending to produce more of the trans isomer due to its lower energy state.
Le Châtelier's Principle
Le Châtelier's Principle is a fundamental concept in chemistry that helps predict the behavior of a reaction upon a disturbance. It asserts that if a dynamic equilibrium is disturbed by changing conditions such as concentration, pressure, or temperature, the system will adjust itself to counteract the effect of the disturbance.
This principle is crucial in determining how the system will behave when conditions change. For instance, in the isomerization of cis-2-pentene and trans-2-pentene, if the concentration of trans-2-pentene is increased by adding more to the reaction vessel, the system will try to minimize this increase. It will shift the equilibrium towards the formation of more cis-2-pentene, in order to restore balance.
Equilibrium Shift
An equilibrium shift refers to the movement of a reaction system's balance point in response to changes in conditions. This can occur due to adjustments in concentration, temperature, or pressure, in accordance with Le Châtelier's Principle. The equilibrium can shift in the forward direction (towards more products) or the reverse direction (towards more reactants).
In the context of the isomerization reaction between cis- and trans-2-pentene, adding more trans-2-pentene causes a shift in equilibrium. This shift occurs because the system reacts to the increase in trans-2-pentene by favoring the production of more cis-2-pentene. As a result, the position of equilibrium moves to the left, increasing the concentration of the less initially abundant cis-2-pentene.
Standard State Conditions
Standard state conditions serve as a reference point for scientists, ensuring that measurements are consistent and comparable. These conditions typically involve a pressure of 1 bar, a temperature of 298 K, and each substance at a concentration of 1 M. However, it's important to note that in our exercise, the temperature is specified as 400 K rather than the usual 298 K.
Gibbs free energy changes, such as the \( \Delta G^{\circ} \) given for the reaction, are calculated under these standard conditions. They are essential for understanding the spontaneity and direction of a reaction. A negative Gibbs free energy change, as observed here at 400 K, suggests that the isomerization from cis-2-pentene to trans-2-pentene is spontaneous under these conditions.

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

Find the value of \(\Delta \mathrm{H}_{\text {Reaction }}^{\circ}\) for the reaction given below by using the given data. \(3 \mathrm{CO}+2 \mathrm{O}_{2} \rightarrow \mathrm{CO}_{3} \mathrm{O}_{4}\) \(2 \mathrm{CO}+\mathrm{O}_{2} \rightarrow \underset{\mathrm{g}}{2 \mathrm{COO}} \Delta \mathrm{H}_{1}^{0}=-475.8 \mathrm{~kJ}\) \(6 \mathrm{COO}+\mathrm{O}_{2} \rightarrow 2 \mathrm{CO}_{3} \mathrm{O}_{4} \Delta \mathrm{H}_{2}^{0}=-355 \mathrm{~kJ}\) (a) \(-445.6 \mathrm{~kJ}\) (b) \(+891.2 \mathrm{~kJ}\) (c) \(+445.6 \mathrm{~kJ}\) (d) \(-891.2 \mathrm{~kJ}\)

The dissociation energies of \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) to convert them into gaseous atoms are 360 and \(620 \mathrm{kcal}\) mol respectively. The bond energy of \(\mathrm{C}-\mathrm{C}\) bond is: (a) \(280 \mathrm{kcal} \mathrm{mol}^{-1}\) (b) \(240 \mathrm{kcal} \mathrm{mol}^{-1}\) (c) \(160 \mathrm{kcal} \mathrm{mol}^{-1}\) (d) \(80 \mathrm{kcal} \mathrm{mol}^{-1}\)

The enthalpy change for a reaction does not depend upon the: (a) Physical state of reactants and products (b) Use of different reactants for the same product (c) Nature of intermediate reaction steps (d) Difference in initial or final temperatures of involved substances

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

For which of the following changes \(\Delta \mathrm{H} \neq \Delta \mathrm{E} ?\) (a) \(\mathrm{H}_{2}+\mathrm{I}_{2} \rightleftharpoons 2 \mathrm{HI}\) (b) \(\mathrm{HCl}+\mathrm{NaOH} \longrightarrow \mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})\) (d) \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \longrightarrow 2 \mathrm{NH}_{3}\)

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.

Sign-up for free