Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 17: Problem 23

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Short Answer

Expert verified
In conclusion, the difference in spontaneity and reaction rates for the two methods of producing monochloroethane can be explained by the activation energy barriers associated with each reaction. The first reaction has a more negative ΔG°, indicating a higher spontaneity, but requires light as an energy source to overcome the activation energy barrier and proceed. The second reaction, with a less negative ΔG°, can occur rapidly without a catalyst or additional energy input due to its lower activation energy barrier, resulting in a high yield of monochloroethane.

Step by step solution

01

Understanding the Reactions

We need to understand the two reactions given: 1. Direct reaction of ethane gas (C2H6) with chlorine gas to produce monochloroethane (C2H5Cl). 2. Reaction of ethylene gas (C2H4) with hydrogen chloride gas to produce monochloroethane (C2H5Cl). The first reaction needs light as an energy source, while the second reaction doesn't and produces a nearly 100% yield of C2H5Cl without catalysis.
02

ΔG° vs Reaction Rate

Keep in mind that ΔG° (Gibbs Free Energy change in standard conditions) is used to determine the spontaneity of a reaction. If ΔG° is negative, a reaction is spontaneous, whereas if it's positive, it's non-spontaneous. However, this does not take into account the reaction rate or how fast the reaction will occur.
03

Light as an Energy Source

The first reaction has a more negative ΔG°, implying that it's more spontaneous than the second reaction. However, in terms of reaction rate, it needs light as an energy source to occur. This indicates that the activation energy of the first reaction might be high, and light provides the necessary extra energy to overcome this barrier.
04

Yield and Reaction Rate

On the other hand, the second reaction has a less negative ΔG°, but it doesn't require a catalyst and occurs rapidly, producing a nearly 100% yield of monochloroethane (C2H5Cl). This suggests that the activation energy for the second reaction is comparatively lower, and the reaction can proceed quickly once the reactants are brought together.
05

Conclusion

In conclusion, the difference in ΔG° values and the need for external energy sources or catalysts for the two reactions can be explained by considering the activation energy barriers and the overall reaction rate. A more negative ΔG° indicates a more spontaneous reaction, but it doesn't necessarily mean a faster reaction rate, which is evident in the first reaction needing light to proceed. Conversely, the second reaction, despite a less negative ΔG° value, has a lower activation energy, allowing it to occur rapidly and produce a high yield of monochloroethane without additional energy input or catalysts.

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

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{~atm}\), it freezes at \(54.5 \mathrm{~K}\) to form solid \(\mathrm{I}\). At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{~J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Consider a weak acid, HX. If a \(0.10-M\) solution of HX has a \(\mathrm{pH}\) of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C} ?\)

Consider the dissociation of a weak acid HA \(\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$ \mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\).

Calculate \(\Delta G^{\circ}\) for \(\mathrm{H}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g)\) at \(600 . \mathrm{K}\), using the following data: \(\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6}\) at 600. \(\mathrm{K}\) \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37}\) at \(600 . \mathrm{K}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

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