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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
The difference in \(\Delta G^\circ\) values for the two reactions can be explained by the relationship between thermodynamics and kinetics. The first reaction is more thermodynamically favorable with a more negative \(\Delta G^\circ\), but it has kinetic barriers that require light as an energy source to overcome, resulting in a slower reaction rate. The second reaction is less thermodynamically favorable with a less negative \(\Delta G^\circ\), but it has a low activation energy and proceeds rapidly without needing any catalyst or external energy source.

Step by step solution

01

Understand the given reactions

We are given two reactions: 1. The direct reaction of ethane gas \((C_2H_6)\) with chlorine gas to produce monochloroethane \((C_2H_5Cl)\). 2. The reaction of ethylene gas \((C_2H_4)\) with hydrogen chloride gas to produce monochloroethane \((C_2H_5Cl)\). We are informed that the second reaction gives almost a 100% yield of pure \(C_2H_5Cl\) at a rapid rate without catalysis, while the first reaction requires light as an energy source. However, the \(\Delta G^\circ\) for the first reaction is more negative than the \(\Delta G^\circ\) for the second reaction.
02

Understand the concepts of thermodynamics and kinetics

To make sense of the problem, we need to understand the difference between thermodynamics and kinetics: Thermodynamics tells us the potential energy change in a reaction (like \(\Delta G^\circ\)), which determines if the reaction is spontaneous or not. A negative \(\Delta G^\circ\) implies that the reaction is spontaneous. Kinetics, on the other hand, deals with the reaction rate, which determines how fast a reaction occurs. A reaction can be thermodynamically favorable (negative \(\Delta G^\circ\)) but may have a slow rate due to kinetic barriers.
03

Discuss the role of catalysts and energy sources in reactions

Catalysts and energy sources can affect the reaction rate (kinetics) without changing the thermodynamics of a reaction. Catalysts work by providing an alternative reaction pathway with a lower activation energy, speeding up the reaction rate. In the first reaction, light acts as the energy source which helps to speed up the reaction rate. The second reaction, however, does not require a catalyst or external energy source, implying that it has a low activation energy and proceeds rapidly on its own.
04

Explain the difference in \(\Delta G^\circ\) values

Having discussed thermodynamics, kinetics, and the role of catalysts, we can now explain why \(\Delta G^\circ\) for the first reaction is more negative despite the second reaction being faster and having a higher yield. The first reaction has a more negative \(\Delta G^\circ\), which means it is more thermodynamically favorable. However, due to kinetic barriers, the reaction rate is slow and requires an energy source (light) to overcome these barriers. The second reaction, on the other hand, is less thermodynamically favorable (less negative \(\Delta G^\circ\)), but it has a low activation energy and proceeds rapidly without needing any catalyst or external energy source. In conclusion, the difference in \(\Delta G^\circ\) values can be explained by looking at the relationship between thermodynamics and kinetics. A reaction can have more negative \(\Delta G^\circ\), but still not proceed rapidly unless the kinetic barriers are overcome using catalysts or external energy sources.

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Key Concepts

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

Thermodynamics
In the realm of chemical reactions, thermodynamics provides insight into the potential energy changes that occur during a reaction. It helps us determine if a reaction is spontaneous or not. This is where the term \(\Delta G^\circ\) comes into play. Simply put, \(\Delta G^\circ\) represents the Gibbs free energy change under standard conditions.
- A negative \(\Delta G^\circ\) indicates that a reaction is spontaneous, meaning it can occur without external energy input.- On the flip side, a positive \(\Delta G^\circ\) suggests that the reaction requires energy to proceed.
In our given exercise, even though Reaction 1 involving ethane and chlorine has a more negative \(\Delta G^\circ\), it still requires light to proceed. This highlights a critical misunderstanding: thermodynamics tells us *if* a reaction can go, but not *how fast* it goes. That's the kinetic part, which often surprises students. Imagine thermodynamics like a book on a shelf; it tells you the book is there, but not if it’ll fall quickly when nudged.
Activation Energy
Activation energy is the energy barrier that must be overcome for a reaction to proceed. Picture it as a hill a molecule must climb before it can descend into products.
In the first reaction, light provides the energy needed to overcome this barrier. This means that, without sufficient energy (like photons from light), the reaction would not proceed, despite being thermodynamically favorable.
  • Energy sources such as light help to initiate reactions by providing the activation energy.
  • A high activation energy implies a slower reaction rate unless an external energy source or catalyst is involved.
The second reaction, involving ethylene and hydrogen chloride, proceeds rapidly without additional energy, suggesting a much lower activation energy. This is why it doesn't require light or any catalyst to achieve a near 100% yield rapidly.
Ultimately, activation energy serves as a bridge between thermodynamics and kinetics, defining not just the possibility of a reaction but its practicability at a given condition.
Reaction Mechanism
The reaction mechanism is a detailed sequence of steps that describes how reactants are transformed into products. It offers a deeper understanding of the path and timeline a chemical reaction follows.
In our case:
  • The first reaction has a complex mechanism that involves multiple steps and energy barriers. It needs light to generate reactive intermediates that can eventually lead to the formation of monochloroethane.
  • The second reaction, however, is simpler, suggesting fewer intermediates and a more straightforward path. This makes it highly efficient and rapid, without needing extra energy or a catalyst.
Understanding the mechanism is like reading the entire story of a reaction from start to finish. It not only aids in identifying each step's energetic demand but also illustrates how modifications, like adding a catalyst, might alter the pace or yield of a reaction. Grasping mechanisms is crucial in designing industrial processes or laboratory reactions to ensure efficiency and economy.

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Most popular questions from this chapter

Consider two perfectly insulated vessels. Vessel 1 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and water at \(0^{\circ} \mathrm{C}\) . Vessel 2 initially contains an ice cube at \(0^{\circ} \mathrm{C}\) and a saltwater solution at \(0^{\circ} \mathrm{C}\) . Consider the process \(\mathrm{H}_{2} \mathrm{O}(s) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)\) a. Determine the sign of \(\Delta S, \Delta S_{\text { sum }}\) and \(\Delta S_{\text { univ }}\) for the process in vessel 1 . b. Determine the sign of \(\Delta S, \Delta S_{\text { sum }},\) and \(\Delta S_{\text { univ }}\) for the process in vessel \(2 .\) (Hint: Think about the effect that a salt has on the freezing point of a solvent.)

For the reaction at 298 K, $$2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{kJ}\) and \(-176.6 \mathrm{J} / \mathrm{K}\) , respectively. What is the value of \(\Delta G^{\circ}\) at 298 \(\mathrm{K}\) ? Assuming that \(\Delta H^{p}\) and \(\Delta S^{\circ}\) do not depend on temperature, at what temperature is \(\Delta G^{\circ}=0 ?\) Is \(\Delta G^{\circ}\) negative above or below this temperature?

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) . Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M .\)

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving NaCl in water

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}\) . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

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