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Chapter 16: 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 first reaction of ethane and chlorine, even though it is more thermodynamically favorable (ΔG° is more negative), has a large activation energy and requires an external energy source, such as light, to proceed. The second reaction involving ethylene and hydrogen chloride has a smaller activation energy, and thus, it can proceed rapidly without any catalyst, resulting in almost a 100% yield of pure monochloroethane (C2H5Cl).

Step by step solution

01

Understand the reactions and given information

The first reaction is the direct reaction of ethane gas (C2H6) with chlorine gas to produce monochloroethane (C2H5Cl): \[ C_2H_6 + Cl_2 \rightarrow C_2H_5Cl + HCl \] This reaction has a more negative ΔG° but needs light as an energy source. The second reaction involves ethylene gas (C2H4) reacting with hydrogen chloride gas to produce monochloroethane (C2H5Cl): \[ C_2H_4 + HCl \rightarrow C_2H_5Cl \] This reaction gives almost a 100% yield of pure C2H5Cl at a rapid rate without any catalysis. Now, let's recall the concepts of Gibbs free energy, reaction spontaneity, and activation energy.
02

Recall Gibbs free energy and reaction spontaneity

Gibbs free energy change (ΔG) is an important parameter to determine whether a reaction will be spontaneous or not. If ΔG < 0, the reaction is spontaneous, whereas if ΔG > 0, the reaction is non-spontaneous. However, a negative ΔG only indicates that the reaction is thermodynamically favorable. It does not say anything about the reaction rate or whether a catalyst or energy input is required for the reaction to occur.
03

Recall activation energy

Activation energy (Ea) is the minimum energy required for a reaction to proceed. It is related to the reaction rate, which determines how fast the reaction happens. If the activation energy is high, the reaction rate will be slow and the reaction may need an external energy source, such as heat or light, or a catalyst to lower the activation energy and speed up the reaction.
04

Examine the role of Gibbs free energy and activation energy in these reactions

In both given reactions, the products are the same, i.e., C2H5Cl. However, the first reaction has a more negative ΔG°, which means it is more thermodynamically favorable. Still, it has a large activation energy. The large activation energy is causing the first reaction to be slow, and it requires an external energy source, which is light in this case, to make the reaction proceed. On the other hand, the second reaction, despite having a less negative ΔG°, has a small activation energy in comparison to the first reaction. Since the activation energy is small, this reaction proceeds rapidly and does not need a catalyst.
05

Conclusion

In conclusion, the first reaction of ethane and chlorine, even though it is more thermodynamically favorable (ΔG° is more negative), has a large activation energy and requires an external energy source, such as light, to proceed. The second reaction involving ethylene and hydrogen chloride has a smaller activation energy, and thus, it can proceed rapidly without any catalyst, resulting in almost a 100% yield of pure monochloroethane (C2H5Cl).

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Most popular questions from this chapter

Sodium chloride is added to water (at \(25^{\circ} \mathrm{C}\) ) until it is saturated. Calculate the \(\mathrm{Cl}^{-}\) concentration in such a solution.

Consider the reaction $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ $$2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{S}_{\text {thombic }}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C},\) predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{atm}\) c. \(P_{\mathrm{NO}_{2}}=0.29 \mathrm{atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.6 \mathrm{atm}\)

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: $$\begin{aligned} &\mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}_{2}(g) \quad K=2.3 \times 10^{6} \text { at } 600 . \mathrm{K}\\\ &2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g) \quad K=1.8 \times 10^{37} \mathrm{at} 600 . \mathrm{K} \end{aligned}$$

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ} ?\) a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Is \(\Delta S_{\text {surr favorable or unfavorable for exothermic reactions? }}\) Endothermic reactions? Explain.
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