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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
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

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{J} / \mathrm{K} \cdot\) mol. What is the normal boiling point of mercury?

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ},\) for a reaction at \(25^{\circ} \mathrm{C} .\) How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C} ?\) What assumptions are made?

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}},\) where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right).\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction.”

You have a 1.00 -L sample of hot water \(\left(90.0^{\circ} \mathrm{C}\right)\) sitting open in a \(25.0^{\circ} \mathrm{C}\) room. Eventually the water cools to \(25.0^{\circ} \mathrm{C}\) while the temperature of the room remains unchanged. Calculate \(\Delta S_{\text {surr }}\) for this process. Assume the density of water is \(1.00 \mathrm{g} / \mathrm{cm}^{3}\) over this temperature range, and the heat capacity of water is constant over this temperature range and equal to \(75.4 \mathrm{J} / \mathrm{K} \cdot\) mol.

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of ±1 in all quantities): a. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{atm}\) b. \(T=298 \mathrm{K}, P_{\mathrm{N}_{2}}=200 \mathrm{atm}, P_{\mathrm{H}_{2}}=600 \mathrm{atm}\) \(P_{\mathrm{NH}_{3}}=200 \mathrm{atm}\)

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