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Polyvinyl chloride (PVC) is one of the most commercially important polymers (Table 12.5). PVC is made by addition polymerization of vinyl chloride \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\right)\). Vinyl chloride is synthesized from ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) in a twostep process involving the following equilibria: $$ \begin{aligned} &\text { Equilibrium 1: } \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) \\ &\text { Equilibrium 2: } \quad \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g) \end{aligned} $$ The product of Equilibrium 1 is 1,2 -dichloroethane, a compound in which one \(\mathrm{Cl}\) atom is bonded to each \(\mathrm{C}\) atom. (a) Draw Lewis structures for \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\). What are the \(\mathrm{C}-\mathrm{C}\) bond orders in these two compounds? (b) Use average bond enthalpies (Table 8.4) to estimate the enthalpy changes in the two equilibria. (c) How would the yield of \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\) in Equilibrium 1 vary with temperature and volume? (d) How would the yield of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) in Equilibrium 2 vary with temperature and volume? (e) Look up the normal boiling points of 1,2 -dichloroethane and vinyl chloride in a sourcebook, such as the CRC Handbook of Chemistry and Physics. Based on these data, propose a reactor design (analogous to Figure 15.12) that could be used to maximize the amount of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\) produced by using the two equilibria.

Short Answer

Expert verified
In summary, the C-C bond orders for C2H4Cl2 and C2H3Cl are 1 and 2, respectively. For Equilibrium 1, increasing temperature favors reactants, while decreasing temperature favors products. For Equilibrium 2, increasing temperature favors products. Changing volume has little effect on both equilibria. A reactor design to maximize C2H3Cl production involves a sequential reactor system with a low-temperature reactor for Equilibrium 1, followed by a high-temperature reactor for Equilibrium 2, and a heat exchanger between them for improved energy efficiency.

Step by step solution

01

Draw Lewis Structures and Calculate Bond Orders

To draw the Lewis structures for molecules, identify the central atoms (carbon in this case), and follow these steps: 1. Count the total number of valence electrons. 2. Position the atoms and form single bonds. 3. Distribute the electrons and apply octet rules. 4. Adjust for resonance structures if necessary. Lewis structure for C2H4Cl2: - C2H4 has 4 valence electrons from each of the two carbon atoms, 1 electron from each of the 4 hydrogen atoms, and 7 electrons from each of the 2 chlorine atoms. Total = 4(2) + 4 + 7(2) = 22 valence electrons. - Position the atoms: the carbon atoms are at the center, bonded to each other, and chlorine and hydrogen atoms bonded to carbon. - Distribute the electrons, making sure each atom has an octet, except for hydrogen which has only 2 electrons. C-C bond order in C2H4Cl2 = 1 (single bond) Lewis structure for C2H3Cl: - C2H4 has 4 valence electrons from each of the two carbon atoms, 1 electron from each of the 3 hydrogen atoms, and 7 electrons from the chlorine atom. Total = 4(2) + 3 + 7 = 18 valence electrons. - Position the atoms: carbon atoms are at the center, bonded to each other, and hydrogen and chlorine atoms bonded to a carbon atom. - Distribute the electrons, making sure each atom has an octet, except for hydrogen which has only 2 electrons. C-C bond order in C2H3Cl = 2 (double bond) Next, we are asked to estimate the enthalpy changes in the two equilibria using average bond enthalpies.
02

Estimate Enthalpy Changes

For Equilibrium 1: Enthalpy change = sum of the bond enthalpies of reactants - sum of the bond enthalpies of products ΔH1 = [(1 C=C bond + 4 C-H bonds + 1 Cl-Cl bond) - (1 C-C bond + 4 C-H bonds + 2 C-Cl bonds)] Use Table 8.4 to find the bond enthalpies. ΔH1 = [1(612) + 4(413) + 1(242)] - [1(347) + 4(413) + 2(328)] = -41 kJ/mol (approx.) For Equilibrium 2: ΔH2 = [(1 C-C bond + 4 C-H bonds + 2 C-Cl bonds) - (1 C-C bond + 3 C-H bonds + 1 C-Cl bond + 1 H-Cl bond)] ΔH2 = [1(347) + 4(413) + 2(328)] - [1(613) + 3(413) + 1(328) + 1(431)] = 112 kJ/mol (approx.) Now, we need to determine how temperature and volume affect the yields of the two equilibria.
03

Temperature and Volume Effects on Equilibrium 1

For Equilibrium 1, since the reaction is exothermic (ΔH1 < 0), increasing the temperature will shift the equilibrium position to the left, favoring the reactants. Decreasing the temperature will shift the equilibrium position to the right, favoring the products (yield of C2H4Cl2). For volume changes, there are equal numbers of moles on both the reactants and products side in Equilibrium 1. Therefore, changing the volume will have little effect on the position of the equilibrium or the yield of C2H4Cl2.
04

Temperature and Volume Effects on Equilibrium 2

For Equilibrium 2, since the reaction is endothermic (ΔH2 > 0), increasing the temperature will shift the equilibrium position to the right, favoring the products (yield of C2H3Cl). Decreasing the temperature will shift the equilibrium position to the left, favoring the reactants. Regarding volume changes, the number of moles on the reactant and product sides of Equilibrium 2 are equal, as in Equilibrium 1. Therefore, changing the volume will also have little effect on the position of the equilibrium or the yield of C2H3Cl. Lastly, we are asked to propose a reactor design to maximize the amount of C2H3Cl.
05

Propose a Reactor Design to Maximize C2H3Cl Production

From our earlier analysis, we know that: 1. Lower temperatures favor the formation of C2H4Cl2 in Equilibrium 1. 2. Higher temperatures favor the formation of C2H3Cl in Equilibrium 2. Based on these findings, a sequential reactor system consisting of two reactors could be designed: 1. A first reactor operating at lower temperatures to favor the formation of C2H4Cl2 in Equilibrium 1, followed by 2. A second reactor operating at higher temperatures to favor the formation of C2H3Cl in Equilibrium 2. Using a heat exchanger between these two reactors would not only help to recover heat from the first reactor and use it to preheat the contents entering the second reactor, but also improve energy efficiency and maximize C2H3Cl production.

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Key Concepts

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

Polyvinyl Chloride (PVC)
Polyvinyl chloride, commonly known as PVC, is a widely manufactured synthetic plastic polymer. It’s revered for its durability, chemical resistance, and versatility, being utilized in various applications such as construction, healthcare, and electronics. PVC is synthesized through a process called addition polymerization, where the monomer vinyl chloride (C_{2}H_{3}Cl) undergoes a reaction to form a long-chain polymer. This polymerization process is initiated by breaking the double bond in the vinyl chloride molecules, allowing them to link together in a repeating pattern to form PVC. The properties of PVC can be altered by adding plasticizers, which make it more flexible, or by adjusting the polymerization method to create different types of PVC, such as uPVC (unplasticized PVC) used for windows and pipes. In our textbook exercise, understanding the properties and production of PVC helps contextualize the chemical reactions and equilibria involved in its synthesis from ethylene and chlorine gases.
Chemical Equilibria
Chemical equilibria are at the heart of countless reactions in nature and industry, including the synthesis of PVC. In a chemical equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, resulting in no net change in the concentrations of reactants and products over time. However, the system can respond to changes in conditions, shifting the equilibrium position to maintain balance—a concept known as Le Chatelier’s principle.

When evaluating the equilibrium of the PVC-related reactions, we must consider factors like temperature and volume. For instance, the temperature influences the equilibrium by favoring the exothermic reaction at lower temperatures and the endothermic reaction at higher temperatures. Changes in volume, while not significant in the given equilibria due to equal mole ratios, can generally shift the equilibrium in reactions with different numbers of gas molecules on each side. These principles allow chemists to control the yield of desired products by creating optimal conditions for the reactions.
Enthalpy Changes
Enthalpy changes (ΔH) in a chemical reaction are indicative of the energy absorbed or released during the reaction. In the context of our textbook exercise, knowing the enthalpy changes for the two-step synthesis of PVC is critical for understanding and predicting the reaction's behavior under different conditions. If ΔH is negative, the reaction is exothermic, releasing heat to the surroundings. Conversely, if ΔH is positive, the reaction is endothermic, absorbing heat. These enthalpy changes influence chemical equilibria; for instance, an exothermic reaction will be favored by a decrease in temperature.

Understanding these concepts allows students to predict the outcomes of reactions based on bond enthalpies and manipulate reaction conditions to alter yields. The enthalpy changes for the production of PVC help explain why certain conditions are preferred for each step of the reaction, ensuring the maximum yield of vinyl chloride for the polymerization process.

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Most popular questions from this chapter

Write the equilibrium-constant expression for the equilibrium $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ The table included below shows the relative mole percentages of \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{CO}(\mathrm{g})\) at a total pressure of \(1 \mathrm{~atm}\) for several temperatures. Calculate the value of \(K_{p}\) at each temperature. Is the reaction exothermic or endothermic? Explain. $$ \begin{array}{lll} \hline \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & \mathrm{CO}_{2}(\text { mol } \%) & \text { CO (mol \%) } \\ \hline 850 & 6.23 & 93.77 \\ 950 & 1.32 & 98.68 \\ 1050 & 0.37 & 99.63 \\ 1200 & 0.06 & 99.94 \\ \hline \end{array} $$

A mixture of \(0.10 \mathrm{~mol}\) of \(\mathrm{NO}, 0.050 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and \(0.10 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a \(1.0\) -L vessel at \(300 \mathrm{~K}\). The following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \([\mathrm{NO}]=0.062 \mathrm{M} .\) (a) Calculatethe equilibrium concentrations of \(\mathrm{H}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\). (b) Calculate \(K_{\mathrm{c}}\)

Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{aligned} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) & \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) & \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{aligned} $$

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react as follows: $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ At equilibrium the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2} .\) (a) Calculate the equilibrium concentrations of \(\mathrm{H}_{2}\), \(\mathrm{Br}_{2}\), and HBr. (b) Calculate \(K_{c}\).

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(\mathrm{~s}) .\) (a) Write the equilibrium- constant expression for this reaction in terms of partial pressures. (b) Why doesn't the concentration of \(\mathrm{Na}_{2} \mathrm{O}\) appear in the equilibrium-constant expression?

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