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Acrylonitrile is the starting material used in the manufacture of acrylic fibers (U.S. annual production capacity is more than two million pounds). Three industrial processes for the production of acrylonitrile are given below. Using data from Appendix 4, calculate \(\Delta S^{\circ}, \Delta H^{p},\) and \(\Delta G^{\circ}\) for each process. For part a, assume that \(T=25^{\circ} \mathrm{C} ;\) for part \(\mathrm{b}, T=70 .^{\circ} \mathrm{C} ;\) and for part \(\mathrm{c}, T=700 .^{\circ} \mathrm{C}\) Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short Answer

Expert verified
Using the data from Appendix 4 and the balanced chemical equations, calculate the change in entropy (\(\Delta S^{\circ}\)) and change in enthalpy (\(\Delta H^{\circ}\)) for each process using the formulas: \[\Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}}\] \[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\] Then, convert the given temperatures from Celsius to Kelvin and calculate the change in Gibbs free energy (\(\Delta G^{\circ}\)) for each process using the equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]

Step by step solution

01

Find \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for each process

Using the Appendix 4, we need to find the standard enthalpies of formation and standard entropies of all reactants and products involved in the reactions for processes a, b, and c. Once we have found these values, we can calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for each process using the following formulas: \[\Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}}\] \[\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\]
02

Calculate \(\Delta G^{\circ}\) for each process

Now that we have the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for each process at the given temperatures, we can calculate \(\Delta G^{\circ}\) using the following equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] where \(T\) is the temperature in Kelvin. Remember to convert the given temperatures in Celsius to Kelvin by adding 273.15 to the Celsius temperatures before plugging them into the equation. Calculate \(\Delta G^{\circ}\) for each process with their respective temperatures: a) For part a, \(T = 25^{\circ}\mathrm{C} \Rightarrow T = 298.15 \mathrm{K}\): \[\Delta G^{\circ}_a = \Delta H^{\circ}_a - (298.15 \mathrm{K})(\Delta S^{\circ}_a)\] b) For part b, \(T = 70^{\circ}\mathrm{C} \Rightarrow T = 343.15 \mathrm{K}\): \[\Delta G^{\circ}_b = \Delta H^{\circ}_b - (343.15 \mathrm{K})(\Delta S^{\circ}_b)\] c) For part c, \(T = 700^{\circ}\mathrm{C} \Rightarrow T = 973.15 \mathrm{K}\): \[\Delta G^{\circ}_c = \Delta H^{\circ}_c - (973.15 \mathrm{K})(\Delta S^{\circ}_c)\] With these equations, we can find the \(\Delta G^{\circ}\) values for each of the three processes.

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Key Concepts

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

Enthalpy
Enthalpy is a fundamental concept in thermodynamics that represents the total heat content of a system. It is denoted by the symbol \(H\) and is measured in joules or kilojoules. It gives us insight into how much energy is absorbed or released during a chemical reaction. In terms of chemical reactions, enthalpy change (\(\Delta H\)) can be used to determine if a reaction is exothermic (releases heat) or endothermic (absorbs heat). A positive \(\Delta H\) indicates that the reaction requires heat, meaning it's endothermic. A negative \(\Delta H\) suggests that the reaction gives off heat, categorizing it as exothermic. In calculations, the enthalpy change for a reaction can be found using the standard enthalpies of formation for the products and reactants:
  • \(\Delta H^{\circ} = \sum H^{\circ}_{\text{products}} - \sum H^{\circ}_{\text{reactants}}\)
This equation allows us to determine the heat change under standard conditions (1 bar and 298 K). Understanding enthalpy helps predict the energy requirements or releases in industrial processes, like the production of acrylonitrile.
Entropy
Entropy, symbolized as \(S\), is a measure of the disorder or randomness in a system. In chemical thermodynamics, it's crucial for understanding how energy spreads in a process or reaction. The second law of thermodynamics states that in any spontaneous process, the total entropy of the system and its surroundings always increases.When examining reactions, the change in entropy (\(\Delta S\)) tells us how the disorder of the system changes:
  • \(\Delta S^{\circ} = \sum S^{\circ}_{\text{products}} - \sum S^{\circ}_{\text{reactants}}\)
A positive \(\Delta S\) indicates that the products are more disordered than the reactants, which is typically favorable for spontaneity. Conversely, a negative \(\Delta S\) suggests greater order in the products.Entropy change is a significant factor when predicting whether reactions occur spontaneously, as its interplay with enthalpy will determine the reaction's overall Gibbs free energy change.
Gibbs Free Energy
Gibbs free energy, represented by \(G\), combines enthalpy and entropy to determine the spontaneity of a reaction. It is a critical concept providing insight into whether a reaction can proceed on its own under constant pressure and temperature. The change in Gibbs free energy, \(\Delta G\), defines this spontaneity:
  • \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\)
Here, \(T\) is the temperature in Kelvin. A negative \(\Delta G\) indicates a spontaneous process, while a positive \(\Delta G\) suggests a non-spontaneous one. When \(\Delta G = 0\), the system is in equilibrium.The clever marriage of enthalpy and entropy through Gibbs free energy helps chemists and engineers predict and control chemical reactions, especially significant in deciding the viability of industrial processes like synthesizing acrylonitrile at different temperatures.
Temperature Conversion
Temperature conversion is essential in thermodynamics for consistent and accurate calculations, especially when determining properties like Gibbs free energy that require absolute temperature scales. The standard conversion from Celsius to Kelvin is simple and necessary:
  • To convert from Celsius to Kelvin, add 273.15 to the Celsius temperature.
  • For example, \(25^{\circ}C\) converts to \(298.15 K\), \(70^{\circ}C\) to \(343.15 K\), and \(700^{\circ}C\) to \(973.15 K\).
Using Kelvin, a unit explained by absolute zero where all thermal motion stops, ensures that all thermodynamic equations, including \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), are correctly applicable without negative temperatures that complicate calculations. Understanding and accurately converting temperatures is fundamental to reliable thermodynamic assessments and calculations.

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Most popular questions from this chapter

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}\) Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

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.

Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{f} / k_{\mathrm{r}},\) 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 \(\left(k=A e^{-E_{2} / R T}\right)\) 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."

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from 99.90% to 99.99% purity by the Mond process. The primary reaction involved in the Mond process is $$\mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g)$$ a. Without referring to Appendix \(4,\) predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperature-dependent. Predict the sign of \(\Delta S_{\text { sum }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}\) at 298 \(\mathrm{K}\) . Using these values and data in Appendix 4 calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C} .\) The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .^{\circ} \mathrm{C}\) f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\) . The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\) . g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\) . Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\) . The boiling point for Nic CO) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is 29.0 \(\mathrm{kJ} / \mathrm{mol} .\) [Hint: The phase change reaction and the corresponding equilibrium expression are \(\mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{NiCO} 4}\) greater than the \(K\) value. \(]\)

The melting point for carbon diselenide \(\left(\mathrm{CSe}_{2}\right)\) is \(-46^{\circ} \mathrm{C}\) . At a temperature of \(-75^{\circ} \mathrm{C},\) predict the signs for \(\Delta S_{\mathrm{surr}}\) and \(\Delta S_{\mathrm{univ}}\) for the following process: \(\operatorname{CSe}_{2}(l) \rightarrow \operatorname{CSe}_{2}(s)\)

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