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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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?

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate $\Delta G^{\circ}$ for this reaction. b. Is this reaction spontaneous under standard conditions at 298 $\mathrm{K} ?$ c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Predict the sign of \(\Delta S_{\text { surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

Calculate the entropy change for the vaporization of liquid methane and liquid hexane using the following data. $$\begin{array}{lll} {\text { }} & {\text { Boiling Point (1 atm)}} & { \Delta H_{\mathrm{vap}} } \\ \hline {\text { Methane }} & \quad\quad\quad {112 \mathrm{K}} & {8.20 \mathrm{kJ} / \mathrm{mol}} \\ {\text { Hexane }} & \quad\quad\quad {342 \mathrm{K}} & {28.9 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Compare the molar volume of gaseous methane at 112 \(\mathrm{K}\) with that of gaseous hexane at 342 \(\mathrm{K}\) . How do the differences in molar volume affect the values of \(\Delta S_{\mathrm{vap}}\) for these liquids?

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. $\mathrm{K}(s)+\frac{1}{2} \mathrm{Br}_{2}(g) \longrightarrow \mathrm{KBr}(s)$ b. $\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$ c. \(\mathrm{KBr}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{Br}^{-}(a q)\) d. \(\mathrm{KBr}(s) \longrightarrow \mathrm{KBr}(l)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free