Microstates of Entropy Chapter 19 cont. 19.3 and 19.4 http://ocw.mit.edu/resources/res-tll-004-stem-concept-videos-fall-2013/videos/governing-rules/entropy/ Microstates of Entropy

Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample. © 2012 Pearson Education, Inc.

Entropy on the Molecular Scale Molecules exhibit several types of motion: Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule about an axis or rotation about  bonds.  http://www.youtube.com/watch?v=3RqEIr8NtMI and disc from book: D:\Chapter_19\C_Media\molecular-motions\_molecular-motions.html © 2012 Pearson Education, Inc.

Entropy on the Molecular Scale Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules. He referred to this sampling as a microstate of the thermodynamic system.  © 2012 Pearson Education, Inc.

Entropy on the Molecular Scale Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy is S = k ln W where k is the Boltzmann constant, 1.38  1023 J/K.  © 2012 Pearson Education, Inc.

Entropy on the Molecular Scale The change in entropy for a process, then, is S = k ln Wfinal  k ln Winitial Wfinal Winitial S = k ln Entropy increases with the number of microstates in the system. © 2012 Pearson Education, Inc.

Entropy on the Molecular Scale The number of microstates and, therefore, the entropy, tends to increase with increases in Temperature Volume The number of independently moving molecules. © 2012 Pearson Education, Inc.

Entropy and Physical States Entropy increases with the freedom of motion of molecules. Therefore, S(g) > S(l) > S(s) © 2012 Pearson Education, Inc.

In which phase are water molecules least able to have rotational motion? A. Vapor B. Liquid C. Ice Answer: C

© 2012 Pearson Education, Inc. Solutions Generally, when a solid is dissolved in a solvent, entropy increases. However, water becomes more organized. Why? © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Entropy Changes In general, entropy increases when Gases are formed from liquids and solids; Liquids or solutions are formed from solids; The number of gas molecules increases; The number of moles increases. © 2012 Pearson Education, Inc.

What major factor leads to a decrease in entropy as the reaction shown takes place? A. Increase in number of molecules during change B. Decrease in number of molecules during change C. Change in pressure of system D. Change in internal energy of system Answer: B

Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. © 2012 Pearson Education, Inc.

Why does the plot show vertical jumps at the melting and boiling points? A. During melting or boiling at constant temperature entropy dramatically increases because energy is removed from the system during the change. B. During melting or boiling at constant temperature entropy changes significantly because more microstates exist for the final phase compared to the initial phase. C. During melting or boiling at constant temperature entropy changes significantly because fewer microstates exist for the final phase compared to the initial phase. D. During melting or boiling at constant temperature entropy changes significantly because no change in the number of microstates occurs during the phase change. Answer: B

© 2012 Pearson Education, Inc. Standard Entropies These are molar entropy values of substances in their standard states. Standard entropies tend to increase with increasing molar mass. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Standard Entropies Larger and more complex molecules have greater entropies. © 2012 Pearson Education, Inc.

What might you expect for the value of S° for butane, C4H10? A. 270 J/mol-K to 300 J/mole-K B. 280 J/mol-K to 305 J/mole-K C. 310 J/mol-K to 315 J/mole-K D. 320 J/mol-K to 340 J/mole-K Answer: C

Entropy Changes S = nS(products) — mS(reactants) Entropy changes for a reaction can be estimated in a manner analogous to that by which H is estimated: S = nS(products) — mS(reactants) where n and m are the coefficients in the balanced chemical equation. © 2012 Pearson Education, Inc.

Entropy Changes in Surroundings Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: Ssurr = qsys T At constant pressure, qsys is simply H for the system. © 2012 Pearson Education, Inc.

Entropy Change in the Universe The universe is composed of the system and the surroundings. Therefore, Suniverse = Ssystem + Ssurroundings For spontaneous processes Suniverse > 0 © 2012 Pearson Education, Inc.

Entropy Change in the Universe Since Ssurroundings = and qsystem = Hsystem This becomes: Suniverse = Ssystem + Multiplying both sides by T, we get TSuniverse = Hsystem  TSsystem qsystem T Hsystem T © 2012 Pearson Education, Inc.

Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g) Sample Exercise 19.5 Calculating ΔS° from Tabulated Entropies Calculate the change in the standard entropy of the system, ΔS° , for the synthesis of ammonia from N2(g) and H2(g) at 298 K: N2(g) + 3 H2(g)  2 NH3(g) Solution Analyze We are asked to calculate the standard entropy change for the synthesis of NH3(g) from its constituent elements. Plan We can make this calculation using Equation 19.8 and the standard molar entropy values in Table 19.1 and Appendix C. Solve Using Equation 19.8, we have ΔS° = 2S°(NH3)–[S°(N2) + 3S°(H2)] Substituting the appropriate values from Table 19.1 yields ΔS° = (2 mol)(192.5 J/mol–K)–[(1 mol)(191.5 J/mol–K) + (3 mol)(130.6 J/mol–K)] = –198.3 J/K Check: The value for ΔS° is negative, in agreement with our qualitative prediction based on the decrease in the number of molecules of gas during the reaction. Practice Exercise Using the standard molar entropies in Appendix C, calculate the standard entropy change, ΔS°, for the following reaction at 298 K: Al2O3(s) + 3 H2(g)  2 Al(s) + 3 H2O(g) Answer: 180.39 J/K 26

Plan today Gibbs Free energy WS Graph problems from WS Problems from book Reminder: Quest is a quiz grade, next text is Thursday after break!

Whiteboard Group Problem 1 2,7 2 3,8 3 4,7 4 5,8 5 6,7 6 7 8