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Miscellaneous Notes

Miscellaneous Notes. The end is near – don’t get behind. All Excuses must be taken to 233 Loomis before 4:15, Monday, April 30. The PHYS 213 final exam times are * 8-10 AM, Monday, May 7

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Miscellaneous Notes

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  1. Miscellaneous Notes • The end is near – don’t get behind. • All Excuses must be taken to 233 Loomis before 4:15, Monday, April 30. • The PHYS 213 final exam times are * 8-10 AM, Monday, May 7 • * 8-10 AM, Tuesday, May 8 and* 1:30-3:30 PM, Friday, May 11. The deadline for changing your final exam time is 10pm, Monday, April 30. • Homework 6 is due Tuesday, May 1 at 8 am. (NO late turn-in). • Course Survey = 2 bonus points (now on SmartPhysics – due Wed. May 2)

  2. Lecture 19: Chemical Equilibria, Surfaces,and Phase Transitions Chemical equilibria - Law of mass action Surface chemistry Phase equilibria and chemical potentialsVapor pressure of a solid Reading for this Lecture: Elements Ch 13

  3. Chemical Equilibrium “Chemical” is a bit of a misnomer. We’re describing any process in which things combine (or rearrange) to form new things. These problems involve reactions like: aA +bB  cC,where A, B, and C are the particle types and a, b, and c are integers. In equilibrium the total free energy, F, is a minimum.We must have DF = 0 when the reaction is in equilibrium, for any reaction that takes us away from equilibrium: Therefore: amA + bmB = cmC

  4. Internal energy per molecule Chemical Equilibrium (2) Treating the components as ideal gases or solutes: Plug these chemical potentials into the equilibrium condition, amA + bmB = cmC, and solve for the density ratios: K(T) is called the “equilibrium constant”. It depends on D’s and T, but not on densities. This equilibrium condition is a more general version of the law of mass action that you saw before for electrons and holes. The exact form of the equilibrium condition (how many things in the numerator and denominator, and the exponents) depends on the reaction formula: aA +bB  cCRHS  numerator LHS denominator

  5. Examples of Chemical Equilibrium ProcessReactionEquilibrium condition Dissociation of H2 molecules H2 2H mH2 = 2mHIonization of H atoms H e + p mH = me + mpSynthesize ammonia N2+ 3H2 2NH3mN2 + 3mH2 = mNH3General reaction aA+ bB … cC + dD … amA + bmB … = cmC + dmD … For the monatomic gases (circled) you can use nT = nQ. The others are more complicated, and we won’t deal with it here.However, remember that nT often cancels, so it won’t be a problem. Ideal solutions follow the same general form, but m isn’t close to the ideal monatomic gas value, because interactions in a liquid can be strong, modifying both U and S. Units and notation: Chemists measure density using units of moles per liter, and write the law of mass action like this:

  6. Interactions between the particles (e.g., molecules):In addition to simple PE terms from external fields, there are usually PE terms from interactions between particles (which are not usually ideal gases). Interactions between the molecules can often be neglected. That is, we’ll treat the molecules as ideal gases. Internal energy of each particle (e.g., molecule):Atoms can combine in any of several molecular forms, each of which has a different binding energy. The U term in F includes all those binding energies (which we’ll call D’s) , so they must be included in the m’s. (dF/dN) The reaction will NOT proceed to completion in either direction, because m depends on n for each type of molecule. As any one type becomes rare, its m drops until equilibrium is reached, with some of each type present. (Just as not all air molecules settle into the lower atmosphere.) Chemical Equilibrium (3)

  7. ACT 1: Equilibrium in the Ammonia Reaction Consider a reaction that is essential to agriculture: the synthesis of ammonia from nitrogen and hydrogen: N2 + 3 H2  2 NH3 1) Insert the correct superscripts and subscripts in the equilibrium equation: 2) Suppose the reaction is in equilibrium. Now double the number of N2 molecules.What will happen? A) Make more NH3. B) Dissociate more NH3. C) Nothing.

  8. Solution Consider a reaction that is essential to agriculture: the synthesis of ammonia from nitrogen and hydrogen: N2 + 3 H2  2 NH3 1) Insert the correct superscripts and subscripts in the equilibrium equation: 2) Suppose the reaction is in equilibrium. Now double the number of N2 molecules.What will happen? A) Make more NH3. B) Dissociate more NH3. C) Nothing. X 2 NH3 Of course, you could write the wholething upside down, with K’(T) = 1/K(T). 1 3 H2 N2

  9. Solution Consider a reaction that is essential to agriculture: the synthesis of ammonia from nitrogen and hydrogen: N2 + 3 H2  2 NH3 1) Insert the correct superscripts and subscripts in the equilibrium equation: 2) Suppose the reaction is in equilibrium. Now double the number of N2 molecules.What will happen? A) Make more NH3. B) Dissociate more NH3. C) Nothing. X 2 NH3 Of course, you could write the wholething upside down, with K’(T) = 1/K(T). 1 3 H2 N2 You’ve decreased the density ratio. To restore it, nNH3 must increase and/or nH2 must decrease. Some of the new N2 reacts with some of the H2, (decreasing nH2), producing more NH3 (increasing nNH3). There’s still some of the new N2, i.e., nN2 still increases somewhat.

  10. D = H2 binding energy = 4 eV T = 1000 K Non-monatomic Gases Formation of H2 from hydrogen atoms:H2 2H, so mH2 = 2mH.Equilibrium condition:We can use nQH, because it’s monatomic. We don’t know how to calculate nQH2, because it is diatomic and has extra U and S.However, we saw last time the effect that H2 being diatomic has on mH2, namely, it reduces the chemical potential.

  11. Act 3: Formation of H2 We can also write it like this: We have: 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease

  12. Solution We have: 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease Since nH √nH2, decreasing nH2 decreases nH. Makes sense: The overall density is reduced.

  13. Solution We have: 1) What happens to nH if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease 2) What happens to nH/nH2 if we decrease nH2? A) Decrease B) Increase C) Increase, then decrease Since nH √nH2, decreasing nH2 decreases nH. Makes sense: The overall density is reduced. nH/nH2 1/√nH2, so decreasing nH2 increases the fraction of free atoms. 2H H2 requires that two atoms meet, whileH2 2H only requires a single molecule.At low density, the rate of the second process is higher, shifting equilibrium to more H. At a given T, the fraction of atoms increases at lower molecule density! There are more H atoms in outer space than H2 molecules. Why? Two particles (H + H) have more entropy than one particle (H2). Entropy maximization dominates the tendency of atoms to bind!!!

  14. Phase Transitions Roadmap: We’ll start by looking at a simple model of atoms on surfaces, and discover that, depending on the temperature, the atoms prefer to be bound or to be flying free. This is related to the common observation that materials can be found in distinct phases: E.g., solid, liquid, gas. We’ll learn how equilibria between these phases work. Then we’ll go back and try to understand why distinct phases exist in the first place.

  15. Adsorption of Atoms on a Surface M = # (single occupancy) binding sites on the surface D = binding energy of an atom on site Ns = number of bound atoms Fs = Free energy of bound atoms Calculate the chemical potentials: Bound atoms: Atoms in the gas: Equilibrium: We could solve this equation for Ns, but …

  16. f 1 T1 T2 T3 0.5 T3 > T2 > T1 0 p p0(T2) Adsorption of Atoms (2) … usually we want to know what fraction of the surface sites are occupied, for a given gas pressure p and temperature T: Using our result: We obtain a simple relation for the fraction of occupied sites: More atoms go onto the surface at highpressure, becausemgas increases with pressure. po(T) is the characteristic pressure at which half the surface sites are occupied. It increases with temperature due to theBoltzmann factor.

  17. Act 4: Adsorption 1) At 10 atm, half the sites are occupied. What fraction are occupied at 0.1 atm? A) 1% B) 11% C) 90% 2) Keep the pressure constant, but increase T. What happens to f? A) Decrease B) No effect C) Increase

  18. Solution 1) At 10 atm, half the sites are occupied. What fraction are occupied at 0.1 atm? A) 1% B) 11% C) 90% 2) Keep the pressure constant, but increase T. What happens to f? A) Decrease B) No effect C) Increase At lower pressure, gas atoms hit the surface less often.

  19. Solution 1) At 10 atm, half the sites are occupied. What fraction are occupied at 0.1 atm? A) 1% B) 11% C) 90% 2) Keep the pressure constant, but increase T. What happens to f? A) Decrease B) No effect C) Increase At lower pressure, gas atoms hit the surface less often. Higher T: higher pQ and e-D/kT p0 increases f decreases Makes sense? More atoms have enough thermal energy to leave.

  20. Solids, Liquids and Gases Solid Liquid Gas Solids have fixed relationships among the atoms (or molecules) Liquids have looser relationships among atoms. The liquid has more entropy, because there are more ways to arrange atoms in the liquid. In liquids there are still some correlations between atoms, but in gases there are essentially none. In most situations the entropy of the gas is vastly larger than that of the liquid or solid.

  21. Phase Transitions and Entropy ice > 0 < 0 water If Sgas > Sliguid > Ssolid, why are different phases stable at different temperatures? The answer is that we must also consider the entropy of the environment. That’s what free energy and chemical potential do for us. For example: At low enough temperatures a substance like water is a solid. Its entropy is lower than that of the liquid so it must give up enough energy to its environment to make the total entropy increase when ice forms: Liquid H2O  Solid:DStot = DSL+DSS + DSenv 0 In order for this to work, enough heat must be given to theenvironment to make DStot 0. OK. So, why is liquid H2O favored at higher temperatures? The relative sizes of the DS terms must be different. Let’s look at the problem more quantitatively.

  22. Solid-gas equilibrium: vapor pressure V,T Consider solid-gas equilibrium at constant volume and temperature. Some substances (e.g., CO2) don’t exist as liquids at atmospheric pressure. The solid has negligible entropy (compared to the gas), so we’ll ignore it. In that case: Fs = Us -TSs = -NDms = -D The gas: mg = kT ln(n/nQ) Set them equal and solve for the equilibrium gas pressure:The equilibrium pressure is called the “vapor pressure” of the solid at temperature T. If p < pvapor, atoms will leave the solid until pgas = pvapor.This is called sublimation. For liquids, it’s called evaporation. Examples: Si (28 g/mol): pvapor = (4.04104 atm)(28)3/2 e-3eV/.026eV = 510-44 atmCO2 (44 g/mol): pvapor = (4.04104 atm)(44)3/2 e-0.26eV/.026eV = 535 atm Some solids don’t sublimate. Some do.Question: Does water ice sublimate? Binding energy of an atom in the solid Note the different D’s

  23. Next time Phase diagrams Latent heats Phase-transition fun Freezing point depression/Boiling point elevation Superheated/cooled water

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