Advanced Physical Chemistry Problems (VI), Free Energies

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University of Connecticut DigitalCommons@UConn Chemistry Education Materials Department of Chemistry June 2008 Advanced Physical Chemistry Problems (VI), Free Energies Carl W.

1 University of Connecticut Chemistry Education Materials Department of Chemistry June 2008 Advanced Physical Chemistry Problems (VI), Free Energies Carl W. David University of Connecticut, Follow this and additional works at: Recommended Citation David, Carl W., "Advanced Physical Chemistry Problems (VI), Free Energies" (2008). Chemistry Education Materials

2 Problems for the Advanced Physical Chemistry Student Part 6, Free Energies C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut (Dated: June 9, 2008) I. SYNOPSIS his is a set of problems that were used near the turn of the century and which will be lost when the web site they were on disappears with my demise. Because these problems are being taken from the web and are being edited, their statements and the hints/answers offered are subject to the typical editorial errors that ensue when such work is undertaken in the vacuum of a non-teaching situation. herefore, I claim any errors for myself, and hate to note that there most likely is no point in contacting me about them for obvious reans. II. GIBBS AND HELMHOLZ FREE ENERGIES. he molar volume of benzene (liquid) is 88.9 cc at 20 o C and atm pressure. Assuming the volume to be constant, find G (Change in Gibbs Free Energy) for compression of mole of the liquid from to 00 atm (in calories/mole). his problem asks you to remember (or derive) the pressure dependence of the Gibbs Free Energy. Here is a micro derivation: de = ds pdv da = ds Sdt + ds pdv dg = Sdt pdv + pdv + V dp and integrating, we have p=00 atm p= atm dg = 00 V dp where V = l iters (per mole) G(p = 00 atm) G(p = atm) = (00 ) where the r.h.s. is in liter-atm, which can be converted to any alternate desired units. 2. wo moles of an ideal gas are compressed ithermally and reversibly at 00 o C from a pressure of 0 to 25 atm. Find the change in the Helmholtz Free Energy (in calories) for this process. ( ) A = p V 25 R da = 2 0 V dv and the rest is left to the interested reader. 3. hree moles of an ideal gas are allowed to expand freely at 300 o K from a volume of 00 to 000 liters. What is the change in Gibbs Free Energy for this process? implying dg = Sdt + V dp ( ) G = V p Electronic address: Carl.David@uconn.edu his is done exactly the way the previous problem was done, except the r.h.s. integral is over dp. 3R 000 dg = 3 3R 00 R ypeset by REVEX

3 2 4. Calculate the magnitude of the difference between the Gibbs and the Helmholtz free energies at 25 o C for the reaction H 2 (g, atm) + 2 O 2(g, atm) > H 2 O(liq, atm) of liquid water. 6. Given the reaction at 25 o C: H 2 O(l, mm Hg) H 2 O(gas, mm Hg) what is G o (25 o C) (in kj/mole) for the reaction H 2 O(l, atm) H 2 O(gas, atm) at the same temperature. (he vapor pressure of liquid water is mm Hg at 25 o C.) Assume that the Gibbs free energy of liquid water is independent of pressure. and G = A + (pv ) (pv ) = nr where n is the change in the number of moles of gaseous materials (liquids and lids generally don t count). herefore liquid, atm G=? gas, atm (pv ) = (3/2) R which is what was asked for, depending on the units desired for R. 5. For a certain process the Gibbs Free Energy change G = 3, ln( ) liquid, VP G=0 gas, VP Find S for the process at 25 o C., ( ) G p = S S = 6.ln his is an appropriate time to discuss units in thermodynamics. When you see mething like ln you know there s mething wrong. One can t have a unit inside a logarithm. Instead, we should have had ln std, and this goes for sines, cosines, error functions, etc., i.e., the argument should have been rendered unit-free by use of me standard state of standard comparative value which established the correct unit to use (via cancellation)! FIG. : Raising the pressure on a liquid to standard state, evaporating it there, and lowering the pressure on the resultant gas (vapor) back to the vapor pressure value. We have G(l, V P l, atm) = and G(g, atm g, V P ) = grams mole grams dp cc R which can be combined to form the G we desire. 7. At the melting point of KF, the standard molar entropy of the liquid is J/(moleK), and that of the lid is (in the same units). he molar heat of fusion at 3 o K is kj/mol Assuming that the entropy of both liquid and lid do not change appreciably with temperature, calculate the molar Gibbs free energy of fusion for KF at 3 o K. (in Joules).

4 3 G = 27, 96 3(63 39) 8. At 25 o C the standard (p = atm) Gibbs free energy of formation of H 2 S(g) is kJ/mole. Calculate the pressure (in atm) at which the Gibbs free energy of the reaction H 2 (g, atm) + S(s) H 2 S(g, P ) would be zero. 0. he standard molar Gibbs free Energy of reaction for a certain substance s liquid to be transformed into vapor at atm pressure, when the temperature is 25 o C, is 4.8 kj/mole. What is the vapor pressure of this substance (in mm Hg) at this temperature? A(l, atm) A(vap, atm); G = G o = 4.8kJ/mole A(l, atm) A(l, v p); G 0 A(l, v p) > A(vap, v p); G =? H 2 (g, atm) + S(s) H 2 S(g, atm); G o = 33.56kJ 0 G = G o + R ln p atm As a continuation of the comments on units, notice that here I ve explicitly made the argument of the logarithm unit-free by specifying the unit system and standard state ( atm) (Of course, we could have used any other unit system and the standard states are just conventions we ve all agreed to). 9. he density of graphite is 2.25 gm/(cc) and that of diamond is 3.5 gm/(cc). If the standard Gibbs free energy of transformation from graphite to diamond (at 25 o C) is 2.9 kj/mole, determine the pressure (in atm) at which graphite and diamond are in equilibrium. G(g, atm g, V P ) = p R per mole, and then (See Figure ), we have G = G o = 4.8kJ and which can be combined to form the G we desire.. At 25 o C, H o = 92, 300 Joule mole, and Go = 95, 200 Joules mol. Assuming that H o is constant, i.e., independent of temperature over the temperature range in this question, calculate G o at 50 o C. Give your answer in kj/mole. Here, we have mething of the likes of 2900 J = p 8.34J 2900 J 0.082l atm = V dp; p (l atm) ( ) dp 3.5 At this point, I always tease my students about diamonds are forever, and ask the women whether or not they believe the I will love you forever statements of beaus. If diamonds decay to graphite, to what does love decay? We know that G o ( ) Go ( std ) std = std H o 2 d the infamous Gibbs-Hlemholtz equation. Knowing the H o is constant over the temperature domain in question helps in integrating the r.h.s. of the G-H equation, leading to G o ( ) Go ( std ) std = Ho std where std = 298 o K and = o K. he rest is an exercise.

5 4 It is interesting to ask what happens if H o is not constant over the temperature domain? he answer leads to nested integrals, i.e., G o ( ) Go ( std ) = std Ho ( std) + C std pd std 2 d which we re-write in a manner less confusing: G o ( ) Go ( std ) = std y Ho ( std) + C std pdz std y 2 dy We note in passing that the internal nested integral concerning C ps can be more complicated if, in fact, these heat capacities are themselves temperature dependant, i.e., functions of, i.e., C p ( ) represented herein as C p (z). III. EPILOGUE After editing this material for weeks, and continuously finding errors, me small, me huge, I have to wrap it up and send this off. If, in the years or, you come across an error, and you me, I will try to have it corrected. But since this material is written in LaeX there is me doubt whether or not I ll have access to a Linux machine, and access to the digitialcommons site. You can try; we ll see what happens, if anything. hanks to all the students over the last 45 years who ve taught me Physical Chemistry.

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