Problem Set #7, Chem 340, Fall 2013 Due Friday, Oct 18, 2013 To hand in :

Transcription

1 1. Atkns 3.11(b) Problem Set #7, Chem 34, Fall 213 Due Frday, Oct 18, 213 To hand n : 2. Atkns 3.18 (b) Ths seems to have a sgn error n formula for V but answer s rght, hgh pressure should reduce volume. Problem s defnton of mssed sgn: =-(1/V)(dV/dP)

2 3. Atkns 3.19 (b) Chap. 3 dd not defne but t equals G m, Was n lecture 4. Atkns 3.16 Ths s formula from book and lecture

3 5. Atkns 3.26

4 6. Atkns 3.38

5 7. Atkns 3.4

6 8. Engel P5.36 P5.36) Plottng the data for crystallne glycne as Cp 12 C p,m versus T gves: T K To obtan the molar entropy of crystallne glycne at 3 K we calculate the area under the curve for each temperature ncrement, dvde the area by the upper temperature of the ncrement, and addng that number for all temperature ncrements. The area for each temperature ncrement can be obtaned by calculatng the areas under the red and blue step functons, addng the two, and dvdng by two. For example the molar entropy for the ncrement (lned area): 1 K 8 K 43.2 J mol K 1 K 8K 35.2 J mol K Sm (1 K 8 K) 2 1 K Sm (1 K 8 K) 7.84 J mol K Dong ths calculaton for all ncrements and addng yelds: S (glycne, 3 K) 96.4 J mol K m

7 9. Engel P5.45 P5.45 The entropy of aturaton at 34 K can be calculated as: ΔH 641 J mol ΔS 34 K J K mol T 34 K The entropy of aturaton at 31 K s: 34K ΔCp ΔS 31 K ΔS 34 K dt ΔS 34 K ΔC T ΔS p 31K 31 K 31 K J K mol 8.37 kj K mol ln The enthalpy of aturaton at 31 K s: ΔH 31 K T ΔS 31 K J K 34 K 31K 34K 1 T dt J K mol kj K mol 1. Engel P6.13 mol 3 ΔGH O 1 ΔG NH aq ΔGreacton ΔG kj mol kj mol kj mol The equlbrum constant s then: ΔG reacton kj mol kj mol K p Exp Exp R T J K mol 298 K Equlbrum constant goes past the Wednes. lecture, can skp -49

8 11. Engel P6.2 P6.2) Calculate the Gbbs energy change for the proten aturaton descrbed n Problem 5.45 at T = 31.K and T = 34.K. From P5.34 we have: 31 K J K mol 31 K kj K mol 31 K ΔS ΔH ΔG s then: ΔG 31 K ΔH 31 K T ΔS 31 K kj mol 298 K J K mol 13.3 kj mol At 34 K we obtan: ΔG reacton kj mol 133 J mol 34 K 34 K 31 K kj mol 1 34 K 1 31 K

9 1. Atkns 3.12 (b) Extra problems 2. Atkns 3.16 (b) 3. Atkns 3.17 (b)

10 4. Atkns 3.21 (b) 5. Atkns 3.24 s 6. Atkns 3.25 The double dervatve method used n class s much clearer than ths. It seems steps are mssng

11 7. Atkns 3.28

12 8. Atkns Engel P5.27 The standard entropy for ths reacton s: ΔS reacton (298 K) ν S f, J K mol J K mol J K mol 1 J K mol J K mol 44.2 J K mol

13 1. Engel P5.2 The standard entropy and enthalpy and heat capacty for ths reacton are: ΔS (298 K) ν S reacton f, J K mol J K mol J K mol J K mol J K mol ΔH reacton (298 K) kj mol 6 kj mol kj mol kj mol kj mol C ν H f, 1 1 p,reacton (298 K) J K mol J K mol J K mol J K mol J K mol At 31 K the entropy and enthalpy for ths reacton are then: Tf ΔSreacton (31 K) ΔSreacton (298 K) Cp,reacton ln T 33 K J K mol J K mol ln 29.7 J K K mol ΔH reacton (31 K) ΔH reacton (298 K) C reacton kj mol J K mol 33 K K kj mol T f T And the entropes for the surroundngs and unverse: -dq H kj mol reacton ΔSsurroundngs J mol K T T 33 K unverse reacton surroundngs ΔS ΔS ΔS J K mol J mol K J mol K

14 11. Engel P5.41 S corresponds to the area under the trs C p curve (see fgure P4.31). Dvdng the area nto small fragments, addng them up, and multplyng by the molecular weght of the proten yelds: ΔS 78 J K mol Calculatng the enthalpy of aturaton n the same way the entropy of aturaton was calculated n P5.41 yelds: ΔH kj mol kj mol The entropy of aturaton s then: ΔS ΔH T kj mol 34 K J K mol

15 Engel P Engel P6.4 a) for the sothermal reversble path P f Pf V G VdP nrt ln nrt ln P V P 1. L 5. L Vf Vf A PdV nrt ln V mol J mol K 298 K ln J V f L mol J mol K 298 K ln J 1. L b) Because A and G are state functons, the answers are the same as to part a) because the systems go between the same ntal and fnal states, T,V T,V f. G A = H U = PV) = nrt). Therefore, G = A for an deal gas f T s constant.

16 13. Engel P6.9 We frst need to know how much energy s generated n the metabolsm of glucose: 6 ΔG 6 ΔG 6 ΔG 1 ΔG ΔG reacton ΔG H2O CO2 O2 C6H12O kj mol kj mol 6. kj mol kj mol kj mol Now we need to convert the work performed by the horse: w horse m g h lb.4536 kg lb 1 ft.348 m ft 9.81m s mn 6 mn h kj h 1 That means that the horse needs the followng amount of glucose: m glucose w w reacton glucose M glucose kj h kj mol g mol 168.1g