Physics - Molecular Arrangement Calculation

Physics - Molecular Arrangement Calculation Student name___________________________ TA name__________________________ Section_______________________________     Calculating arrangements due to distribution of molecules: (𝑎+𝑏+𝑐+...)! Wpos(a,b,c,….) = 𝑎!∙𝑏!∙𝑐!… ; Where a, b, c, ...— number of different particles Calculating arrangements due to distribution of quanta (q) of energy among particles (m) or with (p) partitions(where p=m−1): (𝑞 + 𝑝)! (𝑞 + (𝑚 − 1))! W(m,q) = 𝑞! ∙ 𝑝! 𝑞! ∙ (𝑚 − 1)! Entropy S: S= kB ln∙W where kB = 1.381 × 10−23 J/K (Boltzmann constant) J Gas constant R(mol∙K) = NA∙ kB where NA = 6.022140857 × 1023 mol−1 i. Stotal = S1 + S2 = R ∙lnW object1 + R ∙lnW object2 = R ∙ln(W1∙W2) Wtotal= Wobject1∙Wobject2 W ii. ΔS = Sfinal – Sinitial = R ∙lnWfinal - R ∙lnWinitial = R∙ ln W final 1. ln W final > 0; ΔS < 0 process is not spontaneous: ln W initial Wfinal initial < 0; Wfinal > 1; Winitial Wfinal < 1; Winitial Wf > Win Wf < Win For a reaction 2A(aq) + B(aq)  3C(aq) at 298K; E = 4.00 V and E° = – 2.00V and ne = 2.00mol circle everything that must be true: Q=1 Q = Keq 2. initial W ΔS > 0 process is spontaneous: Q>1 Q > Keq Q<1 Keq> 1 Keq <1 Q < Keq Keq=1 This question concerns a concentration cell with positive voltage at 25 oC, constructed from chlorine half cells of different chlorine ion concentration and sealed so that their Cl2 gas pressure can be controlled. In one half-cell the chlorine ion concentration is 2.0 M and the Cl2 gas pressure is 0.45 atm, and in the other half cell the chlorine ion concentration is 0.015 M and the Cl2 gas pressure is 10.00 atm. Platinum electrodes are used to connect the cell to an external circuit. Anode half-reaction: Cathode half-reaction: NET reaction: Write the line notation of the cell. Indicate for each species and its concentration or pressure. Q= Eocell = Calculate the cell voltage Ecell at 25°C 1. Consider the 3 systems below: a. Calculate the entropy difference (ΔS) between the two systems A and B. Express your answer in the correct units and to the correct number of significant figures. WparticleA = WparticleB = ΔS = b. Two quanta of energy are added to the molecules of system B. Calculate the total number of unique arrangements of system B. Hint: WtotalB =WparticleB· WenergyB c. System A, with no energy quanta, and system B, with two energy quanta, are combined into system C, shown above. Energy quanta can pass between the two subsystems, but particles cannot. Calculate the total number of unique energy arrangements of system C. d. The barrier between the two parts of system C is removed so that particles can move throughout the combined system. Calculate the total number of unique arrangements Considering arrangements of both particles and energy. 2. Consider the following two transformations. Transformation I: Add 1 quantum to a 10-atom solid that contains 20 quanta Transformation II: Add 1 quantum to a 10-atom solid that contains 10 quanta a. Without doing any calculations, circle which transformation results in the larger increase in the number of energy arrangements. Transformation I Transformation II b. Without doing any calculations, circle which transformation results in the larger increase in the entropy(∆S). Transformation I Transformation II W c. Entropy change is proportional to the logarithm of the ratio of energy arrangements, W final . initial Calculate this ratio for the transformation of adding 1 quantum of energy to a 3-atom solid that already contains 3 quanta. Express your answer as a decimal to two significant figures. 3. Consider 4 “X” particles and 3 quanta of energy in 5 boxes (shown below). If one of the particles and J 𝐽 one quantum of energy removed, calculate the change in entropy in (K) and in (𝑚𝑜𝑙∙𝐾 )to one significant figure. Is this process spontaneous? X X X X 4. For the following examples predict if ΔSsys is positive, negative or more information needed. Hint: Remember that ΔSsys is dominated by positional entropy Sgas > Sliquid > Ssolid and beware of liquids that contain ions in aqueous solutions. a. CaCO3(s) → CaO(s) + CO2(g) positive negative ΔSsys more information needed b. NH3(g) → NH3(l) positive negative more information needed c. N2(g) + 3H2(g) → 2 NH3(g) positive negative more information needed d. AlCl3(s) → Al3+(aq) + 3 Cl−(aq) positive negative more information needed 5. At −1 °C, liquid water spontaneously freezes. H2O(l)  H2O(s) a. For this process, the entropy change of the system, ΔSsys is: < 0 =0 >0 b. During the process of freezing are H−bond broken or formed? c. During the process of freezing ΔHsys is: < 0 =0 >0 d. During the process of freezing ΔHsur is: < 0 =0 >0 e. At −1 °C, liquid water spontaneously freezes. For this process, the entropy change of the surroundings, ΔSsurr is: <0 =0 >0 f. At −1 °C, liquid water spontaneously freezes. For this process, in comparison to the magnitude of the entropy change of the system, the magnitude of the entropy change of the surroundings is… │ΔSsur│< │ΔSsys│ │ΔSsur│=│ΔSsys│ │ΔSsur│>│ΔSsys│ g. At −1°C ΔSuniv of the process will be: < 0 =0 >0 h. At +1°C ΔSuniv of the process will be: < 0 =0 >0 i. At 0°C ΔSuniv of the process will be: =0 >0 j. If ΔHfus(water) = 6.000kJ/mol, calculate ΔSsys , ΔSsur and ΔSuniv at −1.00°C and at 1.00°C. <0 6. Consider the system of two identical four-atom solids that are in contact with each other, one having six quanta of energy, and the other having ten quanta. a. Calculate Wsystem = W1· W2 for the system described W1(4 particles, 6 quanta) = W1(4 particles, 10 quanta) = Wsystem = b. Allowing quanta to pass from one system to another calculate Wsystem that will give you the maximum number of arrangements. What is special about this new system? Hint: what will be new W1 and W2? 7. Consider 4 identical particles are confined to the left half of a 12-position lattice gas by a nonpermeable barrier. The barrier is then removed and the 4 particles distribute throughout the 12 𝑊 lattice positions. What is the ratio of 𝑊 𝑓𝑖𝑛𝑎𝑙 ? 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 8. Consider the process of adding 1 quantum of energy to a system: System A: Winitial(4atoms, 10quanta)  Wfinal(4 atoms, 11quanta) System B: Winitial(4 atoms, 4quanta)  Wfinal(4 atoms, 5quanta) Which process has the greater entropy change (∆S)? Hint: Comparing ΔS of different systems: 𝑊 ΔSa vs. ΔSb is the same as comparing (𝑊 𝑓𝑖𝑛𝑎𝑙 ). Prove your answer : W What is final (for system A)? Winitial W What is W final initial 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 (for system B)? 9. The membranes of red blood cells are semipermeable, allowing water to flow in and out but not other solutes. Water flowing out of the cell causes the cells to shrivel (crenation) and water flowing into the cell causes the cell to swell and rupture (hemolysis). The diagrams below model a red blood cell and a surrounding sucrose solution. The cell membrane (vertical heavy line) is impermeable to the round particles (sucrose) but permeable to the square particles (water). The empty squares are spaces inbetween molecules. The concentration of sucrose in a healthy red blood cell is 2%. Π=MRT a. Which diagram corresponds to red blood cells in a 2% sucrose solution? How do you know? b. Which diagram corresponds to red blood cells in a solution that would cause crenation? c. Which diagram corresponds to red blood cells in a solution that would cause hemolysis? In preparation for next week, useful information:  Second law of Thermodynamics: ΔSuniv= ΔSsystem+ ΔSsurrounding i. ΔSuniv = ΔSsystem + ΔSsurrounding > 0 process is spontaneous ii. ΔSuniv = ΔSsystem + ΔSsurrounding < 0 process is not spontaneous iii. ΔSuniv = 0 process is at Equilibrium T = Tequilibrium ∆H ∆H  ∆Ssurroundings = surr = − sys Tsurr  Tsurr i. ΔSsur is thermal entropy due to distribution of quanta (q) of energy among fixed particles (m). Increase in number of quanta (q) of energy will increase the number of distinguishable arrangements. More quanta (q) of energy correspond to higher temperature. ° ∆Ssys = ∆Sreaction = ∑products nj (∆S ° ) − ∑reactants nj (∆S ° ) = Sfinal−Sinitial i. ΔSsystem is positional entropy due to distribution of molecules. The more volume per molecule, the more distinguishable arrangements. ii. S° is standard entropy of formation. Standard entropy of formation for elements in the standard state does not equal zero iii. Sgas > Ssolution > Ssolvent > Ssolid iv. ΔSsys = T ∆Hsys equilibrium 10. Is the following expression ° ∆Hsys = ∆Hreaction = Always true ∑ nj (∆Hf° ) − products never true only true at T=Teq nj (∆Hf° ) ∑ reactants only true at TTeq only true at TTeq a. Is ΔHsys temperature dependent? ∆Hsys 11. Is the following expression ΔSsys = Always true never true T only true at T=Teq a. Is ΔSsys temperature dependent? 12. Is the following expression ° ∆Ssys = ∆Sreaction = Always true never true ∑ nj (∆S ° ) − products only true at T=Teq ∑ nj (∆S ° ) reactants only true at TTeq only true at TTeq a. Is ΔSsys temperature dependent? ∆H 13. Is the following expression ΔSsurr =− Tsys Always true never true only true at T=Teq a. Is ΔSsur temperature dependent? 14. Given the entropy diagram for freezing point lowering of water, answer the following questions: ∆Sfr = ∆H(H2 O(l)→H2O(s)) Tfp ΔHfr – represents heat transfer when H2O(l)→H2O(s) it is the same in both cases. Ssolution ((H2O+solute)) Ssolvent (H2O(l)) ∆S(l→s) = ∆S(solution→solid) = ∆H(H2 O(l)→H2 O(s)) Tfp(solution) ∆H(H2O(l)→H2 O(s)) Tfp(solvent) Ssolid(H2O(s)) Ssolid(H2O(s)) a. The ΔS on the diagram represents (choose one): ΔSsystem b. S (water, (l)) compared to S (solution, (l)) is… Greater ΔSsurrounding ΔSuniverse smaller the same c. The magnitude of ΔS (l  s) compared to the magnitude of ΔS (solution  s) is ... Greater smaller the same d. ΔHfp (water (l)→ water (s)) compared to ΔHfp (solution, l → solid) is… Greater smaller the same e. Tfp (solution) compared to Tfp (solvent) is ... Greater smaller the same 15. Compare the entropy change when a pure solvent freezes and when a solution freezes: ΔSsoln > ΔSpure ΔSsoln < ΔSpure ΔSsoln = ΔSpure 16. Compare the magnitudes of entropy change when a pure solvent freezes and a solution freezes: │ΔSsoln│ >│ΔSpure│ │ΔSsoln │< │ΔSpure│ ΔSsoln = ΔSpur 17. I like making “sweet tea” and thought making ice cubes of sugar water would be useful. However, when I put my sugar water in my freezer at 0°C, the sugar water did not freeze. Circle the correct statements and correct any incorrect statements for the freezing of sugar water. Hint: You might use the diagram above to assist you. a. ΔHfus,soln > ΔHfus,water e. │ΔSfreeze,soln│ < │ ΔS freeze,water│ b. ΔSfreeze,soln < ΔSfreeze,water f. Sliquid,water > Ssol c. ΔSfreeze,soln > ΔS freeze, water g. ΔHmelt,soln is temperature dependent d. │ΔSfreeze,soln│ > │ ΔS freeze,water│ 18. For the following reaction: AlCl3(s)  Al3+(aq) + 3 Cl−(aq), AgCl3 is observed to be highly soluble at room temperature. For the reaction above at 300.K, by using the information given, calculate ΔHsys, ΔSsys, ΔSsur, ΔSuniv and the temperature at which process will cease to be spontaneous. kJ AlCl3(s) Al3+ (aq) Cl− (aq) J ΔH°f (mol) S° (mol∙K) −700. −550. −150. 100. −300. 50. Review: Use your notes and your book to complete the following: 19. State the Second Law of Thermodynamics: 20. Define ΔSuniverse(ΔSnet) 21. What is the relationship between ΔSsystem and ΔSreaction? 22. What equation relates ΔSsystem and ΔHsystem when the system is at equilibrium? 23. What is the relationship between ΔSsurrounding and ΔSreaction? 24. What equation relates ΔSsurrounding and ΔHsurroundings? 25. What equation relates ΔSsurrounding and ΔHsystem? 26. Fill the table by using the following possible answers (Teq is the temperature at equilibrium): a. Always spontaneous process b. Only spontaneous at T < Teq ΔHsys ΔSsys Exothermic Positive Exothermic Negative Endothermic Positive Endothermic Negative c. Only spontaneous T >Teq d. Never spontaneous ΔSnet(specify at which temperature) Numerical Answers: b. 27225; equlibrium 1. 2. 1. a. b. c. d. 7. 8. 9. 0 24 10 2800 a. Diagram A b. Diagram B c. Diagram C 2. a. Transformation I b. Transformation II c. 1.5 3. 4. -0.7∙10-23 J/K ; not spontaneous a. b. c. d. Positive Negative Negative more information a. b. c. d. e. f. g. h. i. j. <0 Formed <0 >0 >0 >│ΔSsys│ >0 <0 =0 (@ -1°C): ΔSsys = -21.97 J/mol∙K ΔSsurr = 22.05 J/mol∙K ΔSuniv = 0.08067 J/mol∙K (@ +1°C) ΔSsys = -21.97 J/mol∙K ΔSsurr = 21.89 J/mol∙K ΔSuniv = -0.08012 J/mol∙K 5. 6. a. 24024 33; spontaneous 1.6>1.3  ΔSsystem B > ΔSsystem A 10. Always true a. Not temperature dependent 11. only true at T=Teq a. Not temperature dependent 12. Always true a. Not temperature dependent 13. Always true a. temperature dependent 14. a. b. c. d. e. ΔSsys smaller smaller the same smaller 15. ΔSsoln < ΔSpure 16. │ΔSsoln│ >│ΔSpure│ 17. a. b. c. d. e. f. g. ΔHmelt,soln = ΔHmelt,water ΔSfreeze,soln < ΔSfreeze,water ΔSfreeze,soln <ΔSfreeze,water │ΔSfreeze,soln│ > │ ΔS freeze,water│ │ΔSfreeze,soln│> │ ΔS freeze,water│ Sliquid,water < Ssoln ΔHmelt,soln is NOT temperature dependent 18. ∆Hsys = -300 kJ/mol; ∆Ssys = -0.25 kJ/mol; ∆Ssurr = 1 kJ/mol; ∆Suniv = 0.75 kJ/mol; 1200K