Chemistry 5 Final. Name(printed): Note: 10 points will be assigned if you print your name so I can read it effortlessly. Print legibly!

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1 Chemistry 5 Final ame(printed): ote: 10 points will be assigned if you print your name so can read it effortlessly. Print legibly! You must show your work to receive credit PERODC TABLE OF TE ELEMETS 1A 8A A 3A 4A 5A 6A 7A Li Be B C O F a K Rb Cs Fr (223) 12 Mg B 4B 5B 6B 7B 8B 1B 2B 20 Ca Sr Ba Ra Sc Y La Ac Ti Zr f Rf (261) 23 V b Ta Db (262) 24 Cr Mo W Sg (263) 25 Mn Tc (98) 75 Re Bh (264) 26 Fe Ru Os s (265) 27 Co Rh r Mt (268) 28 i Pd Pt Cu Ag Au Zn Cd g Al Ga n Tl Si Ge Sn Pb P As Sb Bi metals 16 S Se Te Po (209) 17 Cl Br At (210) 2 e e Ar Kr Xe Rn (222) nonmetals 58 Ce Th Pr Pa d U Pm (145) 93 p Sm Pu (244) 63 Eu Am (243) 64 Gd Cm (247) 65 Tb Bk (247) 66 Dy Cf (251) 67 o Es (252) 68 Er Fm (257) 69 Tm Md (258) 70 Yb o (259) 71 Lu Lr (260) A few Constants Avogadro s umber, = 6.022*10 23 Gas constant, R = L atm/k mole, 1 L atm = J 1 Faraday = 96,485 coulombs 1 coulomb = 1 amp-sec work = - coulombs*e max 1 joule = 1 volt-amp 1 watt = 1 joule/second

2 ame PLEASE PRT CLEARLY (5 of the 10 points depend on it) 1. Calculate the entropy change when 3.00 mole of benzene vaporizes reversibly at its normal boiling point of 80.1 o C. o Vap = 30.8 kj at this temperature. What is G for this process? S o vap = o vap/t B = 30.8 kj/ K = kJ/K-mole = 87.2 J/K-mole S = 3mole benzene* S o vap = 3mole*87.2 J/k-mole = 262 J/K At equilibrium G = X. Total:

3 mole of an ideal monatomic gas has an initial pressure of 1.00 atm and a T = 300 K is expanded reversibly and adiabatically until the volume has doubled.; then it is expanded irreversibly and isothermally into a vacuum until the volume has doubled again; next it is heated reversibly at constant volume to 400 K. Finally, it is compressed reversibly and isothermally to a final state with P = 1.00 atm and T = 400 K. Calculate S Sys for this process. int: there are 2 ways to solve this problem an easy way and a hard way. Remember, S is a state function, it depends only on the initial and final states and S = n(cp+r)lnt2/t 1 = 1mole*[5/2*8.31*ln(400/300)]J/K-mol = 5.98 J/K ard Way find S for each step and sum them The answer is: S = 1mole*( )J/K mole = 5.98 J/K

4 3. The square planar compound, Pt( 3 ) 2 2 exists in two forms, the cis and the trans, which differ in their molecular structure. o f is and kj/mol and G o f is and respectively for the cis and trans compounds. A. What is S for the cis to trans transformation at 25 o C? B. What are reasonable structures (3D drawings) for these two compounds? (int: Platinum is the central atom in each structure.) Pt Pt Cis Trans o f kj/mole G o f kj/mole G o = o T S o, or S o cis-trans = ( o cis-trans G o cis-trans)/t S o cis-trans = 940 J/K

5 4. A i i 2+ Ag + Ag galvanic cell is constructed in which the standard cell voltage is 1.03V and the cell volumes are 100.ml each. A. Calculate the free energy change at 25 o C when 1.00 gram of silver plates out if all concentrations remain at 1.00 M throughout the process. B. What is the maximum electrical work done during the process? Suppose the 99.9% of the silver ion is reduced. C. What will the new nickel concentration be? D. What will the new cell voltage be? i i 2+ Ag + Ag All concentrations = 1.00M, 100ml each a). 1.00g of Ag is 1/107.9 = moles = 9.26*10-3 moles. G E = - G = 1.03J/C *9.26*10-3 moles* C/mol = J nf b) so, the electrical work will be W max = G = 920 J. Suppose that 99.9% silver Ag + reduced. f this accounts for b), then the total work would be 920J * = 919J c) the remaining Ag is 0.001M moles of Ag + consumed, moles of Ag + remain. i = i e - 2Ag + + 2e - = 2Ag Moles on i dissolved = /2 = Total # of moles i = (from the initial concentration) = moles in 100mL of solution. The i 2+ concentration will be M = 1.49M with significant figures. Some students may attempt to solve this part with respect to one gram of Ag +, saying that after 99.9% reaction, the remaining Ag + is 9.26*10-5 moles which is close to d) Using the ernst equation for 25C: E = E o (0.059/n)log(red/ox) = /2*log(1.49/0.001) 2 = *6.173 =0.848 V

6 5. An aqueous solution of gold() is electrolyzed and gold metal is deposited at the cathode while oxygen is evolved at the anode. A. Draw a diagram of the cell, label all components and show the reaction(s) taking place in the cell. A. What volume of oxygen at standard pressure and 25 o C is liberated when 450. milligrams of gold is deposited? B. What is the current used if the electrolysis of the 450 milligrams of gold is completed in 15 minutes? Oxidation Reaction (Anode): 2 2 O O 2 (g) + 4e - Reduction Reaction (Cathode): Au e - Au (s) Overall: 4Au O 4Au (s) + 3O 2 (g) O O 2 (g) + 4e - Au e - Au (s) 5b) 450 mg Au.45g/(197g/mol) = mol Au in overall reaction: Au = 4 = mol X =.0017 mol O 2 O 2 3 X PV = nrt V = nrt =.0017mol ( L atm) (298K) = ml O 2 p 1 atm mol K 5c) mol Au in 15 min mol ( 3mol e - ) (1 farad) (96,485 coulombs) ( 1min) =.733 coulomb = amp 15 min (1 mol Au) (1mol e - ) ( 1 farad) (60 sec) sec

7 10 points EXTRA CREDT (no partial credit) 1. name: a, O 2 nitrous acid b. MnCl 2 manganese() chloride c. P 2 O 3 phosphorus trioxide or diphosphorus trioxide 2. write a formula a. sodium chlorate aclo 3 b. ammonia 3 c. acetic acid C 2 3 O 2 3. f one wished to create the battery, Mg Mg 2+ ClO 1- Cl 2, one would have to use a basic solution for the electrolyte (else the magnesium would react with proton, releasing hydrogen, which might explode destroying the battery and those near it. Complete and balance the reaction used to power the battery. (not the hydrogen generating reaction or the explosion ;> ). See your lecture notes from Friday, March 7, 2003

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