CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question Credit

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1 Chemistry 11 Name Final Exam May 7, 1998 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question Credit 1(15) 5(0) (0) 6(0) 3(0) 7(5) 4(40) 8(40) Total Total TOTAL POINTS = R = J/mol-K = C/mol e - C = amp x sec Δ = Δ /n log Q ΔG = ΔG + RT ln Q ΔG = -RTlnKeq ΔG = -nδ PV = nrt R = L-atm/mol-K ln(ai/af) = kδt 1/Af - 1/Ai = kδt ln(k 1 /k ) = Ea/R (1/T - 1/T 1 ) 1)Write down the equilibrium that occurs when the following compounds are mixed. a) NaOH + NaBenz b) NH 4 OH + NH 4 Cl c) NaCN + HCl General Chemistry II Lab Manual 07

2 ) Last night I poured myself a glass of water from the tap and took a drink. The water was not cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water tasted better. If my cup had 50 ml of water at C, and I put two ice cubes into the water that weighed 0 grams each, what was the final temperature of the water if the ice cubes were initially at 0 C? ΔH fus = 6.01 kj/mol, Cp(H O) = 75. J/molK 3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100 C. If the activation energy of the cooking of an egg is 75 kj/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80 C? 08 General Chemistry II Lab Manual

3 4a) Calculate the ph at each of the following points for the titration of 100 ml of.00 M Phthalic acid with 5 M NaOH INITIAL ph H Phthal H + + HPhthal - pk A1 =.89 HPhthal - H + + Phthal - pk A = ml NaOH 40 ml NaOH 50 ml NaOH 80 ml NaOH 100 ml NaOH 4b) How many milliliters of 5 M NaOH must you add to 100 ml of M Phthalic acid to make a buffer of ph = 3.5? General Chemistry II Lab Manual 09

4 5) At low temperatures, the rate law for the reaction, CO(g) + NO (g) CO (g) + NO(g) can be determined by the following data; Initial Rate/10-4 [NO]/10-3 M [CO]/10-3 M [NO ]/10-3 M Write a rate law in agreement with the data. Work out the rate law for each of the following mechanisms (show your work) and then circle the one that is consistent with the rate law determined above. (Circle A, B, or C) A) CO + NO CO + NO slow B) NO NO + O equil. O + CO CO slow C) NO NO + O equil. CO + O CO + O slow O + NO NO fast 10 General Chemistry II Lab Manual

5 6) What is the solubility of Barium Carbonate in a solution saturated with H CO 3 at a ph of 10? [H CO 3 ] sat'd = 0.034M. BaCO 3 Ba CO 3 H CO 3 H CO 3 Ksp = 8.1x10-9 Keq =.0x ) A cell has the following cell diagram; Ag / AgCl / 1M HCl // 1M AgNO 3 / Ag 7a) What is the reduction half reaction? 7b) What is the overall reaction? 7c) Based on the cell diagram please draw the electrochemical cell. General Chemistry II Lab Manual 11

6 7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are 0.01 M? 8) Please name or draw the following compounds. OH CH 3 OH OH H N C C C O C C C C C C Cl C O C C C C Cl C CH 3 C C butanone 3,3-dichloro butanoic acid trans 1,3-cyclopentadiol o-methylaniline 1 General Chemistry II Lab Manual

7 Chemistry 11 Name Answer Key Final Exam May 7, 1998 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question Credit 1(15) 5(0) (0) 6(0) 3(0) 7(5) 4(40) 8(40) Total Total TOTAL POINTS = R = J/mol-K = C/mol e - C = amp x sec Δ = Δ /n log Q ΔG = ΔG + RT ln Q ΔG = -RTlnKeq ΔG = -nδ PV = nrt R = L-atm/mol-K ln(ai/af) = kδt 1/Af - 1/Ai = kδt ln(k 1 /k ) = Ea/R (1/T - 1/T 1 ) 1)Write down the equilibrium that occurs when the following compounds are mixed. a) NaOH + NaBenz Benz - + H O HBenz + OH - b) NH 4 OH + NH 4 Cl NH 4 OH NH OH - c) NaCN + HCl HCN H + + CN - General Chemistry II Lab Manual 13

8 ) Last night I poured myself a glass of water from the tap and took a drink. The water was not cold and didn't taste good so I put a couple of ice cubes in it. The ice cubes melted and the water tasted better. If my cup had 50 ml of water at C, and I put two ice cubes into the water that weighed 0 grams each, what was the final temperature of the water if the ice cubes were initially at 0 C? ΔH fus = 6.01 kj/mol, Cp(H O) = 75. J/molK First, use the C water to melt the ice. The result will be colder water and melted ice. ncpt = nh Fus 50 g/18 g/mol (75. J/molK)(T F - C) = 40g/18 (6010 J/mol) T F = 9.1 C Now, mix this colder water (9.1 C) with the melted ice (0 C), -ncpt = ncpt -[50 g/18 g/mol (75. J/molK)(T F - C)] = 40 g/18 g/mol (75. J/molK)(T F - 0 C) T F = 7.94 C 3) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100 C. If the activation energy of the cooking of an egg is 75 kj/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80 C? t 1 = C and 80 C t ln 3 min 75, t = min 14 General Chemistry II Lab Manual

9 4a) Calculate the ph at each of the following points for the titration of 100 ml of.00 M Phthalic acid with 5 M NaOH H Phthal H + + HPhthal - pk A1 =.89 K A1 =.88x10-3 HPhthal - H + + Phthal - pk A = 5.41 K A = 3.89x10-6 INITIAL ph x x.88x10 3 ph = 1.9 M 1V 1 = M V (M)((0.100L) = (5 M)(V ) V = 40 ml for each region Also (M)(0.100L) = 0. mol H Phthal 0 ml NaOH ph = ml NaOH pka1 pka ph ph = ml NaOH 10 ml into nd Region (5 M)(0.010 L) = 0.05 mol OH ph 5.41 log ph = ml NaOH [Phthal mol 1x10 ] 1.111M and Kb.57x L 3.89x10 x.57x x 9 x = 5.3x10-5 M OH - ph = ml NaOH (0.00 L)(5 M) = 0.10 mol OH mol OH 0.00 L 0.5 M OH ph = b) How many milliliters of 5 M NaOH must you add to 100 ml of M Phthalic acid to make a buffer of ph = 3.5? The ph is nearest the 1 st pka so the buffer is in Region I. You must add, (x) log x mol NaOH = (5 M)(V) V = 3.14 ml NaOH (0.0 - x) General Chemistry II Lab Manual 15

10 5) At low temperatures, the rate law for the reaction, CO(g) + NO (g) CO (g) + NO(g) can be determined by the following data; Initial Rate/10-4 [NO]/10-3 M [CO]/10-3 M [NO ]/10-3 M Write a rate law in agreement with the data. [CO][NO] rate k [NO] Work out the rate law for each of the following mechanisms (show your work) and then circle the one that is consistent with the rate law determined above. (Circle A, B, or C) A) CO + NO CO + NO slow rate = k [CO] [NO ] No, doesn t match B) NO NO + O equil. O + CO CO slow [NO][O] [NO] rate = k [O] [CO] and Keq so, [O] Keq [NO] [NO] [CO][NO] [CO][NO] Substituting, we get rate kkeq or rate k' [NO] [NO] This matches our experimentally determined rate law so it is a possible mechanism. C) NO NO + O equil. CO + O CO + O slow O + NO NO fast rate = k [O ] [CO] The only reason to use the equilibrium is to get rid of something that is not in the overall reaction. Since the rate law does not contain anything that is not already in the overall reaction (no intermediates) there is no reason to continue. This does not match. 16 General Chemistry II Lab Manual

11 6) What is the solubility of Barium Carbonate in a solution saturated with H CO 3 at a ph of 10? [H CO 3 ] sat'd = 0.034M. BaCO 3 Ba CO 3 H CO 3 H CO 3 Ksp = 8.1x10-9 Keq =.0x [H ] [CO3 ] (1x10 M) (CO3 ) 17.0x10 [CO3 [H CO ] 0.034M 3 [Ba + ] [CO - 3 ] = [Ba + ] [68.68 M] = 8.1x10-9 [Ba + ] = 1.18x10-10 M 7) A cell has the following cell diagram; Ag / AgCl / 1M HCl // 1M AgNO 3 / Ag 7a) What is the reduction half reaction? Ag + + e - Ag 7b) What is the overall reaction? Ag + + Cl - AgCl 7c) Based on the cell diagram please draw the electrochemical cell. ] M General Chemistry II Lab Manual 17

12 7d) If the Ksp for this cell is 1.8x10-10 then what is the cell voltage when all the ionic species are 0.01 M? Δ = 0.059/n log Ksp = 0.059/(1) log (1/1.8x10-10 ) = volts Δ = Δ /n log Q = Δ /n log [1/([Ag + ][Cl - ])] Δ = V /(1) log 1/[(0.01)(0.01)] Δ = volts 8) Please name or draw the following compounds. OH CH 3 OH OH H N C C C phenol para methylaniline 1,-propadiol O C C C C C C O C C cyclopropyl ethyl ether H 3 C Cl C Cl CH 3 C C C H CH 3 3-methyl--pentanone cis,trans-,4-dichloro-3-methyl--,4-hexadiene butanone 3,3-dichloro butanoic acid O O Cl C C C C HO C C C Cl C trans 1,3-cyclopentadiol OH o-methylaniline NH CH 3 OH 18 General Chemistry II Lab Manual

13 Chemistry 11 Name Final Exam May 6, 000 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question Credit 1(15) 6(0) (15) 7(15) 3(15) 8(30) 4(15) 9(0) 5(5) 10(30 ) Total Total TOTAL POINTS = R = J/mol-K = C/mol e - C = amp x sec = /n log Q ΔG = ΔG + RT ln Q ΔG= -RTlnKeq ΔG = -n PV = nrt R = L-atm/mol-K ln(ai/af) = kδt 1/Af - 1/Ai = kδt 1) Which of the following compounds is the most soluble in water? Show your work. BaSO 4 Ksp = 1.98x10-10 Fe(OH) Ksp = 1.64x10-14 Ag 3 PO 4 Ksp = 1.x10-15 ) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HF Na 3 Asp + HCl HCl + KOH General Chemistry II Lab Manual 19

14 3) The bromination of acetone is acid catalyzed; CH 3 COCH 3 + Br CH 3 COCH Br + H + + Br - The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine and hydrogen ions; Rate [Acetone] [Br ] [H + ] 6x x x x x a) What is the forward rate law for this reaction? b) What is the value and units of the forward rate constant? 4) What is the solubility of Cu S in a saturated solution of 0.10M H S that is buffered at ph=10? Cu S Cu + + S - Ksp = x10-47 H S H + + S - Keq = 1.1x General Chemistry II Lab Manual

15 5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO. Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be; HgO + H O Hg + + OH - Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 5b) Draw the cell diagram for the above cell. 5c) If the = volts, what is the voltage of the cell when concentration of the ionic species in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0. M? General Chemistry II Lab Manual 1

16 6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100 C. If the activation energy of the cooking of an egg is 75 kj/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80 C? 6b) If the rate of a reaction doubles with a 15 C change in temperature, what is it's activation energy? State your assumptions. 7) A block of copper intially at 500 C was placed on top of a 100 gram block of ice at 0 C. All of the ice melted and turned into water. The water and the block of copper ended up at 0 C. How much did the block of copper weigh? ΔH fus = 6.01 kj/mol, Cp(Cu) = 4.5 J/mol-K, Cp(H O) = 75. J/mol-K, FWT (Cu) = g/mol, FWT (H O)= 18 g/mol General Chemistry II Lab Manual

17 8) Calculate the ph at each of the following points for the titration of 100 ml of 3 M Phthalic acid with 5 M NaOH H Phthal H + + HPhthal - pk A1 =.89 HPhthal - H + + Phthal - pk A = 5.41 INITIAL ph 0 ml NaOH 60 ml NaOH 75 ml NaOH 90 ml NaOH 8b) How much 5 M NaOH would you add to 100 ml of 3 M H Phthal to make a buffer of ph = 5? General Chemistry II Lab Manual 3

18 9) How much heat must be added to 100 g of ice at -5 C to turn it into 60 grams of water and 40 grams of steam, both at 100 C? Cp H O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1 J/mol-K, ΔHfus = 6.01 kj/mol, ΔHvap = 40.7 kj/mol 4 General Chemistry II Lab Manual

19 Chemistry 11 Name Answer Key Final Exam May 6, 000 CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit. You may use a calculator. Question Credit Question Credit 1(15) 6(0) (15) 7(15) 3(15) 8(30) 4(15) 9(0) 5(5) 10(30 ) Total Total TOTAL POINTS = R = J/mol-K = C/mol e - C = amp x sec = /n log Q ΔG = ΔG + RT ln Q ΔG= -RTlnKeq ΔG = -n PV = nrt R = L-atm/mol-K ln(ai/af) = kδt 1/Af - 1/Ai = kδt 1) Which of the following compounds is the most soluble in water? Show your work. BaSO 4 Ksp = 1.98x10-10 (x)(x) = x = 1.98x10-10 x = 1.41x10-5 Fe(OH) Ksp = 1.64x10-14 (x)(x) = 4x 3 = 1.64x10-14 x = 1.6x10-5 Ag 3 PO 4 Ksp = 1.x10-15 (3x) 3 (x) = 7x 4 = 1.x10-15 x = 8.16x10-5 ) Please write the equilibrium reaction that will occur when the following compounds are mixed. Write all reactions as dissociations where appropriate. NaOH + HF F - + H O HF + OH - Na 3 Asp + HCl HAsp - H + + Asp 3- HCl + KOH H O H + + OH - General Chemistry II Lab Manual 5

20 3) The bromination of acetone is acid catalyzed; CH 3 COCH 3 + Br CH 3 COCH Br + H + + Br - The rate of disappearance of bromine was measured for several different concentrations of acetone, bromine and hydrogen ions; Rate [Acetone] [Br ] [H + ] 6x x x x x a) What is the forward rate law for this reaction? rate = k [Acetone] [Br ] [H + ] b) What is the value and units of the forward rate constant? 6x10-5 M/s = k [0.30 M] [0.050 M] [0.05 M] k = 1.6 1/M 3 s 4) What is the solubility of Cu S in a saturated solution of 0.10M H S that is buffered at ph=10? [H + ] [S - ]/[H S] = 1x10-0 Cu S Cu + + S - Ksp = x10-47 H S H + + S - Keq = 1.1x10-0 [S - ] = 1x10-0 [H S]/[H + ] = 1x10-0 [0.10]/[1x10-10 ] = 0.10 M S - Substitute this result into the following equation, [Cu + ] [S - ] = x10-47 [Cu ] x 7 /[S - ] x x10 4 [Cu S] 7.07x10 M Cu S /[0.10] 1.41x10 3 M Cu 6 General Chemistry II Lab Manual

21 5a) Set up a WORKING electrochemical cell that will determine the solubility product for HgO. Note that this reaction is spontaneous in the reverse direction. Your overall reaction should be; HgO + H O Hg + + OH - The half-reactions; Hg + + e - Hg Hg + OH - HgO + H O + e - Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of all ionic species. Indicate the composition of the electrodes. 5b) Draw the cell diagram for the above cell. Pt / Hg(l) / Hg O(s) / 1M NaOH // 1M Hg(NO 3 ) / Hg / Pt 5c) If the = volts, what is the voltage of the cell when concentration of the ionic species in the reduction cell are 0.5 M and the ionic species in the oxidation cell are 0. M? = /n log 1/[Hg + ][OH - ] = V / log 1/[0.5M][0.M] = Volts General Chemistry II Lab Manual 7

22 6a) Assume that the cooking of an egg is a first order process. Assume further that it takes 3 minutes to half-cook (soft boil) an egg at 100 C. If the activation energy of the cooking of an egg is 75 kj/mole, how long would it take to soft boil an egg at the top of Mt. Everest where water boils at 80 C? t 1 = C and 80 C t ln 3 min 75, t = min 6b) If the rate of a reaction doubles with a 15 C change in temperature, what is its activation energy? State your assumptions. Assume k 98 K and k 313 K k ln k 1 1 Ea Ea = 35,834 J/mol 7) A block of copper intially at 500 C was placed on top of a 100 gram block of ice at 0 C. All of the ice melted and turned into water. The water and the block of copper ended up at 0 C. How much did the block of copper weigh? ΔH fus = 6.01 kj/mol, Cp(Cu) = 4.5 J/mol-K, Cp(H O) = 75. J/mol-K, FWT (Cu) = g/mol, FWT (H O)= 18 g/mol ncpt Cu = n fus + ncpt HO x g J molk x g = 5.55 g Cu 100 g 18 J 100 g mol 18 J molk 0C 500C C 8 General Chemistry II Lab Manual

23 8) Calculate the ph at each of the following points for the titration of 100 ml of 3 M Phthalic acid with 5 M NaOH H Phthal H + + HPhthal - pk A1 =.89 HPhthal - H + + Phthal - pk A = 5.41 M 1V 1 = M V (3 M)(0.100 L) = (5 M)(V ) V = 60 ml Also, (3 M)(0.1-0 L) = mol H Phthal x 3 INITIAL ph 1.88x10 x M H ph = x ml NaOH ph.89 log ph = pka1 pka ml NaOH ph ph = ml NaOH ph 5.41 log ph = ml NaOH ph = b) How much 5 M NaOH would you add to 100 ml of 3 M H Phthal to make a buffer of ph = 5? The ph is nearest the nd pka so the buffer is in Region II. You must add, (3 M)(0.100 L) = (5 M)(V) V = 60 ml to get to the starting point. (x) log x mol NaOH= (5 M)(V) V = 16.8 ml ( x) Total volume = 16.8 ml + 60 ml = 76.8 ml NaOH 9) How much heat must be added to 100 g of ice at -5 C to turn it into 60 grams of water and 40 grams of steam, both at 100 C? Cp H O = 75.7 J/mol-K, Cp Ice = 37.7 J/mol-K, Cp Steam = 35.1 J/mol-K, ΔHfus = 6.01 kj/mol, ΔHvap = 40.7 kj/mol 100g J molk 100g 18 J 100g mol 18 J molk 40g 18 J mol 5C C 40, ,658J General Chemistry II Lab Manual 9