CHEM 1332 SPRING 2007 VERSION 1 NO CHEATING!

1 CHEM 1332 SPRING 2007 VERSION 1 NO CHEATING! 1. How many kilojoules of heat are required to raise the temperature of 484 grams of ethanol (mw = 46) from 0 C to 25 C and then vaporize it completely? (specific heat = 2.46 J/(g C), Hvap = 43.3 kj/mol) (A) 485 (B) 456(C) 73.1 (D) 30,200 (E) More information is needed to answer this question. From solid at 0 C to solid at 25 C q = mc T = 484g x 2.46J/g C x 25 C = 29.766 kj from solid 25 C to vapor at 25 C heat = n Hvap = 484g/46 g/mol x 43.3 = 455.591 kj total heat = 29.766 kj + 455.591 kj = 485 kj 2. Which of the following compounds is predicted to have the highest enthalpy of vaporization at room temperature? (A) H 2 O (B) CH 4 (C) CCl 4 (D) C 6 H 14 (E) CO 2 This is because of hydrogen bonding 3. Which solution would have the largest osmotic pressure? (A) 0.1 M (CH 3 ) 2 CO (acetone) at 25 C (B) 0.1 M C 6 H 12 O 6 at 30 C (C) 0.1 M NH 4 NO 3 at 25 C (D) 0.05 M KBr at 25 C (E) 0.05 M CaCl 2 at 25 C The most concentrated would have the largest osmotic pressure C contains 2 ions which is multiplied by 0.1 and the resultant is 0.2. This is higher than 0.1 in A and B since A and B do not dissociate and 0.1 and 0.15 for D and E respectively. 4. When 0.750 g sample of an unknown substance is dissolved in 20.0 g of benzene, the freezing point of the solution is 4.53 o C. What is the molar mass of the substance? (The normal freezing point of benzene is 5.53 o C and Kf is 5.12 o Cm -1.) (A) 180 (B) 870 (C) 7.32 (D) 192 (E) 341 MM = g of solute x K f /( f p x kg of solven)t = (0.750 x 5.12)/(5.53-4.53)(20 x 10-3 kg) = 192gmol -1 5. Which of the following pairs is most likely to form a heterogeneous mixture?

2 (A) methanol, CH 3 OH, and water, H 2 O (B) hexane, CH 3 (CH 2 ) 4 CH 3, and octane, CH 3 (CH 2 ) 6 CH 3 (C) potassium chloride, KCl, and carbon tetrachloride, CCl 4 (D) acetic acid, CH 3 COOH, and water, H 2 O (E) an allotrope of iodine, I7, and benzene, C 6 H 6 Like dissolves like. A and D are all polar and would form a homogeneous mixture, also B and E are non polar and would form a homogenous mixture, whereas in C, KCl is polar and CCl 4 is non polar and would form a heterogeneous mixture. 6. Nitrogen dioxide decomposes to nitrogen and oxygen with a rate constant of 12.5 M -1 s -1. What is the half-life for the reaction in seconds if [NO 2 ] 0 = 0.00260 M? (A) 10.6 (B) 22.8 (C) 30.8 (D) 42.4 (E) 68.4 From the rate constant unit, the reaction is second order T 1/2 = 1/K[A] 0 = 1/ 12.5M -1 s -1 x 0.00260 M = 30.8 s -1 7. Determine the rate law for the reaction, NH + 4 + NO - 2 N 2 + 2 H 2 O given the following data: [NH 4 ] (M) [NO - 2 ] (M) Rate (M/s) 0.010 0.020 0.020 0.015 0.020 0.030 0.010 0.010 0.005 (A) Rate = k[nh + 4 ][NO - 2 ] (B) Rate = k[nh + 4 ] 2 [NO - 2 ] (C) Rate = k[nh + 4 ][NO - 2 ] 2 (D) Rate = k[nh + 4 ]2[NO - 2 ] 2 (E) None of the above 0.01[NH 4 ]/0.01[NH 4 ] x 0.02[NO 2 - ]/0.01[NO 2 - ] = 0.020/0.005 2[NO 2 - ] = 4 or 2 2, therefore [NO 2 - ] is second order 0.015[NH 4 ]/0.01[NH 4 ] x 0.02[NO 2 - ]/0.02[NO 2 - ] = 0.03/0.02 1.5[NH 4 ] = 1.5, therefore [NH 4 ] is first order Rate = k[nh 4 + ][NO 2 - ] 2 8. The activation energy of a reaction is 37.6 kj/mol and the rate constant is 5.4 x 10-3 s -1 at 45 C. What is the rate constant at 145 C? (A) 0.16 (B) 5.5 x 10-3 (C) 8.4 x 10-3 (D) 0.38 (E) 0.56 solution ln = =. /. / = 3.4 = e 3.4 = 30.03

3 k 2 = 30.03 x 5.4 x 10-3 = 0.16 9. Based on a reaction profile, the enthalpy of reaction is 430 kj and the energy of activation for the backward reaction is 156 kj. What is the energy of activation for the forward reaction? (A) 586 kj (B) 274 kj (C) 274 kj (D) 137 kj (E) 293 kj E a = 430 + 156 = 586 kj 10. Which plot would not have a slope from which the indicated quantity for the reaction A Product(s) could be determined? (A) ln [A] vs. t: rate constant for a first order reaction (B) 1/[A] vs. t: rate constant for a second order reaction (C) ln k vs. 1/T (K): activation energy (D) ln k vs. 1/T (K): frequency factor (E) All of these plots would have a slope from which the indicated quantity could be determined. 11. Consider the following mechanism of the oxidation of bromide ions by hydrogen peroxide in acid solution: H + + H 2 O 2 H 2 O-OH + Fast 1 H 2 O-OH + + Br - HOBr + H 2 O Slow 2 HOBr + H + + Br - Br 2 + H 2 O Fast 3 Which of the following rate laws is consistent with this mechanism? (A) Rate = k[h 2 O 2 ][Br - ][H+] 2 (B) Rate = k[h 2 O 2 ][Br-][H+] (C) Rate = k[h 2 O-OH+][Br - ] (D) Rate = k[hobr][h+][br - ][H 2 O 2 ] (E) Rate = k[h 2 O 2 ][Br - ] 2 [H + ] 2 The slow step determines the rate law and H 2 O-OH + is an intermediate which is used up as it is produced. H + + H 2 O 2 + Br - HOBr + H 2 O from 1 and 2 12. 230 Th decays to form 214 Po. If two beta particles are released, how many alpha particles were also released? (A) 1 (B) 2 (C) 3 (D) 4 (E) 6 230 Th 2 0-1β + 4 4 2α + 214 Po 13. K for the system A (g) + B (g) 2C (g) is 4.0. In HOW MANY of the following experiments will equilibrium be realized through a net reaction forming C from A and B? Experiment [A] [B] [C] 1 1.0 0.5 0 2 1.0 1.0 1.0 3 1.0 1.0 2.5 4 0 1.0 1.0 (A) none (B) 1 (C) 2 (D) 3 (E) 4

4 solution for equilibrium to be realized, K should not be zero K = [C] 2 /[A][B] In experiment 1 and 4 K will be zero and infinity respectively. 14. K P for the reaction CaCO 3 (s) CaO (s) + CO 2 (g) is 0.88 at 600 o C. If a 15.0-g sample of CaCO 3 (molar mass = 100 g/mol) is put into a 10.0 L container and heated to 600 o C, what percent of the CaCO 3 reacts? (A) 100 (B) 81.9 (C) 36.1 (D) 18.1 (E) 13.1 CaCO 3 (s) CaO (s) + CO 2 (g) K p = P CO2 = 0.88 PV = nrt Moles (n) = PV/RT = 0.88 atm x 10L/ (0.0821atm L.K -1 x 873K ) = 0.1228 moles of CO 2 produced. This is moles of CaCO 3 reacted. Mass of CaCO 3 reacted = 0.1228 mol x 100g/mol = 12.28 g Percentage reacted = (12.28g/15g) x 100 = 81.9. This is also a nice one. You just need to play with algebra and some reasoning. 15. An equilibrium mixture of NO(g), O 2 (g) and NO 2 (g) is allowed to expand from 1.0 to 2.0 L at a constant temperature. Given that 2 NO (g) + O 2 (g) 2 NO 2 (g) which of the following statements is correct? (A) The concentrations of all three gases are unchanged. (B) The value of K P would decrease. (C) The number of moles of NO 2 would increase. (D) The number of moles of O 2 would increase. (E) The number of moles of all three gases are unchanged. Increasing the volume, thereby decreasing the pressure, the reaction would try to increase the pressure by increasing the number of gas molecules, goes to the side with more gas moles. Le Chatelier s principle If you mess up a reaction, it tries to unmess itself Dr. Bott 16. What is the [OH - ] in an aqueous solution of orange juice with [H 3 O + ] = 3.4 x 10-4 M? (A) 1.00 x 10-14 (B) 10.53 (C) 3.47 (D) 3.4 x 10-4 (E) 2.9 x 10-11

5 ph = -log [H 3 O + ] = -log 3.4 x 10-4 = 3.47 poh = 14 3.47 = 10.53 poh = - log [OH - ] [OH - ] = 10-10.53 = 2.9 x 10-11 M 17. Calculate the ph of 1 x 10-8 M HCl. (A) 8 (B) 6 (C) slightly less than 7 (D) slightly more than 7 (E) 7 HCl is a strong acid so it s ph cannot be greater than 7 [H + ] = 10-8 + 1x 10-7 = 1.1 x 10-7 ph = - log [H + ] = - log 1.1 x 10-7 = 6.96 18. What molarity of a CH 3 COONa solution has a ph = 9.35? (Ka (CH 3 COOH) = 1.8 x10-5 ) (A) 2.68 (B) 0.75 (C) 0.37 (D) 0.90 (E) 1.10 ph = 9.35 poh = 14 9.35 = 4.65 = - log [OH - ] [OH - ] = 10-4.65 = 2.24 x 10-5 CH 3 COO - + H 2 O <==> CH 3 COOH + OH - Kb = [CH 3 COOH][ OH - ]/[ CH 3 COO - ] 1 x 10-14 / 1.8 x 10-5 = (2.24 x 10 5 ) 2 / [ CH 3 COO - ] [ CH 3 COO - ] = 0.90 M 19. The Ka of HSO 4 - (aq) is 1.2 x 10-2. Which of the following is the BEST approximation of the ph of a 0.0500 M solution of H 2 SO 4 (aq) to 2 significant figures? (A) ph is exactly 1.3 (B) ph is exactly 1.0 (C) ph is exactly 1.6 (D) ph is less than 1.0 (E) ph is between 1.0 and 1.3 ph of 0.1M of H 2 SO 4 = 1.0 ; and that of 0.05 M = 1.3 Therefore the ph should be between these two numbers 20. Which of the following statements is true? (A) HAt should be a strong acid. It is is below iodine on the periodic table. Acidity increases down the group. (B) Oxy acids get weaker as the number of oxygen atoms increases (C) HCN is a weak acid so CN - is a strong base. (D) OH - is a weaker base than SH -. (E) All metal hydroxides are strong bases.

6 21. In the titration of a solution of the extremely weak acid, HCN (Ka = 6.2 x 10-10 ), by a solution of sodium hydroxide of equal concentration, which of the following statements is true? (A) The equivalence point will occur at an acidic ph, and a suitable indicator would be one like alizarin that changes color between ph of 6 and about 7 (B) The equivalence point will occur at a neutral ph, and a suitable indicator would be one like bromthymol blue that changes color between ph of 6 and about 7.5 (C) The equivalence point will occur at a basic ph, and a suitable indicator would be one like bromthymol blue that changes color between ph of 6 and about 7.5 (D) The equivalence point will occur at a basic ph, and a suitable indicator would be one like phenol red that changes color between ph of 7 and about 8 (E) The equivalence point will occur at a basic ph, and a suitable indicator would be one like phenolphthalein that changes color between ph of 8 and about 11. NaOH is a strong base and it s ph should be between 8 and 11 22. Calculate the ph of a solution prepared by adding 20.0 ml of 0.100 M HCl to 80.0 ml of a buffer that is comprised of 0.25 M NH 3 and 0.25 M NH 4 Cl. Kb of NH 3 = 1.8 x 10-5. (A) 9.17 (B) 4.83 (C) 9.67 (D) 9.34 (E) 1 New [H + ] = 0.1 M x 20 ml /100 ml = 0.02 New [NH + 3 ] = 0.25 x 80 /100 = 0.2 = new [NH + 4 ] NH 3 + H + + <===> NH 4 S 0.2 0.02 0.2 C -0.02-0.02 + 0.02 AN 0.18 0 0.22 NH 3 + + H 2 O <===> NH 4 + OH - S 0.18 0.22 0 C - x x x E 0.18 x 0.22 + x x Kb = [NH 4 + ][OH - ]/[NH 4 + ] 1.8 x 10-5 = (0.22 +x)(x)/(0.18-x) ; let s assume x is very small, the 1.8 x 10-5 = (0.22 )(x)/(0.18) x = 1.47 x 10-5 = [OH - ] poh = - log[oh - ] = - log 1.47 x 10-5 = 4.83 ph = 14 4.83 = 9.17 23. What is the molar solubility of AgCl in 0.15 M HCl? Ksp of AgCl is 1.6 x 10-10 (A) 0.15 (B) 0.075 (C) 1.1 x 10-9 (D) 1.3 x10-5 (E) 4.3 x 10-6 AgCl Ag + + Cl - Ksp = [Ag + ][Cl - ]

7 1.6 x10-10 = s (s +0.15) ; s is very small compared to 0.15 1.6 x 10-10 = 0.15s s = 1.1 x 10-9 24. A solution is 0.10 M AgNO 3 and 0.10 M Ba(NO 3 ) 2. If solid Na 2 SO 4 is added to the solution, what is [Ba 2+ ] when Ag 2 SO 4 begins to precipitate? (Ksp BaSO4 = 1.1 x 10-10 ; Ag 2 SO 4 = 1.1 x 10-5 ) (A) 2.4 x 10-6 (B) 1.0 x 10-7 (C) 2.4 x 10-7 (D) 3.2 x 10-6 (E) 5.1 x 10-6 As Ag 2 SO 4 precipitates, BaSO 4 Ba 2+ 2- + SO 4 ksp = [Ba 2+ 2- ][SO 4 ] Ag 2 SO 4 2Ag + 2- + SO 4 ksp = [Ag + ] 2 [SO 2-4 ] Same [SO 2-4 ] 1.1 x 10-5 = (0.1) 2 [SO 2-4 ] [SO 2-4 ] = 1.3 x 10-3 1.1 x 10-10 = [Ba 2+ ][1.3 x 10-3 ] [Ba 2+ ] = 1.0 x 10-7 M 25. For a certain reaction the standard free energy change is 80.0 kj at 300 K and 40.0 kj at 600 K. For this reaction (A) H is positive and S is positive. (B) H is positive and S is negative. (C) H is negative and S is positive. (D) H is negative and S is negative. (E) impossible to tell. G = H - T S -80 = H - 300 S (1) -40 = H 600 S (2) Equation (1)- (2) -40 = 300 S S is negative ( -0.133 kj) We substitute S in either equation 1 or 2 to get H I chose equation 1-80 = H 300 (-0.133) and H = -80 40 = -120

8 26. For a reaction the standard free energy change is 200 kj at 450 K and 500 kj at 900 K. What is H for this reaction in kj? (A) -200 (B) +500 (C) +200 (D) +100 (E) 500 G = H - T S -200 = H - 450 S (1) -500 = H - 900 S (2) (1) (2) 300 = 450 S S = 0.667 From (1) -200 = H 450(0.667) H = +100 27. Consider the reaction, NH 3 (g) + 2 O 2 (g) HNO 3 (l) + H 2 O, calculate K and Sº given the values below (all at 25 ºC): 1. 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O K = 81 Sº = -124 J/mol K 2. 2 NO 2 (g) 2 NO(g) + O 2 (g) K = 16 Sº = 120 J/mol K 3. 3 NO 2 (g) + H 2 O 2 HNO 3 (l) + NO(g) K = 64 Sº = 200 J/mol K (A) K = 3, Sº = -21 J/K (B) K = 24, Sº = -21 J/K (C) K = 3, Sº = 196 J/K (D) K = 24, Sº = -221 J/K (E) K = 3, Sº = -221 J/K Multiply equation (1) by 4 Reverse equation (2) and multiply by ¾ Multiply equation (3) by ½ (1)/4 = NH 3 + 5/4O 2 NO + 3/2 H 2 O K = (81) 1/4 = 3 S = -124/4 = -31 (2) -1 x ¾ = 3/2 NO + ¾ O 3/2 NO 2 K = (1/16) 3/4 = 0.125 S = -120 x ¾ = -90 (3) x ½ = 3/2 NO 2 + ½ O 2 HNO 3 + ½ NO K = (64) 1/2 = 8 S = 200 x ½ = 100 NH 3 (g) + 2 O 2 (g) HNO 3 (l) + H 2 O after cancelling out the rest K = 3 x 0.125 x 8 = 3 and S = -31+ (-90) + 100 = -21 J/K This is another nice piece. It took me a little while to figure out, so have fun with it!!! 28. The standard free energy of formation of water is 237.18 kj/mol. Determine the value of Keq at 25 o C for the reaction: 2H 2 (g) + O 2 (g) 2 H2O (l). (A) 3.76 x 10 41 (B) 1.10 (C) 1.21 (D) 0.826 (E) 1.41 x 10 83 solution G 0 = -RTlnK eq K eq = e - G0/RT = e -(2 x -237.18)/8.3145 x 10^-3 x 298 = 1.41 x 10 83

9 29. Balance the following equation in basic solution, using the lowest possible integer coefficients. What is the coefficient of water? H 2 O 2 (aq) + Cr(OH) 3 (s) CrO 4 2- (aq) + H 2 O(l) (A) 1 (B) 2 (C) 4 (D) 6 (E) 8 solution -1 +3 +6-2 2- H 2 O 2 + Cr(OH) 3 CrO 4 + H 2 O H 2 O 2 + 2e - H 2 O x 3 2- Cr(OH) 3 CrO 4 + 3e - x 2 2-3H 2 O 2 +2 Cr(OH) 3 2 CrO 4 + 2H 2 O To balance the 4 negative charges on the right, add 4OH - on the left, and balance the number of O and H with H 2 O 3H 2 O 2 + 2Cr(OH) 3 + 4OH - 2CrO 4 2- + 8 H 2 O 30. If K is much less than 1 for a given reaction then, (A) Gº is a large negative number and Eº is a positive number (B) Gº is a large negative number and Eº is a negative number (C) Gº is a large positive number and Eº is a positive number (D) Gº and Eº are both small negative numbers (E) Gº is a large positive number and Eº is a negative number G = -RTlnK eq G = -nfe If K is much less than 1, lnk will be and multiply the negative signe infront of the equation. We have a positive number E = G /-nfe = negative 31. A voltaic cell consists of a strip of lead metal in a solution of Pb(NO 3 ) 2 in one beaker, and in the other beaker a platinum electrode immersed in an NaCl solution, with Cl 2 gas bubbled around the electrode. The two beakers are connected with a salt bridge. (I) Which metal is at the anode? (II) Which electrode loses mass? (III) What is the cell potential under standard conditions? Pb 2+( aq) + 2e - Pb(s) Eºred = -0.126 V Cl 2 (g) + 2e - 2Cl - (aq) Eºred = +1.359 V (A) (I) Pb, (II) Pt, (III) +1.233V (B) (I) Pt, (II) Pt, (III) -1.485 V (C) (I) Pt, (II) Pb, (III) 1.233 V (D) (I) Pb, (II) Pb, (III) +1.485 V (E) None of the above

10 Oxidation takes place at the anode, Pb will be oxidized to Pb Pb(s) Pb 2+ + 2e - Pb will lose mass since it has been oxidized to Pb 2+ which is smaller than Pb The reaction is Pb(s) + Cl 2 (g) Pb 2+ (aq) + 2Cl - (aq) = 0.126 + 1.359 = 1.485 V 32. A voltaic cell consists of a Cu 2+/ Cu electrode (Eºred = 0.34 V) and an Au3 + /Au electrode (Eºred 1.50 V). Calculate the [Au 3+( aq)] if [Cu2+(aq)] = 1.20 M and Ecell = 1.13 V. (A) 0.001 M (B) 0.002 M (C) 0.04 M (D) 0.2 M (E) 5.0 M Cu(s) + Au 3+ (aq) Cu 2+ (aq) + Au(s) E ell = 1.5 = 0.34 = 1.16 V E = E - (0.059/n) log [Cu 2+ ]/[ Au 3+ ] 1.13 = (1.16 0.059/6)log(1.2) 3 ]/(Au 3+ ) 2 [Au 3+ ] = 0.04 M 33. Which of the following elements can be isolated by electrolysis in the manner described? (A) Oxygen at the cathode when a solution of potassium nitrate is electrolyzed. (B) Hydrogen at the cathode when a solution of nickel(ii) nitrate is electrolyzed. (C) Potassium at the cathode when a solution of potassium nitrate is electrolyzed. (D) Nitrogen at the anode when a solution of potassium nitrate is electrolyzed. (E) Oxygen at the anode when a solution of potassium nitrate is electrolyzed. 34. 5.9 g of cobalt metal is deposited from a solution of a rare cobalt compound when a solution of that compound is electrolyzed for 4.0 hours by a current of 2.0 amps. What is the charge of the cobalt in the original compound? (A) +1 (B) +2 (C) +3 (D) +4 (E) -1 Charge = current x time = 2.0 amps x 4.0 hrs x 60 mins x 60 secs = 28800 Charge = nef = mass/mm x e x F n = 5.9/58.93 = 0.1001 moles e = Charge/nF = 28800/0.1001 x 96500C = 3 e = number of electrons and F= Faraday constant.