Chemistry 10401, Sections H*, Spring 2015 Prof. T. Lazaridis Final examination, May 18, Last Name: First Name:

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1 The City College of New York Chemistry Department Chemistry 10401, Sections H*, Spring 2015 Prof. T. Lazaridis Final examination, May 18, 2015 Last Name: First Name: Instructions: There are 13 questions in this exam, check both sides of the sheets * Do not cheat. You will automatically fail and may face further action. * You may tear off the last sheet, which contains the periodic table, and use it as scrap paper. Anything written on it will not be counted during grading. * Use a calculator. No cellphones or devices for /texting. * Please do not leave the room during the exam. If you must go to the bathroom, please give the exam to the Professor. * Place your student ID on the desk in front you. To guarantee full marks, you must SHOW WORK/REASONING, even for multiple choice questions. No work or reasoning means no points. Constants: Speed of light : X 10 8 m/s Planck's constant : X Js Nav: 6.022X10 23 F=96485 C/mol e MeV/u R= L atm/mol K = bar L/mol K = J/mol K = L torr/mol K Data: Standard enthalpies of formation (ΔH f o, kj/mol): SF 4(g): - 763, CF 4 (g): - 925, SO 2(g): CO 2(g): Absolute entropies at 298 K (S o, J/mol K) : SF 4(g): 299.6, CF 4 (g): 261.6, SO 2(g): CO 2(g): O 2 (g) + 4H + (aq) + 4e - - > 2H 2O (l) E o = O 2 (g) + 2H + (aq) + 2e - - > H 2O 2 (l) E o = I 2(s) + 2 e - - > 2 I - (aq) E o = Spectrochemical series: CN - >NO 2 - >en>py~nh3 >EDTA 4- >SCN - >H 2 O>ONO - >ox 2- >OH - >F - >SCN - >Cl - >Br - >I - Equations: ΔT f=- K fxm ΔT b=k bxm K p=k c(rt) Δn ΔG o =- RTlnK ΔG= ΔG o +RTlnQ 1 st order: t 1/2= ln2/k ln(a/a o)=- kt 2 nd order: 1/A=1/A o+kt k=aexp(- E a/rt) ph=pka + log [base]/[acid] E o cell = ( V/z ) ln K = (0.0592V/z) log K E cell = E o cell (0.0592V/z) log Q ΔG o =- zfe o Score: Total:

2 1. (8) Arrange the following substances in the expected order of increasing boiling point: H 2 O, NH 3, CH 4, CH 3 CH 3. Explain your rationale. CH 4 < CH 3 CH 3 < NH 3 < H 2 O The first two are nonpolar, but ethane is larger. The last two make h bonds, but those in water are stronger and of a greater number (there are more h bonds per molecule in water than in ammonia). 2. (8) How many grams of naphthalene (C 10 H 8 ) would you add to 50.0 g of benzene (C 6 H 6 ) to produce a solution with the same freezing point as pure water? (T f of pure benzene is 5.53 o C and K f = 5.12 o C/m) ΔT f = = o C ΔT f = -K f Xm => m= 5.53/5.12 = 1.08 m Moles of C 10 H 8 : 1.08 moles/1000 g benzene X 50.0 g benzene= Molar mass of C 10 H 8 : 10* *1.008= g/mol g of C 10 H 8 : X 128.2= 6.92 g 3. (8) Determine the order of the following reaction with respect to each reactant and the reaction rate constant using the given initial rate data. 2-2 HgCl 2 + C 2 O 4 à 2 Cl CO 2 (g) + Hg 2 Cl 2 (s) Exper. [HgCl 2 ], M [C 2 O 2-4 ], M Initial rate, mol/(l min) X X X 10-5 From 1 & 2: 7.1/1.8=3.94 and 0.30/0.15=2 => order with respect to C 2 O 4 2- is 2. From 2 & 3: 7.1/3.5=2.03 and 0.105/0.052=2.02 => order with respect to HgCl 2 is X 10-5 =k X X => k= 7.6 X 10-3 M -2 min (7) The reaction A + 2 B à C + 2 D has been found to be first order in A and first order in B. Write a plausible mechanism that explains the observed rate law. A + B à I + D I + B à C + D slow fast

3 5. (8) 5.25 g of I 2 and 2.15 g of Br 2 are mixed in a 3.15-L flask at 115 o C. At equilibrium, 1.98 g of I 2 is present. What is K c at 115 o C for the reaction I 2 (g) + Br 2 (g) 2 IBr (g) Initial concentrations: [I 2 ] o = 5.25 g/3.15 L X 1mol/253.8g = 6.57 X 10-3 M [Br 2 ] o = 2.15 g/3.15 L X 1mol/159.8g = 4.27 X 10-3 M I 2 (g) + Br 2 (g) 2 IBr (g) I X 10-3 M 4.27 X 10-3 M 0 C. -x -x +2x E X x 4.27 X x 2x Equilibrium concentrations: [I 2 ] eq = 1.98 g/3.15 L X 1mol/253.8g = 2.48 X 10-3 M = 6.57 X x =>x=4.09x10-3 [Br 2 ] eq = 4.27 X X10-3 =0.18X10-3 [IBr] eq = 2X 4.09X10-3 = 8.18X10-3 K c = [IBr] 2 /[I 2 ] [Br 2 ]= (8.18X10-3 ) 2 / (2.48 X 10-3 )( 0.18X10-3 )= 1.5X (8) Calculate the ph of a 1-L aqueous solution containing 32.9 g formic acid, HCOOH (Ka = 1.8 X 10-4 ). Initial concentration: 32.9 g/l X 1mol/46.03g = M HCOOH (aq) + H 2 O HCOO - (aq) + H 3 O + (aq) I C. -x x x E x x x K a =[H 3 O + ][ HCOO - ]/[ HCOOH] = x 2 /(0.715-x) x 2 /0.715 => x=[h 3 O + ]=1.1X10-2 M ph=-log[1.1x10-2 ] = (7) What is the ph of a solution that contains M HCOOH and M of the salt HCOONa? Use data from the previous problem. ph=pka + log [base]/[acid] = - log(1.8 X 10-4 ) + log(0.326/0.405) = 3.65

4 8. (8) A mL portion of a 0.50 M AgNO 3 (aq) solution is added to ml of a solution that is M in Cl - a) Will AgCl (s) (K sp = 1.8X10-10 ) precipitate from this solution? If so, how many moles will precipitate and what will be the concentrations of the ions after precipitation? Volume of new solution: ml [Ag + ] = 10.00mLX0.50M / ml = 4.5 X10-2 M [Cl - ]= 100.0mLX0.010M / ml = 9.1 X 10-3 M Q sp =[Ag + ] [Cl - ]= 4.1X10-4 > K sp, so precipitation will occur Ag + (aq) + Cl - (aq) à AgCl (s) I. 4.5 X10-2 M 9.1 X 10-3 M - C. x -x +x E. 4.5 X10-2 -x 9.1 X x x K sp = (4.5 X10-2 -x)(9.1 X x) => x ~ 9.1X10-3 (x must be 9.1X10-3 ) So, 9.1X10-3 M X 110X10-3 L = 1.0X10-3 moles of AgCl will precipitate. After precipitation: [Ag + ]= 4.5 X X10-3 =3.6X10-2 [Cl - ]= K sp / [Ag + ] = 5.0X10-9 (much smaller than [Ag + ] but not quite zero) 9. (6) For each of the following reactions indicate whether you would expect the entropy of the system to increase or decrease and why. If you cannot tell, also state why. a) 2 KClO 3 (s) à 2 KCl (s) + 3 O 2 (g) S increases (gas is produced from solid) b) CH 3 COOH (l) à CH 3 COOH (s) S decreases (liquid becomes solid) c) N 2 (g) + O 2 (g) à 2 NO (g) Cannot tell, 2 gas molecules on both sides 10. (8) Use thermodynamic data given to determine K eq for the following reaction at 45 o C. CO 2 (g) + SF 4 (g) CF 4 (g) + SO 2 (g) ΔH o = (-393.5)-(-763) = kj ΔS o = = -3.5 J/K Assuming these are independent of temperature: ΔG o (318K) = ΔH o -318 X ΔS o = kj ΔG o = -RTlnKeq => Keq= 3.5X10 10

5 11. (8) Using data given, determine the values of E o cell and ΔG o for the following reaction: O 2 (g) + 4 I - (aq) + 4H + (aq) à 2 H 2 O (l) + 2 I 2 (s) O 2 (g) + 4H + (aq) + 4e - - > 2H 2O (l) E o = I 2(s) + 2 e - - > 2 I - (aq) E o = Flip the second, multiply by 2 and add to the first to get the given equation. E o cell = = V ΔG o =- zfe o =- 4X96485X0.694 V = -268 kj 12. (8) Draw orbital diagrams showing the possible distributions of d electrons in the central metal of the ion [PtCl 6 ] 2-. If more than one distribution is possible, indicate which one is low spin and which one high spin. Given information in your data sheet, which one is more likely? Pt 4+ : 5d 6 low spin high spin Cl - is a weak-field ligand, so the high-spin state is more likely 13. (8) Radium-224 decays by α-particle emission. a) (4) Write the complete nuclear equation (include atomic numbers) b) (4) Determine the energy in MeV associated with this decay given the following nuclear masses: 224 Ra: u; 220 Rn: ; 4 He: u. a) Ra - > Rn + 4 2He b) Δm= =-0062u X MeV/u = -5.8 MeV

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