CHEM 1412 Problem Set #3A

CHEM 1412 Problem Set #3a Dr. Ya-Ping Huang Solubility Equilibrium o 1. Calculate the molar solubility of CaF2 at 25 C in: a. pure water (2.1 x 10-4 M) -5 b. in 0.010M Ca(NO3)2 solution (3.1 x 10 M) c. in 0.010M KF solution (3.9 x 10-7 M) [Ans]: The 1st step is to write the equation for dissolution, and find the relationship between Ksp and molar solubility.

2+ - . a. R CaF2(s)  Ca (aq) + 2F (aq) Let S represent the molar solubility E S  S 2S 2+ - 2 2 3 -11 Ksp = [Ca ]eq [F ]eq = (s)(2s) = 4S = 3.9x10 K 3.9x1011 S  3 sp  3  3 9.75x1012  2.14x104 M 4 4

2+ b. In 0.010M Ca(NO3)2 solution, there is the common –ion effect. The extra Ca from Ca(NO3)2 will

reduce the solubility of CaF2(s) (Le Chatelier’s principle) –the common-ion effect

2+ - R CaF2(s)  Ca (aq) + 2F (aq) Let S represent the molar solubility E S  S 2S +0.010 2+ - [Ca ]eq = 0.010+ S [F ]eq = 2S

2+ - 2 2 2 2 -11 Ksp = [Ca ]eq [F ]eq = (S+0.010)(2S)  (0.010)(2S) = 0.04S = 3.9x10 (s is dropped from s+0.010, because when there is common-ion effect , the solubility will be severely reduced such that the s term is negligible compared to the concentration of commen-ion) Ksp 3.9x1011 S    3.12x105 M 0.04 0.04

2+ - 2 2 3 -11 Ksp = [Ca ]eq [F ]eq = (s)(2s) = 4S = 3.9x10

- c. in 0.010M KF solution: there is the common –ion effect. The extra F from KF will reduce the

solubility of CaF2(s) through common-ion effect (Le Chatelier’s principle)

2+ - R CaF2(s)  Ca (aq) + 2F (aq) Let S represent the molar solubility E S  S 2S +0.010 2+ - [Ca ]eq = S [F ]eq = 0.010+ 2S

2+ - 2 2 2 -4 -11 Ksp = [Ca ]eq [F ]eq = (S)(0.01+2S)  (S)(0.01) = 1.0x10 S = 3.9x10 S = Ksp/1.0x10-4 = 3.9x10-7 M

0f44a222470a951b7682d36e3b1558bf.doc p. 1 4/4/2018 o 2. Calculate the molar solubility of BaSO4 at 25 C: a. in pure water

b. in 0.10M Na2SO4 solution

c. Grams of BaSO4 dissolved in 2.0 Liters of water 2+ 2+ e. LD50 of Ba for rat is around 20 mg/Kg body weight of rats. Would the Ba in part (d) dangerous if ingested? (1.05 x 10-5M, 1.1 x 10-9 M, 4.9 x 10-3 g, 2.16 mg) [Ans]: a. In pure water (no common-ion or other effects) 2+ 2- BaSO4(s)  Ba (aq) + SO4 (aq) 2+ 2- Let S represent the molar solubility, then [Ba ] = S and [SO4 ] = S

2+ 2- 2 -10 10 5 Ksp = [Ba ] [SO4 ] = (S)( S)= S = 1.1x10 S= 1.1x10  1.05x10 M 2- b. in 0.10M Na2SO4 solution, the additional SO4 ions from Na2SO4 will reduce solubility of BaSO4 through common-ion effect 2+ 2- -10 -9 Ksp = [Ba ] [SO4 ] = (S)(0.10+S)= S(0.10) = 1.1x10 S=1.1x10 M 2+ d. mg of Ba in 1.5 L of saturated BaSO4 solution. 5  2+ 1.05x10 mol 137g 1000mg  mg of Ba = (M)(V)(MM) =  x(1.5L)( )   2.16mg  L  1mol.Ba  1g 

e. LD50 refers to lethal dose for 50% population exposed. Assume a typical adult weighs 50kg, then

LD50 = (20mg/Kg body weight)(50 kg) = 1000mg. So 2.16 mg is quite safe. 3. How would the solubility of the following substances be affected by  pH of a solution?

a. Ni(OH)2 b. CaCO3 c. BaSO4 d. AgCl [Ans]: The key to see how does  pH of a solution affect the solubility is to see if additional H+ will affect the dissolution process

Dissolution reaction Rexn w/ H+ dissolution solubility

2+ - + - A Ni(OH)2 (s) Ni(OH)2 (s)  Ni (aq) + 2OH (aq) H + OH  H2O  

2+ 2- + 2- B CaCO3 CaCO3 Ca (aq) + CO3 2H +CO3    H2O + CO2

2+ 2- + 2- - C BaSO4 BaSO4(s)  Ba (aq) + SO4 (aq) H +SO4 HSO4 Slightly barely favored

+ - d AgCl AgCl Ag (aq) + Cl (aq) No reaction Not affected same

4. A solution is 0.10 M in and the pH is adjusted to 8.0. Would Mg(OH)2 precipitate? (No)

[Ans]: Calculate Qsp of Mg(NO3)2

2+ -11 Mg(OH)2(s) Mg + 2OH Ksp = 1.5x10 Qsp = [Mg2+][OH-]2 = (0.10)(1.0x10-6)2 = 1.0x 10-13 Qsp < Ksp, no ppt

- 5. A solution is 0.10M in Mg(NO3)2. What concentration of OH is required to just start precipitation of

0f44a222470a951b7682d36e3b1558bf.doc p. 2 4/4/2018 + Mg(OH)2? If NH3-NH4 buffer is used to control the pH, and [NH3] = 0.10 M. What concentration

+ -5 of NH4 is required to prevent the precipitation of Mg(OH)2? (1.2 x 10 M, >0.15 M) [Ans]: To find out conditions for ppt, compare the requirements for different situation: The easiest way is to calculate conditions for saturated solution and modify condition for different situation Saturated soln, Unsaturated soln, ppt equilibrium no ppt

2+ - 2 -11 Qsp = [Mg ][OH ] = Ksp = 1.5x10 < Ksp > Ksp [OH-] 1.2x10-5 M < 1.2x10-5 M > 1.2x10-5 M

+ [NH4 ] 0.15M > 0.15 M, need more CA < 0.15M, need less + + (NH4 ] to keep solution CA (NH4 ] to keep acidic solution basic

2+ - 2 - 2 -11 For saturated solution, Qsp = [Mg ] [OH ] (0.10) [OH ] = 1.5x10 1.5x1011 [OH-] = 1.2x105 M 0.10

- If [OH ] is maintained by buffer: you can either use Handerson-Hasselbalch equation or Kb of NH3

Kb of NH3 will be more straight forward: [NH  ][OH  ]  5 + - 4 5 [NH 4 ][1.2x10 ] NH3 + H2O  NH4 + OH Kb= 1.8x10  [NH3 ] 0.10 (1.8x105 )(0.10) [NH  ]   0.15M 4 1.2x105

+ 7. Solid AgNO3 is slowly added to a solution that is 0.0010M in NaCl, NaBr and NaI. Calculate the [Ag ] required to initiate the ppt of each silver salt. Assume the solution volume does not change in the process. (AgCl, requires [Ag+] = 1.8 x 10-7M, AgBr: [Ag+] = 3.3 x 10-10M, AgI: [Ag+] = 1.5 x 10-13M) [Ans]; For each ppt, 1st calculate the [Ag+] required for saturation. All 3 ppts (AgCl, AgBr, AgI) follow the same dissolution process:

+ - AgX(s)  Ag (aq) + X (aq) (X= Cl, Br or I) K In saturated solutions, Q = [Ag+] [X-] = K [Ag+] = sp sp eq eq sp eq [X  ]

K K sp K sp X sp [Ag+] =  Order of ppt eq [X  ] 0.001

AgCl Cl 1.8x10-10 1.8x10-7 M 3rd, requires most amt of Ag+ AgBr Br 3.3x10-13 3.3x10-10 M 2nd, requires 2nd amt of Ag+ AgI I 1.5x10-16 1.5x10-13 M 1st, requires least amt of Ag+

** Since all three ppts have the same ion ratio, the Ksp indicates the relative molar solubility

st AgI has lowest Ksp, therefore the lowest molar solubility, will be the 1 to ppt

0f44a222470a951b7682d36e3b1558bf.doc p. 3 4/4/2018 AgCl has highest Ksp, therefore the highest molar solubility, will be the last to ppt

8. Solid AgNO3 is slowly added to a solution that is 0.0010M in NaCl and Na2CrO4. Which one will ppt + first, AgCl or Ag2CrO4? Calculate the [Ag ] required to initiate the ppt of each silver salt. st + -7 + -5 (AgCl 1 , requires [Ag ] = 1.8 x 10 , Ag2CrO4 requires [Ag ] = 9.5 x 10 ) [Ans]:

AgCl and Ag2CrO4 have different ion ratio, values of Ksp does not indicates the relative molar solubility

+ Ksp Ksp and conc [Ag ]eq= + Ion [Ag ]eq = Order of ppt K K ratio sp  sp [X  ] 0.001

-10 + - K 10 AgCl 1:1=1 1.8x10 Ksp= [Ag ]eq [Cl ]eq sp 1.8x10 7 1st, requires   1.8x10 [Cl ] 0.001 less Ag+

-12 + 2 2- K 9.0x1012 Ag2CrO4 2:1 = 2 9.0x10 Ksp=[Ag ] eq [CrO4 ]eq sp   9.5x105 M 2nd, requires [CrO2 ] 0.001 4 more Ag+

9. Refer to problem 7, what is [Cl-] and [I-] when AgBr just starts to precipitate? ([Cl-] = 0.001, [I-] = 4.5 x 10-7 when AgBr starts to ppt. [Ans]: When AgBr starts to ppt, [Ag+ ] = 3.3x10-10 M. AgCl has not ppt’d yet. All Cl- remains in solution. [Cl-] = 0.001M AgI ppt’d before AgBr starts to ppt. Once pptn starts, the Qsp of the compound that has ppt’d always equals to its Ksp. K 1.5x1016 I   sp,AgI   4.5x107 M [Ag  ] 3.3x1010

+ 10. A 0.010 M solution of AgNO3 is made 0.50 M in NH3 and Ag(NH3)2 complex forms. + + 7 Ag (aq) + 2 NH3(aq)  Ag(NH3)2 (aq) Kf = 1.7 x 10 . a. What is the equilibrium concentration of Ag+ in the solution? (2.56 x 10-9 M)

+ -5 b. What % of the total silver is in the form of Ag (aq)? (2.56 x 10 %) [Ans]: A, use RICE + + 7 R Ag (aq) + 2 NH3(aq)  Ag(NH3)2 (aq) Kf = 1.7 x 10 . I 0.010 0.50 0 C -x -2x +x E (0.01-x) 0.5-2x x

 [Ag(NH 3 ) 2 ] x K f   2  2 [Ag ][NH 3 ] (0.01 x)(0.5  2x)

You can either solve the math equation or make some assumption to simplify the solution.

0f44a222470a951b7682d36e3b1558bf.doc p. 4 4/4/2018 Since K is very large, we can assume reaction is near 100%. All limiting reagent Ag+ will be used up. Let x = 0.01 and reset the RICE: + + 7 R Ag (aq) + 2 NH3(aq)  Ag(NH3)2 (aq) Kf = 1.7 x 10 . I 0.010 0.50 0 C -0.01 -0.02 +0.01 E 0 0.48 0.01 New equation to solve is [Ag(NH )  ] 3 2 0.01 7 + -9 K f   2  2  1.7x10 [Ag ] = x= 2.56x10 M [Ag ][NH 3 ] (x)(0.48) 2.56x109  x100  2.56x107% + b. % of the total silver is in the form of Ag (aq) 0.010 Since complex formation reduces % Ag+ to extremely low level, solubility of AgCl will be significantly increased

Problem Set 3b. Thermodynamics

Q1. 1.000g C2H5OH(ℓ) burned in bomb calorimeter, heat capacity = 2.71kJ/C

[Ans]: C2H5OH(ℓ) + 3 O2(g)  2CO2(g) + 3H2O(ℓ) n = 2-3 = -1

o o qH2O = (SH)H2O(m)H2O(ΔT)H2O = (4.184kJ/kg C.)(3.000kg)(1.941 C)= 24.36 kJ

o o qcal = Ccal(ΔT)cal = Ccal(ΔT) H2O = (2.71 kJ/ C)(1.941 C) = 5.26 kJ

qrxn= qethanol = nethanol (ΔE) ethanol = (1.000g/46.07) (ΔE) ethanol

qtotal = qH2O + qcal + qrxn = 24.36 kJ + 5.26 kJ + qrxn

qrxn = -29.62 kJ = (0.02171 mol) (ΔE) ethanol (ΔE) ethanol = -1364.3 kJ/mol H = E + nRT = -1364.3 kJ/mol + (-1)(8.314E-3)(298.15K)= -1364.3 -2.48 = -1366.8 kJ/mol)

Q2. A 50.0 mL sample of 0.400M copper (II) sulfate solution at 23.35 oC is mixed with 50.0 mL sample of 0.600M NaOH solution, also at 23.35 oC in a coffee cup calorimeter (heat capacity = 24.0J/oC). After the reaction, the temperature of the resulting mixture is measured to be 26.65 oC. The density of the final solution is 1.02 g/mL. Calculate the amount of heat evolved and H. (1.49 x 103J, -99.2 kJ/mol rxn) Write the thermochemical equation for the reaction)

[Ans]: CuSO4(aq) + 2NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq) H= -99.2 kJ/mol rxn)

o o qH2O = (4.184 J/g C)(100 mL)(1.02 g/mL)(26.65 – 23.35 C) = 1408 J

o o qCal = (24.0 J/ C)(26.65 – 23.35 C) = 79.2 J

(Total heat evolved = qH2O + qCal = 1487 J)

qtotal= qH2O+ qCal+ qrexn = 0

0f44a222470a951b7682d36e3b1558bf.doc p. 5 4/4/2018 qrexn = -1487 J = nH To find n, you have to find moles of each reagent and find the limiting reagent. 1molrexn mol of CuSO4 = (0.400M)(0.050L) = 0.020 mol x  0.0200molrexn 1molCuSO4 1molrexn mol of NaOH = (0.600M)(0.0500L) = 0.030 mol x  0.0150molrexn 2molNaOH So NaOH is limiting reagent, and n = 0.0150 mole reaction

qrexn = -1487 J = nH = (0.0150 mole reaction) H H = -1487 J/(0.0150 mole reaction) = - 99133 J/mole rexn = -99.13 kJ/mole rexn

Complete thermochemical equation includes balanced chemical equation and H

CuSO4(aq) + 2NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq) H = -99.13 kJ/mol rxn

Hess’s law

(3). Find H for the rxn: 2HCl(aq) + F2(g)  2HF(aq) + Cl2(g) (-988.8 kJ/mol rxn) Given the following enthalpies of reaction:

4HCl(aq) + O2(g)  2H2O(ℓ) + 2 Cl2(g) H= -148.4 kJ/mol rxn)

HF(ℓ)  ½ H2(g) + ½ F2(g) H= +600.0 kJ/mol rxn

H2(g) + ½ O2(g)  H2O(ℓ) H= -285.4 kJ/mol rxn

Q4. The thermite reaction, used for welding iron, is as following:

8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)

A. To calculate H, the best method is using Hf:

8 Al(s) + 3Fe3O4(s)  9 Fe(s) + 4Al2O3(s)

H = 9Hf, Fe(s) + 4 Hf,Al2O3(s) – 3 Hf,Fe3O4(s) –8 Hf, Al(s) = 4(-1676) – 3(-1118) = -3350 kJ/mol rxn

B. To find the energy released, given both reactants, it’s a limiting reagent type of problem. The approach is to calculate the energy released by the reaction of each reactant. The smaller value is the answer. a. H based on 8.0 g Al(s): 1molAl 1molrexn  3350kJ 8.0gAlx x x  124kJ 27gAl 8molAl 1molrexn

b. H based on 20.0 g Fe3O4

1molFe3O4 1molrexn  3350kJ 20.0gFe3O4 x x x  96.3kJ 232gFe3O4 3molFe3O4 1molrexn

The answer is therefore 96.3 kJ released. And Fe3O4(s) is the limiting reagent.

c. Grams of Fe produced is based on the limiting reagent Fe3O4(s)

0f44a222470a951b7682d36e3b1558bf.doc p. 6 4/4/2018 1molFe3O4 9molFe 55.8gFe 20.0g Fe3O4 x x x  14.4gFe 232gFe3O4 3molFe3O4 1molFe Q5. Use the standard enthalpy of formation and bond energy to calculate H for burning of 1mole of

ethyl alcohol, C2H5OH(ℓ). (- 1367, -1022 kJ/mol rxn )

[Ans]: C2H5OH(ℓ) + 3 O2(g)  2CO2(g) + 3H2O(ℓ)

H° = Hf°,RHS - Hf°,LHS = 2Hf°,CO2(g) + 3Hf°,H2O(g) – (Hf°,C2H5OH(l) + 3Hf°,O2(g)) = 2 (-393.5) + 3(-285.8) – (-277.7 + 3x0) = -1644.4 – (-277.7)= -1366.7

To calculate with bond energy, you need to write the Lewis structure

H-C-C-O-H + 3 O=O  2 O=C=O + 3 H-O-H ( + 4 missing C-H bonds from C2H5OH)

H° = BE of LHS - BE of RHS = 5 C-H + C-C + C-O + O-H + 3 O=O –(4 C=O + 6 O-H) = 5 (414)+ 347 + 351 + 464 + 3x498 –(4x741 + 6x464) = 4726 – 5748 = -1022 kJ/mol rxn

Q7. Estimate G° and determine if the following reaction is spontaneous at 25, 1000 and 2000°C?

N2(g) + O2(g) 2NO(g) [Ans]: G° = H° - TS°

H° = Hf°,RHS - Hf°,LHS = 2Hf°,NO(g) – (Hf°N2(g) + Hf°,O2(g)) = 2(90.25) – (0+0) = 180.5 kJ/mol rxn

S° = S°,RHS - S°,LHS = 2S°,NO(g) – (Hf°N2(g) + Hf°,O2(g)) = 2(210.7) – (191.5 + 205) = 24.9 J/K.mol rxn = 24.9x10-3 kJ/K.mol rxn G° = H° - TS° At 25°C (298.15 K) G° = 180.5 kJ/mol rexn –(298.15)( 24.9x10-3 kJ/K.mol rxn) = 173.08 kJ/mol rxn (non-spontaneous) At 1000°C (1273 K) G° = 180.5 kJ/mol rexn –(1273)( 24.9x10-3 kJ/K.mol rxn) = 148.80 kJ/mol rxn (non-spontaneous) At 2000°C (2273 K) G° = 180.5 kJ/mol rexn –(2273)( 24.9x10-3 kJ/K.mol rxn) = 123.90 kJ/mol rxn (non-spontaneous)

At 25°C (298.15 K) G° can be calculated by GHf°

G° = Gf°,RHS - Gf°,LHS = 2Gf°,NO(g) – (Gf°N2(g) + Gf°,O2(g)) = 2(86.57) –(0+0) = 173.14 kJ/mol rxn, almost identical to the one calculated by G° = H° - TS°

H 180.5kJ / mol.rxn Transition temperature, Ttransition =   7249K  6976C S 24.9x103 kJ / K.mol.rxn

CHEM 1412 Solubility and Precipitation Reaction Dr. Ya-Ping Huang

1. Mg(OH)2 2. Ag3PO4 3. Fe4[Fe(CN)6]3 2+ - + 3- 3+ a. reaction for Mg(OH)2(s) Mg + 2OH Ag3PO4(s)  3Ag + PO4 Fe4[Fe(CN)6]3 4Fe dissolving + 3 [Fe(CN) ] 4- (dissolution) 6 b. Algebraic Ksp = S(2S)2=4S3= 1.5x10-11 10 Ksp = (3S)3S=27S4 = 11 Ksp = (4S)4(3S)3 = 6912

0f44a222470a951b7682d36e3b1558bf.doc p. 7 4/4/2018 expression of S= molar solubility 1.3x10-20 S7 Ksp and (S) = 3.0 E-41 c. Molar S= S= S= solubility 11 20 41 Ksp 1.5x10 4 Ksp 1.3x10 6 Ksp 3.0x10 3  3  1.55x10 M 4  4  4.68x10 M7  7  4.59x10 7 M 4 4 27 27 6912 6912 d. Molar mass 58.33 418.58 859.24 e. Solubility, 1.55x10-4M)(58.33g/mol)= 9.04 1.96 E-3 3.95 E-4 g/L x10-3 g/L f. g in 2.50 L M.V.MM = 0.0049 g 9.9 x10-4g sat’d solution (1.55x10-4M)(2.5L)(58.33g/mol)= 0.0226g g. given 0.10 M HCl in 0.1 M AgNO3 skip solution Possible WB + SA, 100 % neutralization Common-ion (Ag+) effect on reaction or + 2+ solubility of Ag PO Mg(OH)2 + 2H Mg + 2H2O 3 4(s) effect Molar ½ (.10MH+)=.05M (solubility Ksp = (3S+0.1)3 S skip solubility determined by the neutralization) = 1x 10-3 S S= 1.3 E- 17 M

Q1g, WB-SA neutralization between HCl and Mg(OH)2 would be 100%. So the moles of Mg(OH)2 neutralized by HCl is the molar solubility. Since HCl is the limiting reagent in the

neutralization, moles of Mg(OH)2 dissolved is ½ of mole of HCl used.

Given the combination of solutions,: (25.0 mL 0.010M of 1st solution and 50.0 mL 0.020 M 2nd solution) After dilution, [Solution1] = 0.010M(25.0mL)/75.0mL = 0.0033M, [Solution2] = 0.020M(50.0mL)/75.0mL = 0.013M Solution combination 4. Cu(NO3)2 + NaOH 5. CaCl2 + Na3PO4 6. AgNO3 + K2CrO4

reaction Cu(NO3)2 + 2 NaOH  3CaCl2 +2 Na3PO4  2AgNO3 + K2CrO4 

Cu(OH)2 + 2 NaNO3 Ca3(PO4)2(s) + 3 NaCl Ag2CrO4(s) +2 KNO3 a. formula of ppt Cu(OH)2 Ca3(PO4)2 Ag2CrO4 b. Ksp of the ppt 1.6x10-19 1.0x10-25 9.0x10-12 c. Qsp in solution (0.0033)(0.013)2 =5.57x10-7 (0.0033)3(0.013)2 (0.0033)2(0.0133) = 6.07x10-12 =1.41x10-7 d. Will ppt form? Yes yes yes

0f44a222470a951b7682d36e3b1558bf.doc p. 8 4/4/2018 Thermodynamics Worksheet Dr. Ya-Ping Huang

Reactions 1. Decomposition of a 2. Habor process 3. Ionization of water fertilizer

+ - 2NH4NO3(s)  N2(g) + 3H2(g)  2NH3(g) H2O(ℓ)  H (aq) + OH (aq) 2N2(g) + 4H2O(g) + O2(g) a H kJ/ mol rxn -236 -92.22 /mol rxn 55.8 kJ/mol rxn

b qp for 500 g (kJ) NH4NO3(s) -737.5 NH3(g) -1353 kJ H2O(ℓ): 1550 kJ c n 7 -2 0 d E kJ/ mol rxn -253.35 -87.26 kJ/mol rxn 55.8kJ/mol rxn e S kJ/mol.K 1.0406 -0.1987 kJ/mol.K -0.0806 kJ/mol.K f Transition Temp Does not exist = 464.1 K = 190.9 C Does not exist

25 C g G kJ/mol rxn -546.1 **a. -40.0 (by H & S) 79.82 kJ

b. -32.38 (Gf)

95 7 -14 h Keq 4.53 x 10 a.1.0 x 10 1.04 x 10 b. 4.7 x 105 i Spontaneous? yes yes pH value for pure water: 6.99 300 C 90 C j G kJ/mol rxn -832.47 21.64 kJ/mol rxn 85.06 kJ

75 -13 k Keq 7.08 x 10 0.0107 5.78 x 10 l Spontaneous? yes No pH value for pure water; 6.12

** There is discrepancy in G calculated by Gf and by G = H -T S for this reaction (2) at 25 C

(1). 2NH4NO3(s)  2N2(g) + 4H2O(g) + O2(g) always spontaneous

H = 2HfN2(g) + 4HfH2O(g) + HfO2(g) - 2HfNH4NO3(s) = 2(0) +4(-241.8) + 0 –2(-365.6) = -236 kJ/mol rxn n = [500g/(80)]/2 = 3.125 mole reaction

0f44a222470a951b7682d36e3b1558bf.doc p. 9 4/4/2018 q = 3.125 mol rxn x (-236 kJ/mol rxn) = -737.5 kJ H = E +nRT, E = H -nRT = -236 kJ – (7)(8.314E-3)(298.15) = -737.5 kJ –17.35 = -253.35 kJ

Ttran = H/S = -236 kJ/1.0406kJ/mol.K = -227 K, does not existent,(Transition temperature does not exist if T< 0 K.) (In this reaction, reaction is always spontaneous, favored by lower enthalpy, H < 0 and increased entropy, S> 0) 25 C :G = H -T S = -236kJ – 298(1.0406) = -546.1 kJ/mol rxn G = -5.709logK logK = G /( -5.709) = -546.1/-5.709 = +95.66, K = 10exp(+95.66) = 4.53 x 1095 300 C :G = H -T S = -236 – (300+273)(1.0406) = -832.47 kJ G = -2.303RTlogK logK = G /-2.303(8.314E-3)(300+273.15)= -832.47/-10.975 = +75.85, K = 10exp(+75.85) = 7.08 x 1075

+ - (3) H2O(ℓ)  H (aq) + OH (aq) always nonspontaneous

+ - H = HfH (aq) + HfOH (aq) - HfH2O(l) = 0 + (-230.0) –(-285.8) = 55.8kJ/mol rxn q = [500g/(18)]mol x 55.8 = 1550 kJ 90 C: G = H -T S =55.8kJ/mol - 363(-0.0806) = 85.06 kJ logK = G /-2.303(8.314E-3)(90+273.15)= 85.06/-2.303(8.314E-3)(363) = -12.24, K = 10exp(-12.24) = 5.78 x 10-13

K = [H+][OH-] = [H+]2, [H+] = 5.78x1013 = 7.5 x 10-7, pH = 6.12

CHEM 1412 Sample Test 3

1. Calculate the molar solubility for each compound and compare

Ion ratio Dissolution reaction Ksp S (molar solubi;ity) a. AgCl 1/1 =1 AgCl(s) Ag+ + Cl- 1.8E-10 = S2 1.34E-5 M

2+ 2- 2 b. BaCO3 1/1 = 1 BaCO3 (s) Ba + CO3 8.1E-9 = S 9.0E-5 M

2+ - 3 c. Cd(OH)2(s) 2/1 = 2 Cd(OH)2(s) Cd +2OH 1.2E-14 = 4S 1.44E-5 Ans: b > c > a

2. a. qv = nE -27.03 kJ = (1.000/44.05) E E = (-27.03)(44.05) = - 1190.8 kJ/mol rxn

b. C2H4O (g) + 5/2 O2(g)  2CO2(g) + 2H2O (ℓ) n = 2 – 3.5 = -1.5 (Since the question asked about 1mol of ethanal, the coefficient for ethanal is fixed at one in balanced equation, instead of using equation of minimal integral coefficients) c. H = E + nRT = - 1190.8 kJ/mol + (-1.5)(8.314E-3kJ/k.mol)(298.15K) = -1194.5 kJ/mol rxn

0f44a222470a951b7682d36e3b1558bf.doc p. 10 4/4/2018 o o o o d. H = 2 Hf , CO2(g) + 2 Hf , H2O(l) – (Hf ,C2H4O(g) + 2.5Hf , O2(g) )

o -1194.5 = 2 (-393.5) + 2(-285.8 ) -(Hf ,C2H4O(g) + 2.5 x 0 )

o Hf ,C2H4O(g) = +2 (-393.5) + 2(-285.8 ) –(-1194.5) = -164.1 kJ/mol

2+ - 3. Mg(OH)2(s)  Mg +2OH At pH 3, high conc of H+ will neutralize OH-, make rxn , increase the solubility At pH 10, high conc of OH- will increase OH-, make rxn  (CI effect), decrease the solubility

4. 2NO(g) + O2(g)  2NO2(g) At 25C, G < 0 (see following calculation), reaction is spontaneous

o o o G = 2 Gf , NO2(g) - (2 Gf , NO(g) + Gf , O2(g) ) = 2 (51.30) – 2(86.57) = -70.54 kJ/mol rxn G = -5.709 logK, -70.54 = -5.709 LogK logK = -70.54/-5.709 = 12.36, K = 1012.36 = 2.29E12

o o o H = 2 Hf , NO2(g) - (2 Hf , NO(g) + Hf , O2(g) ) = 2 (33.2) – 2(90.25) = -114.1 kJ/mol rxn

o o o S = 2 S , NO2(g) - (2 S , NO(g) + S , O2(g) ) = 2 (240.0) – [2(210.7)+ 205.0] = -146.4 J/K.mol rxn At 600C, G > 0 (see following calculation), reaction is non-spontaneous G = H - TS = -114.1 kJ/mol rxn – (600+273)(-0.1464 kJ/K.mol rxn) = 13.71 kJ/mol rxn G = -2.303RTlogK 13.71 = -2.303(8.314E-3)(873)logK logK = -0.82 K = 10-0.82 = 0.15

5. a. melting of ice : solid to liquid, entropy  b. dissolving of NaCl in water: solid NaCl dissolved in liquid, entropy 

c. formation of H2O(ℓ) from H2(g) and O2(g): gas to liquid, entropy 

o o 6. CH3OH(ℓ)  CH3OH(g)  H = +37.4kJ and S = 111J/K b.p. =  Ho/So = 37.4 kJ/0.111kJ/K = 336.9K = 63.8 C

To calculate Keq at 25C, use G = -5.709log K G = H - TS = 37.4 kJ – (298K)(0.111kJ/K) = 4.32 kJ G = -5.709log K 4.32 = -5.709logK logK = -0.757 K = 0.175 Based on G> 0, K < 1, evaporation is non-spontaneous. However, non-spontaneous Is not the same as no reaction. It just means reaction is less likely to go forward than reverse

7. [Ans]: C6H12O6(s) + 6O2(g)  6CO2(g) + 6H2O(ℓ) n = 6-6 = 0

o o qH2O = (SH)H2O(m)H2O(ΔT)H2O = (4.184kJ/kg C.)(0.950kg)(3.15 C)= 12.52 kJ

o o qcal = Ccal(ΔT)cal = Ccal(ΔT) H2O = (2.25 kJ/ C)(3.15 C) = 7.09 kJ

0f44a222470a951b7682d36e3b1558bf.doc p. 11 4/4/2018 qrxn= qglucose = nglucose (ΔE) glucose = (1.25g/180.16) (ΔE) glucose

qtotal = qH2O + qcal + qrxn = 12.52 kJ + 7.09 kJ + qrxn

qrxn = -19.61 kJ = (0.00694 mol) (ΔE) ethanol (ΔE) ethanol = -2826 kJ/mol

Heat released is 19.61 kJ ( a positive term, since heat released implies exothermic)

H = E + nRT = -2826 kJ/mol + (0)(8.314E-3)(298.15K)= -2826 kJ/mol) When n = 0, H = E

8. Given a. 2NF3(g) + 2NO(g)  N2F4(g) + 2ONF(g) H = -82.9kJ b. NO(g) + 1/2 F2(g)  ONF(g) H = -156.9kJ c. Cu(s) + F2(g)  CuF2(s) H = -530.1kJ

Target rxn: 2NF3(g) + Cu(s)  N2F4(g) + CuF2(s), start from left to right

2NF3(g) need rxn (a): 2NF3(g) + 2NO(g)  N2F4(g) + 2ONF(g) Ha = -82.9kJ

Cu(s) need rxn (c): Cu(s) + F2(g)  CuF2(s) H = -530.1kJ

N2F4(g)needs rxn (a), but already used, skip to next

CuF2(s)needs rxn (c), but already used, stop to see what do we have now Add rxns (a) and (c),

2NF3(g) + 2NO(g) + Cu(s) + F2(g)  N2F4(g) + 2ONF(g) + CuF2(s) H = -82.9+ (-530.1) = -613 kJ

Compared with target rxn, there are extra terms (2NO, F2 and 2ONF) need to be eliminated All the extra terms exist in rxn (b) If subtract rxn (b)x2, we can get rid of all the terms:

2NF3(g) + 2NO(g) + Cu(s) + F2(g)  N2F4(g) + 2ONF(g) + CuF2(s) H = -613kJ

-2(b) 2NO(g) + F2 (g)  2ONF(g)  H = (-156.9kJ)x2

Final 2NF3(g) + Cu(s)  N2F4(g) + CuF2(s) H = (-613) – (-156.9)x2 = -299.2 kJ

9. CH4 + 4 Cl2  CCl4 + 4 HCl H Cl H C H + 4 Cl-Cl Cl C Cl + 4 H-Cl H Cl H = 4 C-H + 4 Cl-Cl – ( 4 C-Cl + 4 H-Cl) = 4 x 414 + 4 x 243 – (4 x 330 + 4 x 431) = -416 kJ

o o o o 10. H = Hf , CCl4(g) + 4 Hf , HCl(g) - (Hf , CH4(g) + 4 Hf , Cl2(g) ) = (-103) + 4 x (-92.31) – [(-74.81) + 4 x 0] = - 397.43 kJ/mol rxn

o The value obtained by Hf ( -397.43 in this Q10) is more accurate than the one by bond energy

0f44a222470a951b7682d36e3b1558bf.doc p. 12 4/4/2018 (-416 kJ in Q9). Because BE ‘s are given as average values and specifically for gaseous rxns. 11. a. Estimate the enthalpy of neutralization of 1.0 M HCl(aq) and 1.0 M NaOH(aq)

Based on complete molecular equation: HCl (aq) + NaOH(aq)  NaCl(aq) + H2O (ℓ)

o o o o H = Hf , NaCl(aq) + Hf , H2O(ℓ) - (Hf , HCl(aq) + 4 Hf , NaOH(aq) ) = (-407.1) + (-285.8) –[(-167.4) + (-469.6)] = -692.9 + 637 = - 55.9 kJ/mol rxn

+ - Based on net ionic equation: H (aq) + OH (aq)  H2O (ℓ)

o o o H = Hf , H2O(ℓ) - (Hf , H+(aq) + Hf , OH-(aq) ) = (-285.8) – (0 + (-230.0)) = -55.8 kJ/mol rxn

b. heat measured using coffee cup calorimeter(qp) or bomb calorimeter (qv) will be the same

qp = nH qV = nE H = E + nRT

since n = 0, H = E and qp = qV c. the enthalpy of neutralization of 1.0 M HOAc(aq) and 1.0 M NaOH(aq) will be smaller compare to the enthalpy of neutralization of 1.0 M HCl (aq) and 1.0 M NaOH(aq). HOAc is a WA, the neutralization involves the breaking bond (endothermic, energy required) between H and acetate; as a result less energy will be released.

- - HOAc (aq) + OH (aq)  H2O (ℓ) + OAc (aq) 12. EDTA O O CH2 C OH HO C CH2 N-CH2-CH2-N CH2 HO C CH2 C OH O O a. the four acidic H are those bonded to oxygen in the acetate groups b. as shown in the Lewis structure, there are 10 atoms with lone paired electrons ( 2 N and 2 oxygens of each of the 4 acetate group) c. the bond formation between EDTA and metal is an example of Lewis acid-base reaction EDTA with many pairs of lone pair electron acts as Lewis base, donating electron pairs to (sharing with) electron deficient metal (Lewis acid) d. EDTA forms only 6 bonds with metal because the geometry (space limitation) such that only one oxygen from each acetate group can bond to metal

0f44a222470a951b7682d36e3b1558bf.doc p. 13 4/4/2018