CHEM 1412 Problem Set #3a
Dr. Ya-Ping Huang
Solubility Equilibrium
1. Calculate the molar solubility of CaF2 at 25oC in:
a. pure water
(2.1 x 10-4 M)
b. in 0.010M Ca(NO3)2 solution
(3.1 x 10-5 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.
. a.
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

S
Let S represent the molar solubility
2S
Ksp = [Ca2+]eq [F-]eq2 = (s)(2s)2 = 4S3 = 3.9x10-11
S3
K sp
4
3
3.9 x1011 3
 9.75x1012  2.14 x10 4 M
4
b. In 0.010M Ca(NO3)2 solution, there is the common –ion effect. The extra Ca2+ from Ca(NO3)2 will
reduce the solubility of CaF2(s) (Le Chatelier’s principle) –the common-ion effect
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

S
Let S represent the molar solubility
2S
+0.010
[Ca2+]eq = 0.010+ S
[F-]eq = 2S
Ksp = [Ca2+]eq [F-]eq2 = (S+0.010)(2S)2  (0.010)(2S)2 = 0.04S2 = 3.9x10-11
(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)
S
Ksp
3.9 x1011

 3.12 x10 5 M
0.04
0.04
Ksp = [Ca2+]eq [F-]eq2 = (s)(2s)2 = 4S3 = 3.9x10-11
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)
R
CaF2(s)  Ca2+(aq) + 2F-(aq)
E
S

S
Let S represent the molar solubility
2S
+0.010
[Ca2+]eq = S
[F-]eq = 0.010+ 2S
Ksp = [Ca2+]eq [F-]eq2 = (S)(0.01+2S)2  (S)(0.01)2 = 1.0x10-4S = 3.9x10-11
S = Ksp/1.0x10-4 = 3.9x10-7 M
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2. Calculate the molar solubility of BaSO4 at 25oC:
a. in pure water
b. in 0.10M Na2SO4 solution
c. Grams of BaSO4 dissolved in 2.0 Liters of water
e. LD50 of Ba2+ for rat is around 20 mg/Kg body weight of rats. Would the Ba2+ 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)
BaSO4(s)  Ba2+(aq) + SO42-(aq)
Let S represent the molar solubility, then [Ba2+] = S and [SO42-] = S
Ksp = [Ba2+] [SO42-] = (S)( S)= S2 = 1.1x10-10
S= 1.1x1010  1.05x105 M
b. in 0.10M Na2SO4 solution, the additional SO42- ions from Na2SO4 will reduce solubility of BaSO4
through common-ion effect
Ksp = [Ba2+] [SO42-] = (S)(0.10+S)= S(0.10) = 1.1x10-10
S=1.1x10-9 M
d. mg of Ba2+ in 1.5 L of saturated BaSO4 solution.
 1.05x105 mol 
137 g  1000mg 
mg of Ba2+ = (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
A
Ni(OH)2 (s)
Ni(OH)2 (s)  Ni2+(aq) + 2OH-(aq)
H+ + OH-  H2O


B
CaCO3
CaCO3 Ca2+(aq) + CO32-
2H++CO3 2-
H2O + CO2


C
BaSO4
BaSO4(s)  Ba2+(aq) + SO42-(aq)
H++SO42HSO4-
Slightly
favored
barely
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
Mg(OH)2(s) Mg2+ + 2OH
Ksp = 1.5x10-11
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
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Mg(OH)2? If NH3-NH4+ buffer is used to control the pH, and [NH3] = 0.10 M. What concentration
of NH4+ is required to prevent the precipitation of Mg(OH)2?
(1.2 x 10-5M, >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,
equilibrium
Unsaturated soln,
Qsp = [Mg2+][OH-]2
= Ksp = 1.5x10-11
< Ksp
> Ksp
[OH-]
1.2x10-5 M
< 1.2x10-5 M
> 1.2x10-5 M
[NH4+]
0.15M
> 0.15 M, need more CA
(NH4+] to keep solution
acidic
< 0.15M, need less
CA (NH4+] to keep
solution basic
For saturated solution, Qsp = [Mg2+] [OH-]2
ppt
no ppt
(0.10) [OH-]2 = 1.5x10-11
1.5 x10 11
 1.2 x10 5 M
0.10
[OH-] =
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:
NH3 + H2O  NH4+ + OH-


Kb= [ NH 4 ][OH ]
[ NH 3 ]
1.8 x10 5 
[ NH 4 ][1.2 x10 5 ]
0.10
(1.8 x105 )(0.10)
 0.15M
1.2 x105
[ NH 4 ] 
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)
In saturated solutions, Qsp = [Ag+]eq [X-]eq = Ksp
X
Ksp
[Ag+]eq = K sp
[X ]
[Ag+]eq= K sp  K sp

Order of ppt
[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
AgI has lowest Ksp, therefore the lowest molar solubility, will be the 1st to ppt
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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.
(AgCl 1st, requires [Ag+] = 1.8 x 10-7, Ag2CrO4 requires [Ag+] = 9.5 x 10-5)
[Ans]:
AgCl and Ag2CrO4 have different ion ratio, values of Ksp does not indicates the relative molar solubility
AgCl
Ion
ratio
Ksp
1:1=1
1.8x10-
Ksp and conc
[Ag+]eq= K sp  K sp
[X  ]
9.0x1012
Order of ppt
0.001
Ksp= [Ag+]eq [Cl-]eq
10
Ag2CrO4 2:1 = 2
[Ag+]eq =
Ksp=[Ag+]2eq [CrO42]eq
K sp

[Cl  ]
1.8 x10 10
 1.8 x10 7
0.001
K sp
2
4
[CrO ]

9.0 x10 12
 9.5 x10 5 M
0.001
1st, requires
less Ag+
2nd, requires
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.
I   K[ Ag

sp, AgI

]

1.5 x1016
 4.5 x107 M
3.3x1010
10. A 0.010 M solution of AgNO3 is made 0.50 M in NH3 and Ag(NH3)2+ complex forms.
Ag+(aq) + 2 NH3(aq) 
Ag(NH3)2+(aq)
Kf = 1.7 x 107.
a. What is the equilibrium concentration of Ag+ in the solution?
(2.56 x 10-9 M)
b. What % of the total silver is in the form of Ag+ (aq)?
(2.56 x 10-5 %)
[Ans]:
A, use RICE
R
Ag+(aq) + 2 NH3(aq) 
I
0.010
0.50
0
C
-x
-2x
+x
E
(0.01-x) 0.5-2x
Kf 
Ag(NH3)2+(aq)
Kf = 1.7 x 107.
x
[ Ag ( NH 3 ) 2 ]
x


2
[ Ag ][ NH 3 ]
(0.01  x)(0.5  2 x) 2
You can either solve the math equation or make some assumption to simplify the solution.
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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:
R
Ag+(aq) + 2 NH3(aq) 
I
0.010
0.50
0
C
-0.01
-0.02
+0.01
E
0
0.48
0.01
Ag(NH3)2+(aq)
Kf = 1.7 x 107.
New equation to solve is
Kf 
[ Ag ( NH 3 ) 2 ]
0.01

 1.7 x10 7

2
2
[ Ag ][ NH 3 ]
( x)(0.48)
[Ag +] = x= 2.56x10-9M
b. % of the total silver is in the form of Ag+ (aq) 
2.56 x10 9
x100  2.56 x10 7 %
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
qH2O = (SH)H2O(m)H2O(ΔT)H2O = (4.184kJ/kg oC.)(3.000kg)(1.941 oC)= 24.36 kJ
qcal = Ccal(ΔT)cal = Ccal(ΔT) H2O = (2.71 kJ/ oC)(1.941 oC) = 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)
qH2O = (4.184 J/goC)(100 mL)(1.02 g/mL)(26.65 – 23.35oC) = 1408 J
qCal = (24.0 J/oC)(26.65 – 23.35oC) = 79.2 J
(Total heat evolved = qH2O + qCal = 1487 J)
qtotal= qH2O+ qCal+ qrexn = 0
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qrexn = -1487 J = nH
To find n, you have to find moles of each reagent and find the limiting reagent.
mol of CuSO4 = (0.400M)(0.050L) = 0.020 mol x
1molrexn
 0.0200molrexn
1molCuSO4
mol of NaOH = (0.600M)(0.0500L) = 0.030 mol x
1molrexn
 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)
Given the following enthalpies of reaction:
(-988.8 kJ/mol rxn)
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):
8.0 gAlx
b.
1molAl 1molrexn  3350kJ
x
x
 124kJ
27 gAl
8molAl 1molrexn
H based on 20.0 g Fe3O4
1molFe3O 4
1molrexn  3350kJ
20.0 gFe3O 4 x
x
x
 96.3kJ
232 gFe3O 4 3molFe3O 4 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)
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1molFe3O4
9molFe
55.8gFe
x
x
 14.4 gFe
232 gFe3O4 3molFe3O4 1molFe
20.0 g Fe3O4 x
Q5. Use the standard enthalpy of formation and bond energy to calculate H for burning of 1mole of
ethyl alcohol, C2H5OH(ℓ).
[Ans]:
(- 1367, -1022 kJ/mol rxn )
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°
Transition temperature, Ttransition =
H 
180.5kJ / mol.rxn

 7249 K  6976C
S  24.9 x10 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
a. reaction for
dissolving
(dissolution)
Mg(OH)2(s) Mg2+ + 2OH-
Ag3PO4(s)  3Ag+ + PO43-
Fe4[Fe(CN)6]3 4Fe3+
b. Algebraic
Ksp = S(2S)2=4S3= 1.5x10-11
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+ 3 [Fe(CN) 6] 410 Ksp = (3S)3S=27S4 =
p. 7
11 Ksp = (4S)4(3S)3 = 6912
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S= molar solubility
expression of
Ksp and (S)
c. Molar
S7
= 3.0 E-41
1.3x10-20
S=
Ksp 3 1.5 x10 11
3

 1.55 x10 4 M
4
4
S=
d. Molar mass
58.33
418.58
859.24
e. Solubility,
g/L
1.55x10-4M)(58.33g/mol)=
1.96 E-3
3.95 E-4
f. g in 2.50 L
sat’d solution
M.V.MM =
(1.55x10-4M)(2.5L)(58.33g/mol)=
0.0226g
0.0049 g
9.9 x10-4g
g.
given
solution
0.10 M HCl
in 0.1 M AgNO3
skip
Possible
reaction or
effect
WB + SA, 100 % neutralization
Common-ion (Ag+) effect on
solubility of Ag3PO4(s)
solubility
Molar
solubility
9.04
4
S=
Ksp 4 1.3x10

27
27
20
 4.68 x10 6 M7
Ksp 7 3.0 x10 41

 4.59 x10 7 M
6912
6912
x10-3 g/L
Mg(OH)2 +
2H+Mg2++
2H2O
½ (.10MH+)=.05M (solubility
determined by the neutralization)
Ksp = (3S+0.1)3 S
= 1x 10-3 S
S= 1.3 E17 M
skip
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
Cu(NO3)2 + 2 NaOH 
Cu(OH)2 + 2 NaNO3
3CaCl2 +2 Na3PO4 
Ca3(PO4)2(s) + 3 NaCl
2AgNO3 + K2CrO4 
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
= 6.07x10-12
(0.0033)2(0.0133)
=1.41x10-7
d. Will ppt form?
Yes
yes
yes
reaction
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Thermodynamics Worksheet
Reactions
Dr. Ya-Ping Huang
1. Decomposition of a
2. Habor process
3. Ionization of water
N2(g) + 3H2(g)  2NH3(g)
H2O(ℓ)  H+ (aq) + OH-
fertilizer
2NH4NO3(s) 
(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
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
G
-546.1
**a. -40.0 (by H & S)
79.82 kJ
kJ/mol.K
25 C
g
kJ/mol rxn
b. -32.38 (Gf)
h
Keq
4.53 x 1095
a.1.0 x 107
b. 4.7 x 105
1.04 x 10-14
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
k
Keq
7.08 x 1075
0.0107
5.78 x 10-13
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
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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+] =
CHEM 1412
5.78x10 13 = 7.5 x 10-7, pH = 6.12
Sample Test 3
1. Calculate the molar solubility for each compound and compare
Ion ratio
Dissolution reaction
Ksp
S (molar solubi;ity)
S2
a. AgCl
1/1 =1
AgCl(s) Ag+ + Cl-
1.8E-10 =
1.34E-5 M
b. BaCO3
1/1 = 1
BaCO3 (s) Ba2+ + CO32-
8.1E-9 = S2
9.0E-5 M
c. Cd(OH)2(s)
2/1 = 2
Cd(OH)2(s) Cd2+ +2OH-
1.2E-14 = 4S3
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
d. H = 2 Hfo, CO2(g) + 2 Hfo, H2O(l) – (Hfo,C2H4O(g) + 2.5Hfo, O2(g) )
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-1194.5 = 2 (-393.5) + 2(-285.8 ) -(Hfo,C2H4O(g) + 2.5 x 0 )
Hfo,C2H4O(g) = +2 (-393.5) + 2(-285.8 ) –(-1194.5) = -164.1 kJ/mol
3. Mg(OH)2(s)  Mg2+ +2OHAt 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
G = 2 Gfo, NO2(g) - (2 Gfo, NO(g) + Gfo, O2(g) ) = 2 (51.30) – 2(86.57) = -70.54 kJ/mol rxn
G = -5.709 logK, -70.54 = -5.709 LogK
K=
1012.36
logK = -70.54/-5.709 = 12.36,
= 2.29E12
H = 2 Hfo, NO2(g) - (2 Hfo, NO(g) + Hfo, O2(g) ) = 2 (33.2) – 2(90.25) = -114.1 kJ/mol rxn
S = 2 So, NO2(g) - (2 So, NO(g) + So, 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
logK = -0.82
13.71 = -2.303(8.314E-3)(873)logK
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 
6.
CH3OH(ℓ)  CH3OH(g)  Ho = +37.4kJ and So = 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
qH2O = (SH)H2O(m)H2O(ΔT)H2O = (4.184kJ/kg oC.)(0.950kg)(3.15 oC)= 12.52 kJ
qcal = Ccal(ΔT)cal = Ccal(ΔT) H2O = (2.25 kJ/ oC)(3.15 oC) = 7.09 kJ
qrxn= qglucose = nglucose (ΔE) glucose = (1.25g/180.16) (ΔE) glucose
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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)
Cu(s)
need rxn (c): Cu(s) + F2(g)  CuF2(s)
Ha = -82.9kJ
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)
-2(b)
2NO(g) + F2(g)  2ONF(g)
H = -613kJ
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
H C H
H
Cl
+ 4 Cl-Cl
Cl C Cl
+ 4 H-Cl
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
10. H = Hfo, CCl4(g) + 4 Hfo, HCl(g) - (Hfo, CH4(g) + 4 Hfo, Cl2(g) )
= (-103) + 4 x (-92.31) – [(-74.81) + 4 x 0] = - 397.43 kJ/mol rxn
The value obtained by Hfo ( -397.43 in this Q10) is more accurate than the one by bond energy
(-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)
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Based on complete molecular equation: HCl (aq) + NaOH(aq)  NaCl(aq) + H2O (ℓ)
H = Hfo, NaCl(aq) + Hfo, H2O(ℓ) - (Hfo, HCl(aq) + 4 Hfo, 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)
H = Hfo, H2O(ℓ) - (Hfo, H+(aq) + Hfo, 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
H = E + nRT
qV = nE
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
HO C
CH2
HO C
CH2
CH2
C OH
N-CH2-CH2-N
O
CH2
C OH
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
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