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 email/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: 8
-­‐34
Speed of light : 2.9979 X 10 m/s Planck's constant : 6.626 X 10 Js Nav: 6.022X1023 F=96485 C/mol e 931.5 MeV/u R= 0.0821 L atm/mol K = 0.083145 bar L/mol K =8.3145 J/mol K = 62.364 L torr/mol K Data: Standard enthalpies of formation (ΔHfo, kJ/mol): SF4(g): -­‐763, CF4 (g): -­‐925, SO2(g): -­‐296.8 CO2(g): -­‐393.5 Absolute entropies at 298 K (So, J/mol K) : SF4(g): 299.6, CF4 (g): 261.6, SO2(g): 248.1 CO2(g): 213.6 O2 (g) + 4H+ (aq) + 4e-­‐ -­‐> 2H2O (l) Eo=+1.229 -­‐
+
O2 (g) + 2H (aq) + 2e -­‐> H2O2 (l) Eo=+0.695 I2(s) + 2 e-­‐ -­‐> 2 I-­‐ (aq) Eo= +0.535 Spectrochemical series: CN-­‐>NO2-­‐>en>py~NH3>EDTA4-­‐>SCN-­‐>H2O>ONO-­‐>ox2-­‐>OH-­‐>F-­‐>SCN-­‐>Cl-­‐>Br-­‐>I-­‐ Equations: ΔTf=-­‐KfXm ΔTb=KbXm Kp=Kc(RT)Δn ΔGo=-­‐RTlnK ΔG= ΔGo +RTlnQ 1st order: t1/2= ln2/k ln(A/Ao)=-­‐kt 2nd order: 1/A=1/Ao+kt k=Aexp(-­‐Ea/RT) pH=pKa + log [base]/[acid] Eocell = (0.025693 V/z ) ln K = (0.0592V/z) log K Ecell = Eocell – (0.0592V/z) log Q ΔGo=-­‐zFEo Score: 1._____ 2._____ 3._____ 4._____ 5._____ 6._____ 7._____ 8._____ 9._____ 10._____
11.____ 12._____ 13._____
Total: _____
1. (8) Arrange the following substances in the expected order of increasing boiling point:
H2O, NH3, CH4, CH3CH3. Explain your rationale.
CH4 < CH3CH3 < NH3 < H2O
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 (C10H8) would you add to 50.0 g of benzene
(C6H6) to produce a solution with the same freezing point as pure water? (Tf of pure
benzene is 5.53 oC and Kf = 5.12 oC/m)
ΔTf=0.00-5.53= -5.53 oC
ΔTf = -KfXm => m= 5.53/5.12 = 1.08 m
Moles of C10H8: 1.08 moles/1000 g benzene X 50.0 g benzene= 0.054
Molar mass of C10H8: 10*12.01+8*1.008= 128.2 g/mol
g of C10H8: 0.054 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 HgCl2 + C2O42- à 2 Cl + 2 CO2 (g) + Hg2Cl2 (s)
Exper.
[HgCl2], M
[C2O42-], M
1
0.105
0.15
2
0.105
0.30
3
0.052
0.30
Initial rate, mol/(L min)
1.8 X 10-5
7.1 X 10-5
3.5 X 10-5
From 1 & 2: 7.1/1.8=3.94 and 0.30/0.15=2 => order with respect to C2O42- is 2.
From 2 & 3: 7.1/3.5=2.03 and 0.105/0.052=2.02 => order with respect to HgCl2 is 1.
1.8 X 10-5 =k X 0.105 X 0.152 => k= 7.6 X 10-3 M-2 min-1
4. (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
5. (8) 5.25 g of I2 and 2.15 g of Br2 are mixed in a 3.15-L flask at 115 oC. At equilibrium,
1.98 g of I2 is present. What is Kc at 115 oC for the reaction
I2 (g) + Br2 (g) ⇌ 2 IBr (g)
Initial concentrations:
[I2]o = 5.25 g/3.15 L X 1mol/253.8g = 6.57 X 10-3 M
[Br2]o = 2.15 g/3.15 L X 1mol/159.8g = 4.27 X 10-3 M
I.
C.
E.
I2 (g)
+
Br2 (g)
⇌ 2 IBr (g)
6.57 X 10-3 M 4.27 X 10-3 M
0
-x
-x
+2x
6.57 X 10-3-x 4.27 X 10-3-x
2x
Equilibrium concentrations:
[I2]eq = 1.98 g/3.15 L X 1mol/253.8g = 2.48 X 10-3 M = 6.57 X 10-3-x =>x=4.09X10-3
[Br2]eq = 4.27 X 10-3-4.09X10-3=0.18X10-3
[IBr]eq = 2X 4.09X10-3= 8.18X10-3
Kc= [IBr]2/[I2] [Br2]= (8.18X10-3)2 / (2.48 X 10-3)( 0.18X10-3)= 1.5X102
6. (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 = 0.715 M
I.
C.
E.
HCOOH (aq) + H2O ⇌ HCOO- (aq) + H3O+ (aq)
0.715
0
0
-x
x
x
0.715-x
x
x
Ka=[H3O+][ HCOO -]/[ HCOOH] = x2/(0.715-x) ≈ x2/0.715 =>
x=[H3O+]=1.1X10-2 M
pH=-log[1.1X10-2] = 1.95
7. (7) What is the pH of a solution that contains 0.405 M HCOOH and 0.326 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
8. (8) A 10.00-mL portion of a 0.50 M AgNO3 (aq) solution is added to 100.0 mL of a
solution that is 0.010 M in Cl a) Will AgCl (s) (Ksp = 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: 110.0 mL
[Ag+] = 10.00mLX0.50M / 110.0 mL = 4.5 X10-2 M
[Cl-]= 100.0mLX0.010M / 110.0 mL = 9.1 X 10-3 M
Qsp=[Ag+] [Cl-]= 4.1X10-4 > Ksp , so precipitation will occur
-
Ag+ (aq) + Cl (aq)
à AgCl (s)
-2
-3
I. 4.5 X10 M 9.1 X 10 M
C. –x
-x
+x
E. 4.5 X10-2-x 9.1 X 10-3-x
x
Ksp= (4.5 X10-2-x)(9.1 X 10-3-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 X10-2-9.1X10-3=3.6X10-2
[Cl ]= Ksp/ [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 KClO3 (s) à 2 KCl (s) + 3 O2 (g)
S increases (gas is produced from solid)
b) CH3COOH (l) à CH3COOH (s)
S decreases (liquid becomes solid)
c) N2 (g) + O2 (g) à 2 NO (g)
Cannot tell, 2 gas molecules on both sides
10. (8) Use thermodynamic data given to determine Keq for the following reaction at 45
o
C. CO2 (g) + SF4 (g) ⇌ CF4 (g) + SO2 (g)
ΔHo = -925 -296.8 -(-393.5)-(-763) = -65.3 kJ
ΔSo = 261.6 + 248.1-213.6-299.6 = -3.5 J/K
Assuming these are independent of temperature:
ΔGo(318K) = ΔHo -318 X ΔSo = -64.2 kJ
ΔGo = -RTlnKeq => Keq= 3.5X1010
11. (8) Using data given, determine the values of Eocell and ΔGo for the following
reaction: O2(g) + 4 I (aq) + 4H+ (aq) à 2 H2O (l) + 2 I2 (s)
O2 (g) + 4H+ (aq) + 4e-­‐ -­‐> 2H2O (l) I2(s) + 2 e-­‐ -­‐> 2 I-­‐ (aq) Eo=+1.229 Eo= +0.535 Flip the second, multiply by 2 and add to the first to get the given equation.
Eocell = 1.229-0.535 = 0.694 V
ΔGo=-­‐zFEo =-­‐4X96485X0.694 V = -268 kJ
12. (8) Draw orbital diagrams showing the possible distributions of d electrons in the
central metal of the ion [PtCl6]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?
Pt4+: 5d6
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: 224Ra: 223.9719 u; 220Rn: 219.9642 ; 4He:
4.00150 u.
a) 22488Ra -­‐> 22086Rn + 42He
b) Δm= 219.9642+4.00150-223.9719=-0062u X 931.5 MeV/u = -5.8 MeV