Exam 1 Spots
CH 142
Practice problems from Chapt. 12: (recommended additional problems, do not turn in)
29, 31, 33*, 51, 53, 55, 57, 59, 61*, 63*, 65, 67, 69**, 71, 73, 75**, 77**, 79**, 81**, 83**, 85, 85, 89, 99, 105, 111, 117**
(**highly recommended)
Practice problems from Chapt. 13: (recommended additional problems, do not turn in)
25, 27, 33, 35, 39**, 41**, 43, 45, 47, 49, 55**, 59, 63, 67, 69**, 71**, 73, 77, 85**, 95**, 97, 105 (**highly recommended)
1. In each group of substances pick the one that has the property listed:
(a) isotonic with 0.1 M NaCl:
0.1 M BaCl2 (aq) 0.2 M C6H12O6 (aq)
(b) lowest boiling point:
.1 M KCl (aq)
(c) least volatile:
0.1 M Cs3(PO4)2 (aq)
.1 M Pb(NO3)2 (aq)
0.05 M Na2SO4 (aq)
.1 M NH4F (aq)
0.1M LiBr(aq)
0.1M C6H12O6 (aq)
(d) the most polar bond:
NH3
HI
PH3
(e) lowest vapor pressure:
H2O
CH3OH
Ba(OH)2
(f) highest melting point:
01.M NaBr (aq),
NH4Cl
Na2O
0.1m LiOH (aq)
0.1m CsCl (aq)
(g) highest boiling point:
0.1 m CuCl2 (aq)
(h) highest osmotic pressure:
0.1M CaI2 (aq)
0.1 M NaNO3 (aq)
(i) Lowest boiling point:
H2O
0.5 M CsOH (aq)
0.5 M glucose (C6H12O6) (aq)
0.1 M CaF2 (aq)
0.1 M C6H12O6 (aq)
0.1 m CH3CH2OH (aq)
0.1m CH3OH (aq)
0.1m KOH (aq)
H2O
CH3(CH2)3CH3
CH2Cl2
HOCH2CH2OH
CH3OH
CsOH
(j) Largest osmotic pressure: 0.1 M NaCl (aq)
H2O (l)
(k) Least volatile
(l) Highest Boiling point:
(m) Highest Vapor pressure
1.
2
For each of the sets, circle the one with the property listed. Notice the presence of pure compounds and
solutions. (25 pt)
(a) Lowest freezing pt:
H2O
0.1 M KCl (aq)
CH4
0.1 M NH4Br (aq)
0.1 M HOCH2CH2OH (aq)
0.2 M LiOH (aq)
0.1 M NH4Cl (aq)
0.1 M LiOH (aq)
H2O
0.5 M NaOH (aq)
0.5 M HOCH2CH2OH (aq)
0.3 M Pb(NO3)2 (aq)
0.3 M Na3PO4 (aq)
0.3 M NaI (aq)
(b) Largest osmotic pressure: 0.1 M BaBr2 (aq)
(c) Least volatile
(d) Highest boiling point:
(e) Highest Vapor pressure
2. Iodide ion can be used to decompose aqueous hydrogren peroxide in acidic aqueous solution.
2 H+ (aq)
+ 2 I- (aq)
+
H2O2 (aq) 
I2 (aq) + 2 H2O
A proposed rate law for this reaction is as follows:
rate = k [I-]2[H2O2]
Based on the proposed rate law, indicate what happens to the specific quantity (increase, decrease, no change)
the rate constant
the rate
(a) KI is added
_______________
___________________
(b) H2O is added
_______________
___________________
(c) The temperature is increased
_______________
___________________
(d) [H+] concentration is increased
_______________
___________________
3.
3
Provide brief answers for the following questions. (20 pt)
(a) Often concentrations are written as ppm. What does the symbol ppm stand for and when would it be used?
(b) Using collision theory, briefly explain what happens to the rate constant of a reaction if the temperature is
increase?
(c)
What is the advantage of a heterogeneous catalyst versus a homogenous catalyst?
(d) Give an example of a solution that is isotonic with respect to a 0.1 M NH4Cl (aq) solution
4
2.
Concentrated nitric acid, HNO3 (molar mass: 63.013 g/mol), available in the laboratory is 71% nitric acid
by mass. The other component is water (molar mass: 18.02 g/mol, density: 0.9971 g/ml, kf = 0.512C/m,
kb = 1.858C/m). The density of the concentrated nitric acid solution is 1.418 g/ml at 20°C.
(a) Draw the molecular geometry of HNO3. (15 pts)
Just practice
molecular geometry around:
Bond angles
Hybridization
N: _______________
N: _______________
N: _______________
O: _______________
O: _______________
O: _______________
Polarity: ____________
Predominate IMF: _______________
(b) Determine the molarity of the solution. (10 pt)
(c) Determine the molality of the solution. (10 pt)
3.
5
Concentrated nitric acid, HNO3 (molar mass: 63.013 g/mol), available in the laboratory is 71% nitric acid
by mass. The other component is water (molar mass: 18.02 g/mol, density: 0.9971 g/ml, kf = 0.512C/m,
kb = 1.858C/m). The density of the concentrated nitric acid solution is 1.418 g/ml at 20°C.
(continued)
(d) What is the expected boiling point of this solution? Assume that HNO3 is non-volatile. (10 pt)
6.
Determine the vapor pressure of 100 mL of pure water at 20°C? Additional data is located on the last page.
skip. And look answer up in text for next problem
(b)
What is the vapor pressure after 15.0 grams of NaCl (s) has been added to the water?
(c)
Sketch a vapor pressure plot that shows the total vapor pressure of the solution as a function of
concentration.
6
4. Hemocyanin is an oxygen transport protein that is found in the blood of crabs such as the Horseshoe Crab
found along the coast of Cape Cod. It is an example of a metalloprotein. Oxygen molecules reversibly bind
simultaneously to two copper atoms covalently bonded within the protein.
An aqueous solution of hemocyanin was prepared by dissolving 1.50 g of the protein in enough water (Kf =
1.86 °C Kg/mol, Kb = 0.51 °C Kg/mol) to bring the total solution volume to 250 mL. The resulting solution
had an osmotic pressure of 2.60 mm Hg at 4°C. What is molar mass of hemocyanin? (20 pt)
7
7. Urease is an enzyme that catalyzes the hydrolysis of urea, C(O)(NH2)2, to ammonium ion and bicarbonate
ion. The reaction is shown below
(NH2)2CO (aq)
+ 2 H2O (aq)
+
H+ (aq) 
2 NH4+ (aq)
+
HCO3- (aq)
k (uncatalyzed) = 3 x 10-10 s-1 , k (catalyzed with urease) = 3 x 104 s-1
The enzyme is incredibly efficient and selective. It acts only on urea and provides a rate enhancement of 14
orders of magnitude (1014)!
In an effort to determine its molar mass, an aqueous solution of Urease was prepared by dissolving 0.450 g
of the protein in enough water (Kf = 1.86 °C Kg/mol, Kb = 0.51 °C Kg/mol) to bring the total solution
volume to 0.015 L. The resulting solution had an osmotic pressure of 1.084 torr at 5°C. What is molar mass
of Urease? (25 pt)
8
5. Rhombic and Monoclinic sulfur, the two stable solid phases of sulfur, do not exist as individual sulfur atoms
in the solid state but rather as a circular chains of sulfur atoms. The addition of 0.24 g of sulfur to 100 g of
carbon tetrachloride (CCl4, MW = 153.82, kf = 29.8C/m) lowers the latter's freezing point by 0.28°C.
What is the molar mass and molecular formula of Rhombic and Monoclinic sulfur?
6. 2.37 g of a particular protein is dissolved in a small amount of water. The sample is then diluted to 100.0 mL
and its osmotic pressure was measured. The osmotic pressure of the solution was found to be 24.244 torr at
20C. What is the molar mass of this protein?
9
7. The density of a 1.06 M aqueous sucrose solution (C12H22O11, MW = 342.3) is 1.14 g/cm3 at 25°C. At
25°C the vapor pressure of water (l) is 23.76 Torr and its density is 0.9971 g/cm3
(a) What is the molality of the solution?
(b) What is the mass percent of the solution?
(c) What is the mole fraction of sucrose?
10
8. The density of a 1.06 M aqueous sucrose solution (C12H22O11, MW = 342.3) is 1.14 g/cm3 at 25°C. At
25°C the vapor pressure of water (l) (kf = 1.86C/m) is 23.76 Torr and its density is 0.9971 g/cm3
Continued
(d) What is the vapor pressure of H2O above the solution?
(e) What is the freezing point of this solution
(f) What is the osmotic pressure of the solution
4.
11
Aqueous solutions of sodium hydroxide (NaOH, mol. wt. 40.00 g/mol) can be purchased that are 50%
NaOH (aq) by mass and have a density of 1.515 g/ml at 25C. The other component is water (molar mass:
18.02 g/mol, density: 0.9971 g/ml, kb = 0.512C/m, kf = 1.858C/m).
(a) Determine the molarity of the solution. (15 pt)
(b) Determine the molality of the solution. (15 pt)
(c) What is the expected freezing point of this solution in C? (10 pts)
9.
12
Provide brief answers for the following questions. (20 pt)
(a) A particular reaction takes 10 min complete when the temperature is 20C. When the temperature is raised
to 50C the reaction only takes 3 minutes. At which temperature is the rate constant k larger? Explain
(b) In class we discussed the relative stabilities of Cdiamond and C graphite at room temperature and pressure.
Cdiamond is said to be kinetically stable but thermodynamically unstable. What does this statement mean?
(c) In order for a reaction to take place, what three things must take place at the molecular level.
13
5. Consider the following hypothetical mechanism, where k1 <<< k2 and Hrxn = 56.3 KJ/mo
(a)
step (1)
2 A (g) + B (g) 
step (2)
D (g) + A (g) 
C (g) + D (g) k1
E (g) + B (g)
What is the overall reaction? (10 pts)
(b)
Identify the following in the above mechanism: (12 pts)
Reactants:
Products:
Intermediates:
Catalysts:
(c)
Sketch an accurate reaction profile. Label the following on it: (10 pts):
(i) both axes (ii) reactants and products, (iii) any transition states
(iv) Ea for the rate determining step (v) Hrxn




(d) Based on the above mechanism, what would be the expected rate law? (10 pts)








(e) If the reaction was experimentally found to be 3nd order overall, would this
(i) support, (ii) prove or (iii) disprove
the proposed mechanism? (3 pts)
k2
14
6. Coordination complexes often undergo the replacement of the coordinated water molecules for an anion in a
reaction referred to as anation. Below is an anation reaction of aqueous Mn2+ with chloride ion:
Mn(H2O)62+ (aq) + Cl- (aq)  Mn(H2O)5Cl+ (aq) +
H2 O
The rate of anation can be determined by monitoring the change in the absorption of the Mn2+ ion. The
following data were obtained:
[Mn(H2O)62+] (M) [Cl-] (M)
Initial Rate (mol/L • min)
__________________________________________________________
0.075
0.025
0.0032
0.100
0.050
0.0113
0.150
0.025
0.0128
0.150
0.050
0.0256
0.200
0.025
0.0057
__________________________________________________________
(a) Write the rate expression for the reaction. (14 pts)
(b) Answer the following concerning order of the reaction? (6 pts)
(i) Order with respect to [Mn(H2O)62+]
(ii) Order with respect to [Cl-]
(iii) Overall order
15
10.
Consider the following mechanism
step (1)
Ce4+ (aq) + Mn2+ (aq) 
Ce3+ (aq) + Mn3+ (aq)
step (2)
Ce4+ (aq) + Mn3+ (aq) 
Ce3+ (aq) + Mn4+ (aq)
Mn4+ (aq) + Tl+ (aq) 
Mn2+ (aq) + Tl3+ (aq)
step (3)
(a)
What is the overall reaction? Mind those charges! (10 pts)
(b)
Identify the following in the above mechanism: (12 pts)
Reactants:
Products:
Intermediates:
Catalysts:
(c)
Label the proposed reaction profile for this mechanism, which is shown below (9 pts):
g.
h.
f.
e.
a.
c.
i.
d.
b.
(d) Using the profile above, is the reaction exothermic or endothermic? (3 pts)
(e) Using the profile above determine which step is rate determining? (3 pts)
(f) Using the profile above, which step has the largest rate constant, k? (3 pts)
16
Consider the following mechanism
11.
CONTINUED
step (1)
Ce4+ (aq) + Mn2+ (aq) 
Ce3+ (aq) + Mn3+ (aq)
step (2)
Ce4+ (aq) + Mn3+ (aq) 
Ce3+ (aq) + Mn4+ (aq)
Mn4+ (aq) + Tl+ (aq) 
Mn2+ (aq) + Tl3+ (aq)
step (3)
(g) Determine the rate law expected for the proposed mechanism (15 pts)
(h) If the reaction was experimentally found to be 3rd order overall, would this
(i) support, (ii) prove or (iii) disprove
the proposed mechanism? (5 pts)
17
12. Boron trifluoride, BF3, is archetypal example of a Lewis Acid. Its empty p-orbital readily accepts an
electron pair from a variety of electron donors. One such reaction with ammonia is shown below:
BF3 (g) + NH3 (g)  F3BNH3 (g)
The rate of the reaction is determined by measuring the appearance of F3BNH3. The following data are
obtained:
[BF3] (M)
[NH3] (M)
Initial Rate (mol/L • s)
__________________________________________________________
0.250
0.250
0.2130
0.250
0.125
0.1065
0.200
0.100
0.0682
0.350
0.100
0.1193
0.175
0.100
0.0596
__________________________________________________________
(a) Write the rate expression for the reaction. (15 pts)
(b) Answer the following concerning order of the reaction? (5 pts)
(i) Order with respect to [BF3]
(ii) Order with respect to [NH3]
(iii) Overall order
18
Useful Constants and Equations
Planck's constant
h = 6.626 x 10-34 J•s
Speed of light
Atomic mass unit
Boltzmann constant
Gas Constant
c = 2.9979 x 108 m/s
amu = 1.66054 x 10-27 kg
k = 1.38066 x 10-23 J / K
R = 8.31451 J / K
R = 0.08206 L • atm / K mol
F = 96,485 C/mole e-
Faraday
Kf [Ag(NH3)2]+ = 1.7 x 107
kb (H2O) = 0.512C/m
Enthalpy of vaporization for H2O
Enthalpy of fusion for H2O
kf (H2O) = 1.858C/m
40.7 KJ / mol
6.02 KJ / mol
E = 2.31 x 10-19 J • nm (q1 q2 / r)
[A] = -kt + [A]0; t1/2 = [A]0/2k
(KE)avg = 3/2 RT
ln [A] = -kt + ln [A]0; t1/2 = ln 2/k
urms = (3RT / M)1/2
[A]-1 = kt + [A]-1; t1/2 = 1/k[A]0
ln(P1/P2) = H/R (1/T2 – 1/T1)
k = Ae-Ea/RT
sc: l = 2r;
ln(k1/k2) = Ea/R (1/T2 – 1/T1)
bcc: l = 4r / (3)1/2
T = i K m
fcc l = r (8)1/2
=iMRT
G = H - TS
P =  P
G = G° + RTlnQ
-b ± b2 - 4ac
2a
x=
 = ° - (RT/nF)lnQ
 = ° - (0.0592/n)logQ
E = mc2
E = h
 = c
PMm = dRT