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Formulas and Constants
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NA = 6.022x1023 mol-1
h = 6.626x10-34 J s
c = 2.998x108 m s-1
-19
-31
e = 1.602x10 C
me = 9.109x10 kg
1 Å = 1x10-10 m = 100 pm
-1
-1
1 atm = 760 torr
R = 0.08206 L atm K mol = 8.314 J K-1 mol-1
1
hc
~
~
ν =
∆E = = hcν
λ
λ
[gas(aq)]eq = KH (Pgas)eq
∆Tf = Kf m
vpsolution = Xsolvent vppure solvent
∆Tb = Kb m
Π V = nRT
 [A] 
ln
 = – kt
[A]o
[A] = [A]o e–kt
t½ = ln 2/k = 0.6931/k
1
t½ = [A] k
o
∆Ea  1 1 
k2
–
ln  = –
R T2 T1
k1
1
IA
1A
2
IIA
2A
3
4
IIIB IVB
3B 4B
5
VB
5B
1
1
–
= kt
[A] [A]o
∆E = Ea,f – Ea,r
k1 ∆Ea  1 1 
ln  =
–
R T2 T1
k2
6
7
VIB VIIB
6B 7B
8
-----
9
VIII
8
Π = MRT
10
-----
k = A e–Ea/RT
11
IB
1B
12
IIB
2B
13
14
IIIA IVA
3A 4A
15
VA
5A
16
17
18
VIA VIIA VIIIA
6A 7A 8A
1
2
H
He
1.008
4.003
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.941 9.012
11
10.81 12.01 14.01 16.00 19.00 20.18
12
13
14
15
16
17
18
Na Mg
Al
Si
P
S
Cl
Ar
22.99 24.31
26.98 28.09 30.97 32.07 35.45 39.95
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 69.72 72.59 74.92 78.96 79.90 83.80
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.47 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55
56
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba La* Hf
57
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 190.2 195.1 197.0 200.5 204.4 207.2 209.0 (210) (210) (222)
CH145
Chemistry 145
Prof. Shattuck
2
Practice Test 2
R = 8.314 J mol-1 K-1 = 0.08206 L atm mol-1 K-1
Name_______________________________
T(0°C) = 273.15 K
Part 1: Answer 8 of the following 10 questions. If you answer more than 8 cross out the one
you wish not to be graded, otherwise only the first 8 will be graded. 8 points each.
1. In the following pairs, which substance has the higher boiling point:
a. C2H6 or C6H14 ? more electrons
b. H2S or H2O ?
hydrogen-bonding
c. CO2 or NO2 ? NO2 has a dipole moment
2. Which of these statements about benzene is true?
__ A. All carbon atoms in benzene are sp3 hybridized.
__ B. Benzene contains only π-bonds between C atoms.
_* C. The bond order of each C—C bond in benzene is 1.5.
__ D. Benzene is an example of a molecule that displays ionic bonding.
__ E. All of these statements are false.
3. From the following information calculate the Keq for reaction 3.
Reaction 1 Ag+ + NH3 = Ag(NH3)+
Reaction 2 Ag(NH3)+ + NH3 = Ag(NH3)2+
Reaction 3 Ag+ + 2 NH3 = Ag(NH3)2+
K1 = 1.7x103 M-1
K2 = 1.1x107 M-1
Keq = K1 K2 = 1.87x1010
4. Give the hybridization for NH3 __sp3______ and for SO2 _sp2______.
5. For the reaction NO2 + CO → NO + CO2 the experimental rate law is:
rate = k [NO2]2
The following mechanism has been proposed:
2 NO2 → NO3 + NO
NO3 + CO → NO2 + CO2
a. Which step is the rate determining step? Slow step is the rate detemrinining step—step 1
b. Is there a reactive intermediate in this reaction? If so what is it?
NO3 (not a reactant or product)
6. For a first order reaction, the concentration of the reactant dropped from 0.200 M to 0.100 M
in 6.00 minutes. How long does it take for the concentration to drop from 0.200 M to 0.0100 M?
Answer: The first time interval is for [A] = [A]o/2, that is one half-time, t½= 0.6931/k = 6.00
min. The rate constant is then k = 0.1155 min-1. The second time interval is:
ln([A]/[A]o) = ln(0.0100 M/0.200 M) = -2.996 = – (0.1155 min-1) t or t = 25.9 min
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3
7. The rate law for a third order reaction is
1
1
[A]2 – [A]o2 = 2kt
To make a straight line plot to verify third order behavior,
a. what do you plot on the vertical axis?__1/[A]2___________.
b. what do you plot on the horizontal axis? ______t (time)____________.
c. what is the slope equal? _____slope = 2 k_________________
8. The osmotic pressure of a solution of a protein in water is 1.54 torr at 25.0°. The solution
contained 0.700 g of protein per liter of solution. Calculate the molar mass of the protein.
Answer: In atmospheres, Π = 1.54 torr/760 torr/atm = 2.026x10-3 atm.
From Π = MRT the molar concentration is M = Π/RT = 2.026x10-3 atm/0.08206 L atm K-1
mol-1/298.15 K = 8.281x10-5 mol L-1. The molar mass is then MM = 0.700 g/8.281x10-5 mol =
8.45x103 g mol-1.
9. Which molecule has a stronger bond, CF or F2? Explain your answer for credit.
CF
F2
E
––
–
+
σ*u (2pz)
+
+
–
–
+
+
+
+
↑↓
+
+
–
–
+
–
+
+
σg (2pz) 2pF
↑↓
↑↓
↑↓
↑↓
Bond order:
C
BO =
σg (2pz) 2pF
↑↓
↑↓
πu (2px) πu (2py)
πu (2px) πu (2p
↑↓y)
σ*u (2s)
2sC
C––––F
↑↓
π*g (2px) π*g (2py)
2pF
–
+
↑↓
π*g (2px) π*g (2py)
2pC
–
σ*u (2pz)
↑
–
+
E
↑↓
σ*u (2s)
↑↓
σg (2s)
F–F
F
2sC
2sN
↑↓
σg (2s)
C–F
F
F
8–3
= 2½
2
BO =
2sN
8–6
=1
2
stronger bond in CF (greater bond order)
10. Determine the rate law for the reaction 2NO (g) + Br2(g) → 2NOBr(g) from the following
initial rate study (you don’t need to get a numerical value for k, just leave it as “k”):
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4
[NO] (mol L-1)
1.00
2.00
1.00
[Br2] (mol L-1)
1.00
1.00
2.00
rate (L mol-1 sec-1)
1.00x10-6
2.00x10-6
4.00x10-6
Answer: Experiments 1&2: when [NO] doubles the rate doubles—the order with respect to NO
is one. Experiments 1&3: when [Br2] doubles the rate quadruples –the order with respect to Br2
is two: rate = k [NO][Br2]2
If it were requested: For experiment 1, to get the rate constant rate = k [NO][Br2]2 or:
1.00x10-6 l mol-1 s-1 = k (1.00 mol L-1)(1.00 mol L-1)2 giving k = 1.00x10-6 L3 mol-3 s-1
Part 2. Answer 3 of the following 5 questions. If you answer more than 3 cross out the ones
you wish not to be graded, otherwise only the first 3 will be graded. 12 points each.
11. (a.) Draw an energy level diagram for the molecular orbitals for the CN- ion. (b.) Label each
orbital with the type of orbital, sigma, or pi and bonding or anti-bonding. (c.) Fill the levels with
the proper number of electrons, and (d.) calculate the bond order.
CN–
E
––
–
+
σ*u (2pz)
+
+
–
–
+
–
+
π*g (2px) π*g (2py)
2pC
–
+
+
–
+
+
↑↓
+
+
–
–
+
–
+
+
σg (2pz) 2pN
↑↓
↑↓
πu (2px) πu (2p
↑↓y)
↑↓
σ*u (2s)
↑↓
σg (2s)
2sC
C––––N
Bond order:
C
BO =
C–N
2sN
N
8–2
=3
2
12. NOBr decomposes according to
NOBr (g) →
← NO (g) + ½ Br2 (g)
With Kp = 0.15. If 1.0 atm of NOBr, 0.8 atm of NO, and 0.4 atm of Br2 are mixed, will any
reaction occur? If a net reaction is observed, will NO be formed or consumed?
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5
PBr2½ PNO (0.4 atm)½ (0.8 atm)
=
= 0.506 with Q > Kp the reaction runs in the
1 atm
PNOBr
reverse direction.
Answer: Q =
13. What is the molarity of 50.0% by weight NaOH solution? The density of the solution is 1.53
g mL-1. (Molar Mass(NaOH) = 40.0 g mol-1 )
Answer: Assume 100 g of solution, which then contains 50.0 g of NaOH or 1.25 moles of
NaOH. The volume of the solution is Vsoln = 100.0 g/dsoln = 100. g/1.53 g mL-1 = 65.36 mL or
0.06536 L. The final molarity is c = (moles solute)/Vsoln = 19.1 mol L-1
14. The rate constant for the following reaction is 1.05x10-4 sec-1 at 786 K . The rate constant
increases to 7.88x10-4 sec-1 at 834 K. Calculate the activation energy for the reaction:
C4H8O → C4H6 + H2O
∆Ea  1 1 
k2
–
giving
Answer: ln  = –
R T2 T1
k1
-4
∆Ea
1 
7.88x10 
 1
ln1.05x10-4 = – 8.314 J K-1 mol-1834. K – 786. K




-1
-1
-5
2.015 = – ∆Ea/8.314 J K mol (-7.322x10 )
∆Ea = 229. kJ mol-1
15. NOBr decomposes according to
NOBr2 (g) →
← NO (g) + Br2 (g)
With Kp = 2.15. If 1.00 atm of NOBr is placed in a constant volume container, calculate the
equilibrium pressure of Br2.
Answer: NOBr2 (g) →
← NO (g) + Br2 (g)
initial
1.00 atm
0
0
change
–x
+x
+x
equil
1.000 – x +x
+x
Kp =
PNO PBr2
x2
=
= 2.15
PNOBr2 1 – x
cross multiplying gives: x2 = 2.15 – 2.15 x
solving:
x=
or
x2 + 2.15 x – 2.15 = 0
–b ± b2 – 4ac –2.15 ± (2.15)2 – 4(-2.15) –2.15 ± 3.636
=
=
= 0.743 atm
2a
2
2