1
Chem 1000A Midterm 2 – Spring 2004 - Solutions
A
March 12th, 2004: 8:00 to 8.50 am
Your name _______________
Instructor: Dr. M. Gerken
Student ID _______________
Time: 50 min
No. of pages: 5 +1
Report all your answers using significant figures. Use SI units and show all units and their conversions
throughout your calculations.
Q1
Q2
Q3
Q4
Q5
Total Percent
6.5
13.5
8
10.5
1.5
40
100 %
Question 1 (6.5 Marks)
Balance the following redox reaction in a basic aqueous solution and show that the final equation has the
correct electron, material, and charge balance. Show as many steps as possible.
(note: N2H4 is hydrazine: H2N-NH2)
MnO4- + N2H4
→
MnO2 + N2
Mn+VIIO-II - + N-II2H+I4
→
Mn+IVO-II2 + N02
oxidation half-reaction (unbalanced): N2H4 → N2
reduction half-reaction (unbalanced): MnO4- → MnO2
principle element balance:
oxidation half-reaction (unbalanced): N2H4 → N2
reduction half-reaction (unbalanced): MnO4- → MnO2
oxygen balance:
oxidation half-reaction (unbalanced): N2H4 → N2
reduction half-reaction (unbalanced): MnO4- → MnO2 + 2H2O
hydrogen balance (basic solution):
oxidation half-reaction (unbalanced): 4OH- + N2H4 → N2+ 4H2O
reduction half-reaction (unbalanced): 4 H2O + MnO4- → MnO2 + 2H2O + 4OHInserting electrons:
oxidation half-reaction (balanced): 4OH- + N-II2H4 → N02+ 4H2O + 4e- |×3
reduction half-reaction (balanced):3e- +4 H2O + Mn+VIIO4- → Mn+IVO2 + 2H2O + 4OH-|×4
electron balanced
oxidation half-reaction (balanced): 12OH- + 3N2H4 → 3N2+ 12H2O + 12ereduction half-reaction (balanced): 12e- +16 H2O + 4MnO4- → 4MnO2 + 8H2O + 16OH combination:
12OH- + 3N2H4 +16 H2O + 4MnO4- +12e- → 4MnO2 + 8H2O + 16OH – +3N2+ 12H2O + 12esimplification:
3N2H4 + 4MnO4- → 4MnO2 + 4H2O + 4OH- +3N2
material balance:
12H, 16O, 6N, 4Mn |4Mn, 6N, 16O, 12H
charge balance:
(4-)
|4correct electron, material, and charge balance!
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Question 2 (13.5 Marks)
In a complete combustion of 1.569 g of a diole (CxHyOz), 2.225 g of CO2 and 1.366 g of H2O is produced.
(a) Write the balanced reaction equation.
CxHyOz + (2x + y/2 – z)/2 O2 → xCO2 + y/2 H2O
Or
2C2H6O2 + 5O2 → 4CO2 + 6H2O
(b) Determine the empirical formula for this diole.
M(CO2) = 44.0098 g/ mol; M(H2O) = 18.01534 g/mol
n(H2O) = 1.366 g/(18.01534 g/mol) =0.07582 mol
n(CO2) = 2.225 g/(44.0098 g/ mol) = 0.05056 mol
n(H) = 0.07582 mol H2O × 2 mol H/1 mol H2O = 0.1516 mol H
n(C) = 0.05056 mol CO2× 1 mol C/1 mol CO2 =0.05056 mol C
m(H) = 0.1516 mol ×1.0079 g/mol =0.1528 g
m(C) = 0.05056 mol × 12.011 g/mol =0.6073 g
m(O) = m (acid = CxHyOz) – m(H) – m(C) = 1.569 g -0.1528 g – 0.6073 g = 0.809 g
n(O) = 0.809 g/(15.999 g/mol) = 0.0506 mol
n(C) : n(H) : n(O) = 0.05056 mol: 0.1516 mol : 0.0506 mol
= 1 : 2.998 : 1.000
≈1:3:1
empirical formula: CH3O
(c) The molar mass of the diole has been determined to be 62.1 g/mol. What is the molecular formula of
this diole.
molar mass of CH3O: M(CH3O) = 31.034 g mol-1
experimental molar mass: M(diole) = 62.1 g mol-1
molecular formula of the diole: C2H6O2
(d) You burn (assume complete combustion) 2.152 g of this diole. What is the volume (in m3 and L) of
CO2(g) that is produced at 683 Torr and 25.23 °C?
M(C2H6O2) = 62.067 g/mol; n(C2H6O2) = 2.152 g/(62.067 g mol-1) = 0.03467 mol
n(CO2) = 0.03467 mol diole × 4mol of CO2/2mol of diole = 0.06934 mol CO2
CO2 is a gas, therefore, you can use pV = nRT;
p = 683 Torr × (101325 Pa/760Torr) =91100 Pa; T = (25.23 +273.15)K = 298.38 K
V = nRT/p = 0.06934 mol × 8.3145 J mol-1 K-1 ×298.38 K/( 91100 Pa) = 1.89 ×10-3 J/Pa
= 1.89 ×10-3 (kg m2 s-2 )(kg-1m s2) = 1.89 ×10-3 m3= 1.89 L
3
Name _________________________
Student ID _________________
Question 3 (8 Marks)
Determine the oxidation states in the following compounds. Indicate the oxidation states of all elements in
the Lewis structure.
(a) S2F2
..
:S
..
:F
..
S.. :
..
F..:
F-I-S+I-S+I-F-I
(b) I3.. .. ..
: I ..I I :
.. _.. ..
I0-I-I-I0
(c) CH3NO2
H
..
O:
H C N .. _
+ O:
..
H
H
.. _
O
.. :
H C N
+ O:
..
H
(H+I)3C-II-N+III(O-II)2
Question 4 (10.5 Marks)
NO reacts with oxygen to NO2.
NO(g) + O2(g) → NO2(g)
You fill 792 Torr of NO in a 1.523 L vessel at 25.35 °C. You add 0.0456 mol of O2 to the vessel.
(a) Balance the reaction equation.
2NO(g) + O2(g) → 2NO2(g)
(b) What type of reaction is this reaction.
This reaction is a redox reaction.
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(c) Calculate the partial pressures (in Pa and Torr) of NO and O2 at the beginning of the reaction.
The partial pressure of NO is 792 Torr since p(NO) does not change upon addition of a second gas.
p(NO) = 792 Torr × 101325 Pa/760 Torr = 106000 Pa
For oxygen: pV=nRT; T = (25.35 + 273.15)K= 298.50 K; V = 1.523 L × 1 m3/(1000 L) = 1.523 ×10-3m3
p(O2) = nRT/V = 0.0456 mol ×8.3145 J mol-1 K-1 ×298.50 K/(1.523 ×10-3m3) = 74300 J m-3
= 74300 kg m2 s-2 m-3= 74300 kg m-1 s-2= 74300 Pa × 760 Torr/101325Pa = 557 Torr
(d) Which reactant is the limiting reagent?
T and V are constant (25.35 °C and 1.523 L), therefore, p and n are proportional to each other.
p(NO)/(O2) = 792 Torr/557Torr = 1.42
The reaction stoichiometry specifies that two molecules of NO are required for each O2 molecule; in this
experiment only 1.42 molecules are present for every O2 molecule. Therefore, NO is the limiting reagent.
(e) Calculate the pressure of NO2 (in Torr) after the complete reaction.
You generated the same number of moles of NO2 as the number of moles of NO consumed. The pressure
of NO2 after the complete reaction is the same as the partial pressure of NO.
106000 Pa and 792 Torr
or
pV=nRT;
p(NO) = 106000 Pa; T = 298.50 K; V = 1.523 ×10-3m3
n(NO)=pV/RT = 106000 Pa×1.523 ×10-3m3/(8.3145 J mol-1 K-1 ×298.50 K) = 0.0650 Pa m3 J-1 mol
= 0.0650 kg m s-2 m3 kg-1 m-2 s2 mol= 0.0650 mol
n(NO2) = 0.0650 mol NO × 1 mol NO2/1mol NO = 0.0650 mol NO2
p(NO2) = nRT/V = 0.0650 mol ×8.3145 J mol-1 K-1 ×298.50 K/(1.523 ×10-3m3) = 106000 J m-3
= 106000 kg m2 s-2 m-3= 106000 kg m-1 s-2= 106000 Pa × 760 Torr/101325Pa = 792 Torr
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Name _________________________
Student ID _________________
Question 5 (1.5 Marks)
Define the temperature of a gas with respect to the energy of the gas particles.
In a gas at equilibrium, the gas particles have a distribution of energies, i.e., Boltzmann distribution. The
maximum of a Boltzmann distribution curve/the average kinetic energy of the gas particles is a measure
of the temperature of a gas.
6
Name _________________________
Student ID _________________
Prefixes
Pico, p 10-12; nano, n 10-9; micro, µ 10-6 ;milli, m 10-3; centi, c 10-2; deci, d
Fundamental Constants
Avogadro's number 6.022 × 1023 mol-1
Proton mass
-19
Neutron mass
Elementary charge, e 1.6022 × 10 C
Electron mass
9.1095 × 10-28 g
molar volume of gases at STP
-1
-1
Gas constant, R
8.314 J K mol
m
m
3RT
n=
; ρ = ; pV = nRT ; Ptot = ∑ p i ; pi = X i ⋅ Ptot ; u 2 = u rms =
V
M
M
i
Physical quantity
Unit
Symbol
Definition
Frequency, f or ν
Energy , W or E
Force, F
Pressure, p
hertz
joule
newton
pascal
10-1; kilo, k
103
1.67252 × 10-24 g
1.6749 × 10-24 g
22.414 L/mol
s-1
kg m2 s-2
J m-1 = kg m s-2
N m-2 = kg m-1 s-2
Hz
J
N
Pa
Temperature: 0 K = -273.15 °C; 0 °C = 273.15 K
Pressure: 1 atm = 760 Torr = 760 mmHg = 1.01325 bar = 101325 Pa; 1 bar = 105 Pa
Volume: 1 mL = 1 cm3; 1 L = 1000 cm3 = 1 dm3 = 0.001 m3
Chem 1000 Standard Periodic Table
Electronegativities in [ ]
1
1.0079
1H
18
4.0026
2He
hydrogen
[2.1]
6.941
2
13
14
15
16
17
helium
9.0122
10.811
12.011
14.0067
15.9994
18.9984
20.1797
3Li
4Be
5B
6C
7N
8O
9F
10Ne
11Na
12Mg
magnesium
lithium berrylium
[1.0]
[1.5]
22.9898 24.3050
sodium
[1.0]
39.0983
19K
[1.2]
40.078
3
4
5
6
7
8
9
10
11
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
20Ca
21Sc
22Ti
23V
24Cr
25Mn
26Fe
27Co
28Ni
29Cu
potassium calcium scandium titanium vanadium chromium manganese
iron
[1.6]
[0.9]
[1.0]
[1.3]
[1.4]
[1.5]
[1.6]
[1.7]
87.62
88.9059 91.224 92.9064
95.94
(98)
85.4678
101.07
37Rb
38Sr
rubidium strontium
[0.9]
[1.0]
132.905 137.327
55Cs
cesium
[0.8]
(223)
87Fr
francium
[0.8]
39Y
40Zr
41Nb
yttrium zirconium niobium
[1.3]
[1.4]
[1.5]
178.49 180.948
56Ba
La-Lu
88Ra
Ac-Lr
barium
[1.0]
(226.025)
72Hf
42Mo
[1.6]
183.85
73Ta
74W
hafnnium tantalum tungsten
[1.3]
[1.4]
[1.5]
(266)
(262)
(261)
104Rf
rutherfordium
radium
[1.0]
43Tc
molybdenum technetium
34Se
35Br
36Kr
51Sb
52Te
83Bi
silicon
[1.8]
72.61
30Zn
31Ga
32Ge
germanium
49In
75Re
rhenium
[1.7]
(264)
76Os
78Pt
79Au
77Ir
osmium
[1.9]
(269)
iridium
[1.9]
(268)
150.36
151.965
lanthanum cerium
(227.028) (232.038) (231.036) (238.029) (237.048)
protactinium
33As
65.39
aluminum
[1.5]
69.723
48Cd
(145)
92U
chlorine
[3.0]
79.904
18Ar
[2.1]
74.9216
sulfur
[2.5]
78.96
17Cl
47Ag
platinum
[1.8]
(271)
93Np
80Hg
[1.9]
118.710
50Sn
tin
[1.8]
207.19
arsenic selenium bromine
[2.1]
[2.4]
[2.8]
121.757 127.60 126.905
antimony tellurium
[1.9]
[2.1]
208.980
(210)
53I
82Pb
158.925
162.50
164.930
167.26
168.934
173.04
174.967
66Dy
67Ho
dysprosium holmium
68Er
69Tm
70Yb
71Lu
(244)
(243)
(247)
94Pu
95Am
96Cm
97Bk
uranium neptunium plutonium americium curium
(251)
(252)
erbium
(257)
54Xe
86Rn
81Tl
terbium
(247)
krypton
[2.9]
131.29
85At
thallium
[1.6]
65Tb
83.80
84Po
mercury
[1.7]
bismuth polonium astatine
[1.8]
[1.9]
[2.1]
110Ds
157.25
argon
xenon
[2.3]
(222)
gold
[1.9]
lead
[1.7]
neon
iodine
[2.5]
(210)
darmstadtium
60Nd
61Pm
62Sm
63Eu
64Gd
praesodymium neodymium promethium samarium europium gadolinium
91Pa
16S
46Pd
144.24
90Th
15P
phosphorus
45Rh
109Mt
89Ac
14Si
ruthenium rhodium palladium silver cadmium indium
[1.6]
[1.7]
[1.7]
[1.8]
[1.8]
[1.6]
[1.6]
186.207
190.2
192.22
195.08 196.967 200.59 204.383
108Hs
59Pr
39.948
13Al
gallium
[1.7]
114.82
hassium meitnerium
140.908
fluorine
[4.0]
35.4527
zinc
[1.6]
112.411
107Bh
58Ce
oxygen
[3.5]
32.066
copper
[1.8]
107.868
106Sg
140.115
nitrogen
[3.0]
30.9738
nickel
[1.8]
106.42
105Db
57La
carbon
[2.5]
28.0855
cobalt
[1.7]
102.906
dubnium seaborgium bohrium
138.906
actinium9 thorium
44Ru
12
boron
[2.0]
26.9815
thulium ytterbium lutetium
(258)
(262)
(259)
98Cf
99Es
100Fm 101Md
102No
103Lr
berkelium californium einsteinium fermium mendelevium nobelium lawrencium
radon