Download DEPARTMENT OF CHEMISTRY, CFS, IIUM

SHE 1315
CHEMICAL BONDING
TOPIC 1: MATTER AND MEASUREMENT
1.
Classify the following as a substance or a mixture. If it is a substance choose either
element or compound. If it is a mixture choose either heterogeneous or homogeneous.
(1c)
Substance
Mixture
Type of Matter
Element
a. Chlorine
Compound
Homogeneous
√
b. Water
√
c. Soil
√
d. Sugar Water
√
e. Oxygen
Heterogeneous
√
f. Carbon Dioxide
√
g. Chocolate chip ice
cream
√
h. Pure Air
i. Iron
2.
√
√
Classify the following properties as either chemical or physical by putting a check in the
appropriate column.
(1c)
Physical Property
a. Blue Color
√
b. Density
√
c. Flammability
√
d. Solubility
√
e. Reacts with acid to form H2.
Chemical Property
√
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CHEMICAL BONDING
√
f. Supports combustion
g. Sour Taste
√
h. Melting Point
√
i. Reacts with water to form a gas
√
j. Reacts with a base to form water
√
k. Hardness
√
l. Boiling Point
√
√
m. Can neutralize a base
3.
n. Luster
√
o. Odor
√
Classify the following as an element, a compound, homogeneous or heterogenous mixture.
(1a)
a)
gold chain - homogeneous
b)
human tears - homogeneous
c)
chunky peanut butter- heterogeneous
d)
ink - homogeneous
e)
sand - heterogeneous
f)
gasoline- homogeneous
g)
distilled water -compound
h)
steel (an alloy of several metals) - homogeneous
i)
Helium – element
j)
mud - heterogeneous
k)
sugar -compound
SHE 1315
4.
5.
CHEMICAL BONDING
Identify the following properties as intensive or extensive.
(1c)
a)
Mass – extensive
e)
Melting point– intensive
b)
Density – intensive
f)
Length – extensive
c)
Color – intensive
g)
Temperature– extensive
d)
Volume– extensive
h)
Ductility - intensive
Choose words from the list to fill in the blanks in the paragraphs.
(1a, 1b, 1c)
chemical property
intensive property
compound
mixture
element
physical property
extensive property
property
heterogeneous mixture
substance
homogenous mixture
Matter has uniform characteristics throughout is called homogeneous mixture. Matter that has
parts with different characteristics is called heterogeneous mixture.
A characteristic by which a
variety of matter is recognized is called a property. A characteristic that depends upon the amount
of matter in the sample is called an extensive property. A characteristic that does not depend upon
the amount of matter is called an intensive property. A characteristic that can be observed without
producing new kinds of matter is called a physical property. A characteristic that depends on how
a kind of matter changes suring interactions with other kinds of matter is called chemical property.
Matter can also be classified according to the basic types of matter it contains. A simple
substance that cannot be broken down into other substances by chemical means is called element.
A chemical combination of simple substances is called compound.
A physical combination of
different substances that retain their individual properties is called a mixture. Either an element or
a compound may be referred to as a substance.
6.
State the number of significant digits in each measurement.
a)
2804 m
4
d)
0.003068 m
g)
75,000.0 m
4
6
b)
2.84 km 3
e)
4.6 x 105 m
h) 75 m
2
2
(1d)
c)
5.029 m 4
f)
4.06 x 10-5 m
i)
75.00 m
4
3
SHE 1315
7.
CHEMICAL BONDING
Round the following numbers as indicated:
a)
To four significant figures:
i) 3.682417
3.682
b)
iii) 375.6523
375.7
iv) 112.511
112.5
ii) 2.473
2.5
iii) 5.687524
5.7
iv) 7.555
7.6
iii) 0.03062
0.03
iv) 3.4125
3.41
To two decimal places:
i) 22.494
22.50
8.
ii) 21.860051
21.86
To one decimal place:
i) 1.3511
1.4
c)
(1d)
ii) 79.2588
79.26
The world’s tallest waterfall is Angel Falls. It is 4210 ft high. What is the height of Angel
Falls:
[1 mi = 5280 ft = 1.609 km]
(1e)
a)
in miles?
4210 ft x 1 miles = 0. 7973 miles
5280 ft
b)
in km?
4210 ft x 1.609 km = 1.2829km
5280 ft
c)
in m?
4210 ft x 1.609 km x 1000 m = 1282.9 m
5280 ft
1 km
9.
A slow jogger runs a mile in 13 minutes. Calculate the speed in:
[1 mi = 1609 m ; 1 in = 2.54 cm]
a)
in miles?
1 mile x 1609 m x 100 cm x 1in
13 min 1 mile
1m
2.54 cm
b)
(1e)
x 1min
60 s
in km?
1 mile x 1609 m = 123. 7692 m/min
13 min 1 mile
c)
in m?
=
81. 2134 in/s
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CHEMICAL BONDING
1 mile x 1609 m x 1 km x
13 min 1 mile 1000 m
10.
60 min
1 hr
7.4262 km/hr
Liquid nitrogen is obtained from liquefied air. It is used in preparation of frozen goods and in
low-temperature research. The density of the liquid at its boiling point (-196oC or 77 K) is
0.608 g/cm3. Convert the density to units of kg/m3.
(1e)
0.608 g X 1 kg X 100 cm3
cm3 1000 g 1 m3
11.
=
= 0.0608 kg/m3
Convert the average speed of H2 molecules at 25oC (1.70 x 103 m/s) to miles/hr, keeping
track of significant figures.
[1 mi = 1.6093 km]
(1e)
1.70 X 103 m X 1 km X 1 mi
X 60 s X 60 m = 3.803 X103 mi/hr
s
1000 m 1.6093 km
1m
1 h
12.
Traveling at 65 miles/hour, how many feet can you travel in 22 minutes? (1 mile = 5280
feet).
(1e)
65 miles x
1 hour
13.
5280 feet
1 mile
x
1 hour
x 22 min = 125840 feet
60 min
Carry out the following operations and give your answers to the correct number of
significant figures.
a)
6.201 cm + 7.4 cm + 0.68 cm +12.0 cm = 26.3 cm
b)
8.264 g - 7.8 g =
c)
10.4168 m - 6.0 m = 4.4 m
d)
1.31 cm x 2.3 cm =
e)
0.5 g
3.0 cm2
5.7621 m x 6.201 m = 35.73 m2
f)
20.2 cm ÷ 7.41 s =
2.73 cm/s
h)
40.002 g ÷ 13.000005 g = 3.077
(1d)
SHE 1315
CHEMICAL BONDING
TOPIC 2: ATOMS, MOLECULES AND IONS
1.
a)
Complete the following table on sub-atomic particles
Sub-atom
particles
Absolute mass
(g)
Relative mass
(amu)
Relative
Charge
Electron
9.10939 x 10-24
0.000548579
-1
Proton
1.67262 x 10-24
1.007276
+1
Neutron
1.67493 x 10-24
1.008665
0
(2b)
b)
Draw the structure of an atom and show the location of proton, electron and neutron.
State the charges for each.
Nucleus : Proton(+ve) and neutron (neutral)
Electron(-ve)
(2a)
2.
Define:
a)
Atomic number, Z
Atomic number is the number of proton in the nucleus of an atom.
b)
Nucleon number, A
Mass number is the sum of the number of protons and neutrons in the
nucleus of an atom.
(2c)
A
3.
Write the Z X notation or the number of proton, neutron and electron at its right position
for each of the given diagram.
.
a)
18p, 20n
18e
38
SHE 1315
CHEMICAL BONDING
18Ar
b)
25p, 30n
55
23e
Mn2+
25
17p, 18n
c)
18e
35
Br-
17
(2d)
4.
Three naturally occurring isotopes of Pb are,
204
Pb,
Pb, and
206
Pb.
208
a) Define isotope.
An atom with the same number of protons but differ in number of neutrons.
(2e)
b) Calculate the number of protons, neutrons and electrons present in each of the isotope.
protons
neutrons
electrons
204
Pb
82
122
82
206
Pb
82
124
82
208
Pb
82
126
82
(2f)
5.
An atom has been chosen as a reference in determining the mass of elements. State the
atom.
Carbon-12
(2g)
6.
Define:
SHE 1315
CHEMICAL BONDING
a)
Relative atomic mass, Ar
A mass exactly equal to 1/12 the mass of a carbon-12 atom.
b)
Relative molecular mass, Mr
The sum of all the relative atomic masses of the atoms in a molecule; the
ratio of the average mass per molecule of a specified isotopic composition of
a substance to one-twelfth the mass of an atom of carbon-12.
c)
Relative isotopic mass
It is the ratio of the mass of one atom of that isotope to a twelfth of the
mass of one atom of carbon-12
(2h)
7.
Vanadium has two naturally occurring isotopes, 50V (isotopic mass 49.947 amu, abundance
0.25%) and 51V (isotopic mass 50.944 amu , abundance 99.75%). Calculate the atomic
mass of vanadium.
Atomic mass of V= (% abund x mass)
V + (% abund x mass)
50
V
51
= ( 49.947 amu)(0.25) + ( 50.944 amu)(99.75)
(100)
(100)
=
50.942 amu
(2i)
8.
Copper has two naturally occurring isotopes, 63Cu (isotopic mass 62.93 amu) and 65Cu
(isotopic mass 64.93 amu). Calculate the percent abundance of each isotope if the atomic
mass of Copper is 63.55 amu.
Atomic mass of Cu
63.55 amu
63.55 amu
2.00y
y
= (% abund x mass) 63Cu + (% abund x mass)
= y(62.93 amu) + (1-y)(64.93 amu)
= -2.00y + 64.93 amu
= 1.38
= 0.69
Cu
65
So, percent abundance for
63
Cu = y x 100% = 69%
65
Cu = (1 – y) x 100% = 100% - 69% = 31%
(2i)
9.
Given isotopes of hydrogen are 1H and 2H and isotopes of bromine are 79Br and 81Br. When
HBr is vaporized and analyzed in mass spectrometer, the following spectrum is found:
SHE 1315
CHEMICAL BONDING
P
R
S
Percent
abundance
T
Q
1
U
V
2
79 80
81 82 83
m/e
Identify the species of peaks P, Q, S and V
P =
H+ ;
1
Q =
H+ ;
S = (1H79Br)+ ;
2
V = (2H81Br)+
(2j)
10.
Both N and O have two isotopes each; that are 14N, 15N, 16O and 18O. If a sample of NO is
injected into a mass spectrophotometer, draw the complete mass spectrum obtained
considering fragmentation does occur.
N+
O+
14
(14N 16O)+
16
(15N 16O)+
Percent
abundance
(14N 18O)+
N
O+
15
14 15
(15N 18O)+
18
16
18
30 31 32 33
(2j)
11.
The table below gives the m/e values and the peak size of a sample atom X.
Use periodic table to answer the following questions:
m/e
32
33
34
36
SHE 1315
CHEMICAL BONDING
Relative number
71
1
3
1
a)
Calculate the atomic mass of X, in amu.
X= (71 x 32) + (1 x 33) + (3 x 34) + (1 x 36)
76
= 32.14 amu
b)
Based on your answer to the question above, suggest the name for atom X.
Sulphur
(2i)
12.
Define mole
A mole is defined as the amount of substance which contains equal number
of particles (atoms / molecules / ions) as there are atoms in exactly
12.000g of carbon-12.
(2k)
13.
a)
Molar mass
Molar mass is defined as the mass per one mole of a substance.
b)
Atomic mass
The average of the masses of the naturally occurring
element weighted according to their abundances.
isotopes
c)
Formula mass
The sum (in amu) of the atomic masses of a formula unit of a compound.
d)
Molecular mass
The sum (in amu) of the atomic masses of the elements in a
a compound.
molecule
(2l)
14.
of
a)
Calculate the molecular mass and molar mass of glucose, C6H12O6.
Molecular mass : (6 x 12.01) + (12 x 1.008) + (6 x 16.00)
= 180.156 g
an
of
SHE 1315
CHEMICAL BONDING
Molar mass : 180.156 g/mol
(2m)
b)
By using Factor Label Method/Dimensional Analysis, calculate the following number
of atoms or molecules in 150 g of C6H12O6.
(i) C atoms
=150 g C6H12O6 x
1 mole C6H12O6 x
180.156 g C6H12O6
6 mole atoms C
1 mole C6H12O6
x 6.022 x 1023 atoms C
1 mole atom C
= 3.008 x 1024 atoms C
(ii) H atoms
=150 g C6H12O6 x
1 mole C6H12O6 x
180.156 g C6H12O6
12 mole atoms H
1 mole C6H12O6
x 6.022 x 1023 atoms H
1 mole atom H
= 6.0168 x 1024 atoms H
(iii) C6H12O6 molecules
=150 g C6H12O6 x 1 mole C6H12O6 x 6.022 x 1023 molecules
180.156 g C6H12O6
1 mole C6H12O6
= 5.0140 x 1023 molecules C6H12O6
(2n)
15.
a)
Calculate the formula mass and molar mass of CaF2
Formula mass : (40.08) + (2 x 19.00)
= 78.08 g
Molar mass : 78.08 g/mol
(2m)
b)
Given 27.5 g of CaF2, calculate the number of:
(i)
cations
= 27.5 g CaF2 x 1 mole CaF2 x 1 mole Ca2+ x 6.022 X 1023 Ca2+ ion
78.08 g
1 mole CaF2
1 mole Ca2+
= 2.121 x 1023 Ca2+ ion.
SHE 1315
CHEMICAL BONDING
(ii)
anions
27.5 g CaF2 x 1 mole CaF2 x
78.08 g
2 moles F- x 6.022 X 1023 F- ions
1 mole CaF2
1 mole F-
= 4.2419 x 1023 F- ion
(iii) ions
27.5 g CaF2 x 1 mole CaF2 x 3 moles ions x 6.022 X 1023 ions
78.08 g
1 mole CaF2
1 mole ions
= 6.3629 x 1023 ions
(2n)
16.
a)
Calculate the amount in grams of KHCO3 is equivalent to 0.0725 mol KHCO3
Molar mass,KHCO3 = 39.10 + 1.008 + 12.01 + (3 x 16)=100.118 g/mol
0.0725 mol KHCO3 x 100.118 g KHCO3 = 7.26 g KHCO3
1 mole KHCO3
b)
Calculate the mass, in g for 6.022 X 1023 atoms of Cl in Cl2.
=6.022 X 1023 Cl atoms x
1 mole Cl2
2 x 6.022 X 1023 Cl atoms
x
71 g
1 mole Cl2
= 35.5 g
c)
Calculate the mass of oxygen, O2 which contains the same number of molecules as in
4.4 g of carbon monoxide, CO.
4.4g CO x 1 mole CO x 32 g O2 = 5.03 g N2
28 g CO
1 mole O2
(2n)
17.
a)
Define percentage composition by mass. Explain how it is calculated.
Mass percentage the percentage of mass of each element in a compound. It
is calculated as the mass of a component divided by the total mass of the
mixture, multiplied by 100%
(2o)
b)
Calculate the percentage composition by mass of each element in 1 mole ammonium
phosphate, (NH4)3PO4.
Molar mass [(NH4)3PO4] = 149.096 g/mol
%N = (42.03) ÷ (149.096) x 100% = 28.19%
%H = (12.096) ÷ (149.096) x 100% = 8.11%
SHE 1315
CHEMICAL BONDING
%P = (30.97) ÷ (149.096) x 100% = 20.77%
%O = (64.00) ÷ (149.096) x 100% = 42.93%
(2p)
c)
18.
Calculate the mass of each element in 74.548 g ammonium phosphate, (NH4)3PO4.
Mass N = 28.19% x 74.548 g =21.015 g
Mass H = 8.11% x 74.548 g = 6.046 g
Mass P = 20.77% x 74.548 g = 15.484 g
Mass O = 42.93% x 74.548 g = 32.003 g
(2p)
Define:
a)
Empirical formula
Empirical formula is a chemical formula that shows the simplest ratio of all
elements in a molecule.
b)
Molecular formula
Molecular formula is a formula that shows the actual number of atoms of
each element in a molecule
(2q)
19.
State the empirical formulae for the following molecular formulae.
a)
C8H18 :C4H9
c)
Na2C2O4 :NaCO2
b)
H2O2
d)
C7H12 :C7H12
:HO
(2q)
20.
Determine the empirical formula for a sample containing 5.39 g Al and 4.81 g O.
Element
Al
Mass (g)
Moles (mol)
O
5.39
5.39/ 26.98
=0.20
0.20/0.20
=1 x 2 =2
Mole ratio
Empirical Formula
4.81
4.81 /16.00
=0.30
0.30/0.20
=1.5 x 2 =3
Al2O3
(2r)
21.
A substance is decomposed and found to consist of 53.2% C, 11.2% H, and 35.6% O by
mass. Calculate the molecular formula of the unknown if its molar mass is 90.0 g/mol.
Assume 100 g total.
Thus: 53.2 g C, 11.2 g H, and 35.6 g O
Element
Mass (g)
C
53.2
H
11.2
O
35.6
SHE 1315
CHEMICAL BONDING
Moles (mol)
53.2/12.01
11.2/1.01
35.6/16.00
Mole ratio
=4.430
4.430/2.225
=11.09
11.09/2.225
=2.225
2.225/2.225
=1.99  2
=4.98 5
=1
Empirical Formula
C2H5O
Calculate the empirical formula mass: 45.1 g/mol
Molecular Mass = n x empirical formula mass
90.0 = n x 45.1
n=2
Molecular Formula: 2 x (C2H5O) = C4H10O2
(2r)
22.
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and
O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of CO 2 and 0.1159 g of
H2O. Determine the empirical formula of menthol.
Mass of C = 0.2829 g CO2 x 1 mol CO2 x 1 mol C
44.01 g CO2 1 mol CO2
x 12.01 g C = 0.0772 g C
1 mol C
Mass of H = 0.1159 g H2O x 1 mol H2O x 2 mol H
18.02 g H2O 1 mol H2O
x 1.01 g H = 0.0130 g H
1 mol H
Mass of O = 0.1005 g menthol - 0.0772 g C - 0.0130 g H = 0.0103 g O
Element
Mass (g)
Moles (mol)
C
0.0772
0.0772/12.01
H
0.0130
0.0130/1.01
O
0.0103
0.0103/16.00
Mole ratio
=6.43 x 10-3
0.00643/0.000644
=0.0129
0.0129/0.000644
=6.44 x 10-4
0.000644/0.000644
=10
=20
=1
Empirical Formula
C10H20O1
(2r)
SHE 1315
23.
CHEMICAL BONDING
Quinone is an organic compound containing only C, H, and O. Determine the empirical
formula of the compound if 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of
H2O when burned completely. Given a molecular weight of approximately 108 g/mol,
determine its molecular formula.
Mass of C = 0.257 g CO2 x 1 mol CO2 x 1 mol C
44.01 g CO2 1 mol CO2
x 12.01 g C = 0.0701 g C
1 mol C
Mass of H = 0.0350 g H2O x 1 mol H2O x 2 mol H
18.02 g H2O 1 mol H2O
x 1.01 g H = 0.0039 g H
1 mol H
Mass of O = 0.105 g menthol - 0.070 g C - 0.0039 g H = 0.031 g O
Element
C
H
O
0.0701
0.0039
0.031
Mass(g)
Moles (mol)
Mole ratio
0.0701/12.01
0.0039/1.01
0.031/16.00
= 0.0058
= 0.0039
=0.0019
0.0058/0.0019
0.0039/0.0019
0.0019/0.0019
=3
=2
=1
C3H2O1
Empirical Formula
Empirical formula mass = 3(12.01) + 2(1.01) + 1(16.00)
= 54.05 g
SHE 1315
CHEMICAL BONDING
Molecular formula
(C3H2O1)n = 108 g/mol
(54.05)n
n
= 108
= 108/54.05 = 2
Molecular formula = (C3H2O1)2 = C6H4O2
(2r)
SHE 1315
CHEMICAL BONDING
TOPIC 3: REACTIONS AND STOICHIOMETRY
1.
Define;
a)
(3a)
Acid-base, according to Arrhenius and Bronsted-Lowry theory
According to Arrhenius:
Acid is a substance that has H in its formula and dissociates in water to
yield H3O+.
Base is a substance that has OH in its formula and produces OH - in water.
According to Bronsted-Lowry Theory:
Acid is a proton donor, any species that donates an H + ion.
Base is a proton acceptor, any species that accepts an H+ ion.
b)
Redox reaction, in terms of electron transfer and oxidation number
Oxidation is a loss of electrons by a substance.
Reduction is a gain of electrons by a substance.
Oxidation occurs when there is an increase in oxidation number.
Reduction occurs when there is a decrease in oxidation number.
c)
Precipitation reaction
a reaction in which two soluble ionic compounds form an insoluble product,
a precipitate
2.
a)
List three examples for each strong and weak acids.
(3b)
Strong acids: HCl, HClO4 and HNO3
Weak acids: HF, H3PO4 and CH3COOH
b)
List three examples of strong bases from group 2A and one example of weak
base.
Strong bases: Ca(OH)2, Sr(OH)2 and Ba(OH)2
Weak base: NH3
3.
Using solubility guidelines, predict whether each of the following compounds is soluble or
insoluble in water.
(3b)
(a)
NiCl2
soluble
(c)
Cs3PO4
soluble
(b)
Ag2S
insoluble
(d)
SrCO3
insoluble
4.
Identify the type of reaction for the following reactions.
(3c)
SHE 1315
a)
b)
c)
d)
5.
CHEMICAL BONDING
Cu(OH)2 (s) + 2 HNO3 (aq)  Cu(NO3)2 (aq) + 2 H2O (l)
Fe2O3 (s) + 3 CO (g)  2 Fe (s) + 3 CO2 (g)
Sr(NO3)2 (aq) + H2SO4 (aq)  SrSO4 (s) + 2 HNO3 (aq)
4 Zn(s) + 10 H+(aq) + 2 NO3-(aq)  4 Zn2+(aq) + N2O(g) + 5H2O
(l)
Acid-base
Redox
Precipitation
Redox
Write a balanced chemical equation for each reaction and identify whether precipitation
occur.
(3c)
a)
Ni(NO3)2 and NaOH
Ni(NO3)2(aq) + 2 NaOH(aq)  Ni(OH)2 (s) + 2 NaNO3 (aq)
b)
NaOH and K2SO4
2NaOH(aq) + K2SO4 (aq)  Na2SO4
c)
(aq)
+ KOH
(s)
Na2S and Cu(C2H3O2)2
Na2S(aq) + Cu(C2H3O2)2(aq)  2 Na(C2H3O2)2(aq) + CuS(s)
6.
Balance the following equations:
a)
N2O5 (g) +
H2O (l) 
HNO3 (aq)
N2O5 (g) + H2O (l)  2HNO3 (aq)
b)
NH4NO3
(s)
2NH4NO3
c)
CH3NH2
(g)
4CH3NH2
7.
(3d)

(s)
O2
+
(g)
 2N2
+
(g)
N2
(g)
(g)
+ 9O2
+ O2

(g)
O2
(g)
(g)
CO2
 4CO2
+
H2O
+ 4H2O
(g)
(g)
(g)
+
H2O
(g)
+ 10H2O
(g)
(g)
+
N2
+ 2N2
(g)
(g)
Write balanced equations for each of the following chemical statements:
(a)
Chuncks of sodium react violently with water to form hydrogen gas and sodium
hydroxide solution
2 Na (s) + 2 H2O (l)
(b)
H2 (g) + 2 NaOH (aq)
Aqueous nitric acid reacts with calcium carbonate to form carbon dioxide, water and
aqueous calcium (II) nitrate
2 HNO3 (aq) + CaCO3 (s)
(c)
3d
Ca(NO3)2 (s)+ CO2 (g) + H2O (l)
Liquid nitroglycerine C3H5N3O9 explodes to produce a mixture of gases-carbon
dioxide, water vapor, nitrogen and oxygen
4C3H5N3O9 (l)
12 CO2 (g) + 10 H2O (g) + 6 N2 (g) + O2 (g)
SHE 1315
8.
CHEMICAL BONDING
Write the molecular, ionic and net ionic equations and identify the spectator ions for the
following reactions :
(3e)
a)
Potassium phosphate solution is mixed with calcium nitrate solution.
Molecular equation
2 K3PO4 (aq) + 3 Ca(NO3)2 (aq)  6KNO3 (aq) + Ca3(PO4)2
Ionic equation
6K+ (aq) + 2PO43-(aq)+ 3Ca2+(aq)+ 6NO3-(aq)  6K+
Net ionic equation
2PO43-(aq) + 3Ca2+(aq)  Ca3(PO4)2
Spectator ions: K+ and NO3-.
b)
(aq)
(s)
+ 6NO3-(aq) + Ca3(PO4)2
(s)
(s)
Mixing Al(NO3)3 solution with NaOH solution.
Molecular equation
Al(NO3)3 (aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
Ionic equation
Al3+(aq) + 3NO3-(aq) + 3Na+(aq) + 3OH-(aq)  Al(OH)3(s) + 3Na+(aq) + 3NO3-(aq)
Net ionic equation
Al3+(aq) + 3OH-(aq)  Al(OH)3(s)
Spectator ions: NO3- and Na+
9.
Consider the following reaction:
(3e)
Ba(NO3)2 (aq) + Na2CO3 (aq) → BaCO3 (s) + 2NaNO3 (aq)
(a)
(b)
Write the net ionic equation for the reaction.
Suggest two (2) pairs of reactants that can react and give the same net ionic
equation as in (a)
1) Molecular: BaCl2(aq) + K2CO3(aq) → BaCO3(s) + 2KCl(aq)
Total ionic: Ba2+(aq) + 2Cl–(aq) + 2K+(aq) + CO32– (aq) → BaCO3(s) +
2K+(aq) + 2Cl– (aq)
SHE 1315
CHEMICAL BONDING
2) Molecular: BaBr2(aq) + (NH4)2CO3(aq) → BaCO3(s) + 2NH4Br(aq)
Total ionic: Ba2+ (aq) + 2Br– (aq) + 2NH4+ (aq) + CO32– (aq) → BaCO3(s) +
2NH4+ (aq) +2Br– (aq)
* any acceptable answers
10.
For the following complete redox reactions:
a)
b)
write the half equations
identify the oxidizing and reducing agents
(i)
2 Sr + O2  2 SrO
(a)
(ii)
Oxidation: 2 Li  2 Li+ + 2 eReduction: H2 + 2 e-  2 H-
(b) Oxidizing agent: H2
Reducing agent: Li
3 Mg + N2  Mg3N2
(a)
11.
(b) Oxidizing agent: O2
Reducing agent: Sr
2 Li + H2  2 LiH
(a)
(iii)
Oxidation: 2 Sr  2 Sr2+ + 4 eReduction: O2 + 4 e-  2 O2-
(3f)
Oxidation: 3 Mg  3 Mg2+ + 6 eReduction: N2 + 6 e-  2 N3-
(b) Oxidizing agent: N2
Reducing agent: Mg
Balance the following redox reactions;
a.
(3f)
ClO3- (aq) + I-(aq) I2(s) + Cl-(aq) [acidic]
ClO3−(aq) + 6 H+(aq) + 6 I−(aq) → Cl−(aq) + 3 H2O(l) + 3 I2(s)
b.
MnO4- (aq) + SO2-3(aq) MnO2(s) + SO42-(aq) [basic]
2 MnO4−(aq) + H2O(l) + 3 SO32−(aq) → 2 MnO2(s) + 3 SO42−(aq) + 2 OH−(aq)
c.
MnO4- (aq) + H2O2(aq) Mn2+(aq) + O2(g) [acidic]
SHE 1315
CHEMICAL BONDING
2 MnO4−(aq) + 6 H+(aq) + 5 H2O2(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)
12.
16.0 g of methane, CH4 react with 48.0 g of oxygen, O2 as shown in the equation below:
(3h)
CH4 + 2 O2
a)
CO2 + 2 H2O
Define limiting reactant and excess reactant
Reactant that limits the amount of the other reactant that can react, and
thus the amount of product that can form/ Reactant that yields the lower
amount of product.
Reactants that remains after a reaction is over
b)
Determine the limiting reactant.
16.0 g CH4
x 1 mol CH4 x 1 mol CO2 x 44.01 g CO2
16.032 g CH4 1 mol CH4
1 mol CO2
= 43.9222 g CO2
48.0 g O2 x 1 mol O2 x 1 mol CO2 x 44.01 g CO2
32.00 g O2 2 mol O2
1 mol CO2
= 33.006 g CO2
O2 is the limiting reactant. O2 form fewer products CO2.
c)
Calculate the mass of CO2 produced, in grams.
Theoretical yield O2 = 33.006 g CO2
13.
Ammonia can be synthesized by the following reaction.
NO (g) + H2 (g)
(3h)
NH3 + H2O (g)
(a)
Write a balanced chemical equation for the above reaction.
2 NO (g) +5 H2 (g)
2 NH3 + 2 H2O (g)
(b)
What is the maximum amount (g) of ammonia in grams that can be synthesized
from 45.8 g of NO and 12.4g of H2?
Mass of NH3 = 45.8 g NO x 1 mol NO x 2 mol NH3 x 17.04 g NH3 = 26.0 g NH3
30.01 g
2 mol NO
1 mol NH3
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CHEMICAL BONDING
Mass of NH3 = 12.4 g H2 x 1 mol H2
x 2mol NH3 x 17.04 g NH3 = 41.8 g
2.02 g
5 mol H2
1 mol NH3
NO is the limiting reactant. It produces less amount of NH3 = 26.0 g NH3
(c)
Calculate the amount (g) of the excess reactant remains after the reaction has
stopped?
Mass of H2 – Mass of H2 react with NO
= 12.4 g – (45.80 g NO x
14.
1 mol NO
x 5 mol H2
30.01 g NO
2 mol NO
x 2.02 g H2) = 4.6929 g
1 mol H2
Define theoretical yield, actual yield and percentage yield.
(3i)
Theoretical yield
Amount of product that can be made
in a chemical reaction based on the
amount of limiting reactant.
Actual yield
Amount of product actually produced
by a chemical reaction
Percentage yield
%yield= actual yield
x100
theoretical yield
15.
The combustion of methane produces carbon dioxide and water. Assume that 2.00 mol of
CH4 burned in the presence of excess air. Calculate the percentage yield if the reaction
produces 87.0 g of CO2.
(3j)
CH4 + 2 O2
CO2 + 2 H2O
Theoretical yield = 2 mol CH4 x 1 mol CO2 x 44 g CO2 = 88g CO2
1mol CH4
% yield = 87.0g/88.0g x 100 = 99%
SHE 1315
16.
CHEMICAL BONDING
70.0 g of manganese dioxide, MnO2 is mixed with 3.50 mol of hydrochloric acid. How many
grams of Cl2 will be produced from this reaction if the % yield for the process is 42.0%?
(3j)
MnO2 + 4 HCl  MnCl2 + 2 H2O + Cl2
70 g MnO2 x 1 mol MnO2 x 1 mol Cl2
x 70.9 g Cl2 = 57.085 g
86.94 g mol-1 1 mol MnO
1 mol Cl2
3.5 mol HCl x 1 mol Cl2 x 70.9 g Cl2 = 62.038 g
4 mol HCl
1 mol Cl2
MnO2 is the limiting reactant because it forms less Cl2.
% yield = actual yield / theoretical yield x 100
Actual yield = 57.085 = 23.98 g Cl2
42%
17.
Complete the following table by giving the definition for each concentration terms.
(3k)
Concentration term
Definition
Amount (Mol) of solute
Molarity (M)
Volume (L) of solution
Amount (mol) of solute
Molality (m)
Percentage by mass
(%w/w)
Mass (kg) of solvent
Mass of solute
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CHEMICAL BONDING
Mass of solution
Volume of solute
Percentage by volume
(%v/v)
18.
Volume of solution
36.25 mL of a standard 0.1750 M NaOH solution is required to neutralize 25.00 mL of H 2SO4
,calculate the molarity of the acid solution.
(3l)
2 NaOH (aq) + H2SO4(aq)  Na2SO4(aq) + 2 H2O(l)
Mol=MV=0.1750M X 0.03625 L= 6.343 X 10-3 mol NaOH
6.343 X 10-3 mol NaOH x 1 H2SO4
= 3.1719 X 10-3 mol H2SO4
2 mol NaOH
M= 3.1719 X 10-3 mol H2SO4 =0.1269 M
0.025
19.
In a laboratory, 6.58 g of barium nitrate Ba(NO3)2 (molar mass = 261.36) is dissolved in
enough water to form 0.850 L of solution. A 0.100 L sample is transferred from this stock
and titrated with 0.051 M solution of K2SO4. What volume of K2SO4 solution is required to
form BaSO4 precipitate?
(3l)
Ba(NO3)2 + K2SO4
BaSO4 + 2 KNO3
Mol Ba(NO3)2 =
6.58 g
261.36 g mol-1
= 0.0252 mol
M Ba(NO3)2 = 0.0252 mol = 0.0296 M
0.850 L
MaVa = a
MbVb
b
= 0.100 L x 0.0296 M = 0.051 M x Vb
Vb = 0.0581L
SHE 1315
CHEMICAL BONDING
TOPIC 4: GASES AND KINETIC MOLECULAR THEORY
1.
a)
State the Avogadro’s Law.
At fixed temperature and pressure, equal volumes of
any ideal gas contain equal no. of particles.
(4a)
b)
Consider a 1.0 L flask containing neon gas and a 1.5 L flask containing xenon gas.
Both gases are at the same pressure and temperature. According to Avogadro’s Law,
what can be said about the ratio of the number of atoms in the two flasks?
V α n (T & P constant)
Since volume of Ne : Xe = 1 : 1.5
So, no. of atoms Ne : Xe = 1: 1.5
2.
Write the correct unit(s) for the given gas constant R :
(4b)
a) 8.314 Jmol-1K-1, m3Pamol-1K-1
b) 0.08206 Latm mol-1K-1
3.
2.50 L of a gas was at an unknown pressure. However, at standard pressure, its volume
was determined to be 8.00 L. What was the unknown pressure?
(4a)
P1V1 = P2V2
P1(2.50 L) = (101.325 kPa) (8.00 L)
P1= (101.325 kPa) (8.00 L) / (2.50 L)
= 324.24 kPa
4.
5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L. What is the new
temperature in order to maintain the same pressure?
(4a)
V1 / T1 = V2 / T2
5.00 L/100 K = 20.0 L/T2
T2 =400 K
SHE 1315
5.
CHEMICAL BONDING
State the temperature (C), pressure (atm) and volume (L) of a gas at Standard
Temperature and Pressure (STP)
(4c)
T=0C
6.
P=1 atm V =22.4 L
Calculate the density of UF6 at STP.
(4c)
d= 352.03 gmol-1 /22.4 Lmol-1 =15.7 g/L
7.
Dry ice is a solid carbon dioxide. A 0.050 g sample of dry ice is placed in an evacuated 4.6 L
vessel at 30oC. Calculate the pressure inside the vessel after all the dry ice has been
converted to CO2 gas.
(4c)
0.050 g X 1 mol
44.01 g
X 0.0821 L atm mol-1 K-1 X (30 + 273)K
4.6 L
P = 6.1 × 10−3 atm
8.
At 741 torr and 440C, 7.10 g of a gas occupy a volume of 5.40 L. Calculate the molar mass
of the gas.
(4c)
T = 44° + 273 = 317 K
P=(741 torr /760 torr) X 1 atm = 0.975 atm
n= 0.975 atm X 5.40 L
= 0.202 mol
0.0821 L atm mol-1 K-1 X 317 K
Ar= 7.10/ 0.202 = 35.15 gmol-1
9.
Calculate the total mass (grams) of oxygen in a room measuring, 4.0 m x 5.0 m x 2.5 m.
Assume that the gas is at STP and the air contains 20.95 % oxygen by volume.
(4c)
3
3
[1m = 1000 dm ]
PV = nRT
; P= 1 atm
T= 273 K
V room= 4.0 m x 5.0 m x 2.5 m = 50 m 3
V O2 = 20.95%x 50 m3 = 10.475 m3
n= PV/RT
=
1 atm (10475 L)
0.0821 Latm mol-1 K-1 (273 K)
= 467.36 mol
SHE 1315
CHEMICAL BONDING
Mass O2 = 467.36 mol x 32.00 g = 1.5 x 104 g
1 mol
10.
One mole of any gas at STP has a volume of 22.4 L. Calculate the densities of the following
gases, in grams per liter at STP.
(4c)
a)
CH4
d= 16.05 gmol-1/22.4 Lmol-1 = 0.717 g/L
b)
CO2
d= 44.01 gmol-1/22.4 Lmol-1 = 1.96 g/L
c)
UF6
d= 352.03 gmol-1/22.4 Lmol-1 = 15.7 g/L
11.
A 0.98478 g of unknown gas is placed in a 1.500 L bulb at a pressure of 356 mmHg and a
temperature of 22.5 0C. Calculate the molar mass of the gas.
(4c)
V= 1.500 L
P= 356 mmHg x
1 atm = 0.468 atm
760 mmHg
T= 295.5 K
m= 0.9847 g
M= mRT/PV
= 0.9847 g (0.0821 L atm mol-1 K-1)(295.5 K)
0.468 atm (1.500 L)
= 34.03 g/mol
12.
Hydrogen gas can be prepared by a reaction of zinc metal with aqueous HCl. (4c)
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CHEMICAL BONDING
Zn
(s)
+ 2 HCl
(aq)
ZnCl2
(aq)
+ H2
(g)
a) Calculate how many liters of H2 would be formed at 742 mmHg and 15.0 0C, if 25.5 g
of zinc was allowed to react.
P= 742 mmHg x
1 atm = 0.976 atm
760 mmHg
T= 15 0C + 273 = 288 K
n Zn= 25.5 g x 1 mol = 0.390 mol Zn
65.39 g
From the balanced equation, 1 mol Zn reacts to produce 1 mol H 2.
Therefore, 0.390 mol Zn reacts to produce 0.390 mol H2.
V H2= nRT/P
= 0.390 mol (0.0821 L atm mol-1 K-1) (288 K)
= 9.45 L
0.976 atm
b)
Calculate how many grams of zinc would you need to produce 5.00 L of
H2 at 350 mmHg and 30.0 0C.
V H2=5.00 L
P = 350 mmHg x
1 atm = 0.461 atm
760 mmHg
T= 30.0 0C + 273 = 303.0 K
n H2 = PV/RT
=
0.461 atm (5.00 L)
SHE 1315
CHEMICAL BONDING
(0.0821 L atm mol-1 K-1)(303K)
= 0.0926 mol H2
From the balanced equation, 1 mol H2 is produced from the reaction of 1
molZn.
Therefore, 0.0926 mol H2 is produced from the reaction of 0.0926 mol Zn.
Mass Zn = 0.0926 mol x 65.39 g = 6.06 g Zn
1 mol
13.
In an industrial lab simulation, 135 mL of N2 gas was collected over water at 250C and 755
torr. Calculate;
(4e)
2 NaN3 (s)
2 Na (s) + 3N2 (g)
(i)
(ii)
the pressure of the nitrogen
mass of azide (NaN3) decomposed (Vapor pressure of the water=23.76 torr)
(i)
PN2= (755-23.76) torr= 731.24 torr
(ii)
nN2= (0.9621 atm) X (0.135 L) = 5.31 X 10-3
0.0821 L atm mol-1 K-1 X 298 K
mass
NaN3=
5.31 X 10-3mol N2 X 2 mol NaN3 X 65.02 g NaN3
3 mol N2
1mol NaN3
=0.2302 g
13. In a laboratory, a sample of benzene was heated and vaporized to 100°C in a flask to reach
a volume of 247.2 mL and drove all air from the flask. When the benzene was condensed to
a liquid, its mass was found to be 0.616 g. The reading of barometric pressure was 742
mmHg. Calculate the molar mass of the gas.
(4c)
V = 247.2 cm3 x 1 L/10 3 cm3 = 0.2472 L
T = (273+ 100) K = 373 K
P = 742 mmHg x 1.00 atm/760 mmHg = 0.976 atm
n = PV/RT = 0.976 atm x 0.2472 L/0.0820Latmmol-1K-1 x 373 K
= 7.89 x 10-3 mol
M = m/n =0.616 g/7.89 x 10-3 mol = 78.0735 gmol-1
SHE 1315
CHEMICAL BONDING
14. A series of measurements are made in order to determine the molar mass of unknown
halogen gas. First, a large flask is evacuated and found to weight 134.567 g. It is then filled
with the gas to a pressure of 735 torr at 31 oC and reweighed; its mass now 137.456 g.
Finally, the flask is filled with water at 31oC and found to weight 1067.9 g. the [density of
water at this temperature is 0.997 g/ml]. Calculate the molar mass of unknown gas. (4c)
Flask=134.56 g
Flask + gas = 137.456 g
Flask + water = 1067.9 g
Mass of unknown gas =137.456 g – 134.567 g = 2.889 g
Mass of water = 1067.9 g – 134.567 g = 933.33 g
Volume of flask;
0.997= 933.34 = 936.15 ml or 0.9361 L
V
T = 31° + 273 = 304 K
P=(735 torr /760 torr) X 1 atm = 0.967 atm
16.
When an evacuated 63.8 mL glass bulb filled with a gas at 220C and 747mmHg,
the bulb gains 0.0725 g in mass. Identify the gas.
(4c)
n= (0.9829 atm) X (0.0638 L) = 2.5892 X 10-3 moles
0.0821 L atm mol-1 K-1 X 295 K
Mr= 0.0725/ 2.5892 X 10-3 = 28.00 gmol-1 N2 gas
17.
a)
State Dalton’s Law of partial pressure.
(4d, 4e)
In a mixture of gases, the total pressure of the mixture is the sum of the
partial pressure of the constituent gases.
PT = PA + PB + PC +…
Where PA, PB , PC = partial pressuere of A, B and C.
PT =total pressure of the mixture.
b)
What do you understand by the term partial pressure?
Is the pressure that each gas would exert if it alone occupy the container
under the same temperature
SHE 1315
CHEMICAL BONDING
(c)
18.
Complete the following table for an ideal gas:
P
V
n
T
2.00 atm
1.00 L
0.500 mol
48.7 K
0.300 atm
0.250 L
3.05 x 10-3 mol
27 0C
650 torr
11.2 dm3
0.333 mol
350 K
1.04 x 106 Pa
585 mL
0.250 mol
295 K
A 250 mL test tube was used to collect hydrogen gas over water. Atmospheric pressure was
102.4 kPa and the temperature of water was 25°C .Calculate
(4e)
i)
ii)
the pressure of the hydrogen
the number of moles of hydrogen for this experiment.
(Vapor pressure of the water=23.76 torr)
P= 23.76 torr= 3.17 kPa
V= 250 mL =0.250 L
(1 mmHg= 7.5 kPa)
T=25 + 273 K = 298 K.
PTotal = PWater + PHydrogen
102.4 kPa = 3.17 kPa + PHydrogen
PHydrogen= 99.23 kPa
PV=nRT
99.2 kPa x 0.250 L = n x 8.31 L-1kPa/mol-1-K x 298 K
n = 99.2 kPa x 0.250 L / 8.31 L-1kPa/mol-1K / 298 K
n = 0.0100 mol or 1.00 x 10-2 mol H2
21.
The stopcock between the two containers is opened and the gases are allowed to
SHE 1315
CHEMICAL BONDING
mix. [Assume no reaction between N2 and O2]
a)
(4e)
Calculate the partial pressures of N2 and O2 after mixing.
P N2= 1.0 atm (2.0L)= 0.4 atm
5.0 L
P O2= 2.0 atm (3.0L)= 1.2 atm
5.0 L
b)
Calculate the total pressure in the container after the gases mix.
PT = PN2 + PO2
= 0.4 atm + 1.2 atm = 1.6 atm
c)
Determine the mole fraction of each gas.
X N2 = 0.4/1.6
= 0.25
X O2 = 1.2/1.6
= 0.75
22.
A closed bulb contains 0.0100 mol of inert helium gas and a solid white ammonium chloride,
NH4Cl. Assume that the volume of the solid NH4Cl is negligible compared to the volume of
the bulb. The pressure of the helium gas at 27.0 °C is found to be 114 mmHg. The bulb is
then heated at 327 °C. All the NH4Cl decomposes according to the equation:
NH4Cl (s)  NH3 (g) + HCl (g)
The final total pressure in the bulb after complete decomposition of the solid is 908 mmHg.
a)
Calculate the pressure of helium gas at 327 °C.
P1 = 114 mmHg
T1 = 27 + 273 K
= 300 K
P2 = P 1 x
T2
T1
= 114 mmHg x
b)
P2 = ?
T2 = 327 + 273 K
= 600 K
600 K =
300 K
228 mmHg
What is the partial pressure of HCl (g)?
From the eq:
Mole of NH3 =
Therefore:
Mole of HCl
SHE 1315
CHEMICAL BONDING
Then:
P (NH3) = P (HCl) = x
Ptotal = P (He) + P (NH 3) + P (HCl)
908 = 228
+ x
+ x
0r
2x
Hence:
c)
= 908 – 228 mmHg
= 680 mmHg
x = 680 mmHg = 340 mmHg
2
Calculate the number moles of HCl (g) produced in the reaction.
Since V and T are the same for all gases (He & HCl)
Hence:
V = n (HCl) = n (He)
RT
P (HCl)
P (He)
Therefore:
n (HCl) = n (He) x P (HCl)
P (He)
= 0.01 mol x 340 mmHg = 0.0149 mol
228 mmHg
d)
24.
How many grams of NH4Cl (s) were in the bulb at 27.0 °C?
Since all of NH4Cl was converted to HCl & NH3,
therefore:
n (NH4Cl) = n (HCl)
= 0.0149 mol
Hence:
Mass of
= 0.0149 mol x 53.5 g/mol = 0.798 g
NH4Cl
State two assumptions of the kinetic theory of gases
(4f)
1. The volume of gas molecules is very small and can be neglected when
compared to the volume of the container.
2. No attractive forces exist between molecules.
25.
State the two conditions will cause the greatest deviation from the ideal gas law.
At high pressure, and low temperature
(4f)
26.
Indicate which of the following statements regarding the kinetic-molecular theory of gases
are correct.
a)
Ideal gas behavior is better achieved at low pressure and high temperature. Correct
b)
The kinetic energy of a molecule is directly proportional to its temperature. Correct
c) The gas molecules are assumed to exert no forces on each other.
Correct
d)
The volume of the gas molecules is negligible in comparison to the total volume in
which the gas is contained.
Correct
SHE 1315
27.
State the two conditions where real gases behave as ideal gases. Explain.
CHEMICAL BONDING
(4g)
At low P, and high T
When the temperature is high the kinetic energy of molecules increases and the
intermolecular
attractions
among
the
atoms
decrease.
The volume of the gas molecules become negligible compared to volume of the
vessel. Therefore the real gases act like ideal at higher temperatures.
At low temperatures volume of the container is larger. Therefore intermolecular
attractive forces are negligible and the volume of the particles also become
negligible compared with the volume of the vessel. Therefore the real gases act
like ideal at lower temperatures.
SHE 1315
CHEMICAL BONDING
TOPIC 5: ATOMIC STRUCTURE
1.
Write a full set of quantum numbers (n, l, ml, ms) for:
a)
(5a)
the outermost electron in a Li atom
n=2 ; l=0 ; ml=0 ; ms=+1/2
b)
the electron gained when a Br atom becomes a Br- ion
n=4 ; l=1 ; ml=+1
c)
3.
;
ms=+1/2
the highest energy electron in the ground- state of B atom
n=2
2.
ms=-1/2
the electron lost when a Cs atom ionizes
n=6 ; l=0 ; ml=0
d)
;
;
l=1
;
ml=-1
;
ms=+1/2
Give all possible ml values for orbital that have each of the following :
a)
l=3
b)
n=2
c)
n=1; l=1
ml= -3,-2,-1,0,+1,+2,+3
l=0 , ml=0
l=1 , ml= -1,0,+1
ml=-1,0,+1
Draw the following orbital with the correct orientation:
a)
(5a)
1s
c)
(5b)
2py
y
x
x
y
z
b)
2s
d)
2px
y
z
x
x
y
SHE 1315
4.
CHEMICAL BONDING
Identify the types of orbital given below:
dx2-y2
5.
(5c)
dz2
Write the condensed ground-state electron configuration and draw orbital diagram showing
valence electrons for:
(5d)
a)
Sr
[Kr]5s2
↑↓
5s
b)
Rh
[Kr] 5s2 4d7
,
↑↓
↑↓
↑↓
↓5s
c)
Cu
↑
↑
4d
[Ar] 4s1 3d5
↑
4s
6.
↑
↑↓
↑
↑↓
↑↓
↑↓
↑↓
3d
Write the full electronic configuration for:
(5d)
SHE 1315
7.
CHEMICAL BONDING
a)
S2-
b)
Ca2+
1s22s22p63s23p6
1s22s22p63s23p6
c)
Ge
1s22s23s23p64s23d104p2
d)
Cr
1s22s2p63s23p64s13d5
How many inner, outer and valence electrons are present in each of the following elements?
(5e)
No.
Elements
Inner electrons
Outer electrons
Valence electrons
8.
a)
Br
28
7
7
Defi
ne:
b)
Cs
54
1
1
c)
Cr
18
1
6
d)
Sr
36
2
2
e)
F
2
7
7
(5h)
a)
Atomic radii
The size of an atom as measured by the distance from the
nucleus to the boundary of the surrounding cloud of electrons.
b)
Ionic radii
The size of an ion as measured by the distance between the
nuclei of adjacent ions in a crystalline ionic compound.
c)
Effective nuclear charge (Zeff)
The nuclear charge an electron actually experiences as a result
of shielding effects due to the presence of other electrons.
9.
a)
Define isoelectronic species.
Atoms or ions with the same electronic configuration.
(5h)
SHE 1315
b)
CHEMICAL BONDING
Given Na+, Mg2+, Al3+, P3- , S2- and Cl- ;
(i)
(5i)
Arrange these isoelectronic ions in the order of increasing size.
P3- > S2- > Cl- > Na+ > Mg2+ > Al3+> Si4+
(ii)
Explain your answer in b(i).
All the cations have ionic radius smaller than the anions. This is
because the anions have higher principle quantum number for
the valence electron (n=3) compared to cations (n=2).
10.
Compare the sizes of cation and anion to their parent atoms.
(5i)
Cation: has lost an electron, therefore electrostatic repulsion between electrons is
decreased and the size of the ion shrinks. The cation of an element is smaller than
the neutral atom of the same element.
Anion: has gained an electron, therefore electrostatic repulsion between electrons
is increased and the size of the ion increases. The anion of an element is larger
than the neutral atom of the same element.
11.
a)
Define first ionization energy, IE1.
(5j)
Ionization energy is the minimum energy needed to remove 1 mol of electron from
1
mole
of
gaseous
atom.
b)
State two factors that influence the IE.
i.
ii.
Principal quantum number, n
Effective nuclear charge, Zeff
SHE 1315
12.
CHEMICAL BONDING
Given sodium and potassium, which would have a larger first ionization energy? Explain.
(5k)
Na has the larger 1st ionization energy because the valence electron is located in
the period 3 compared to K. Valence electron in K is located in period 4. Valence
electron in Na feels stronger nucleus attraction than in the K atom. Thus harder
to remove an electron from Na atom than in K atom.
13.
Explain the difference in first ionization energy between Beryllium, Be and Boron, B.
(5k)
Be : 1s2 2s2
B : 1s2 2s2 2p1
Be have higher 1st ionization energy than B because of the stability of fulfilled
orbital of 2s. The harder to remove the electron and the more energy needed.
14.
Briefly explain the general trend of atomic radii across the period, from left to right.
(5k)
Across the period, atomic radii decrease because Zeff is increasing, nucleus
attraction is stronger, thus the outermost electron in group 8A (18) is more tightly
held than that in group 1A(1). Group 1A(1) atoms are larger than those of group
8A(18) atoms.
15.
Five successive ionization energies (kJ mol -1) for atom M is shown below:
Determine:
IE
1
2
3
4
5
(kJmol-1)
800
1580
3230
4360
16000
(5l, 5m)
SHE 1315
a)
CHEMICAL BONDING
Valence electron configuration for M.
IE 2
1580
 1.98

800
IE1
IE 3
3230
 2.04

1580
IE 2
IE 4
4360
 1.34

3230
IE 3
IE 5
16000
 3.67

4360
IE 4
The first, second, third and fourth electron are removed from the
same energy subshell. The fifth electron is removed from an inner
subshell, hence it requires a higher IE5 (3.67 times) than IE4 since
there is a large jump in the IE5. Since IE5 / IE4 have the highest ratio,
4 valence electrons are present. Valence electron configuration: ns2
np2
b)
Group number of M in the periodic table.
Group 14
16.
The first four successive ionization energies of Period 3 element, A is shown below:
(5f, 5l, 5m)
a)
IE
1
2
3
4
(kJmol-1)
576
1810
2740
11600
In which group does A appear in the periodic table? Explain.
SHE 1315
CHEMICAL BONDING
3A (13). Because this is where the highest jump/significant jump in IE
occurred. Attempting to remove inner core electron requires larger amounts
of energy.
b)
Write the valence electronic configuration for element A.
ns2np1
17
(a)
Explain with reasons the variation in ionization energies of an unknown
element as displayed in the following graph:
(5f, 5l, 5m)
From the 1st to 4th electrons removed, the IE increases gradually. This is
because the remaining electron is more attracted to the more positively
charged ion. Hence, it needs higher energy to remove the electron.
But there is a sharp increase of IE for the 5th electron removed. This shows
that the 5th electron is in the inner shell which is nearer and more attracted
to the nucleus. So it is harder to remove the electron causes very high
energy needed to remove the electron.
Then, to remove the 6th electron, the IE increases gradually since the
electron is in the same shell as the 5th electron., but still the IE is higher due
to the attraction of the electron to the more positively charged ion.
(b)
If the element is located in Period 2, identify the element.
Carbon, C.
(c)
Write the valence electron configuration of the element.
2s2 2p2
SHE 1315
18.
CHEMICAL BONDING
An element X has the following successive ionization energies.
a)
IE1
IE2
IE3
IE4
IE5
738
1450
7732
10 539
13 628
(5f, 5l, 5m)
Write the valence electronic configuration for X.
ns2
b)
What is the group number of X?
2
c)
If X is located in Period 3, write the full electronic configuration for X.
1s2 2s2 2p6 3s2
d)
Name the X.
Mg
19.
a)
Define first and second electron affinity
(5n)
Electron affinity is is the amount of energy change when 1 mole of electron
added to 1 mole of gaseous atom
Electron affinity is is the amount of energy release or absorbed
1 mole of electron added to 1 mole of gaseous ion.
b)
Write a balanced chemical equation to represent the first and second electron
affinity process for element Y.
when
SHE 1315
CHEMICAL BONDING
Y (g) + e¯  Y-(g)
c)
Second electron affinity always has a positive value (an endothermic
process). Explain.
(5o)
When an electron is added to a negative ion (anion), a strong
repulsion is felt, and the energy of the system increases. Thus,
heat
is absorbed to overcome the electrostatic repulsion
between the electron and the negative charge on the
anion.
This phenomenon leads to a positive value for 2 nd
electron
affinity.
SHE 1315
CHEMICAL BONDING
TOPIC 6: CHEMICAL BONDING
1.
Use Lewis electron dot symbols to depict ions formed from the following atoms and predict
the chemical formula of their compound.
(6d)
a) Na and F
b) Ca and O
a)
Na
+
b)
Ca
+ .O
.
Sr
+
c)
F
Br
c) Sr and Br
Na+ [ F ]-
NaF
Ca2+ [ O ]2-
CaO
Sr2+ 2[ Br ]
d) K and O
SrBr2
Br
d)
K
22 [K]+ [ O ]
O
K2O
K
2.
Write the Lewis dot symbols for:
(6c)
a) Ca
b) Ca2+
c) O
d) O2-
2+
Ca
3.
Ca
2-
O
Write the Lewis structures for:
a) AsCl3
b) O3
O
(6g)
c) BI3
d) SeF4
SHE 1315
4.
a)
CHEMICAL BONDING
Define coordinate covalent bond.
(6g)
Coordinate covalent bond : A covalent bond formed when a pair of electron
is contributed by only one of the bonded atoms
b)
State two requirements for the formation of a coordinate covalent bond.
Two requirements:
a) The donor atom must have a lone pair of electrons
b) The acceptor atom must have an empty orbital to accommodate the pair
of electrons.
5.
6.
Boron trifluoride, BF3 accepts an electron pair from ammonia, NH3 to form BF3NH3.
a)
Draw Lewis structures to illustrate the formation of BF3NH3 from boron trifluoride and
ammonia.
b)
Show which of the bond is the coordinate covalent bond.
(6g)
The following species do not obey the octet rule:
a)
ICl2-
b)
NO2
c)
BBr3
Draw a Lewis structure for each species and state the type of octet rule exception. (6l)
SHE 1315
7.
CHEMICAL BONDING
Rank the bonds in each set in order of increasing bond length and bond energy.
a) C-I, C-F, C-Br, C-Cl
b) C-S, C=O, C-O
c) N-H, N-S, N-O
bond length
a) C–F< C–Cl < C–Br<C-I
b) C=O <C–O < C–S
c) N–H < N–O < N–S
8.
(6h)
bond energy
C-I<C-Br<C-Cl<C-F
C–S < C–O < C=O
N–S < N–O < N–H
Match the following bonds with its bond length by using arrows to show your answers.
(6h)
9.
Calculate the formal charges of each atom in the following molecules and determine
whether it represents a neutral molecule or ion.
(6k)
Structure A
Structure B
H : 1-(1/2 2 + 0) = 0
H-O : 6-(1/2 4 + 4) = 0
N=O : 6-(1/2 4 + 4) = 0
N : 5-(1/2 8 + 0) = +1
O : 6-(1/2 2 + 6) = -1
Thus, it is a neutral molecule
10.
Define resonance structures.
H : 1-(1/2 2 + 0) = 0
C : 4-(1/2 8 + 0) = 0
O= : 6-(1/2 4 + 4) = 0
O- : 6-(1/2 2 + 6) = -1
Thus, it is -1 ion.
(6m)
Structures with the same placements of atoms but different locations of bonding
pair and lone pair electrons.
11.
Draw the resonance structures of:
a)
SO3
b)
N3-
SHE 1315
CHEMICAL BONDING
(6n)
12.
A Lewis structure of CS2 usually is written as S=C=S rather than S-CS. Explain in term of
formal charge.
(6k)
Formal charges:
Formal charges:
Sulphur : 6-(1/2 4 + 4) = 0
Carbon : 4-(1/2 8 + 0) = 0
Sulphur (singly bonded) : 6-(1/2 2 + 6) = -1
Sulphur (triply bonded) : 6-(1/2 6 + 2) = +1
Carbon : 4-(1/2 8 + 0) = 0
The structure S=C=S is preferred because each atom has a formal charge.
13.
Use the VSEPR model to predict the geometry of the following molecules and ions.
a) AsH3
d) I3-
g) IF2-
b) OF2
e) C2H4
h) SF6
c) AlCl4-
f) SnCl3-
i) TeF5-
Trigonal pyramidal
linear
octahedral
square pyramidal
(6p)
SHE 1315
14.
CHEMICAL BONDING
Nitrogen and phosphorus are the first two elements in Group 15 of the periodic table.
Nitrogen can form nitrogen trichloride, NCl3 only while phosphorus can forms both
phosphorus trichloride, PCl3 and phosphorus pentachloride, PCl5.
(6q, 6l)
a)
Write the electronic configuration of the nitrogen and phosphorus atom.
N : 1s22s22p3 , P : 1s22s22p63s23p3
b)
Draw the Lewis structures for NCl 3 and PCl5. Predict their shapes.
: :
:
N
:
:
: Cl
Cl :
:Cl :
:
Trigonal pyramidal
c)
Trigonal Bipyramidal
Explain why phosphorus can form the PCl5 molecule, but not nitrogen.
N atom cannot expand its octet, P atom can expand its octet because of the
availability of 3d subshell.
15.
Define electronegativity.
(6i)
Electronegativity is the relative ability of a bonded atom to attract shared
electron/ Is the tendency of a bonded atom to attract electrons.
16.
State the factors that affect the polarity of a molecule.
(6j)
Shape of molecule & bond polarity / the difference in electronegativities between
atoms in a bond
17.
BF3 is a non-polar molecule, whereas PF3 is a polar molecule. Explain
(6r)
The molecular shape of BF3 is trigonal planar which is symmetrical. Although each
bond is polar, the bond dipole moment cancels each other. Therefore, BF3 is non
polar molecule. The molecular of PF 3 is trigonal pyramidal which is not
symmetrical. The bond dipole moments do not cancel each other thus PF3 is a
polar molecule.
SHE 1315
18.
CHEMICAL BONDING
Determine whether these molecules is polar or non-polar.
(6r)
c) PCl5 non polar
a) Br2 non-polar
b) HBr
d) XeF4 non-polar
e) SO2 polar
f) SF6
non polar
g) PH3
h) SO3 non polar
i) ClF3
polar
polar
polar
19.
a)
What is the hybridization of each C and O atom in the molecule? (6t, 6q, 6u)
b)
How many C atoms have a tetrahedral shape?
c)
How many  and π bonds are in the molecule? = 14, π= 3
3
d)
20.
a)
Describe the hybridization process of the central atom in AlI3.
(6s)
b)
Draw the orbital overlapping diagram of AlI3
(6s)
SHE 1315
21.
CHEMICAL BONDING
State the hybrid orbital of the underline atoms and the bond angle for the following
substance:
(6o)
a)
SO32-
c)
sp3, 109.5
b)
CH3—CH=CH-Cl
sp3, 109.5 ; sp2, 120 ; sp2, 120
CH3—CΞN
sp3, 109.5 ; sp, 180
d)
Br—CH2—CH2—COOH
sp3, 109.5 ; sp3, 109.5 ; sp2, 120
22.
Describe the hybridization process of the atoms in N2.
(6s)
23.
Define the term metallic bond.
(6b)
A metallic bond is the electrostatic attraction between the positively charged
metal and the sea of delocalized electron.
24.
The melting point of aluminum is higher than sodium. Explain.
(6e)
Both aluminium and sodium are metallic. The strength of metallic bond depends
on the number of valence electrons. Since aluminium has three valence electrons
and sodium has one valence electron, the metallic bonds in aluminium are
stronger than sodium. Therefore, aluminium has a higher melting point than that
of sodium
25.
All metals are good conductor electricity. Explain.
The free electrons carry the charge through the metals.
(6f)
SHE 1315
CHEMISTRY 1
TOPIC 7: INTERMOLECULAR FORCES, LIQUIDS AND SOLIDS
1.
Water molecules can form hydrogen bonding.
a)
Define hydrogen bonding. 7(a)
Attraction between molecules having a hydrogen atom bonded to a small
and highly electronegative atom (F, O and N), with the lone pair of
electrons.
b)
Explain how hydrogen bonding affects the boiling point of water. 7(b)
More energy is required to break the strong hydrogen bonding between
water molecules. This makes the boiling point of water higher than
expected from its small molecular size.
2.
3.
Name all the intermolecular forces that exist between the particles of the following liquids:
7(b)
a)
trichloromethane, CHCl3
Dipole dipole forces and London dispersion forces
b)
ethanol, CH3CH2OH
hydrogen bonding, dipole dipole forces and London dispersion
c)
aluminium trichloride, AlCl3
London dispersion forces
Explain the following observations in terms of intermolecular forces: 7(e)
a)
The boiling point of water, H2O is higher than, H2S hydrogen sulphide.
H2O : hydrogen bonding
H2S : London Dispersion forces and Dipole dipole forces
Hydrogen bond is stronger.Thus more energy is needed to break the
hydrogen bonds.
b)
The molecular mass of butane and propanone are both 58. However, the boiling
point of butane is 273 K while that of propanone is 330 K.
Butane : Non-polar molecule. Intermolecular force is the weak London
Dispersion forces
Propanone : polar molecule. Intermolecular force is Dipole-dipole forces.
As dipole-dipole forces is stronger than London dispersion forces,
propanone has higher boiling point.
SHE 1315
4.
CHEMISTRY 1
Ethanol, CH3CH2OH is completely soluble with water while phenol, C 6H5OH is partially
soluble with water. Explain. 7(b)
Both ethanol and phenol can form hydrogen bonding with water molecules. But
the presence of a large non-polar hydrocarbon group (C6H5-) makes phenol less
soluble compared to the small C2H5- group in ethanol.
5.
Determine the type of crystalline solids of these following materials, based on their
properties. 7(j)
a)
Forms very hard crystals, very high melting point and usually poor thermal and
electrical conductors.
Network covalent
b)
Hard, brittle, high melting point and dissolves well in water.
Ionic solid
c)
Soft, very low melting point and poor thermal and electrical conductors.
Atomic solid
d)
Soft, malleable, ductile, has low to very high melting point and excellent thermal and
electrical conductors.
Metallic
6.
Which of the two liquids below has greater surface tension? Explain.
7(e)
CH3CH2OH or CH3OCH3
Ethanol (CH3CH2OH) has greater surface tension than dimethyl ether (CH 3OCH3).
Ethanol has stronger intermolecular forces (Hydrogen bond) than dimethyl ether
(dipole dipole forces). Surface tensions tend to increase as the strength of
intermolecular forces increase. Both have identical molar mass so attractions
resulting from London dispersion forces will be equal.
7.
Arrange these substances in the order of increasing boiling point. State your reasons.
7(e)
a)
CH3OCH3, CH3CH2OH, and CH3CH2CH3
SHE 1315
CHEMISTRY 1
CH3CH2CH3 < CH3OCH3< CH3CH2OH
All have similar molar masses. Ethanol can form hydrogen bonds. Dimethyl
ether is polar and has dipole-dipole forces. Propane is nonpolar, so it has
only London dispersion forces. The boiling point increases as the strength of
the intermolecular forces increase:
London dispersion < dipole-dipole forces < hydrogen bond
b)
Br2, Cl2 and I2
Cl2 < Br2 < I2
All are non polar molecules so only London dispersion forces are present.
London dispersion forces get stronger as molar mass increases, thus I 2 has
the highest boiling point.
c)
CaCO3, CH4, CH3OH and CH3OCH3
By using intermolecular forces, we can tell that these compounds will rank:
methane (dispersion forces),< dimethyl ether (dipole-dipole forces)<
methanol (hydrogen bonding) < calcium carbonate (ionic bond) are much
stronger than intermolecular forces).
8.
For each pair of these substance, identify the substance that is likely to have a higher vapor
pressure. Explain your answers. 7(e)
a)
CO2 or SO2
CO2 will have the higher vapor pressure. Vapor pressure tends to increase as
the strength of the intermolecular forces decrease. Carbon dioxide is nonpolar (weak dispersion forces). Sulfur dioxide is polar (dipole-dipole
forces).
b)
CH3OH or CH3-O-CH3
CH3OCH3 will have the higher vapor pressure. Vapor pressure tends to
increase as the strength of the intermolecular forces decrease. CH 3OH can
form hydrogen bond. CH3OCH3 is polar having dipole-dipole forces which is
weaker than hydrogen bonds.
9.
Define “boiling point” and "normal boiling point". 7(d)
Boiling point: The temperature at which the liquid vapor pressure equals to the
external pressure.
SHE 1315
CHEMISTRY 1
Normal boiling point: The temperature at which the liquid vapor pressure equals to
1 atm
10.
State the difference between crystalline solid and amorphous solid. 7(f)
Crystalline solid : the particles are arranged in a well defined shape.
Amorphous solid: the particles are not arranged in a well defined shape
11.
Define the term “lattice” and “unit cell”. 7(g)
Lattice - the points form a regular pattern throughout the crystal.
Unit cell – the smallest portion of the crystal that, if repeated in all three
directions gives the crystal
12.
A
B
a)
Identify element A and B.
A -diamond
B -graphite
b)
Complete the table below for structure A and B.
Physical properties
7(k)
Structure A
Structure B
Very hard
Very soft
High
High
Colorless
Shiny black
Density
2.27
3.51
Electrical conductivity
None
High
Hardness
Melting point
Color
SHE 1315
13.
CHEMISTRY 1
Graphite and diamond are two allotropes of carbon. Graphite is a conductor while diamond
is a non-conductor. Explain their difference in terms of structure and bonding.
7(k)
In graphite, each carbon atom uses only three of its valence electron to form
covalent bonds. The other one electron is delocalized and can move freely. Thus,
graphite is a conductor.
In diamond, all the four valence electrons are used to form covalent bonds with
other carbon atoms. There is no delocalized electron in diamond. Hence, it is not a
conductor.
14.
Draw the crystal structure of simple cubic, body centered cubic and face centered cubic.
Calculate the number of atom/unit cell for each structure. 7(h)
Simple cubic
No atom/unit cell: 1/8(8) = 1
15.
Body centered
1/8 (8) + 1 = 2
Face centered
1/8(8) + ½ (6) = 4
C and Si both react with oxygen to form CO2 and SiO2. Carbon dioxide is a gas at room
conditions, but silicon dioxide is a hard solid. Explain the difference between CO2 and SiO2
in terms of bonding and structure.
7(k)
CO2: non polar covalent molecule with weak dispersion forces between molecules.
SiO2: Network covalent structure with strong covalent bonds holding the atoms
together.
16.
Explain how do intermolecular forces affect surface tension?
7(e)
Surface tension is the energy required to increase the surface area. As
the strength of IMF’s increase, surface tension will also increase. Molecules
are more strongly attracted to each other and will less likely to spread apart.
17
Select which liquid has a higher viscosity. Explain.
(i) HF and HCl
SHE 1315
CHEMISTRY 1
Both are polar. HF has hydrogen bonding, which is a stronger intermolecular
forces, so it will be more viscous.
(ii) CHCl3 and CHBr3
Both are polar and would have dipole-dipole interactions. Although CHCl3 is more
polar but CHBr3 is more massive. In this case, we can compare LDF instead. CHBr3
has more electrons/higher molar mass, so it is more viscous.
(iii) Br2 and ICl
Br2 is nonpolar and has only LDF. ICl is polar and has dipole-dipole interactions, so
it will be more viscous.
18.
Indicate which of the following properties will increase, decrease or remain unaffected by an
increase in the strength of the intermolecular forces?
7(e)
(a) vapor pressure
decreases with increased intermolecular forces
(b) normal boiling point (boiling point at 1 atmosphere pressure)
increases with increased intermolecular forces
(c) surface tension
increases with increased intermolecular
(d) viscosity
increases with increased intermolecular
SHE 1315
CHEMISTRY 1
TOPIC 8: REACTION KINETIC
1.
Define the rate of reaction.
(8a)
The reaction rate is the change in the concentration of reactants or products per
unit time
2.
Explain the difference between average rate, instantaneous rate and initial rate.
(8c)
Average rate is the total change in concentration over a given period of time.
Instantaneous rate is the rate at particular instant during the reaction.
Initial rate is the instantaneous rate at the moment the reactants are mixed or
t=0.
3.
Write the rate expression in terms of changes in concentration for the following reaction.
(8d)
a) 2 X + 3 Y  6 Z
Rate = - ½ ∆[X]/∆t = -1/3 ∆[Y]/∆t = 1/6 ∆[ Z]/∆t
b) E + ½ F  7 G + ¾ H
Rate = -∆[E]/∆t = -2 ∆[F]/∆t = 1/7 ∆[G]/∆t = 4/3 ∆[ H]/∆t
c) 2/3 A + 5 B  1/6 C + D
Rate = -3/2 ∆[A]/∆t = -1/5 ∆[B]/∆t = 6 ∆[C]/∆t = ∆[D]/∆t
4.
The decomposition of N2O5 is shown below,
2 N2O5 (g)
a)
4NO2 (g) + O2(g)
Write the rate expression in terms of changes in concentration.
Rate = - ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t = ∆[ O2]/∆t
(8d)
SHE 1315
b)
CHEMISTRY 1
If the rate of disappearance of N2O5 gas is 1.0 x 102 mol dm-3s-1 , calculate the rate
of formation of NO2 under the same condition.
- ½ ∆[ N2O5]/∆t = 1/4 ∆[ NO2]/∆t
∆[ NO2]/∆t =- 4/2∆[ N2O5]/∆t
= 2.0 x 102 mol dm-3 s-1
5.
Write the chemical equation for the rate expression below.
Rate = - ∆[ CH4]/∆t
= -½ ∆[ O2]/∆t
CH4 + 2 O2
6.
(8d)
= ½ ∆ [ H2O]/∆t = ∆[CO2]/∆t
2 H2O + CO2
The rate of formation of ammonia was measured as 10.0 mol dm −3 s−1.
N2 (g) + 3 H2 (g) → 2 NH3 (g)
Calculate the rate of disappearance of hydrogen.
(8d)
Rate = - ∆[ N2]/∆t = -1/3 ∆[ H2]/∆t = 1/2∆[ NH3]/∆t
∆[ H2]/∆t= 3/2∆[ NH3]/∆t
=3/2 x 10 mol dm−3 s−1
=15 mol dm−3 s−1
7.
2 NO (g) + 2 H2 (g)
2 H2O (g) + N2 (g)
(8e)
The reaction above is second order with respect to NO and first order with respect to H2.
a) Write the rate law and determine the overall order.
Rate = k[NO]2[H2]
Overall order = 3
b) Predict the changes of reaction rate if:
i)
the NO concentration doubles
Rate increases by a factor of 4
ii)
the NO concentration triples
SHE 1315
CHEMISTRY 1
Rate increases by a factor of 9
8.
iii)
the H2 concentration doubles
Rate increases by a factor of 2
iv)
both reactant concentrations are doubles
Rate increases by a factor of 8
Data for the reaction 2 NO(g) + O2 (g)  2 NO2 (g) are given in the table.
(8e, 8f, 8g)
Experiment
[NO] (M)
[O2] (M)
Initial Rate (M/s)
1
3.60 x 10-4
5.20 x 10-3
3.4 x 10-8
2
3.60 x 10-4
1.04 x 10-2
6.8 x 10-8
3
1.80 x 10-4
1.04 x 10-2
1.7 x 10-8
4
1.80 x 10-4
5.2 x 10-3
x
a) Write the rate law for the reaction.
Rate 2
Rate 3
= k[NO]2m [O2]2n
k[NO]3m [O2]3n
6.8 x 10-8 = (3.60 x 10-4 )m (1.04 x 10-2)n
1.7 x 10-8
(1.8 x 10-4)m (1.04 x 10-2)n
m
Rate 2
Rate 1
=2
= [NO]2m [O2]2n
[NO]1m [O2]1n
6.8 x 10-8 = (3.60 x 10-4 )m (1.04 x 10-2)n
3.4 x 10-8
(3.60 x 10-4)m (5.20 x 10-3)n
n
Rate
=1
= k [NO]2 [O2]1
b) Calculate the rate constant of the reaction.
k=
Rate
[NO]2[O2]1
=
3.4 x 10-8
(3.6 x 10-4)2(5.2 x 10-4)
= 50 M-2s-1
SHE 1315
CHEMISTRY 1
c) Determine the initial rate for experiment 4.
Rate
= k [NO]2 [O2]1
= 8.4 x 10-9 Ms-1
9. The rate of reaction between CO and NO2 was studied at 540 K starting with various
concentrations of CO and NO2. The data was collected as shown below.
(8e, 8f, 8g)
Experiment
[CO] (M)
[NO2] (M)
Initial Rate (M/s)
1
5.10 x 10-4
0.350 x 10-4
3.4 x 10-8
2
5.10 x 10-4
0.700 x 10-4
6.8 x 10-8
3
5.10 x 10-4
0.175 x 10-4
1.7 x 10-8
4
1.02 x 10-3
0.350 x 10-4
6.8 x 10-8
5
1.53 x 10-4
0.350 x 10-4
10.2 x 10-8
a)
Predict (without calculation) the reaction order with respect to CO and NO2.
[CO] constant [NO2] doubled, rate doubled so rate order of NO 2 is 1st order
[CO] doubled [NO2] constant, rate doubled so rate order of CO is 1st order
b)
Calculate the value of k.
k=
Rate
[CO]1[NO2]1
=
3.4 x 10-8
= 1.9 M-1s-1
-4
-4
(5.10 x 10 )(0.350 x 10 )
c)
Calculate the rate of reaction if the concentration of CO is 2.1 x 10 -2 M and the
concentration of NO2 is 6.25 x 10-3M.
Rate = k[CO]1 [NO2]1
=(1.9 M-1s-1)(2.1 x 10-2)(6.25 x 10-3 )
=2.49 x 10-4 Ms-1
d)
How does the reaction rate change if the temperature is changed to 480 K, while the
concentration remains constant?
The rate of reaction decreases.
SHE 1315
CHEMISTRY 1
10. Fill in the table below.
(8h)
Conc vs time graph
Reaction Order
Integrated
Rate Law
Unit for Rate
constant (k)
ln[A]
First order
ln [A]0 = kt
time
s-1
[A]t
[A]
Zero order
[A]t - [A]0= -kt
time
mol L-1s-1
1/[A]
1 – 1 = kt
Second order
[A]t [A]0
Lmol-1s-1
time
SHE 1315
CHEMISTRY 1
11. The rate law for the hydrolysis of sucrose, C12H22O11 to produce C6H12O6 is Rate = k
[C12H22O11]. After 2.57 hour the sucrose concentration decreases from 0.0146 molL -1 to
0.0132 molL-1.
(8h, 8i)
a) Find the rate constant.
First order reaction.
ln [A]0 – ln [A]t = kt
k = ln 0.0146 – ln 0.0132
2.57
= 0.0392 hr-1
b) Determine the half-life of the reaction.
t1/2 = ln 2
k
12.
= 0.693
0.0392 hr-1
= 17.7 hr
The rate constant for the decay of
a)
What is the half-life for
Rn is 0.1810 day-1.
222
(8h, 8i)
Rn expressed in days?
222
(since the unit is day-1, so, this is the first order reaction)
t1/2 = ln 2/k
= 3.83 days
b)
Calculate the fraction of a sample of 222Rn decays in a period of exactly one week.
[Assume initial concentration is 100]
ln[A]t = ln[A]0 – kt
ln[A] t = ln[100] – (0.1810 day-1)(7days)
[A] t = 28.1675
Fraction : 28.17/100
13.
The rate constant for the reaction below is 30.0 Lmol -1min-1.
2 HI (g)
(8h)
H2 (g) +2 I2 (g)
a)
State the overall order of the reaction.
Second order
b)
Calculate the time taken for the concentration of HI to drop from 0.010 M to 0.005 M.
SHE 1315
CHEMISTRY 1
1– 1
= kt
[A]t [A]0
1
- 1
= (30.0) t
0.005
0.010
t
= 3.3 min
14. The rate constant for the decomposition of NO2 is 3.40 M-1min-1. Determine the time needed for
the concentration of NO2 to decrease from 2.00 M to 1.50 M.
(8h)
Second order reaction
1– 1
= kt
[A]t [A]0
1
- 1 = (3.4) t
1.50
2.00
t
15.
= 0.049min
The table below represent a reaction with the rate law of Rate = k.
Time (min)
[A]
a)
(M)
0
1
2
3
1.0
0.79
0.58
0.38
(8h)
State the overall order of the reaction.
Zero order reaction.
b)
Calculate the rate constant.
[A]t - [A]0= -kt
k = 1.0 – 0.79
1
k = 0.21 M min-1
16.
Concentration versus time data for the thermal decomposition of ethyl bromide at 700K is
given below.
(8h)
C2H5Br(g)
Time (min)
0
1
C2H4 (g) + HBr (g)
[C2H5Br]
1.00 M
0.82 M
ln[C2H5Br]
0
-0.20
1/[C2H5Br]
1
1.22
SHE 1315
CHEMISTRY 1
2
3
4
0.67 M
0.55 M
0.45 M
-0.40
-0.60
-0.80
1.49
1.82
2.22
a)
Complete the table by filling in the ln [C2H5Br] and 1/[C2H5Br] values.
b)
Determine the order of the reaction by plotting graph.
time
[C2H5Br]
ln
[C2H5Br]
time
1
[C2H5Br]
time
time
time
time
The linear graph is observed from the second graph prove that the reaction is 1 st order
reaction.
c)
Calculate the rate constant from the graph.
Slope = -k
-k = -0.4 – (-0.6)
2-3
= -0.2
k = 0.2 min-1
d)
17.
Calculate the half life of ethyl bromide in this reaction.
t1/2 = ln 2
k
k = ln 2
0.2
= 3.47 min
Define reaction mechanism and molecularity.
(8o)
Reaction mechanism is a sequence of single reaction steps that sum to overall
chemical reaction.
Molecularity is the number of reactant particles in the step.
18.
Refer to the energy profile diagram below.
(8o, 8p, 8q, 8r)
SHE 1315
a)
CHEMISTRY 1
Write the mechanism and overall equation for the reaction.
Step 1
Step 2
Step 3
Overall
I2
H2 + I
H2I + I
H2+I2
I+I
H2I
HI +HI
2HI
b)
State the molecularity for each elementary step.
c)
Step 1
Unimolecular
Step 2
Bimolecular
Step 3
Bimolecular
Define intermediate.
Substances that is formed and used up during the overall reaction.
d)
Identify intermediate(s).
I and H2I
e)
Write the rate law for the reaction.
Rate = k[H2I][I]
19.
The rate law for the endothermic reaction below is Rate = k[H2][N],
2 H2(g) + 2 NO(g)
(8o, 8p, 8q, 8r)
2 H2O(g) + N2 (g)
The following equations represent a proposed mechanism for the reaction.
Step 1:
Step 2:
Step 3:
a)
H2(g) + NO(g)
N(g) + NO(g)
O(g) + H2(g)
Define elementary step.
H2O(g) + N(g)
N2(g) + O(g)
H2O(g)
Ea = 1690 kJ
Ea = 625 kJ
Ea = 436 kJ
SHE 1315
CHEMISTRY 1
The individual steps that make up a reaction mechanism.
b)
State the rate determining step
Step 1
c)
Determine the validity of the above mechanism and state the reason.
Not valid because the given rate law is not tally with the rate law from RDS
in the proposed mechanism above.
20.
State the factors that influence reaction rate.
(8b)
Concentration of reactants, temperature, catalyst and physical state.
21.
Based on Arrhenius equation, state how the rate of reaction can be increased by:
(8k)
a)
increasing the temperature
The higher the temperature,the larger the value of k and leads to the higher
reaction rate.
b)
adding a catalyst
Catalyst causes lower activation energy, the larger the value of k and leads
to the higher reaction rate.
22.
Define catalyst and list all types of catalyst.
(8m)
Catalyst is substances that speed up the reaction without being consumed. Two
type of catalyst are homogenous and heterogeneous catalyst
23.
Explain how the rate of reaction can be affected by:
a)
adding a catalyst. (Use a suitable diagram)
i)
(8n)
A catalyst affects by providing an alternative pathway with lower Ea.
The fraction of reactant molecules with energy same or greater than
Ea increases leads to increasing of reaction rates.
SHE 1315
b)
CHEMISTRY 1
increasing the reactant concentration.
(8j)
As the concentration of reactants increase, the number of reactant particles
increases. Frequency of effective collision increase leads to increasing of
reaction rates.
c)
increasing the temperature. (Use a suitable diagram)
ii)
(8l)
As the temperature increases, the kinetic energy of the reactant
molecules increases, the frequency of effective collision increases.The
fraction of reactant molecules with energy same or greater than Ea
increases leads to increasing of reaction rates.