MINISTRY OF SCIENCE AND TECHNOLOGY
DEPARTMENT OF
TECHNICAL AND VOCATIONAL EDUCATION
E.Ch-01011
ENGINEERING CHEMISTRY
A.G.T.I(First Year)
CHAPTER 1 1. Atoms and Atomic theory 1.1 Dalton's Atomic Theory
1. Each chemical element is composed of minute, indestructible particles called atoms. Atoms can neither be created nor destroyed during a chemical change.
2. All atoms of the elements are alike in mass (weight) and other properties, but the atoms of one element are different from those of all other elements.
3. In each of their compounds, different elements combine in a simple numerical ratio: for example, one atom of A to one of B (AB), or one atom of A to two of B (AB2), 0r...
1.2 The Discovery of Electrons
In passing electricity through evacuated glass tubes, Faraday discovered cathode rays, a type of radiation emitted by the negative terminal or cathode that cross the evacuated tube to the positive terminal or anode.
cathode rays travel in straight lines and have properties that are independent of the cathode material (i.e., whether it is iron, platinum, etc.); cathode rays are invisible and are deflected by electric and magnetic fields in the manner expected for negatively charged particles. Thomson concluded that cathode rays are negatively charged fundamental particles of matter found in all atoms and become known as electrons.
Metal shield with slid green beam of light
Cathod anode
(-) (+)
fluorescent screen
High voltage source
1.3 X-rays and Radioactivity
• X-rays: X-ray is a form of high-energy electromagnetic radiation.
10 10 10
• Radioactivity: Radioactivity is a natural phenomenon that is spontaneous emission of radiation by mineral sources.
• Three types of radiation from radioactive materials:
Alpha rays (∝): are particles carrying two fundamental units of positive charge and having the same mass as He atom.
Beta rays (β): are negatively charged particles produced by changes occurring within the nuclei of radioactive atoms and have the same properties as electrons.
Gamma rays (γ): are electromagnetic radiation of extremely high penetrating power.
1.4 Rutherford's atomic model; (the nuclear atom)
1. Most of the mass and all of the positive charge of an atom are centered in a very small region called the nucleus. The atom is mostly empty space.
2. The magnitude of the positive charge is different for different atoms and is approximately one half the atomic weight of the element.
3. There exist as many electrons outside the nucleus as there are units of positive charge on the nucleus. The atom as a whole is electrically neutral.
z Characteristic properties of fundamental particles of matter Particle Proton (p) Neutron (n) Electron (e-)
Mass 1 amu 1 amu 1/1840 amu Charge Positive (+1) Neutral Negative (-1) 1.5 Isotopes
Two or more atoms having the same atomic number (Z) but different
mass numbers (A) are called isotopes.
For example, 20Ne, 21Ne, 22Ne are the three neon isotopes. z The atomic mass
=
=
The atomic mass (weight) of an element is the average of the isotopic masses, weighted according to the naturally occurring abundances of the isotopes of the element.
Example:
The mass transfer of carbon shows that 98.892% of carbon atoms are carbon-12 with a mass of 12 u, exactly, and 1.108% are carbon-13 with a mass of 13.00335 u.
Example: 1.1 Calculate the atomic mass of carbon. Solution:
contribution to fraction of all at. mass by 12C atoms that are 12C
= 0.9889 x 12.0000 u = 11.867 u contribution to fraction of all
at. mass by 13C atoms that are 13C
= 0.01108 x 13.00335 u = 0.1441 u at. mass of naturally
occurring carbon = (contribution by 12C) + (contribution by13C) = 11.867 u + 0.1441 u
= 12.011 u.
1.6 The Avogadro’s num er (or) Avogadro’s constant
The number of atoms in exactly 12 g of pure carbon-12 (12C) is called Avogadro’s number (or) constant. It’s numeral value is 6.02x1023.
1.7 The concept of mole: (The mole)
The mole is the amount of substance that consists of a number of chemical units equal to the number of atoms in exactly 12 units of pure carbon-12.
The mass of one mole of atoms of a pure element in grams is numerically equal to the atomic weight of that element in amu.
at. mass of an = element
fractional mass of fractional mass of abundance of x isotope + abundance of x isotope + ... isotope (1) (1) isotope (2) (2)
x mass 12C atom
6.023x1023 atoms 1 mol Mg 11.34 g Pb 1 cm3solution 1 mol. Pb 207 g.Pb 6.023x1023 atoms 1 mol Pb
For instance, a pure sample of the metallic element titanium (Ti) shoes atomic weight is 47.88 amu is taken and measure out 47.88 g of it, you will have one mole (or) 6.023x1023 atoms of titanium.
Example: 1.2
How many 24Mg atoms are present in 8.27x10-3 mol Mg ? Solution:
24Mg atoms = 8.27x10-3 mol Mg x = 4.981 x 1021 atoms
Example: 1.3
How many Pb atoms are present in a small piece of lead with a volume of 0.105 cm3 ?
The density of Pb solution = 11.34 g/cm3, (Pb = 207) Solution:
Conv: Vol. g. Pb mol. Pb atoms. Pb Pb atoms = 0.105 cm3 x x = = 3.4645 x 1021 atoms of Pb. CHAPTER 2. 2. Structure of Atom 2.1 Electromagnetic radiation:
z Electromagnetic radiation is a form of energy transmission through a vacuum (empty space) or a medium (such as glass) in which electric and magnetic fields are propagated as waves.
z Wavelength, λ, is the distance between the tops of two successive crests (or the bottoms of two troughs).
z Frequency, ν, is the number of crests or troughs that pass through a given point per unit of time.
z The speed of light, c, is constant velocity of electromagnetic radiation, 2.997925 x 108 ms-1 in a vacuum, and the relationship between this speed and the frequency (ν) and wavelength of electromagnetic radiation (λ) is
c = ν. λ
2.2 The Electromagnetic spectrum z The visible spectrum:
White light consists of a large number of light waves with different wavelengths. When a beam of white light is passed through a transparent medium, the wavelength components are refracted differently. The light is dispersed into a band or spectrum of colors.
z Atomic spectra:
The spectra produced by certain gaseous substances consist of only a limited number of colored lines with dark spaces between them. These discontinuous spectra are called atomic or line spectra.
Balmer's equation, for the wavelengths of atomic spectra lines of
hydrogen, written in a form proposed by J. Rydberg, is
ν = 3.2881 x 1015 s- 1
2.3 Quantum Theory
Planck postulated that the energy of a quantum of electromagnetic radiation is proportional to the frequency of the radiation—the higher the frequency, the greater the energy.
Planck's equation is E = h ν.
The proportionality constant, h, called Planck's constant, has a value of 6.626 x 10-34 J s.
1 _ 1 22 n2
Example:2.1
For radiation of wavelength 242.4 nm, the longest wavelength that will bring about the photodissociation of O2, What is the energy of (a) one photon and (b) a mole of photons of this light?
Solution: c = 2.998 x 108 m s-1 λ = 242.4 nm = 242.4 x 10-9 m (a) ν = = = 1.237 x 1015 s-1 E = h ν = 6.626 x 10-34 J s/photon x 1.237 x 1015 s-1 = 8.196 x 10-19 J/photon
(b) E = 8.196 x 10-19 J/photon x 6.022 x 1023 photons/ mol = 4.936 x 105 J/ mol
2.4 Bohr's atomic model for a hydrogen atom
1. The electron moves in circular orbits about the nucleus, with the motion described by classical physics.
2. The electron has only a fixed set of allowed orbits, called stationary states. The allowed orbits are those in which certain properties of the electron have unique values.
3. An electron can pass only from one allowed orbit to another. In such transitions, fixed discrete quantities of energy (quanta) are involved, in accordance with Planck's equation, E = hν.
When the electron is attracted to the nucleus and confined to the orbit n, energy is emitted. The electron energy becomes negative, with its value lowered to
En =
RH is a numerical constant with a value of 2.179 x 10-18 J.
-RH n2 2.998 x 108 m s-1 242.4 x 10-9 m c λ
6.26 x 10-34 kg m2–1
(9.109 x 10-31 kg) (3.00 x 107 m s-1) Example: 2.2
Is it likely that there is an energy level for the hydrogen atom, En = - 1.00 x 10-20 J?
Solutoin:
n2 = - RH / En
= = =
n = 14.76
Since the value of n is not an integer, this is not an allowed energy level for the hydrogen atom.
z Wave-Particle Duality
According to de Broglie's hypothesis, the wavelength associated with a particle is related to the particle momentum, p, and Planck's constant, h. Momentum is the product of mass, m, and velocity, ν,
λ = = Example: 2.3
What is the wavelength associated with electrons traveling at one-tenth the speed of light? Mass of an electron = 9.109 x 10-31 kg, ν = 3.00 x 107 m s-1 , h = 6.26 x 10-34 J s. Solution: m = 9.109 x 10-31 kg ν = 3.00 x 107 m s-1 h = 6.26 x 10-34 J s = 6.26 x 10-34 kg m2–1 λ = = 2.42 x 10-11 m = 24.2 pm
z Heisenberg uncertainty principle
Heisenberg uncertainty principle is that we cannot measure position and momentum with great precision simultaneously. There must always be
-2.179 x 10-18 J -1.00 x 10-20 J 2.179 x 10 20 n2 h p h mν 217.9
uncertainties in measurement such that the product of the uncertainty in position ∆x, and the uncertainty in momentum, ∆p, is
∆x, ∆p 〈
2.5 Quantum numbers and electron orbitals
• Four sets of quantum numbers: (i) Principal quantum number:
The principal quantum number, n, have only a positive, nonzero integral value.
n = 1, 2, 3, 4, ---
(ii) Orbital (angular-momentum) quantum number:
The orbital quantum number, l, may be zero or a positive integer, but not larger than n - 1 (where n is the principal quantum number).
l = 0, 1, 2, 3, ---, n-1
(iii) Magnetic quantum number:
It, ml, may be a negative or positive integer, including zero, and
ranging from –l to +1 (where l is the orbital quantum number). ml = -l, -l + 1, -l + 2, ---, 0, 1, 2, ---, + l
(iv) Spin quantum number:
The electron spin quantum number, ms, may have a value of + 1/2 (
also denoted by the arrow ) or – 1/2
( denoted by the arrow ) ; the value of ms does not depend on any of the
other quantum numbers. Example: 2.4
Can an orbital have the quantum numbers n = 2, l = 2, ml = 2?
Solution:
h
No, that is due to the followings.
(1) The l quantum number cannot be greater than (n – 1) . Thus, if n =2, l can only be 0 or 1.
(2) If l can only be 0 or 1, ml cannot be 2;
(3) Thus, if l =0, ml must be 0 and,
if l = 1, ml may range from –1 to +1.
Example: 2.5
Give the orbital designation specified by the quantum numbers. (a) n = 4, l = 0 (b) n = 3, l = 1
Solution:
(a) When l =0 the orbitals are s, so that is 4s orbital. ( b) When l =1 the orbitals are p, so that is 4s orbital.
• Principal shells and Subshells
(i) All orbitals with the same value of n are in the same principal electronic shell or principal level, and
(ii) all orbitals with the same n and l values are in the same subshell or sublevel.
Allowed values and orbital designations of quantum numbers. Princi pal Q.No (n) Orbital Q.No ( l) (n-1) Magnetic Q.No (ml) (2l+1) Orbital designation 1 0 0 1s 2 0 1 0 +1, 0, -1 2p 2p 2p 1s
3 0 1 2 0 +1, 0, -1 +2,+1, 0, -1,-2 3s 3d 3d 3d 3d 3d Practice Example:
1. For an orbital with n = 3 and ml = -1, what are the possible vales of l. 2. Write an orbital designation corresponding to the quantum number n
=4, l = 2, ml = 0.
2.6 Electron configuration
The electron configuration of an atom is a designation of how electrons are distributed among various orbitals.
• Rules for assigning electrons to orbitals
( I ) Electrons occupy orbitals in way that minimizes the energy of the atom.
The order in which orbitals fill is:
1s, 2s.2p, 3s.3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p
( ii ) No two electrons in an atom may have all four quantum number alike:
Pauli exclusion principle:
Only two electrons may exist in the same orbital and these electrons must have opposing spins.
Because of this limit of two electrons per orbital, the n subshell consists of one orbital with a capacity of two electrons; the p subshell consists of three orbitals with a capacity of six electrons; and so on.
( iii ) When orbitals of identical energy are available, electrons initially occupy these orbitals singly.
1s 2s 2s 2p 1s 1s Hund's rule:
An atom tends to have as many unpaired electrons as possible.
• The Aufbau Process:
Hypothetically the electron configurations of the atoms can be build up by placing electrons in the lowest available orbitals until the total number of electrons added is equal to the atomic number.
Element: Electron confi: Orbital diagram: (spdf Notation) 1 H 1s1 2 He 2s2 6 C 1s2, 2s2 2p2
Element: Electron confi: Orbital diagram: (spdf Notation)
7 N 1s2, 2s2 2p3
8 O 1s2, 2s2 2p4
Element: Electron confi: Orbital diagram: (spdf Notation) 9 F 1s2, 2s2 2s5 10 Ne 1s2, 2s2 2s6 11 Na [Ne] 3s1 [Ne] 1s 2s 2p 1s 2s 2p 1s 2s 2p 3s 1s 2s 2p
18 Ar [Ne] 3s23p6 [Ne] 21 Sc [Ar] 3d14s2 [Ar] 23 V [Ar] 3d34s2 [Ar] 24 Cr [Ar] 3d54s1 [Ar] 29 Cu [Ar] 3d104s1 [Ar] 30 Zn [Ar] 3d104s2 [Ar] Example 2.6:
Using spdf Notation for an Electron Configuration.
(a) Identify the element having the electron configuration. 1s2, 2s2 2s6, 3s23p5
(b) Write out the electron configuration of Arsenic.
Solution:
a. Superscript numerals (2 + 2 + 6 + 2 + 5) = atomic number 17; the element is chlorine. b. As (Z = 33): [Ar] 3d104s24p3 3s 3p 3d 4s 3d 4s 3d 4s 3d 4s 3d 4s
S
p-block
3. The periodic table and some atomic properties 3.1 Modern periodic tableMost modern periodic tables arrange the elements in 18 groups and the table can be divided into four blocks of elements, according to the subshells involved in the Aufbau process.
• s-block: Group 1A (alkali metals: ns1 electron configuration) and 2A (alkaline earth metals:ns2) as main group or representative elements .
The group number = number of outermost s orbital e-s .
• p-block: Groups 3A, 4A, 5A, 6A, 7A: ns2 npx, and 8A: noble gases: ns2 np6 as representative elements.
The group number = the sum of the ns and np electrons.
• d-block: Groups 3B, 4B, 5B, 6B, 7B, 8B, 1B, and 2B: transition elements :ns, (n-1) d e- configuration.
The group number = the sum of the ns and (n-1) d electrons.
Group number of 1B and 2B correspond to the number of outermost electrons.
• f-block: Inner transition elements: lanthanide and actinide series, having (n-2) f orbitals filled.
Main group element
*Lanthanum
series 58 Ce 59 Pr Nd 60 Pm 61 Sm 62 63 Eu Gd 64 Tb 65 Dy 66 Ho 67 68 Er Tm 69 Yb 70 Lu 71 *Actinum
series 90 Th 91 Pa 92 U Np 93 94 Pu Am 95 Cm 96 Bk 97 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr
3.2 Atomic properties
The sizes of atoms and ions: 1 H 2A 3A 4A 5A 6A 7A He 2 3 Li Be 4 B 5 C 6 N 7 O 8 9 F Ne 10 11 Na 12 Mg d-block Transition element 3B 4B 5B 6B 7B 8B 1B 2B 13 Al 14 Si 15 P 16 S 17 Cl 18 Ar 19 K Ca 20 21 Se 22 Ti 23 V 24 Cr Mn 25 26 Fe Co 27 28 Ni Cu 29 Zn 30 Ga 31 Gr 32 As 33 34 Se 35 Br 36 Kr 37 Rb 38 Sr 39 Y 40 Zr 41 Nb 42 Mo 43 Te 44 Ru 45 Rh 46 Pd 47 Ag 48 Cd 49 In 50 Sn 51 Sb 52 Te 53 I 54 Xe 55 Cs Ba 56 57* La Hf 72 Ta 73 74 W Re 75 Os 76 77 Ir 78 Pt Au 79 Hg 80 81 Tl Pb 82 83 Bi 84 Po 85 At Rn 86 87 Fr 88 Ra 89** Ac 104 105 106 107 108 109 CHAPTER 3
• Covalent radius: One half the distance between the nuclei of two identical atoms joined by a single covalent bond.
• Ionic radius: The distance between the nuclei of ions joined by an ionic bond.
• Metallic radius: One half the distance between the nuclei of two atoms in contact in the crystalline solid metal.
• Unit for atomic radius: 1 A0(angstrom) = 10-10 m SI units: 1 nm (nanometer) = 103pm (picometer) = 1x10-9m
• Variation of atomic radii within a group:
Down a group, atomic radius increases.
Because the more electronic shells in an atom, the larger the atom will be.
• Variation of atomic radii within a period:
From left to right across a period, atomic radius decreases.
As the effective nuclear charge (Zeff ) increase, the core and outer-shell
electrons are attracted more strongly to the nucleus resulting in an overall contraction of the size of the atom.
• Variaton of atomic radii within a transition series:
Atomic radii do not change very much within a transition series.
Due to the fact that additional electrons go into an inner electron shell where they participate in shielding outer-shell electrons from the nucleus; at the same time, the number of electrons in the outer-shell tends to remain constant; thus the outer-shell electrons experience a roughly comparable force of attraction to the nucleus throughout a transition series.
• Ionic radius:
Because there is an excess of nuclear charge over the number of electrons in the resulting cation when a metal atom loses one or more electrons to form a positive ion.
For isoelectronic cation, the more positive the ionic charge is, the
smaller is the ionic radius.
(ii) Anions are larger than the parent atoms. Due to the following facts.
1. When a nonmetal atom gains one or more electrons to form an anion, the nuclear charge remains constant, but Zeff is reduced because of the additional
electron (s), so the electrons are not held as tightly. 2. Repulsion among electrons increase.
Example 3.1:
Determine which is the largest atom: Sc, Ba, Se.
Solution:
21Sc: [Ar] 4s23d1 (Period:4, Group:3B) 56Ba: [Xe] 6s2 (Period:6, Group:2A) 34Se: [Ar] 4s23d10, 4p4 (Period:4, Group:6A)
• In the same period 4, Sc is closer to the left than Se; Zeff : Sc < Se
Atomic size: Sc > Se
• Ba is in the sixth period and so has more electronic shells than either Sc or Se,
∴Ba atom is the largest of the three.
Example 3.2:
Compare the ionic sizes in each of the following pairs. a. K+, Ca2+ b. Cl-, S2-
Solution:
Ca2+: [Ne] 3s23p6
Both are isoelectronic cations; for isoelectronic cations, the higher the charge is, the smaller is the ion and thus,
Ca2+ is smaller than K+ion. b. Cl- : [Ne] 3s23p6
S2- : [Ne] 3s23p6
Both are isoelectronic anions; for isoelectronic anions, the higher the charge is, the larger the ion and so,
S2- is larger than Cl- ion.
Example 3.3:
Arrange the following species in order of increasing size: Ar, K+, Cl-, S2-, Ca2+
Solution:
1. The five species are isoelectronic , having the electron configuration of Ar: 1s2 ,2s22p6 ,3s23p6 .
2. For isoelectronic cations, the higher the charge is, the smaller is the ion.
Thus,
Cationic size: Ca2+ < K+
3. Because K+ has the higher nuclear charge than Ar, size: K+ < Ar
4. Because Ar has higher nuclear charge than Cl--, size: Ar < Cl-
5. For isoelectronic anions, the higher the charge is, the larger is the ion. Thus,
anionic size: Cl- < S2 –
The order of increasing size is Ca2+ < K+ < Ar < Cl- < S2 –
Ionization energy
• The ionization energy, I, is the quantity of energy a gaseous atom must absorb so that an electron is stripped from the neutral atom.
I1
I2
• First ionization energy: the energy required to strip one electron from a neutral gaseous atom.
• Second ionization energy: the energy required to strip one electron from a gaseous ion with a charge of +1.
• Second ion energy: I2 is larger than the first I1:
Mg (g) Mg+(g) + e- Mg+ (g) Mg+2 (g) + e-
This is due to the fact that the attraction between the positive ion and the electron is stronger than between the neutral atom and the electron.
• Variation of ionization energy within a group:
Down a group, ionization energy decreases as atomic radii increases.
Because the farther an electron is from the nucleus, the more easily it can be extracted.
• Variation of ionization energy within a period:
From left to right across a period, ionization energy increase as atomic radii decreases.
Because the larger effective nuclear charge makes removal of the electron more difficult.
Example 3.4:
Predict which member of each pair has the higher ionization energy. a. Al or Mg b. P of S
Solution:
a. 12Mg : [Ne] 3s2 (Period.3, Group.2A) 13Al : [Ne] 3s2 3p1 (Period.3, Group.3A)
More energy is required to strip an electron from the lower energy 3s orbital in Mg than from the 3p orbital in Al.
∴ Ionization energy: Mg > Al. [exception]
b. 15P : [Ne] (Period.3, Group.5A)
16S : [Ne] (Period.3, Group.6A)
3s 3p
I1 for S is slightly lower than for P because repulsion between the pair of
electrons in the filled 3p orbital of an S atom makes it easier to remove one of those electrons than an unpaired electron from a half-filled 3p orbital of a P atom. [exception]
Electron affinity
• Electron affinity, EA, is a measure of the energy change that occurs when a gaseous atom gains an electron.
F(g) + e- F-(g) EA = -322.2 kJ/mol
When an F atom gains an electron, energy is given off. The process is exothermic, the electron affinity is a negative quantity.
• For some atoms the gain of an electron required that energy be absorbed. The process is endothermic, and the EA has a positive value. This is the case for the noble gases, where the added electron enters the empty s orbital of the next electronic shell.
Ne (g) + e- Ne-(g) EA = + 29 kJ/mol (1s22s22p6) (1s22s22p6, 3s1)
• For an element such as oxygen EA1 is negative and EA2 is positive. When the second electron added is approaching a negative ion, a srong repulsion is felt and the energy of the system increase.
O (g) + e- O -(g) EA1 = -141.4 kJ/mol O-(g) + e- O2- (g) EA2 =+141.4 kJ/mol Magnetic properties
• A diamagnetc atom or ion has paired electrons, and these individual magnetic effects cancel out. A diamagnetic species is weakly repelled by a magnetic field. For example,
12 Mg [Ne] 3s2 [Ne]
Magnesium is diamagnetic, corresponding to two 3s electrons. They must be paired , as are all the other electrons.
• A paramagnetic atom or ion has unpaired electrons, and the individual magnetic effects do not cancel out. The unpaired electrons induce a magnetic field that causes the atom or ion to be attracted into an external magnetic field.
The more unpaired electrons present, the stronger the attraction. For example, 11 Na [Ne] 3s1 [Ne]
Sodium is diamagnetic, corresponding to a single 3s unpaired electron outside the Ne core.
3.3 Other period properties of the elements
Physical properties of the elements • Variation within a group
The value of a property often changes uniformly down a group for compounds as well as for elements .
To explain this trend, it is needed for the relationship between melting point and intermolecular forces of attraction: the stronger these forces, the
higher the melting point. And in general, the higher the molecular mass, the stronger the intermolecular forces in a substances.
• Variation within a period
A few properties vary regularly across a period in general.
The ability to conduct heat and the ability to conduct electricity are two that do. Thus among the third period elements, Na, Mg, Al have good thermal and electrical conductivity. The nonmetals P, S, Cl, and Ar have poor thermal and electrical conductivities.
Chemical properties of the elements
Some variations within groups of elements and some variations across a period will be considered as follows.
• Reducing abilities of group 1A and 2A metals
The lower the ionization energy, the better the metal is as a reducing agent and the more vigorous its reaction with water.
The reducing agent makes possibly a reduction half-reaction and itself, by losing electrons, is oxidized. Indeed, reducing ability depends on the ionization energy.
In-group 1A metals, K, for instance, have lower ionization energy than does the next number of the fourth period, Ca. The expectation is that K reacts more vigorously with water than does Ca.
2K (s) + 2 H2O 2 K+(aq) + 2 OH-(aq) + H2 (g) Ca (s) + 2 H2O Ca2+(aq) + 2 OH-(aq) + H2 (g)
With respect to group 2A, Mg and Be do not react with cold water as do the other alkaline earth metals. This is on account of their higher ionization energies compared with the other group 2A metals.
• Oxidizing abilities of the halogen elements
The higher the electron affinity, the better the nonmetal is as an oxidizing agent.
An oxidizing agent makes possibly an oxidation half reaction and itself, by gaining electrons, is reduced. Hence, electron affinity is related to the gain of electrons. We might expect an atom with a strong tendency to gain electrons (a large negative electron affinity) to take electrons away from atoms that lose electrons without much difficulty (low ionization energies).
(i) In these term, it is understandable that active metals (group1A) form ionic compounds with active nonmetals (group 7A).
2 Na (s) + Cl2 (g) 2 NaCl (s)
(ii) In displacement reaction, two halogens, one in molecular form and the other in ionic form, exchange places.
Cl2(g) + 2I -(aq) I2(aq) + 2 Cl-(aq)
(i) Some metal oxides, such as Li2O, react with water to produce basic metal hydroxides, called basic oxides.
Li2O (s) + H2O 2 Li-(aq) + 2 OH-(aq)
(ii) Some nonmetal oxides, like CO2 (g) , produce acidic solutions in water. These nonmetal oxides are acidic oxides.
CO2 (g) + H2O H2CO3 (aq)
(iii) It should be expected that the metal oxides at the left of the period to be basic
CHAPTER. 4
4. Chemical reactions and Stoichiometric Calculation
4.1 Chemical reactions
A chemical reaction is a process in which one set of substances called reactants is converted to a new set of substances called products.
Sometimes products reacts to reform the original reactants that is called reversible reactions. For example, decomposition of silver oxide is illustrated as follow.
2 Ag2O(s) - _4 Ag ( s) + O2 (g) stoichiometric coefficients
Reaction conditions: Temperature, pressure, and catalyst State of matter: (s) = solid; (l) = liquid; (g) = gas;
(aq) = aqueous (water) solution
4.2 Stoichiometry
It means the quantitative (mole) relationships among the reactants and products in a reaction.
N2 (g) + O2 (g) 2 NO(g) Stoichiometry: 1 mole 1 mole 2 moles
Avagadro's constant: 6.02x1023 atoms /molecule For an element, 1 mole of an element contains
6.02x1023 atoms.
For a compound, 1 mole of compound contains 6.02x1023 molecules.
Reactant (A) Product (B)
Solute + solvent solution Stoichiometric calculations:
mass mole mole mass of of of of
substance A substance A substance B substance B *molar mass: (1), (3) ;
*stoichiometric factor: (2)
1 molar mass. = 40g NaOH = 1 mole. NaOH = 6.02x1023 molecules.
1molar mass = 108 g Ag = 1 mol of Ag = 6.02x1023 atoms.
N2 (g) + 3 H2 (g) 2 NH3(g)
• 1 mole of N2 produces 2 moles of NH3 . • 3 mole of H2 produce 2 moles of NH3 .
Hence, stoichiometric factor based on N2 and H2 are 2 and 2/3 respectively.
4.3 Chemical reactions in solutions:
Amount of solutes (moles)
Volume of solution (liters)
n (mol) V (liters)
1mol O2 2 mol H2
The solvent is the component that is present in the greatest quantity in a solution and the solute is less quantity. The solution is a homogeneous mixture of solvent and solute.
Molarity: Molarity = Concentration: C = n = C.V Solution dilution:
When a solution is diluted, the amount of solute remains constant between the initial (i) and final (f) solution.
• ni = nf
• Ci Vi = Cf Vf
1.4 Determining the limiting reagent Limiting reagent:
The reagent or reactant that is completely consumed in a reaction is known as limiting reagent or limiting reactant.
• The quantity of products formed depends on the quantity of limiting
reagent.
For example, 2 H2 (g) + O2 (g) 2 H2O(g) Stoic. Ratio: 2 mol + 1mol 2 mol
(Relationship)
Reactants: [10mol] [7mol]
Assuming that H2 is the limiting reagent:
Required mol of O2 = 10 mol H2 x = 5 mol O2 Unit: mol /L
Actual yield Theoretical yield
[7 mol present – 5 mol consumed = 2 mol O2 in excess] ∴ O2 is in excess; so H2 is the limiting reagent.
Theoretical Yield
The calculated quantity of product in a chemical reaction is called theoretical yield of a reaction.
Actual Yield
The quantity of product that is actually produced is called the actual yield.
Percent Yield
Percent Yield = x 100 %
Consecutive reaction
Reaction that are carried out one after another in sequence to yield a final product are called consecutive reaction .
For example : 2 S (s) + 3 O2 (g) 2 SO3 (g)
SO3 (g) + H2O (l) H2SO4 (l)
Simultaneous reaction
In simultaneous reaction, two or more substances react independently of each other in separate reactions occurring at the same time.
For example: Mg(s) + 2 HCl(aq) Mg Cl2 (aq) + H2 (g) Zn(s) + 2 HCl(aq) Zn Cl2 (aq) + H2 (g)
Solute + Solvent Solution Nonelectrolyte Strong electrolyte Weak electrolyte Water itself is nonconductor Conductor or nonconductor
4.5 Introduction to Reactions in Aqueous solutions.
Characteristic classification of the solute. Non-
electrolyte
A nonelectrolyte is a substance that is not ionized and does not conduct electric current.
Strong electrolyte
A strong electrolyte is completely ionized in water solution and the solution is a good conductor.
HCl (aq) H+(aq) + Cl- (aq)
Weak electrolyte
A weak electrolyte is partially ionized in water solution and the solution is only a fair conductor of electricity.
HC2H3O2 (aq) H+(aq) + C2H3O2- (aq) (CH3COOH)
4.6 Predicting Precipitation Reactions A Set of Solubility Rules:
Soluble salts Insoluble salts (precipitate)
• Group IA metal salts • NH4+ salts
• NO3- Salts
• CO32-, OH- and, S2- Salts • Chlorides, bromides and,
iodides of Pb2+, • Cl-, Br-and, I- Salts • SO42- salts • . • Ag+ and, Hg 2+2 • Sulfates of Sr2+, Ba2+, Pb2+ and, Hg2+2 Molecular equation:
Mg (OH)2 (s) + 2 HCl (aq) Mg Cl2 (aq) + 2 H2O (l) Net ionic equation:
Mg (OH)2 (s) + H+(aq) Mg2+(aq) + 2 H2O(l)
4.7 Acid-Base Reaction
An acid: A substance that provides hydrogen ions (H+) in aqueous solution. (Hydronium ion, H3O+)
Strong acids
Completely ionized in aqueous solutions, such as, HCl, HBr, HI, HCLO4, HNO3, and H2SO4.
HNO3(aq) H+(aq) + NO3-(aq) Hydroxides and sulfides
of Ca2+, Sr2+, Ba2+are slightly Soluble
Weak acids
Partially ionized in aqueous solutions; the ionization is reversible, eg., HCOOH, CH3COOH
HC2H3O2 (aq) H+(aq) + C2H3O2- (aq)
A Base : A base is a substance capable of providing hydroxide ions (OH-) in aqueous solution.
Strong bases
Completely ionized in aqueous solutions, i.e., Hydroxides of Gp.IA and some Gp.IIA metals. NaOH(aq) Na+(aq) + OH -(aq)
Weak bases
Partially ionized in aqueous solutions; the ionization is reversible, e.g., NH3, NH4+
NH3(aq) + H2O NH4+(aq) + OH- (aq)
Neutralization: In a neutralization reaction, an acid and a base react to form water and aqueous solution of an ionic compound called a salt.
HCl(aq) + NaOH(aq) Na Cl(aq) + H2O(l)
4.8 Oxidation-Reduction Reaction:
• Oxidation: Oxidation is a process in which the oxidation state of element increases.
• Reduction: Reduction is a process in which the oxidation state of element decreases.
• Balancing oxidation-reduction reaction: (Half-reaction method)
Balancing the equation for a redox reaction in acid solution is as follow.
SO32-(aq)+ MnO4-(aq)+ H+ SO42-(aq) + Mn2+(aq) + H2O(aq) 1. SO32-(aq) SO42-(aq)
MnO4-(aq) Mn2+(aq)
2. SO32-(aq) + H2O SO42-(aq) MnO4-(aq) Mn2+(aq) + 4H2O
SO32-(aq) + H2O SO42-(aq) + 2 H+(aq) MnO4-(aq) + 8 H+(aq) Mn2+(aq) + 4H2O
3. SO32-(aq) + H2O SO42-(aq) + 2 H+(aq) + 2 e- MnO4-(aq) + 8 H+(aq) + 5 e- Mn2+(aq) + 4H2O 4. 5 SO32-(aq)+ 5 H2O 5 SO42-(aq)+ 10 H+(aq)+ 10 e-
2MnO4-(aq)+ 16H+(aq)+ 10 e- 2Mn2+(aq) + 8H2O 5 SO32-(aq)+ 2MnO4-(aq)+ 6H+(aq)
5 SO42-(aq)+ 2Mn2+(aq) + 3H2O
5.
Practice Example:
2H2O2 (aq) 2H2-2O + O2 (g)
-1 red: oxi:0
Balance the following equations for the redox reactions. 1. SO32-(aq)+ Cr2O72-(aq)+ H+(aq) SO42-(aq)+Cr3+(aq)+H2O 2. CrO42-(aq)+S2-(aq)+OH-(aq) Cr(OH)3(s)+ S (s) + H2O 3. S2O32-(aq) + H+(aq) S(s) + SO2 (g) + H2O
4. H2O2 (aq) + Fe2+(aq)+ H+(aq) H2O + Fe3+(aq)
4.9 Disproportionation Reaction:
Some oxidation-reduction reactions, in which the same substance is both oxidized and reduced, are called disproportionation reactions.
4.10 Oxidizing and Reducing Agents:
• Oxidizing agent; oxidant: The substance that causes some other substance to be oxidized and so itself reduced.
• Reducing agent; reductant: The substance that causes some other substance to be reduced and so itself oxidized.
2 Na (s) + Cl2 (g) 2 NaCl (s) red. agent oxi. agent
For the above reaction, Na makes Cl2 to be reduced to Cl- in NaCl and itself oxidized to Na+in NaCl; Na is reducing agent. Cl2 makes Nato be oxidized to Na+ in NaCl and itself reduced to Cl-in NaCl; Cl2 is oxidizing agent.
1L 1000 ml 0.1mol NaOH 1L 1mol OH -1mol NaOH 3.808x10-3mol HC2H3O2 5.00x10-3L 0.762mol HC2H3O2 1000 ml vinegar 60.05 gHC2H3O2 1 mol HC2H3O2 1 ml vinegar 1.01 g i Example:
Vinegar is a dilute aqueous solution of acetic acid. A 5 ml sample of a particular vinegar was titrated with 38.08 ml of 0.1 M NaOH (aq). The legal minimum A/A content of vinegar is 4% by mass.
(i) Determine the concentration of HC2H3O2. (ii) Does this sample exceed the minimum limit?
(Density of vinegar = 1.01 g/mL)
Solution:
HC2H3O2(aq)+ OH-(aq) C2H3O2-(aq)+ H2O(l) ( i )
mol OH- = 38.08 ml x x x = 3.808 x 10-3 mol OH
-mol HC2H3O2 = mol OH- = 3.808 x 10-3 mol HC2H3O2
Conc.HC2H3O2 =
(ii) % HC2H3O2 = x
x x 100 %
= 4.53 % HC2H3O2
ml NaOH L NaOH mol NaOH mol OH
mol HC2H3O2 conc.HC2H3O2
∴The vinegar sample does exceed the legal minimum limit.
Practice Example:
A 0.235 g sample of a solid that is 92.5% KOH, 7.5% Ba (OH)2 by mass requires 45.62 ml of an HCl (aq) solution for its titration. What is the molarity of the HCl (aq)?
CHAPTER 5.
5. The properties of gases
5.1 The concept of pressure
Pressure is defined as a force per unit area. A force divided by the area over which the force is distributed
SI unit: N/m2 (Pascal: 1P a = 1 N/m2)
Kilopascal: kPa is more commonly used
Liquid pressure
The pressure of a liquid depends only on the height of column and the density of the liquid.
g = Acceleration due togravity:9.80665 m s-2 h = Height of liquid; d = density of liquid.
• Barometric pressure
To measure the atomospheric pressure, barometer is used.
The height of mercury in a barometer,called the barometric pressure, varies with atmospheric conditions and with altitude.
The pressure exerted by a mercury column of exactly 760mm in height is called the standard atmosphere (atm).
1 atm = 760 mmHg; 1 atm = 760 torr (Torricelli).
Example: 5.1
What is the height of a column of water that exerts the same pressure as a column of mercury 76.0 cm (760mm) high?
A F P= d . h . g A d . A . h . g A d . V . g A m . g A W A F P= = = = = =
Solution: Pressure of Hg column = ghHg dHg = g x 76.0 cm x 13.6g/cm3 Pressure of H2O column = ghH2OdH2O = gxhH2Ox1.00g/cm3 3 3 O H x1.00g/cm gx76.0cmx13.6g/cm h x g 2 = m 3 . 10 cm 10 x 03 . 1 cm / g 00 . 1 cm / g 6 . 13 x cm 0 . 76 h 3 3 3 O H2 = = =
We see that the column height is inversely proportional to the liquid density: for a given liquid pressure, the lower the liquid density is the greater is the height of the liquid column.
• Manometers
One can compare the gas pressure and barometric pressure with a manometer.
As long as the gas pressure being measured and the prevailing atmospheric (barometric) pressure are equal, the heights of the Hg column in the two arms of the manometer are equal. A different in height of the two arms signifies a difference between the gas pressure and barometric pressure.
1. If Pgas>Pbar ; ∆P = positive value. 2. If Pgas < Pbar ; ∆P = negative value.
Example: 5.2
What is the gas pressure Pgas, if the conditions, gas pressure less than barometric pressure, are these: the manometer is filled with liquid mercury (d = 131.6 g/cm3); barometric pressure is 748.2 mmHg; and the difference in mercury levels is 8.6 mmHg?
Solution:
Whether the situation corresponds to gas pressure less than barometric pressure, the pressure we seek is Pgas = Pbar + ∆P.
As the gas pressure is less than barometric pressure; ∆P is negative, Pgas = Pbar - ∆P = 748.2 mmHg – 8.6 mmHg = 739.6 mmHg.
5.2 The simple gas law • Boyle’s Law
For a fixed amount of gas at a constant temperature, gas volume is inversely proportional to gas pressure.
2 2 1 1V P V P ) t tan cons ( a PV or V 1 P = = α Example: 5.3
The tank is first evacuated, and then it is connected to a 50.0-L cylinder of compressed nitrogen gas. The gas pressure in the cylinder, originally at 21.5 atm, falls to 1.55 atm after it is connected to the evacuated tank. What is the volume of the tank? Solution: 2 2 1 1V PV P = V2 = V1 x 2 1 P P = 50.0 L x = 694 L
Of this volume, 50.5 L is that of the cylinder. The volume of the tank is 694 L – 50.0 L = 644 L.
• Charle’s Law
The volume of the fixed amount of gas at constant pressureis directly proportional to the Kelvin (absolute) temperature.
V α T or V = bT (where b is constant) Relationship: T (K) = t (°C) + 273.15
STP condition: 0°C = 273.15 K and 1 atm
= 760 mmHg
• Avogadro’s Law
At a fixed temperature and pressure, the volume of the gas is directly proportional to the amount of gas.
atm 5 . 1 atm 215
Molar volume: 1 mol of gas = 22.4 L at STP
= 6.023 x 1023molecules
Example: 5.4
1. What is the mass of 1 L cyclopropane C3H6 at STP?
2. 128 g pieces of solid CO2 (s) sublimates into CO2 (g) . What is the volume of this gas at STP in Liters?
Solution: 1. (?) g C3H8 = 1.875g L 4 . 22 mol 1 x mol 1 g 42 = 2. (?) L CO2 = x128g 65.1636LCO2 g 44 mol 1 x mol 1 L 4 . 22 =
• The Ideal gas equation
From the gas law relationships,
P nT Vα and P nT R V=
PV = nRT ; R = Ideal gas constant
Example: 5.5
What is the volume occupied by 13.7 g Cl2(g) at 45°C and 745 mmHg. Molar mass of Cl2 = 71 g/mol Solution: (?) n of Cl2 = 71g mol 1 x 13.7 g = 0.193 mole P = mmHg 760 atm 1 x 745 mmHg = 0.9803 atm T = 45°C + 273.318 K PV = nRT
V = P nRT = atm 980263 . 0 K 318 x K . mol / atm L 08205 . 0 x mole 193 . 0 = 5.1364 L. Example: 5.6
5.6An oxygen gas is filled into 1L flask at STP. Suppose that. the pressure remains at 1 atm, when the O2 (g) is heated to 100°C, what mass of O2 must release from the flask?
Solution: At STP, (?) mole O2 = x 1L = 0.04464 mole At 100°C, (?) mole O2, PV = nRT or R = nT PV ∴
n1T1 = n2 T2 (as P, V remains constant)
T1 = 0°C +273 = 273 K, and T2 = 100°C +273 = 373 K
n2 =
= 0.0364 mole O2 (in flask)
Thus, Released O2 (mole) = 0.0446 – 0.03264 = 0.012 mole Molar mass of O2 = 32 g/mole,
(g) O2 released = = 0.384 g O2
5.3 Applications of the ideal gas equation • Molar mass determination
L 4 . 22 mole 1 2 2 2 2 1 1 1 1 T n V P T n V P = K 373 K 273 x mole 0446 . 0 mole 012 . 0 x mole 1 g 32
Since the number of moles of gas (n) is equal to the mass of gas (m) divided by the molar mass (M), that is n = m/M, substituting into the ideal gas equation.
• Gas density
To determine the density of gas, d, d =
d =
Example: 5.7
What is the density of oxygen gas at 298°K and 0.987 atm? Solution:
d = =
•
Gases in chemical reactions
Example: 5.8What volume of N2(g) , measured at 735 mmHg and 26C, is produced when 70.0 g NaN3 is decomposed?
2NaN3(g) ∆→ 2 Na(1) + 3 Na2(g) Solution: M mRT PV= M x V n V M x n V m = = T R P M V m = T R P M V m = L / g 29 . 1 K 298 x K . mol / Latm 08206 . 0 atm 987 . 0 x mol / g 00 . 32 =
2NaN3(g) ∆→ 2 Na(1) + 3 Na2(g) ? mol N2 = = 1.62 mol N2 P = 735 mmHg x 1atm/760 mmHg = 0.967 mmHg V = ? n =1.62 mol T = 26OC + 273 = 299 K = 1.62 mol x 0.08206 Lmol-1K-1 x 299K ..-::P 0.967 atm. = 41.1L
• Law of combining volume
Example:5.9
Zinc blend, ZnS, is the most important zinc ore. Roasting (strong heating) of ZnS is the first in the commercial production of zinc.
2ZnS(s) + 3 O2(g) ∆→ 2ZnO(s) + 2S02(g)
What volume of SO(g) forms per liter of O2(g) consumed? Both gases are measured at 25 oC and 745 mmHg.
2ZnS(s) + 3 O2(g) ∆→ 2ZnO(s) + 2S02(g) ? L S02(g) = 1.00 L O2(g) x = 0.667 L S02
• Mixture of gaes
For the nonreactive gases, n = the total amount, in moles, of the gaseous mixture (n tot)
.Example: 5.10
What is the pressure exerted by a mixture of 1.0 g H2 and 5,00 g He when the mixture is confined to a volume of 5.0L at 20°C?
Solution:
n.tot = 1.0 g H2 x
= 0.50 mol H2 + 1.25 mol He = 1.75 mol mol gas
P = nRT/V = 1.75molx0.08206Lmol-1K-1x293K = 8.4 atm
5.0 L
• Daltn’s Law of partial pressures
The total pressure of a mixture of gases is the sum of
the partial pressures of the components of the mixture,
* Ptot = PA + PB + ---
In case of the wet gas collected over water, *Ttot = Tbar = Pgas + H2O
For the composition of gaseous mixtures in % by volume,
* Vtot = VA + VB + —- (Vtot = total vol.of the gas)
• The mole fraction
The mole fraction of a component. nA /n tot , in a mixture is the fraction
of all molecules in the mixture contributed by that component.
B A B A B A V V P P n n = = Example: 5.11
What are the partial pressure of Hz and He in the gaseous mixture described in example 5.10?
From example 5.10, Ptot = 8.4 atm and so,
2 H
P = x Ptot = 0.50 mol x 8.4 atm = 2.4 atm
1.75 mol
He
P = x Ptot = 1.20 mol x 8.4 atm = 6 atm
1.75 mol
Example: 5.12
Collecting a gas over a liquid (water).
In the following reaction 81.2 ml of O2(g) is collected over water at 23°C and barometric pressure 7.51 mmHg. What must have been the mass of Ag2O(s) decomposed?
(Vapor pressure of water at 23°C -= 21.1 mmHg.) 2A2O(s) 4Ag(s) + O2(g) Solution: 1. = 751 mmHg
–
21.1 mmHg = 750 mmHg = (750/760) atm = 0.961 atm V = 81.2 ml = 0.0812 L R = 0.08206 Latm-1K-1 T = 23oC + 273 = 293 K n = ? n = PV/RT = 0.961 atm x 0.0812 L = 0.0082 mol 0.08206 L atm mol-1 K–1 x 296K 2Ag2O(s) 4Ag(s) + O2(g) Ag2O(s) = 0.00321mol O2 x O Ag mol 1 O Ag 231.7g x O 1mol O Ag mol 2 2 2 2 2 = 1.49g Ag 2O5.4 Kinecic molecular theory of gases
1. A gas is composed of a very large number of extremely small particles (molecules or. in some cases, atoms) m
constant, random, straight-line motion.
2. Great distances separate molecules of a gas. The gas is mostly empty space. (The molecules are treated as if they have mass but no volume, so called "point masses.") .
3. Molecules colloid with one another and with the walls of their container. However, these collisions occur very rapidly and most of the time molecules are not
colliding.
4. There are assumed to be no forces between molecules except very briefly during collisions. That is, each molecule acts independently of all the others and is unaffectedly by their presence.
5. Individual molecules may gain otiose energy as a result of collisions. In a collection of molecules at constant temperature, however, the total energy remains constant.
This theory depends on the following facts.
(i) Thefrequencyof molecularcollisions:
Thenumberof collisionper second
The higher this frequency, the greater the total force of these collisions. Collision frequency increases with the number of molecules per unit volume and with molecular speeds.
(ii) Theamount of translational kinetic energy:
That is energy possessed by objects moving through space,
representedasekandhas thevalue
eK = 1/2 mu2
m = themass ofthemoleculeand u =speedof themolecules
The faster molecules move, the greater are their translational kinetic energies and the greater is the force
of their collisions.
(iii) Moleculesmoveinalldirectionandatdifferentspeeds:
When all factors are properly considered, the result
is this expression for the pressure of gas. The basic equation of the kinetic-molecular theory. P = 2 V N 3 1 mu
N = the number of molecules present in the Volume V.
m = mass of the molecules
u2 = the average of the .squares of their speeds • Distributionof Molecularspeed
Root-mean-squarespeed,ursm thee square rootoftheaverageof the squaresof the speedsof allthemoleculesin asample.
Urms =
M 3RT
R = Gas constant, 8.3145 J mol-1 K-1 (or) 8.3145 Kg m2 s-2 mol-1 K-1
Example: 5.13
Which is The greater speed, that of a bullet from a high powered M-16 rifle (21~0 mi/h) or the root-mean-square
speed of Hs molecules at 25C?
Solution:
Urms =
M 3RT
R = Gas constant, 8.3145 J mol-1 K-1 MH2 = 2.016 x 10
Urms = -3 -l -1 -1 -2 2 mol Kg 2.016x10 K 298 x K mol s gm 8.3145k x 3 = 4.29 x 103 mi/h
Comparing the two speeds, the root-mean-square speed of H2
molecules at 25C is greater than the speed of the high-powered rifle bullet.
5.5 Gas properties relating to the kinetic-molecular theory
• Diffusion: is the migration of molecules of different . substances as a result of random molecular motion.
• Effusion: is the escape of gas molecules from their container through a tiny orifice or pinhole.
• Therates at which 'diffusion and effusion occur are
directly proportional to molecular speeds. Thatis, molecules withhigh speedsdiffuseandeffusefasterthan moleculeswithlow speeds.
• Grahm'a law
Theratesof effusion(ordiffusion)oftwo different gasesare inverselyproportionalto thesquarerootof theirmolarmasses.
The result shown in the following equation is a kinetic-theorv statement of Graham's law. B A B A rms rms M M 3RT/M 3RT/M ) (u ) (u B of effusion of rate A of efusion of rate = = = B A
A ratio of effusion-rates, times, distances, .... is equal to the square root of a. ratio of molar masses.
1. molecular speeds
2. rates of effusion
Ratio of 3. effusion times = ratioof twomolar masses 4 .distance traveled
by molecules
5. amount of gas effused Example: 5.14
A sample of Kr(g) escape through a tiny hole in 87.3 s, and unknown gas requires 42.9 s under identical conditions. What is the molar mass of the unknown gas?
Solution:
Again start with qualitative reasoning. Because the unknown gas effuses faster, it. must have a smaller molar mass than Kr. The ratio of molar
masses in the setup below
must be smaller than 1: Munk goes in the numerator.
491 . 0 M M s 87.3 s 42.9 Kr for me efusion ti unknown for me efusion ti Kr unk = = = Munk = (0.491)2 x MKr = (0.491)2 x 83.80 = 20.2 g/mol
5.5 Nonideal (real) gases
Figure 5.1 The behavior of real gas compressibility factor as a function of pressure for three different gases at OoC.
All gases behave ideally at sufficiently low pressures,
say, below I atm, but that deviations set in at increased pressures. At very high pressures the compressibility factor is always greater than 1.
Here is how we might explain nonideal gas behavior: Boyle's law predicts that at very high pressures a gas volume becomes extremely small and approaches zero. This cannot
be, however, because the molecules themselves occupy space and are practically incompressible. The PV product is larger than predicted for an ideal gas and the compressibility factor is greater than 1. We must also allow for the fact that mterrnolecular force exists in gases.
Figure 5.2 Intermolecular forces of
attraction.
Figure 5.2 suggests that because of
attractive forces among the molecules, the
forces of the collisions of gas molecules
with. the container walls is less than
expected for an ideal gas. Intermolecular
forces of attraction account for
compressibility factors less than 1. These
forces become increasingly important at low temperatures, where molecular motion slows down. To summarize
o Gases tend to behave ideally at high temperatures and low pressures.
o Gases tend to behave nonideally at low temperatures and high pressures.
• The van der Waals equation
[ P + (n2 a)/V2 ] (V – n b) = n RT V = the volume of n moles of gas.
(n2a) /V2 is related to intermolecular .forces of attraction. It is added to P because the measured pressure is lower than expected (recall Figure-5.2)
b = the excluded volume per mole; related to the volume of
(V – n b) = free volume within the gas
'a', "b" = specific value for particular gas. varying with temperature and
pressure. Example: 5.13
Use the van der Waal's equation to calculate the pressure exerted by 1.00 mol C12(g) when it is confined to a volume of 2.00 L at 273 K.
(a = 6.49 L2 atm mol-2 b = 0.0562 L mol--1). Solusion:
Solve van der Waal"s equation for P.
2 2 V a n b n -V RT n P= − n = I mol V = 2 mol T = 273 K R = 0.08206 L atm mol-1K-1
n2a = (I mol)2 x 6.49 L2 atm mol-1 = 6.49 L2 atm n b = 1 mol x 0.0562 L mol-1 = 0.0562 L mol-1 P = 2 2 2 -1 -1 xL 2 atm L 6.49 0.0562)L (2 K mol atm L 0.08206 x Imol − −
CHAPTER 6
6. The preoperties of lquids and solids
6.1 Intermolecular forces and some properties of liquids
∗ Surface tension
Surface tension, γ, is the energy or work required to increase the surface area of a liquid.
It has the units of energy per unit area, typically J/m2.
As the temperature—and hence the intensity of molecular motion— increases, intermolecular forces become less effective. Less work is required to extend the surface of a liquid, meaning that surface tension decreases with the increased temperature.
When a drop of liquid spreads into a film across a surface, we say that the liquid wets the surface. Whether a drop of liquid wets a surface or retains its spherical shape and stand on the surface depends on the strengths of two types of intermolecular forces.
Cohesive forces are the intermolecular forces between like molecules, such as within a drop of liquid.
Adheive forces are the intermolecular forces between unlike molecules, molecules of a liquid and of the surface with which it is in contact.
If cohesive forces are strong compared to adhesive forces, a drop maintains its shape. If adhesive forces are enough, on the other hand, the energy requirement for spreading the drop into a film is met through the work done by the collapsing drop.
If the liquid in the tube is water, the interface between the water and the air above it, called a meniscus, is curved upward (concave). Water is drawn slightly up the walls by adhesive forces between water and the glass.
With liquid mercury the meniscus is curved downward (convex). Cohesive forces in mercury, consisting of metallic bonds between Hg atoms, are strong; mercury does not wet glass.
The effect of meniscus formation is greatly magnified in tubes of small diameter, called capillary tubes. In the capillary action the water level inside the capillary is noticeably higher than outside.
Viscosity
Viscosity refers to a liquid’s resistance to flow. Its magnitude depends on intermolecular forces of attraction and, in some cases, on molecular sizes and shapes.
The stronger the intermolecular forces of attraction are, the greater is the viscosity.
When a liquid flows, one portion of the liquid moves with respect to neighboring portions. Cohesive forces within the liquid create an ‘internal friction’ which reduce the rate of flow. The effect is weak in liquid of low viscosity such as ethyl alcohol and water. They flow easily. Liquid such as honey and heavy motor oil flow much more sluggishly. We say that they are viscous.
Vaporization of liquid
The passage of molecules from the surface of a liquid into the gaseous or vapor state is called vaporization; the term evaporation also used.
The tendency for a liquid to evaporate increases with increased temperature and decreased strength of intermolecular forces.
• Enthalpy of vaporization:
The enthalpy of vaporization is the quantity of heat that must be absorbed if a certain quantity of liquid is vaporized at a constant temperature. liquid vapor on vaporizati H H H = − ∆
Vaporization is an endothermic process, ∆Hvap is always positive. In this text enthalpies of vaporization are expressed in terms of ‘one mole of liquid vaporized.
Example: 6.1
To vaporize 1.75 g of acetone, (CH3)2CO, at 298 K, 933 J of heat is required.
What is the enthalpy of vaporization of acetone in kJ/mol?
Solution: vapor H ∆ = CO ) (CH g 08 . 58 CO ) (CH 1mol x CO ) (CH g 75 . 1 J 1000 kJ 1 x J 933 2 3 2 3 2 3 = 31.0 kJ/ mol (CH3)2CO • Condensation:
Condensation is the passage of molecules from the gaseous state to the liquid state.
In turn, the conversion of a gas or vapor to a liquid is called condensation, that is the reverse of vaporization.
vap vapor liquid oon condensati H H H H = − =−∆ ∆
Condensation is an exothermic process. ∆Hcondensatioon is always
negative.
• Vapor pressure:
The pressure exerted by a vapor in dynamic equilibrium with its liquid is called vapor pressure.
Liquid with high vapor pressures are said to be volatile, and those with very low vapor pressures are nonvolatile. The weaker the intermolecular forces, the more volatile the liquid (the higher its vapor pressure).Vapor pressure increases with temperature.
• Boiling and the boiling point:
The pressure exerted by escaping molecules equals that exerted by molecules of the atmosphere, and boiling is said to occur.
• Normal boiling point:
The pressure at which the vapor pressure of a liquid is equal to standard atmospheric pressure
(1 atm = 760 mmHg) is the normal boiling point.
• The critical point:
If a liquid in a glass tube is heated, the followings are observed.
1. The density of the liquid decreases; that of the vapor increases; and eventually the two densities become equal.
2. The surface tension of the liquid approaches zero. The meniscus between the liquid and vapor becomes less distinct and eventually disappears.
The critical point refers to the temperature and pressure where a liquid and its vapor become identical. It is the highest temperature point on the vapor-pressure curve.
The temperature at the critical point is the critical temperature, Tc
and the pressure is the critical pressure, Pc .
The critical point represents the highest temperature at which the liquid can exist. A gas can be liquefied only at temperatures below its critical temperature,Tc. If room temperature is below Tc, this liquefaction
can be accomplished just by applying sufficient pressure. If room temperature is above Tc , added pressure and lowering of temperature to a
value below Tc are required. Example: 6.2
As a result of a chemical reaction, 0.132 g H2O is produced and maintained
at a temperature of 50.0°C in a closed flask of 525 ml volume. Will the water be present as vapor only or as liquid and vapor in equilibrium?
4.0 4.5 5.0 5.5 6.0 6.5 7.0 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.4 3.6 3.8 1/T x 103 ln P
First, calculate the pressure that would exist if the water were present as vapor only.
V
nRT
P
=
The calculated pressure (281 mmHg) exceeds the vapor pressure (92.5 mmHg), some of the vapor must condense to liquid. The water is present as liquid and vapor in equilibrium at 92.5 mmHg.
∗
Clausis-Clapeyron equation: aniline C6H7N Water H2O benzene C6H6 diethyl ether C4H10O mmHg 281 atm 1 mmHg 760 x atm 0.370 P L 0.525 323.2K x 1 K 1 mol Latm Ox0.08206 H 18.02g O H 1mol Ox H g 0.132 P 2 2 2 = = − − =Figure 6.2 Vapor pressure data plotted as in lnP vs. 1/T.
1. A particular common form of a vapor-pressure equation is that shown below, which express the relationship of a straight line plotted for the liquids.
2. To use equation (6.1), we need to have values for the two constants, A and B. ‘A’ is related to the enthalpy of vaporization of the liquid: A = ∆Hvap/R and it is customary to eliminate B by rewriting (6.1) in
a form called Clausius—Clapeyron equation. − ∆ = 2 1 1 2 T 1 T 1 R vap H P P ln
3. For benzene at 60°C, the vapor pressure is 400 mmHg; ln P = ln 400 = 5.99. T = 60 + 273 = 333 K; 1/T = 1/333 = 3.00 x 10-3 ; 1/T x 103 = 3.00.
The point corresponding to (3.00, 5.99) is marked by the arrow ( ).
Example: 6.3
Calculate the vapor of water at 35.0°C with the following data. T2 = 40.0°C, P2 = 55.3 mmHg, ∆Hvap= 44.0 kJ/mol.
Solution:
Let P1 be the unknown vapor pressure at 35.0°C,
T1 = 308.2 K. Substitute the given values into the equation, 1 1 1 1 3 1 K 2 . 313 1 2 . 308 1 K mol J 3145 . 8 mol J 10 x 0 . 44 P mmHg 3 . 55 ln − − − − − = = 5.29 x 103 (0.003245 - 0.003193)
