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Stoichiometry
Chemical Arithmatic
Two aspects of chemistry
Qualitative
‫كيفي أو نوعي‬
Quantitative
‫كمي‬
Stoichiometry in the quantitative aspect of chemical reaction
and composition.
H2O
- Is composed of hydrogen and oxygen. (qualitatively).
- Is composed of two hydrogen atoms and one oxygen atom.
(Quantitatively)
H2 + 1/2 O2
Hydrogen + Oxygen
H2O
Water (Qualitatively)
One mole hydrogen + half mole oxygen
(quantitatively).
1
one mole water
Basic SI units:
Physical Quantity
Mass
Length
Time
Temperature
Amount of substance
Unit
Kilogram
Meter
Second
Kelvin
mole
Symbol
kg
m
s
K
mol
The Mole:
Is formally defined as the amount of substance of a system that
contains as many dementary entities as there are atoms in 0.012
kg of 12C.
Or: It consists of Avogadro's number of objects.
i.e. 6.02 x 1023 object
mole = atomic weight (mass) in grams if the substance consists
of atoms.
1 mole of C = 12.01 g C
2
1 mole = Molecular weight (mass) in grams if the substance is
composed of molecules.
1 mole = Formula weight (mass) in grams if the substance is
composed of ions or molecules.
1 mole of CCl 4
Contains Avogadro's number of CCl4 molecules.
Contains Avogadro's number of C atoms.
Contains 4 moles of Cl atoms.
We need two moles of Cl2 molecules to make 1 mole of CCl4.
Ex: How many gram in 2.25 mol Cu. Cu = 63.5 U amu
1 mol Cu = 63.5 g Cu
moles of Cu = 2.25 mol Cu x 63.5g = 142.88 g
1mol cu
3
E.g: How many moles of Aluminum are required to react with
28.5g of Fluorine to form AlF3
F = 19
1 mole of Al atoms
~
3 moles of F atoms
Number of mol of F atoms = 28.5g F atoms x 1 mol Fatoms
19 g F atoms
= 1.5 mol F
No. of moles of Al = 1.5 mol F x 1 mol Al
3 mol F
= 0.5 mol Al
Molecular Weight (mass):
Sum of atomic weights of the atoms in a molecule is called
molecular weight or mass.
CH4
C =12,
H=1
MW = ( 1 x 12 ) + ( 4 x 1 ) = 16
Formula Weight:
The sum of atomic weights of atoms in a formula unit is called
formula weight or mass.
Ca Cl2
Ca = 40 ,
Cl = 35.5
F.W = ( 1 x 40 ) + ( 2 x 35.5 ) = 111
4
E.x: Calculate the number of carbon atoms and the number of
hydrogen atoms in 600g of propane, C3 H8 C = 12 , H = 1.
MW = ( 3 x 12 ) + ( 8 x 1 ) = 44 amu
Number of moles of propane = 600 g x 1 mol = 13.63 mol
44 g
1 mole of C3 H8 contains 3 mol of C atoms and contains 8 mol
of H atoms.
It contains also Avogadro's number of C3 H8 molecules.
Number of moles of
C atoms = 13.63 mol C3 H8 x 3 mol of C atoms
1 mol C3 H8
= 40.89 mol of C
Number of C atoms = 40.89 mol C atoms x 6.02 x 1023 atoms
1 mol of C atoms
= 2.46 x 1025 C atoms
Number of mol of H atoms = 13.63 mol C3 H8 x 8 mol H atoms
1 mol C3 H8
= 109.04 mol H atoms
Number of H atoms = 109.04 mol H atoms x 6.02 x 1023 atoms
1 mol of H atoms
= 6.56 x 1025 H atoms
5
Sun 24/12/1424
Molar mass = atomic weight or molecular weight in grams.
Percentage Composition:
Percentage by mass contributed by each element present in the
sample.
E.x: What is the percentage composition of CO2.
C = 12
,
C = 16
MW = ( 1 x 12 ) + ( 2 x 16 ) = 44
1 mol = 44g CO2
C% = 12g x 100 = 27.3%
44g
O% = 32g x 100 = 72.7%
44g
Chemical formulas:
1. Empirical formula gives the simplest whole number ratio
between atoms present in the compound.
2. Molecular formula states the actual number of atoms of
each element present in the molecule.
6
E.F.
CH2O
H2O
CH2O
M.F.
C2H4O
H2O
C6H12O6
Acetic acid
Water
Glucose
Derivation of empirical formula:
The E.F doesn't only give the simplest ratio between number of
atoms but also the simplest ratio between moles of atoms.
We can, therefore, find the empirical formula by determining the
number of moles of atoms from their masses present in the
sample. Then divide the number of moles of atoms each by the
smallest number of moles present.
C 0.25
0.25
H 1.5
0.25
PO2.5
P2 O5
C1.33 H2
O 0.5
0.25
CH6 O2
C4 H6
7
Tues 26/12/1424
E.X. What is the empirical formula of a sulfur oxygen
compound, a sample of which contains 2.1g oxygen and
1.4g sulfur. S = 32 , O = 16.
No. of moles of S = 1.4 g S x 1 mol = 0.0437 mol S
32 g S
No. of moles of O = 2.1 gO x 1 mol = 0.131 mol O
16 g O
S0.0437 O0.131
S0.0437
0.0437
O0.131
S1O3
SO3
.0437
Remember:
E.F is a ratio between atoms or moles of atoms, so we use
atomic weight and not molecular weight to calculate number of
moles of atoms.
Empirical formula from percentage composition :
Ex.: The same compound contains 40% S and 60% oxygen.
Determine the empirical formula.
Suppose we have 100g sample, then calculate number of moles.
S = 40 = 1.25 mol S , O = 60 = 3.75 mol O
32
16
S1.25 O3.75
S1.25 O3.75
1.25
1.25
SO3
8
Derivation of Molecular Formula:
Since molecular formula is an integral multiple of empirical
formula, we divide MW by EFW to get molecular formula.
Ex: Find the molecular formula for a compound whose
empirical formula is CH2 and MW = 84.
C = 12 , H = 1
E.F.W = (1 x 12) + (2 x 1) = 14
Number of times of E.F occur in M.F. = 8_4 = 6
14
Molecular formula = C6 H12
Balancing chemical equations:
1. Write all reactants and products with correct formula.
2. Balance by adjusting coefficients that preceed formulas.
Start with elements that are less frequent.
e.g. C4 H10 + O2
C4 H10 + O2
C4 H10 + 13/2 O2
2C4 H10 + 13O2
CO2 + H2O
4CO2 + 5H2O
4CO2 + 5H2O
8CO2 + 10H2O
9
E.X: NH3 + O2
2NH3 + O2
2NH3 + 5/2O2
NO + H2O
2NO + 3H2O
2NO + 3H2O
4NH3 + 5O2
4NO + 6H2O
E.g: Calculations based on chemical equations:
C + O2
CO2
1 mol of C atoms reacts 1 mol of O2 molecules to give 1 mol of
molecules
(requires)
1 atom + 1 molecule
12 g + 32 g
44 g
1 molecule
Ex.: Calculate the mass of carbon required to produce 1 g of
CO2
12g
44 g
Mass of carbon
1
g
1 x 12 = 3 g of C
44
11
10
Limiting Reactant Calculations
If one of the reactants is used in excess of mole ratio
than the other indicated by a balanced equation,
reactant which is not in excess will be used up
completely before the one present in excess.
2 H2 + O 2
2H2O
The amount of product is determined by the reactant
which disappears first. The reactant which is
consumed first is called limiting reactant.
11
Sun 2/1/1425
Ex.: Freon gas, CCl2F2 is prepared by the following reaction:
3CCl4 + 2SbF3
3CCl2F2 + 2SbCl3
If 150g of CCl4 and 100g of SbF3 were used.
a) How many grams of CCl2F2 can be formed.
b) How many grams of which reactant will remain.
C = 12, F = 19, Cl = 35.5, Sb = 122
a)
molar mass of CCl4 = (1 x 12) + (4 x 35.5) = 154 g
molar mass of SbF3 = (1 x 122) + (3 x 19) = 179g
no. of moles of CCl4 = 150 gCCl4 x 1mol CCl4 = 0.974mol SbF3
154g CCl4
no. of moles of SbF3 = 100g SbF3 x 1mol SbF3 = 0.559mol SbF3
179g SbF3
no. of moles of SbF3 required to react completely with CCl4 =
0.974 mol CCl4 x
2mol SbF3 = 0.649mol
3mol CCl4
SbF3
.
0.649 mol is more than 0.559 mol which is available
of SbF3
CCl4 is in excess or SbF3 is the limiting reactant.
12
Another method to determine limiting reactant divide no. of
moles of each reactant by its coefficient. The smallest value is
for the limiting reactant.
0.559 mol SbF3 = 0.279
2
0.974 mol CCl4 = 0.325
3
SbF3 is limiting reactant.
no. of moles of CCl2F2 produced=0.559mol SbF3 x 3mol CCl2F2
2mol SbF3
= 0.839 mol CCl2F2
Mass of CCl2F2= 0.839mol CCl2F2 x 121g Freon=101.5g Freon
1mol Freon
121g is molar mass of CCl2F2
b)
no. of moles of CCl4 consumed = 0.559 x 3 = 0.839 mol CCl4
2
Mass of CCl4 consumed = 0.839 x 154g = 129g CCl4
mass of CCl4 remaining = 150 – 129 = 21g
13
Theoretical Yield
Theoretical yield of a given product is the maximum yield that
can be obtained if the reaction gave only that product.
Percentage Yield
Is a measure of the efficiency of the reaction
% Yield = actual yield x 100
Theoretical Yield
E.g: Calculate the percentage yield of the previous
reaction (in the previous example) if the actual yield
is 76g.
% yield = actual yield x 100
Theoretical yield
= 76g x 100 = 75%
101.5g
E.g: Calculate the actual yield if the percentage yield of
the previous reaction is 80%
actual yield = 101.5g x 80 = 81g
100
14
Molarity
To express the amount of solute in solution, we use
concentration units. Most important unit is molarity. Molarity is
the number of moles of solute in 1L (dm3) solution.
Molarity = no. of mole of solute
Volume solution in litres
Unit: mol L -1
Or: mol dm-3
no. of moles of solute = molarity x volume of solution in litres
Example.: Calculate the molarity of a solution containing
4g of NaOH in 50 ml solution.
Na= 23, H = 1, O = 16
Formula wt = (1 x 23) + (1 x 1) + (1 x 16) = 40 amu
1 mol = 40g
no. of moles of NaOH = 4 g x 1 mol = 0.1 mol
40g
Molrity
=
0.1mol__ = 2.0 M
0.05 L
15
Dilution
On dilution, the number of moles of solute do not change.
No. of moles before dilution = no. of moles after dilution
M1 . V1 = M2 . V2
E.g: How many ml of 1 M HCl must be added to 50 ml of
0.5 M HCl to get a solution whose concentration is 0.6 M.
no. of moles before mixing = no. of moles after mixing
( Y x1) + (50 x 0.5) = (Y + 50) x 0.6
1000
1000
1000
Y + 25 = 0.6Y + 30
Y – 0.6Y = 30 – 25
0.4Y = 5
Y= 5
0.4
Y = 12.5 ml
16
E.g: Calculate the volume of 0.2M H2SO4 required to react
completely with 500ml of 0.1M NaOH.
The equation:
H2SO4 + 2NaOH
Na2SO4 + 2H2O
no. of moles of NaOH = 500 x 0.1 = 0.05 mol NaOH
1000
no. of moles of H2SO4 required = 0.05mol NaOH x 1mol H2SO4
2mol NaOH
= 0.025 = volume x 0.2
Volume = 0.025 = 0.125 L
0.2
Volume = 125 ml
17
Sun 9/1/1425
Oxidation – Reduction Reactions
(Redox Reactions)
Oxidation number is defined as the charge which would the
atom have if electrons of covalent bonds were assigned to the
more electronegative atoms.
-2
O
H
+1
H – Cl
+1 -1
O=C=O
-2 +4 -2
H
+1
Rules for Assigning Oxidation Number
1. Oxidation number of atoms in the elemental form equals
zero, regardless of the number of atoms in the molecules.
He, Ne, Ar
H2, F2, N2, O2, Cl2
Na, Cu, K, Fe
P4
S8
2. Oxidation Number of atoms in simple ions (monatomic
ions) equals charge of the ion.
M2+, X3-, M3+
+2 3- 3+
3. Sum of oxidation number of all atoms in a molecule =
Zero and for polyatomic ion = charge on the ion.
18
PO4-3
y + (4 x -2) = -3
Y – 8 = -3
y = +5
S2O3-2
2y + (3 x -2) = -2
2y – 6 = -2
2y = 4
y = +2
Mn O4y + (4 x -2) = -1
y – 8 = -1
y = +7
S4O6-2
(4 x y) + (6 x -2) = -2
4y = +10
y = 2.5
4. Fluorine in its compounds always has -1 Oxidation
number
MF
MF2
MF3
+1-1
+2-1
+3-1
CF4 PF6+4-1
+5-1
5. Metals of group IA (Li, Na, K, Rb, Cs, Fr) in their
compounds always have +1 oxidation number.
6. Alkaline earth metals group IIA (Be, Mg, Ca, Sr, Ba, Ra)
in their compounds always have +2 O.N.
19
7. Oxygen in its compounds almost always has -2 O. N.
Exceptions:
Peroxides
Super Oxide
Na2O2 ,
H2O2 ,
-1
-1
BaO2
-1
KO2
+1 -1/2
Fluorine Oxide F2O
+2
8. Hydrogen almost always +1 O.N.
Exceptions: metal hydrides Na H, Ca H2, Al H3
+1 -1
+2 -1
-1
KH , LiA H4, Na BH4
-1
-1
9. Halogens in binary halides (Metal + halogen) have -1
O.N.
Fe Cl3
-1
10. For familiar ions the oxidation number can be considered
the charge of the ions.
PO4-3
CO3-2
-2
Na3PO4
NO3
SO4-2
-
-1
-2
20
Ca3(PO4)2
NH4+
+1
Oxidation: An atom, ion or molecule loses electrons or
its oxidation number increases.
Reduction: An atom, ion or molecule accept electrons or
its oxidation number decreases.
Reducing agent: An atom, ion or molecule providing
(losing) electrons.
Oxidizing agent: An atom, ion or molecule accepting
(receiving) electrons.
E.g: Zn + Cu+2
O
+2
Zn
Cu+2
Zn
Cu+2
is
is
is
is
Zn+2 + Cu
+2
O
Oxidized
Reduced
Reducing agent
Oxidizing agent
21
it is Redox reaction
E.g: Which of the following are redox reaction:
H2 + 1/2 O2
O
O
H+ + OH+1
-1
H+ + H2O
+1 +1-2
H2O
+1 -2
H2O
+1 -2
H3O+
H -2
x
neutralization
x
acid-base reaction
Ca CO3 heat Ca O + CO2
decomposition
+2 +1 -2
+2 -2 +4 -2
H2 + F2
O
O
Ag+ + Cl+1 -1
Cu+2 + 4Cl-
Fe+3 + 6 CN-
x
2HF
+1 -1
Ag Cl
+1 -1
x
precipitation
x
complex formation
(Lewis acid – base
Reaction)
Fe (CN)6-3 x
+3
-1
complex formation
(Lewis acid – base
Reaction)
Cu Cl4
+2 -1
22
Balancing of Redox Equations by ion electron method:
We have to remember that the equations
must be balanced in mass and charge.
In the final equation, their should be no
electrons.
This method is convenient for ionic equations. Steps:
1. Divide the equation into two halves (Reduction half and
oxidation half).
2. Balance each half separately for mass and charge.
3. Balance the charge by adding electrons to the more
positive (+ve) side or less negative (-ve) side.
4. Multiply each half by a suitable factor to eliminate
electrons.
And then add:
 in acid solution add a number of H+ to the side deficient in
hydrogen, and to balance oxygen atoms add number of
H2O molecules and to other side add 2H+ for each H2O
added to remove imbalance.
23
Balance in Acid medium
Cl2
Cl- + Cl O-3
Divide then carry out balancing
Cl2
Cl-
Cl2
Cl O-3
I Cl
I- + I O-3 + Cl-
I Cl
I- + Cl-
I Cl
IO-3 + Cl-
Then carry on as usual
Balance in acid medium
CN- + As O-34
As O-2 + CN O-
CN-
,
CN- + H2O
CNO-
As O-34
As O-2
CNO- + 2H+ + 2e-
As O-34 + 4H+ + 2e-
As O-2 + 2H2O
CN- + As O-34 + 2H+
CNO- + As O-2 + H2O
24
E.g: Balance in acid medium:
Zn + NO-3
Zn
NO-3
Zn+2 + NH+4
Zn+2 + 2e- ………….. (1) x4
NH4+
NO-3 + 10H+
NH4+ + 3H2O
NO-3 + 10H+ + 8e-
NH4++ 3H2O ……….. (2) x 1
4Zn+2 + 8e- + NH4+ + 3H2O
4Zn + NO-3 + 10H+ + 8e-
4Zn + NO-3 + 10H+
4Zn+2 + 8e- + NH4+ + 3H2O
25
SUN 16/1/1425
Balancing in Basic medium
The best method is to balance first in acid medium, then carry
out the following steps:
1. Add number of OH- to eliminate H+ to both side.
2. Combine H+ and OH- into H2O.
3. Cancel water if necessary.
Ex.: Balancing in Basic medium:
MnO-4 + C2O4-2
MnO-4
MnO2 + CO3-2
MnO2
MnO2 + 2H2O ………(1) x
MnO-4 + 4H+ + 3eC2O-24
C2O-24 + 2H2O
2
CO-23
2CO-23 + 4H+ + 2e- ………(2)
2MnO-4 + 8H+ + 6e- +3C2O-24 + 6H2O
6CO-23 + 12H+ + 6e2MnO-4 + 3C2O-24 + 2H2O
x
3
2MnO2 + 4H2O +
2MnO2 + 6CO-23 + 4H+
The equation is now balanced in acid medium.
2MnO-4 + 3C2O-24 + 2H2O + 4OH2MnO2 + 6CO-23 +
4H+ + 4OH2MnO-4 + 3C2O-24 + 4OH2MnO2 + 6CO-23 + 2H2O
The equation is now balanced in basic medium.
26
Gases:
The gas occupies the entire volume in which it exists.
Volume of gas = volume of container (vessel)
If you have a mixture of gases, the volume of each
Gas = volume of the container.
Pressure in all directions is equal
Pressure = force
area
. (Nm-2) unit of pressure. It is called Pascal.
Atmospheric pressure:
Measured by barometer.
Pressure exerted by air. It equals the weight of a column of air.
It equals the height of mercury in barometer.
Standard atmosphere:
1 atm = 760 mm Hg = 760 torr = 76 cm Hg = 101325 Nm-2 =
101325 Pascal = 101.325 kpa.
27
E.g: 0.8 atm in torrs
0.8
1
?
760
608 torrs
0.8 atm in Pascal
1.0
0.8
101325
?
81060kPa
0.8 atm in KPa = 81.06 kPa
Boyel’s law:
For a fixed amount of a gas, at constant temperature, the volume
is inversely proportional to pressure.
Vα 1
P
,
V = constant
P
PV = constant
P1 V1 = P2 V2 = P3 V3
Graphically
28
Tues 18/1/1425
At low pressure, the behavior of the gas approaches ideal
behavior
Charle’s law:
At constant pressure, the volume of a given amount of gas is
directly proportional to absolute temperature.
Vα T
T absolute temperature
(Kelvin)
V = constant X T
The volume of a gas changes linearly with Celsius degrees if
you extrapolate lines for gases to volume zero, they will meet at
-273°C. This is the absolute zero or it is zero at Kelvin scale.
29
At high temperature a real gas approaches ideal behaviors.
Real gas approaches ideal behavior at low pressures and high
temperatures.
Amonton’s law (Gay – Lussac’s law):
The pressure of a given quantity of gas is directly proportional
to absolute temperature if the volume is kept constant.
PαT
P = constant
T
P = constant X T or P1 = P2 = ……
T1 T2
Combined gas law:
Pi Vi = Pf Vf
Ti
Tf
i = initial
f = final
At constant T the law reduces to Boyle’s law:
Pi Vi = Pf Vf
At constant pressure  Vi = Vf reduces to Charle’s law
Ti Tf
At constant volume  Pi = Pf reduces to gayLussac’s law
Ti
Ti w
STP standard temperature and pressure:
1 atm
101325 Pa (Nm-2) 760 torr
O°C 273 K
30
Dalton’s law of partial pressure:
The total pressure exerted by a mixture of gases, that do not
interact, equals sum of partial pressures of all gases, if each gas
occupies the container on its own.
P total = P1 + P2 + P3 + …………
Note: if you collect gas above water, the gas will be saturated
with water vapor total pressure.
Pt = Pgas + P H2O
P total for wet gas
Pg for dry gas
Chemical reactions between gases:
Gay-Lussac’s law of combining volumes: the volume of
gaseous reactants and products are in a simple whole number
ratio, if the volumes are measured under the same conditional of
temperature and pressure.
2H2 (g) + O2 (g)
2H2O (g)
2 volumes + 1 volume
2 volumes
2
:
1
:
2
H2 (g) + Cl2 (g)
1
:
1
:
2HCl (g)
2
31
Avogadros hypothesis: Equal volumes of gases contain the
same number of molecules if the volumes were measured
under the same conditions of temperature and pressure.
Since equal number of molecules implies equal number of
moles, then
Vαn
where (n) is the number of moles
It follows that the volume of one mole of any gas is the same
under the same conditions of temperature and pressure. This
volume is called Molar volume.
Molar volume of any gas at STP = 22.4 L
E.g: The molar volume of a gas at temperatures higher that
STP and pressures lower than STP, equal:
a) 22.4 L.
b) Less than 22.4 L.
c) more than 22.4 L. 
d) we can’t tell.
32
The ideal gas law:
Vα 1
P
VαT
Vαn
V α nT
P
V = R nT
P
R: universal gas constant
PV = nRT
state for an ideal gas.
Ideal gas law or equation of
Obeyed by ideal gases and by real gases at normal laboratory
conditions.
Value of R:
For one mole of gas at STP (101325 Pa and 273 K), it occupies
22.4 L.
SI unit
R = PV = 101325 Nm-2 x 0.0224 m3 = 8.314 J mol-1 K-1
nT
1 mol x 273 K
for 1 mol P= 101.325 kPa , V = 22.4L
R = 101325 kPa x 22.4L = 8.314 kPa L mol-1 K-1
1 mol x 273 K
1 mol
1 atm
22.4 L
R = 1 atm x 22.4 L = 0.0821 atm L mol-1 K-1
1 mol x 273K
33
E.g: Calculate the volume of 4g methane, CH4, at 25°C and 80
kPa. C = 12
,
H=1
nCH4 = 4g x 1 mol = 0.25 mol
16g
V = nRT = 0.25 mol x 8.314 Nm mol-1 K-1 x 298K
P
80000 Nm-2
= 7.074 x 10-3 m3
J = N.m
Or
V = 0.25 mol x 8.314 kPa mol-1 K-1 x 298K
80 kPa
= 7.74 L
Or
V = 0.25 mol x 0.0821 atm mol-1 K-1 x 298K
80
atm
101.325
= 7.74 L
For a mixture of gases
P1 V = n1 RT
Divide
Pt V = nt RT
P 1 = n 1 = X1
Pt
nt
mole fraction
P1 = Pt x X1
34
E.g: Calculate total pressure exerted by a mixture of 6g helium
and 40 g oxygen in 10L vessal at 25°C He = 4 , O = 16
T = 273 + 25 = 298°K
nHe = 6g x 1 = 1.5 mo.
49
Inert gases have
monoatomic molecules
nO2 = 40g x 1 = 1.25 mol
32g
PHe = nHe RT , PO2 = no2 RT
V
V
P = 3.67 atm , PO2 = 3.06 atm
He
P total = PHe + Po2 = 3.7 + 3.06 = 6.73 atm
Or
Pt = nt RT
V
Pt = Po2 + PH2o  Wet oxygen (Gas) ‫اذا ذكر السؤال حنسب معه ضغط املاء‬
PO2 only  dry oxygen (gas): ‫أما‬
35
Density of a gas:
PV = n RT
n= m
M
mass of gas
molar mass
P = n RT
V
P = m RT
M V
m =d
V
density
P = d RT
M
PM = dRT
E.g: calculate the density of CO2 at 86.7 kPa and 25°C
C = 12
,
O = 16
86.7 kPa x 44 = d x 8.314 x 298
d = 1.536 gL-1
or you can convert 86.7 into atmospheres and then use R =
0.0821 atm L mol-1 K-1
36
Real Gases:
Gas laws apply perfectly to ideal gases, but real gases deviate
from ideal behavior.
Example: PV = n RT 
‫للغاز احلقيقي‬
‫خاصة عند درجات منخفضة وضغط عايل‬
There are two reasons for deviation of real gases:
1. Molecules of a real gas have a volume. Therefore, the
volume we measure for a real gas is bigger than the
volume in which molecules are free to move.
2. There are attractive forces between molecules of a real gas,
whereas molecules of an ideal gas have no attractive
forces.
Molecules of an ideal gas would be cooled absolute zero without
condensing (Because no attractive forces) in to a liquid.
37
The measured pressure of a real gas is less than that of an
ideal gas because molecules which are about to collide
with the wall are attracted towards interior by other
molecules.
Van der Waals Equation for real gases.
(P + n2a) (V – nb) = nRT
V2
for n mole
(P + a ) (V – b) = RT
V2
for 1 mole
P = measured pressure
V = measured volume
R = Gas constant
T = Absolute temp.
n = number of moles
a and b constants characteristic for each gas.
a related to attractive forces
b related to molecules sizes
38
E.g: Oxygen gas generated by the reaction:
KClO3
KCl + O2 (un balanced)
was collected over water at 30°C in 150 mL vessel until the total
pressure 600 torr.
a. How many grams of O2 were produced?
b. How many grams of KClO3 were consumed?
PH2O=31.8 torr at 300C
S: a)
2KClO3
2KCl + 3O2
PO2 = Pt – PH2O = 600 – 31.8 = 568.2 torr
568.2 x 0.150
nO2 = PV = 760
RT 0.0521 x 303
= 4.5 x 10-3 mol
Mass of O2 = 4.5 x 10-3 x 32 = 0.144g
b) No of moles of KClO3 consumed = 4.5 x 10-3 x 2 = 3 x 10-3 mol
3
-1
Molar mass of KClO3 =122.5g mol
Mass of KClO3 consumed = 3 x 10-3 x 122.5 = 0.369g
39
Change direction of Chemical Thermodynamics
Thermochemistry:
The study of heats of reactions is called thermochemistry.
Virtually every chemical reaction is associated with absorption
or release of energy how this comes out and why?
Bonds are formed and broken. The differences in potential
energies associated with formed and broken. Bonds are the
source of the energy evolved.
Hess’s law of constant heat summation:
The heat change for a process, ΔH, is the same whether change
is carried out in one step or in stepwise manner.
A
C
change in energy is the same
B
Two Steps
It is an application of law of conservation of energy.
For example:
C(graphits) + O2(g)
CO2(g)
; ΔH = -393.5kJ
This is called a thermochemical equation
ΔH = enthalpy or heat change = H products – H reactents
H = heat content or enthalpy
40
41
If ΔH is engative, the reaction is exothermic.
If ΔH is positive, the reaction is endothermic.
H is called a state function. The change in its value depends only
on intial and final state.
In two steps:
C(graphite) + 1/2O2(g)
CO(g) ;ΔH2 = -110.5kJ
CO(gas) + 1/2O2(g)
Add the two equation
CO2(g)
, ΔH3 = -283KJ
C(graphite) + O2(g)
CO2(g)
, ΔH = -393.5KJ
Thermochemical equation can be treated algebraically.
They can be added or subtracted. ΔH is also treated
algebraically.
Enthalpy diagram to illustrate Hess’s law:
C(graphite) + O2(g)
ΔH1
ΔH2
CO(g) + 1/2O2 (g)
ΔH3
CO2(g)
42
ΔH1 = ΔH2 + ΔH3
We can calculate
ΔH of a reaction from
known heats of other
reactions
ΔH is the heat gained or lost by a system under constant
pressure the only work done being due to volume change.
ΔH = qp
System: any thing we focus our study on it. Anything else is
called surroundings. If ΔH for a system is + ve, ΔH for the
surrounding is – ve.
Manipulating thermochemical equations:
How to calculate ΔH of a reaction from known ΔH for other
reactions:
In order to get the desired equation we multiply or divide by a
suitable factor or change direction of reaction if necessary. Do
the same thing for ΔH.
Ex: Given the following:
2H2(g) + O2(g)
2H2O(l) ΔH° = -571.5kJ
N2O5(g) + H2O(l)
½ N(g) + 3/2 O2(g)
2HNO3(l)
HNO3(l), ΔH° = -174kJ
Calculate ΔH for the reaction:
2N2(g) + 5O2(g)
ΔH° = 76.6kJ
2N2O5 (g)
43
1) Multiply equation (3) by 4, multiply equation (2) by 2 and
change direction. And change direction of equation 1.
Do the same thing for ΔH and add.
2N2(g) + 5O2 (g) + 2H2 (g)
4HNO3(l), ΔH° = -696KJ
4HNO3(l)
KJ
2N2O5 (g) + 2H2O(l) ,
2H2O(l)
Add
2H2(g) + O2(g)
2N2(g) + 5O2 (g)
ΔH° = +153.2
ΔH° = +571.5 kJ
2N2O5(g) ,
ΔH° = + 28.7kJ
Heats or enthalpy formation:
Heats or enthalpy change when one mole of substance is formed
from its elements under stated conditions. ΔH° = standard
enthalpy or heat formation, is the enthalpy change when one
mole of pure substance is formed from its elements in their most
stable forms under standard state conditions.
Standard state conditions (°) at 25°C (298K) and 1 atm.
Ex: which of the following has ΔH°f for H2SO4(l) :
SO3(g) = H2O(l)
H2SO4(l)
SO2(g) = ½ O2(g) + H2O(l)
S(s) + H2(g) + 2O2(g)
H2SO4(l)
H2SO4(l)
S(g) + H2(g) + 2O2(g)
stable
H2(s) + Br2(l)
½ H2(g)
H2SO4(l) because S(g) is not most
2HBr(g)
H(g) ΔH°f
44
Example: H2(g) + ½ O2(g)
H2O(l)
ΔH°f (1)
C(graphite) + 2H2 (g)
CH4 (g)
ΔH°(2)
C(graphite) + ½ O2
CO(g)
ΔH°f (3)
1
Reactions like (2) and (3) are difficult to measure. So, we use
calculations to know ΔH°f
E.x: C(graphite) + O2(g)
CO2 (g) ΔH° = -393.5kJ (1)
ΔH° = -285.9kJ (2)
H2(g) + ½ O2(g)
H2O(l)
CH4(g) + 2O2(g)
CO2(g)+2H2O(l)ΔH° = -890.4kJ
Calculate ΔH°f for CH4.
C(graphite) + 2H2(g)
and add to (1)
C(graphite) + O2(g)
CO2(g) (2), multiply (2) by 2
(3)
2H2(g) + O2(g)
CO2(g) ΔH° = -393.5kJ
ΔH° = -571.8kJ
2H2O(l)
CO2(g) + 2H2O(l)
+890.4kJ
CH4(g)+2O2(g)ΔH°=
C(graphite) + 2H2(g)
CH4(g)
C (graphite) + 2O2(g) + 2H2(g)
ΔH1 CO2(g) + 2H2O(l)
ΔHf
ΔH1 = ΔHf + ΔH2
ΔH° = -74.9kJ mol
ΔH2
CH4 + 2O2
ΔH depends on status of reactants and products.
45
Ex:
H2(g) + ½ O2
H2(g) + ½ O2
H2O(l)
H2O(g)
ΔH° = -286kJ (1)
ΔH° = -242 kJ (2)
Calculate ΔH vap
H2O(l)
Change direction of (1) and add to (2)
H2O(l)
H2 + ½ O2
H2 + ½ O2
H2O(g)
H2O(g)
ΔH1
ΔH2
ΔH1 = ΔH2 + ΔHvap
ΔH° = 286 kJ
ΔH° = - 242 kJ
H2O(g)
H2(g) + 1/2 O2(g)
ΔH vap
H2O(g)
ΔH° vap = 44 kJ
H2O(g)
ΔH vap
H2O(l)
ΔHf for many substances are tabulated
ΔH reaction = Σ ΔHf (products) - Σ ΔHf (reactants).
ΔHf for elements in their standard state = o
Ex: CH4(g) +
ΔH = ?
2O2(g)
CO2(g)
+
2H2O(l)
ΔHf (CO2) = -393.5 KJ mol-1
ΔHf (H2O) = -285.9 KJ mol-1
ΔHf (CH4) = -74.9 KJ mol-1
ΔH reaction = Σ ΔHf (products) - Σ ΔHf (reactants)
= Σ (-393.5 + 2 x -285.9) - Σ (-74.9) = -890.9 kJ
46
Heat of combustion:
Heat change when one mole of a substance is burned completely
in oxygen.
Bond energy or Bond enthalpy:
Energy required to break a bond into neutral fragements.
H2 + energy 
2H
Atomization Energy:
Energy required to reduce a gaseous complex molecule into
neutral gaseous atoms. It is the sum of bond energies.
E.x: The structure of propane is
H H
H
H–C–C=C
H
H
ΔH°f (C3H6) = +8 kJ mol-1
Calculate C = C bond energy, you are given the following data:
ΔH°f C (atoms) = 715 kJ mol-1
ΔH°f H (atoms) = 218 kJ mol-1
C – C bond energy = 348 kJ mol-1
C – H bond energy = 418 kJ mol-1
ΔHf = ΔH1 + ΔH2
ΔH1 = 3 x 715 kJ + 6 x 218 kJ = 3453 kJ
ΔH2 is formation
 8 kJ = 3453 kJ + ΔH2
of bonds from
ΔH2 = -3445 kJ
atoms.
ΔH atom = +3445 kJ
it is the reverse of
+ 3445 = 6 x 418 + 348 + C = C bond energy
atomization
C = C bond energy = 607 kJ
47
The first law of thermodynamics:
This is the law of conservation of energy. Change in internal
energy, ΔE, equals heat added to the system plus work done on
the system by surroundings.
ΔE = ΔE final – ΔE initial
, ΔE = q + w
w = -p ΔV = - p (Vfinal - V initial)
E.x: if a system absorbs 442 kj of heat expands from 100L to
445L against 1 atm pressure. Calculate the change in internal
energy.
(445 – 100) L
ΔE =442 kJ – 101325 Nm- 2 x 1000L / m3
1000J / kJ
= 42 kJ – 34.96 kJ
= 7.05 kJ
Measurement of ΔEL:
Bomb calorimeter
48
Heat capacity:
Heat required or released to raise the temperature of the system
1ºC.
= heat compacity x Δt
= heat capcity x (t2 – t1)
Heat released
Then calculate for one mole, ΔE.
ΔE = qv heat at constant volume when ΔV = 0
For pure substances = M x Y x Δt
Y: specific heat, M: mass
Heat change
The relation ship between ΔH and ΔE:
H = E = PV
ΔH = ΔE+ PΔV
ΔH = ΔE + nRT
P ΔV = Δn RT
Δn = number of moles of gases produced
- number of moles of gases consumed
Neglect Δn change in solid and liquids.
E.x: Relation ship between ΔH and ΔE:
1. C(s) + O2(g)
CO2(g)
ΔH = ΔE
because Δn( gas) = O
2. CO(g) +
1/2O2(g)
ΔH = ΔE – 1/2 RT
3. Zn(s) +
CO2(g)
because Δn( gas) = -½
2HCl(aq)
ΔH = ΔE + RT
Zn Cl2(aq) +
because Δn(gas) = 1
49
H2(g)
E.x: if ΔH for the above reaction = -151.5 kJ mol
Calculate ΔE at 27ºC , R = 8.314 J mol-1 K-1
ΔH = ΔE + RT
-151.5 = ΔE + 8.314 x 300
1000
ΔE = -154.44 kJ mol-1
Chemical Bonding
Atoms combine together to form molecules or compounds. The
force of attraction that holds atoms together is the chemical
bond.
Lewis symbols (Lewis structure):
1A 2A 3A 4A 5A
6A 7A 8A
Li Be B
C
N
O
F
Ne
.
.
.
.
..
..
..
X . X . X. . X . . X . . X . ..X . ..X ..
.
.
..
..
..
..
Valence electrons.
The Ionic Bond:
50
Metals tend to react with non-metals to give ionic compounds.
Ionic or electrovalent bonds occurs between positive ions or
cations (atoms losing electrons) and negative ions or anions
(atoms accepting electrons) i.e. it results from attraction between
oppositely charged ions.
Example: Reaction between sodium and fluorine?
11Na
1s
.
.
Na +: F:
2
2s
11Na
2
6
2p
9F
1
3s
:
[Na+] + [: F- : ]
:
51
9F
1s2
2s2
2p5
Noble gas structure is reached. This corresponds, except for
helium, to eight electrons in the valence shell. This is the basis
of octet rule "atoms accept or lose or share electrons until there
are eight electrons in the valence shell". Not all ions obey the
octet rule, transition metal ion and post transition metal ions do
not obey octet rule:
26Fe
24Cr
1s2 2s2 2p6 3s2 3p6 4s2 3d6
[Ar]
[Ar] 4s1 3d5
[Ar] 366
[Ar] 3d3
[Ar] 3d10 pseudo noble gas
structure
The ratio between is not always 1 : 1
.
..
Ca .+ 2 :Cl:
[Ca]+2 + 2 [Cl]- ratio 1 : 2
.
.
2 Li +: O .
2 [Li]+ + [O]-2
ratio 2 : 1
:
30Zn
[Ar] 4s2 3d10
+3
24Cr
+2
26Fe
30Zn
+2
E.x: deducing formula from charges:
Ca+2
(PO4)-3
Al+3 SO4-2
Mg+2 O-2
Ca3 (PO4)2
Al2 (SO4)3
Mg2 O2
52
Mg O
* Foctors influencing the formation of ionic compounds :Q : why compounds are formed ?
A : To reach stability. The system becomes of lower energy.
Lower energy
More Stability .
In the reaction : Li(s) +
1 mole
Li+F-(s)
1mole
1/2 F2(g)
1/2 mole
We can analyze the factors that contribute to the energy change of
this reaction:
- Born Haper cycle :
ΔHf
Li+F- (s)
Li(s) + 1/2 F2 (g)
(1)
(2)
(5)
F(g) (4) + 1eLi(g)
(3) - 1e-
F- (g)
Li+ (g)
Hf = H1 + H2 + H3 + H4 + H5
Li(g); H=155 kJ
endothermic
F(g) ; H= 79 kJ
Step (1) : Vaporization or sublimation. Li(s)
Step (2) : dissociation 1/2 F2 (g)
endothermic
Step (3) : ionization Li(g)
Li+ (g) + 1e- ; H = 520 kJ
endothermic
Step (4) : electron affinity
F(g) + 1eF-(g) ; H = -330 kJ exothermic
Step (5) :Lattice energy: Li+(g)+ F -(g)
Li+F-(s) ; H =- 1016kJ
highly exothermic
53
‫طاقة البناء الشبكي البلوري‬
* Lattice energy : is primarily responsible for the formation and
stability for ionic compounds.
HF0 = 155 + 79 + 520 + (-330) + (-1016) =-592 kJ mol-1
generally ionic compounds are formed from metals and non-metals
particularly group IA and IIA( low ionization energy) with VIIA and
VIA (high electronic affinity).
* The covalent Bond :
When conditions are not favorable for the formation of ionic bond,
a covalent bond is formed.
A covalent bond is formed as a result of sharing a pair of electrons
between the two atoms. Bonding results from attraction between this
pair and both nuclei.
- Formation of H2 molecule :Energy diagram of H2 formation
Energy
Bond
energy
bond distance
or bond length
Inter nuclear distance
54
As the two atoms comes close to each other the energy decreases to
a minimum. At minimum Energy, the distance between nuclei is
called bond length. The depth of this minimum is the bond energy.
Both electrons in bond have opposite spins.
.
.
H+H
H : H or H – H
or H2
Some atoms require to form more than a covalent bond.
H
.
.
..
.C. + 4H
H..C.. H
.
..
H
N + 3H
O + 2H
F+H
..
H..N.. H
H
.. O.. H
H
..Lone pair or free fair or
non- bonding pair
F.. H
Not all atoms obey octet rule
C| .x Be.x C|
F
.x
xB x
F
F
PF5
F
F: : F
S
F: F
F
55
F
F :P: F
F:
F
atoms from third period and beyond can expand their valence shell
to include empty d subshell and thus exceed octet rule.
Some atoms make more than covalent bond :
O=C=O
N≡N
- Drawing Lewis structure for molecules :
After deciding the central atoms, follow the following steps :
1- Count all electrons plus or minus charge.
2- Put a pair of electrons in each bond.
3- Complete 8 electrons to the surrounding atoms.
4- Put any remaining pairs on central atoms.
5- If the central atom didn't reach 8 electrons, make
multiple bond.
* Ex Draw Lewis structure for PF3
1s2 2s2 2p6 / 3s2 3p3
1s2 / 2s2 2s5
1- (1x5) + (3x7) = 26 e2- F:P:F
F
4- F:P:F:
F
3- :F:P:F:
:F:
56
(15P – 9F)
NO3(9N, 8 O)
1- (1x5) + (3 x 6) = 24e
o
2- o:N:o
o
3- o:N:o
4- X :o:
5- :o: :N:
:o:
____________________________________
1)
2)
3)
4)
5)
CO2
4 + (2 x 6) = 16 eO:C:O
o:c:o
X
O:C:O
57
Bond order and Bond properties :
Bond length and bond energy are two characteristic properties
of the covalent bond.
H–H+E
2H.
Bond energy
* Bond order : number of covalent bonds between two atoms.
H
H
H- C-C- H
H
Bond order
between carbon
atoms
C=C
-C≡C-
2
3
H
1
Increase in bond energy
Increase of bond order
decrease in bond length
- Vibrational frequency :Two atoms vibrate towards and away each other along the axis
joining the two atoms.
The factors affecting vibrational frequencies :Increase of mass of atoms decreases vibrational frequency.
Increase of bond order increases vibrational frequency.
58
between
- State of mater and Intermolecular forces :
Intra bonding(intra means inside) is
stronger than intermolecular forces.
a) Dipole - dipole Interaction :
it occurs between polar molecules. It amounts to 1% strength
of the covalent bond.
In solids > liquids > gases
b) hydrogen bonding :
Occurs between molecules when hydrogen is covalently
bonded to one of the three small most electronegative atoms which
are of F, O, N. example : H2O
H
O
H
H-O
H
OO
O
H
OO
O
hydrogen
bond
H
OO
O
H
H
H
H
Its strength amounts to 5-10% the strength of the covalent bond. It is
responsible for exceptionally high boiling point and heat of
vaporization : H2O>H2S
HF >HCl
* Ex: Is there hydrogen bonding in the following :
O
x
H3CF x
HF
H3C
CH3
H
11O
O
N  CH3 – CO-CH3 x CH3- CH x 
59
CH3OH
H
H - N-H 
OH
CH3
H
CH3
C) London Forces :
It plays a significant role in attraction between non-polar
molecules and uncombined atoms.
It results from instantaneous random temporary polarization
due to continuous movement of electrons.
(+)
(-)
(+) (-)
A
Polarization in A induces polarization
in molecule B and attraction results
B
* It occurs between all particles polar, non polar and ionic.
Example : H9,
I2
, AR, Na+CL-
* London force increase with atomic size (atomic number)
I2 > Br2 > d2 > F2
* London force increases with increasing chain length or molecular
weight.
- c- c- c- c- ,
B.P
10
- c- c- c -c- c35c
,
- c- c- c- c- c- c69c
Number of polarization sites in come with chain length
increases resulting in more attraction force.
60
Hydrogen bonding strength between H…F>H…O>H…N because of
electro negativity.
H2O
o
100C
B.P
HF
o
19C
O
H
H2O
o
B.P 100C
NH3
o
-35C
F
H
H
Because water make
more hydrogen bonds
F
N
H
H
CH3O-H
o
68 C
H
CH3 CH2 O-H
o
79C
..
Water makes more hydrogen bonds than CH3-O-H.
..
CH3 CH2 OH has stronger London Force
* Non polar molecules and compounds :
O2, N2, F2, d2, H2 ,…..
F
non polar CO2 O = C = O
compound
H
BF3 F
B
o
120
F
BF3, Bd3, CS2, S = C = S, CC/4, CF4, CBr4, CI4,
PCl5, PF5, SF6, hydrocarbons
Only London forces
Non polar
CH2 Cl2 Polar London forces and dipole – dipole
Cl- I, Br – I : Dipole dipole and London force
61
Definitions of acids
and bases
* Arrhenius definition in modern concepts :
An acid is a substance that increases concentration of
H3O+ in aqueous solution
H3O+ + ClH2CO3
H3O+ + HCO3-
Hcl + H2O
CO2 + H2O
H2CO3+H2O
- Non metal oxides : CO2, N2 O5, P2O5, SO2, SO3
are acid anhydrides
A base : substance that increase concentration of OHin aqueous solution
NaOH(s) + H2O
Na+aq + OH-aq aq= aqueous
NH4+ OH-
NH3 + H2O
- metals oxides : CaO, Na2O , BaO, K2O, M9O
are basic anhydrides
CaO + H2 O
Ca (OH)2
Neutralization : H3O+ + OHor simply :
2H2O
H+ + OH-
H2O
62
* Bronsted – Lowry Definition :
An acid is a proton donor, a base is a proton acceptor
H3O+ + Clacid base
Hcl + H2O
acid base
There are two acids and bases. An acid and abase that differ by only
one proton are called acid-base conjugate pairs.
* E.x :- H2S , S-2
H2O , OH- 
-3
H3PO4, PO4 x
x
H2S , HS- 
+
H3O, OH- x
+
H3 O, H2O 
NH3 + H2O
base
acid
+
NH4 + OH
acid
base
conjugated pairs
* water behaves as an acid and a base, it is amphitricha or
amphiprotic
NH3, CH3 COOH
H2O + H2O
NH3 + NH3
+
H3O + OH
+
NH4 + NH2
63
+
-
CH3COOH + CH3COOH
CH3 COOH2 + CH3 COO
3+
Al (H2O) 6 hydrated ion is acidic
+
+3
Al (H2O)6 + H2O
H3O+ [A13+ (H2O)5OH-]+2
* Bronsted – Lowry definition is not restricted to water.
HCl + NH3
NH4+ + Cl-
Relative strength of acids and bases :
HCl + H2O
H3O+ + Cl100%
+
HF + H2O
H3O + F
only 3%
common
Base
HCl is more able to donate its proton than HF to the same base..:
HCl is stronger acid than HF.
F- is able to accept 97% protons from the same acid.
: F- is stronger base than Cl-.
a strong acid has a weak conjugate base.
as weak acid has a strong conjugate base.
NH3+ OH100%
+
NH4 OH
0.4%
with the same reference acid NH2 is much stronger base than NH3.
NH2 + H2O
NH3 + H2O
with the same reference base, NH4+ is much stronger acid than NH3.
64
strength
of acid
decreases
Acids
HClO4
HCl
bases
ClO4Cl-
Strength
of base
decreases
OH-
H2O
- Relative strength oxy acids :
CClO4 > HClO3 > HClO2 > HClO
H2SO4 > H2SO3
If you increases no of oxygen
HNO3 > HNO2
atom, acid strength increases.
HClO4 > H2SO4 >H3PO4
increases of acid strength
Increases with electro negativity.
P S Cl
Electro negativity
HClO4 > HBrO4 >HIO4
increases of electro negativity,
Increases in acid strength.
- Binary acid :
HI > HBr > Hd > HF
Although polarity of H-F is highest, the strength of the covalent
bond is dominant factor.
NH3 < H2O > HF
Polarity of the bond is dominate factor.
65
Electro negativity
Increases
C
N
O
F
Cl
Br
I
decreases
- Lewis Definition :Lewis acid is a substance that accepts a pair of electrons to
form covalent band.
Lewis base is a substance that donates a pair of electrons to
form covalent bond.
H
+
1
H + O–H
O–H
Acid base
+
..
H + H-N-H
1
H
+
..
H + .. O-H
1
H
H
1
H - N ..
1
H
+
H
H- N- H
1
H
H
1
: O-H
1
H
Cl
1
B-CL
1
Cl
H
1
H - N
1
H
66
+
+
d
1
B-Cl
1
d
Acidity and basicity is associated with a particular electronic
configuration.
‫ترتيب الكرتوين معني‬
An acid has an incomplete valence shell
A base has electron pairs to donate.
Acidity and basicity is not restricted to any element. Proton in an
acid.
+2
NH3
2+
Cu
+ 4 NH3
Lewis
Lewis
Acid
base
H3N
Cu
NH3
Complex ion
complex compound
NH3
H2O
OH2
OH2
Al
3+
Al + 6H2O
Lewis
Lewis
Acid
base
2
HO
OH2
OH2
HO
H2O + SO3
Lewis
Lewis
base
acid
HO - S - O
O
O
2-
O
+
S – O3
O- S–O
with rearrangement
O
67
-
HO
+
O
11
C
HO
with rearrangement
O
C
O
bi coronate HCO3-
HCl + H2O -------
H3O+ + Cl-
H2O stronger base than Cl- and displace it.
Lewis base is also a Bronsted base.
68
Solutions
Homogeneous mixture of two of more components in one phase
composition is uniform (constant).
1) gaseous; such as air
gas in liquid; CO2 in water
2) Liquid
liquid in liquid; acetone in water
solid in liquid; sugar in water
3) Solid
alloys
- Concentration units :
na
mole fraction Xa = ____________
na+nb+nc+…
no of moles of one component over total number of moles of all
components. sum of mole fractions =1
mole percent = mole fraction X 100
mass of one component
mass fraction = ___________________
total mass of solution
mass % = mass fraction X 100
69
- molarity : no. of moles of solute in 1L solution.
- molality : no. of moles of solute in 1kg of solvent.
 Only molarity changes with change in temperature.
* Ex. A solution of 121g of Zn (NO3)2 in 1L solution has a density of
1.107gmL-1. calculate :
a) mass percent
b) mole fraction
c) molarity
d) molality of Zn (NO3)2
Zn = 65.4 , N=14
,
O=16 ,
H=1
molar mass of Zn (NO3)2 = 65.4 + (2X14) + (16X16) =189.4 gmoL-1
a) mass of 1L solution = 1000 mL X 1.107 gmL-1 = 1107g
121
mass percent =____ X 100 = 10.93%
1107
b) mass of water = 1107 – l2l = 986g
no.of moles of Zn (NO3)2 = l2l g x 1mol = 0.639 mol
189.4g
no.of moles of H2O = 986g x 1mol = 54.78 mol
18g
mole fraction =
0.639
. = 0.0115
0.639+54.78
c) molarity =
0.639 mol = 0.639 M or (molL-1)
IL
D) molality =
0.639 mol = 0.648 molkg-1 or m
0.986 kg
70
- Conversions amongst concentration units :
to convert mass percent to or from mol fraction or to molality
we need to know molecular weight. To convert these into molarity
we must know also density of solution.
Ex :An aqueous solution of 1 M H2SO4 has density of 1.05g mL-1
.
calculate its molality.
H=1 ; O = 16 ; S = 32
mass of 1L = 1000 ml X 1.05 gmL-1 = 1050 g
molar mass of H2SO4 = 98 g mol-1
mass of water = 1050 – 98 = 952 g
molality = 1mol = 1.05 mol kg-1
0.952 kg
OH
1
E.X A solution of CH3-CH-CH3 in water has a mole fraction of
alcohol equals 0.25; what is the mass percent of alcohol in solution.
C = 12 ; O = 16 , H = 1
Assume we have 1 mol of solution
no. of moles of alcohol = 0.25 mol
no. of moles of water = 1-0.25 = 0.75 mol
mass of alcohol = 0.25 mol X 60g = 15 g
1mol
mass of water = 0.75 mol X 18g = 13.5 g
1mol
mass of percent of alcohol = 15g x 100= 52.6%
15+135
71
E.X Calculate molality of the same solution above.
Molality = 0.25 mol = 18.5 mol kg-1
0.0135 kg
- Liquid Solution :
* The solution process in liquid solution :
The ease of distribution of solute particles in solvent depends on
relative forces between :
(1) Solvent molecules
(2) Solute particles
(3) Solute – solvent attraction forces
CCl4
+
relatively
weak London
force
O
relatively
weak London
force
These two liquids are miscible. They dissolve in each other
(similar attraction forces).
CCl4
+
relatively
weak London
force
H2O
relatively
strong hydrogen
bonding
they are immiscible. They form two layers.
72
C2H5OH
+ H2O
relatively
relatively
strong hydrogen strong hydrogen
force
bonding
They are miscible in all propotions
Between miscibility and immiscibility, these is partial
miscibility . If you increase length of the hydrocarbon chain,
solubility decreases.
* Solids in liquids :
Attractive forces between particles of a solid is a maximum
because they are closely arranged.
The solute – solvent attractive forces must be relatively high to
get a solution of solid in liquid.
molecular solid dissolve in non polar solvent.
I2(s) dissolve in CCl4
ionic solid lkike NaCl does not dissolve in CCl4.
weak attractive forces can't overcome strong ionic forces between
ions.
ionic solid dissolves in very polar solvents like water.
strong attractive forces between H2O and ions is enough to tear ionic
lattice.
like dissolves like
polar dissolves polar
non polar dissolves non polar
73
- Heat of solution, H soln :Hsoln = Hsolution – Hcomponents
If H soln is – Ve, energy is released (exothermic)
If H soln is +Ve, energy is absorbed (endothermic)
The magnitude of solution depends on relative forces between
various particles that make a solution.
- Energetics of liquid in liquid solution :There are 3 steps, each of which involves heat change :
1- Separation of solvent molecules (endothermic).
2- Separation of solute molecules (endothermic).
3- Bringing solvent and solute molecules together
(exothermic)
H solution is the sum of the energies.
1st case energy diagrams
Energy required
(2)
to expand solute
(3) solventsolute
together
Head solution
CCl4 + benzene
Hsdn = O
Energy required
(1)
To expand solvent
solvent – solvent,
solute – solute =
solvent – solute
Solvent + solute
attractive forces
A-A, B-B=A-B
74
2nd case
acetone / H2O
H-Ve (exothermic)
solute-solute, solvent-solvent <
Solute-solvent
(2)
volume of solution < volume of
solvent + volume of solute
(1)
(3)
Hsolution
3rd case
Hexane+ ethanol
H+Ve (endothermic)
(3) solvent – solvent, solute-solute >
Solute-solvent attractive force
volume of solution > volume of
solvent + volume of solute
(2)
(1)
Hsolution
75
- Energetic of solution of ionic compounds in water :Two processes
K+I-(s)
K+(g) + I-(g)
endothermic equals lattice energy
+
+
K (g) + I (g) + xH2O
Kaq + Iaq hydration energy (exothermic)
K+I-(s)+K+(g) + I-(g) + xH2O
K+(g)+I-(g)+K+aq+I-(aq), Hsoln
+
K+I-(S) + XH2O
Kaq + Iaq ; Hsoln
If lattice energy > hydration energy
Hsoln + Ve (endothermic)
If lattice energy < hydration energy
Hsoln – Ve (exothermic)
Energy diagram
K+(g)
Lattice
Energy
KI(s)
+
I-(a)
hydration energy
+
Kaq + Iaq
Hsdn
endothermic
for LiCl, it is exothermic
76
increasing charge of the ion, increase both lattice energy and hydration
energy.
Decreasing the size of the ion, increases both lattice energy and hydration
energy.
‫التنبؤ‬
It is difficult to predict in advance which phenomenon predominates.
- Solubility and Temperature :If the process is exothermic, solubility decreases with rising temp.
If the process is endothermic, solubility increases with increasing tempt.
Solvent + gas
solution always exothermic
Solubility of a gas decreases with increasing temperature.
That is why gaseous beverages and preferred to be drunk while cold.
- The effect of pressure on solubility :Changing pressure has no effect on solubility of liquid in liquid or solid in
liquid.
But increasing pressure increases solubility of gas in liquid.
77
gas (solute) + Solvent
solution
Two applications :1- carbonated beverages are canned or bottled under high pressure of
CO2.
2- The same phenomenon is responsible for decompression sickness
(bends).
- Henry's Law :The concentration of a gaseous solute in solution is directly proportional to
the partial pressure of the gas above the solution.
Cg = Kg x Pg
Partial
Pressure
of the gas
Conc. of gas
constant
characteristic
for the gas and
liquid
Henry's law applies for relatively low concentrations and pressures, and for
gases that do not interact extensively with solvent.
78
- Vapor pressure of solutions :
When a solute dissolve in a solvent, the vapor pressure of the solvent
in solution decreases. When a non volatile solute dissolves in
a solvent, the vapor pressure of the solution (also the solvent) is
given by Rauole's law.
O
V.P. of solvent
in solution
PA = XA. PA
V.P of pure solvent
mole fraction
of solvent
*
O
PA
PA
0,0
XA
1
* E.X:- A solution prepared from 96.0g of non volatile, non
dissociating solute in 5.25 mol toluene O CH3 has a vapor
pressure of 16.31 kPa at 60C. what is the molecular mass of the
solute. The V.P. of pure toluene at 60 C is 18.63 kPa.
o
PA = XA. PA
16.31 kPa = XA. 18.63 kPa
79
XA = 0.875
0.875 =
nA
__________
nsolute+nA
0.875 =
5.25 mol
__________
nsolute+5.25
n
solute
= 0.754 mol
1mol
0.754 mol = 96 x _______
M
M = 127.3 gmol-1
__________________
* A Solution of two volatile components : In this case the vapor
pressure of solution equals sum of V.P of each components in
solution.
o
PA = XA. P A applying Rauol's law
o
PB = XB. P B
o
o
Ptotal= PA+ P B = XAPA+XB PB
Graphically
V.P. of solution
O
PA
P
O
B
O
1
XA
XO
1
0
80
Volume of solution
<volume of A+
Volume of B
- Ve deviation A-B>A-A+B-B
exothermic H - Ve
rise temp.
O
PA
P
O
B
O
1
XA
XO
1
0
+ Ve deviation endothermic
H + Ve decreased in temp
volume of solution > volume of
A + volume of B
A-B< A-A+B-B
O
PA
P
O
B
O
1
XA
XO
1
0
* E.X Heptance, C7H16, has a vapor pressure of 791 torr at 100c. At
this temp., octane C8H18 has a vapor pressure of 352 torr. What
will vapor pressure in torr of a mixture of 25 g heptane and 35 g
octane assume ideal solution behavior.
Psolution = Pheptane + Poctane
Xheptanex Pleptans +XoctanexPoctane
Xhep =
25/100
25/100+35/114
, Xoctane =
35/114
35/114+25/100
Psolution = 791 X xhep + 352x Xoctane
Psolution = 355.16 torr + 193.9 torr = 549.06 torr
81
‫الصفات التجمعية‬
- Colligative Properties :
These properties depend on the number of solute particle in
solution and not on its nature. These are :
1- Lowering of vapor pressure.
2- Elevation of boiling point and depression (lowering) of
freezing point.
3- Osmotic pressure.
Phase Diagram of Water
Water liquid
for solution
1atm
ice
vapor
tb
Pressure
0C
100C
tf at freezing
t
boiling point
point
___________________________________________
Normal boiling point : the temperature at which vapor pressure of
liquids = 1 atm.
Normal freezing point : the temperature at which 1 atmosphere line
cross solid liquid equilibrium line.
As a result of solution, vapor pressure decreases and causes rise in
boiling point and loweing of freezing point.
82
These properties are used for determination of molecular masses.
tb = Kb molality
Boiling point constant
Characteristic for each liquid
tf = Kf molality
Freezing point constant
Characteristic for each liquid
* For water : Bp = 100 + tb
fp = 0 - tf
* E.X : What is the molecular mass and molecular formula of a nondissociating compound whose empirical formula is C4H2N
if 3.84 g of the compound in 500 g benzene, C6H6, gives a
freezing point dep. 0.307C.
Kf (benzene) = 5.12Cm-1.
tf = Kf. molality
0.307C = 5.12 Cm-1 X molality
Molality = 0.06m or mol Kg-1
0.06 mol Kg-1 nsolute
0.5 Kg
nsolute = 0.03 mol
83
3.845
molar mass = ________ = 128 g mol-1
0.03 mol
E.F. Wt = (4X12) + (2X1) + (1X14) = 64
128
no. of time E.F. in M.F = _____ = 2
64
Molecular formula = C8 H4 N2
3- osmotic pressure :
Osmosis is a process whereby solvent molecules pass through a semi
permeable membrane from dilute solution to a more concentrated
one.
Dialysis : occurs at cell wall permits water and small particles but
restricts large molecules.
There is a net transfer of solvent from dilute to conc. solutions.
11
II & C. T
osmotic
pressure
h
hydrostatic
pressure =
osmotic pressure
absolute temp.
dilute conc.
molarity
II=RCT
Universal gas constant
II = CRT
II = n/v RT
IIV = nRT
volume
84
Van't Hoff
equation
This phenomenon is used for determination molecular masses
particularly high molecular mass like proteins or plastics (polymers).
* E.X : The osmotic pressure found for solution of 5g horse
hemoglobin in IL solution is 1.8X10-3 atm. At 25C, what
is the molar mass of hemoglobin.
IIV = nRT
1.8X10-3 X 1 = n X 0.0821 X 298 K
n = 7.4 X 10-5 mol
7.4 X 10
-5
5g
= _____________
molecular mass
5g
Molar mass =___________ = 68000 g mol-1
7.4X10-5mol
85
-
Solution of electrolytes :
Electrolytes dissociate in solution. The effect of 0.1 M NaCl in
solution is twice the effect 0.1M glucose, because the former
gives twice as many particles as the latter. Because colligative
properties depends on the number of particles of solute in
solution.
Ca Cl2 has 3 times the effect of non electrolytes.
Al2 (SO4)3 has 5 times the effect of non electrolytes.
A weak electrolyte has an effect between a strong electrolyte and
non electrolyte. In case of association like benzoic acid in
benzene, the effect is halved because two particles act as one
particle.
O……H - O
O
-C
C-
O
O – H ….O
- Inter ionic attraction :
These are attractive force between ions in solution, which increase
with increasing concentration. This makes the solute behaves as if
it is not ionizing completely.
L : Vant Hoff factor
tf (measured)
I =_____________ _______________
tf (calculated as non electrolyte
86
It can be used to calculate degree of dissociation.
- Chemical Equilibrium :forward reaction
C+D
E+F
back word reaction
At equilibrium, forward rate = backward rate.
The concentrations of reactants and products remain constant at
equilibrium.
Rate
forward
time
There is relationship between concentrations of reactants and
products at equilibrium called equilibrium law or law of mass action.
[E] . [F]
Kc = _______
[C] . [D]
[ ] = molarity
Kc : equilibrium constant. It is constant at constant temperature. In
general for :-
87
cC + dD
eE + fF
[E]e . [F]f
Kc = _________
[C]c . [D]d
for gases we can use also Kp.
PEe . PFf
Kp = _______
PCc . PDd
P : Partial pressure
If you have large value Kc,
H2 + Cl2
2HCl
Kc = 5 X 1032
The reaction goes to large extent to the right.
For small value of Kc
2H2O
2H2+ + O2
Kc = 1.2 x 10-82
The equilibrium lies to the left.
This is dynamic equilibrium. The reaction does not stop but the
forward rate equals backward rate.
Some important relationship :
H2(g) + 3H2(g)
2NH3 (g)
2NH3 (g)
Kc
N2 (g) + 3H2 (g) K`c
1
Kc = _____
K`c
88
1/2 N2 (g) + 3/2 H2 (g)
K"c = Kc
NH3 (g)
K"c
= Kc 1/2
2NO(g) + O2 (g)
2NO2 (g)
K1
N2 (g) + O2 (g)
2NO (g)
K2
___________________________________
N2 (g) + 2O2 (g)
2NO2
K3
K3 = K1 x K2
[NO2]2
[NO]2
[NO2]2
__________ X __________ = __________
[NO]2 [O2]
[N]2 [O2]
[N2] [O2] 2
K 1 x K2 = K3
- Lechatellier's Principle :
If a stress is applied to a system at equilibrium, the equilibrium
will shift to reduce or cancel this stress.
N2 (g) + 3H2 (g)
2NH3 (g)
* Change of conc. or partial pressure :
Increasing conc. of reactant(s) shifts eq. to the right (more products
are formed). Increasing conc. of product(s) shifts eq. to the left (more
reactants are formed).
89
Change of total pressure (opposite to change in volume) for previous
reaction (NH3 formation) :Increasing total pressure, the reaction shifts to products
to decrease total pressure by decreasing number of moles of gases.
For 2NO2(g) +
2NO (g) + O2(g)
Increasing total pressure, the reaction shifts to the left.
H2(g) + I2 (g)
2HI (g)
Change in total pressure or volume have no effect on equilibrium.
Effect of temperature change
N2(g) + 3H2 (g)
2NH3 (g) + heat or H – Ve
We treat heat of reaction as a reactant (endothermic) or products
(exothermic).
Increasing of temperature for endothermic reaction shift equilibrium
to products.
In this case Kc value increases. For exothermic reaction increasing
temperature reduces value of Kc.
90
4- Addition of Catalyst : has no effect on equilibrium constant or
equilibrium position.
5- Addition of inert gas : (any gas not present in the equation) has no
effect.
2NO2(g)
N2O4(g)
_____________________
- Reaction Quotient :
cC + dD
Ee + Ff
[E]e . [F]f
Q = _________
[C]c . [D]d
at any time before after or at eq
If we start with all reactants and products calculate Q :
If Q > Kc the reaction goes to left (to reactants)
If Q< Kc the reaction goes to right (to products)
If Q = Kc the reaction is at equilibrium
_________________
- Relationship between Kp and Kc :PDV = nDRT
PD = __nD___ RT = [ D ] R T
V
PDd = [ D ]d (RT)d
91
PEe . PFf
[E]e [RT]e. [F]f [RT]f
Kp = _________ = ___________________
PCc. P.Dd
[C]c [RT]c. [D]d [RT]d
[E]e . [F]f
(e+f) – (c+d)
= _________ = (RT)
[C]c. [D]d
KP = KC. (RT)n
An = no. of moles (coefficient)
of gases in products – no. of
moles of gases in reactants
* E.x. For H2 (g) + I2(g)
2HI(g)
In 10L vessel. at eq. there are : 0.1 mol H2
0.1 mol I 2
0.74 mol HI
If we insert 05 mol of HI after equilibrium has been reached calculate
the new equilibrium concentrations.
[HI]2
(0.74)2
KC = _________ = _____________ = 54.76
[H2] [I2]
(0.1) X (0.1)
10
10
H2(g) + I2(g)
2HI
H2(g) + I2(g)
no. of moles at start
0.1
0.1
change in no, of moles +X
+X
no. of moles at eq.
0.1+X 0.1+X
conc.
0.1+X 0.1+X
10
10
2
[HI]
(1.24 – 2x)2
KC = ________ = 54.76 = _____________
[H2][I2]
(0.1+X) (0.1+X)
10
10
92
2HI
0.74+0.5
-2X
1.24-2X
1.24-2X
10
1.24 – 2X
take square root 7.4 = _____________
0.1 + X
X = 0.053 mol
0.1 + 0.053
0.153
[H2] = _________ = _______ = 0.0153
10
10
[I2] = 0.0153
1.24 + 0.106
[HI] = __________ = 0.1134
10
* E.x. Calculate Kp for the reaction :
H2 (g) + I2(g)
at 27 C if Kc = 7.5
2HI(g)
Kp = Kc. (RT)n
n = 0
Kp = Kc = 7.5
heterogeneous Equilibrium
2NaHCO3(s)
Na2CO3(g)+ H2O(g) + CO2(g)
[Na2CO3(s)] [H2O(g)] [CO2(g)]
KC = _________________________________
[NaH CO3(s)]2
concentrations of pure solids and liquids constant and do not change
[NaHCO3(S)]2
__________ __
[Na2 CO3(S)]
= constant
Kc = [H2O] [CO2]
93
Kp = PH2O x PCO2
KP = Kc(RT)2
H2O(L) = H2O (g)
[H2O] = constant
Kp = PH2O (g)
= 3.17 kPa at 25oC
Kc = [H2O(g)]
Kp = 3.17 kpa
Kc = Kp = 3.17 kPa
= 1.28 X 10-3 M
RT 8.31kPaLmol-1K-1x298K
94
95
* Acid – Base Equilibrium in aqueous solution :
Ionic equilibrium for acids and bases.
H3O+ + OH-
Ionization of water : H2O + H2O
[H3O+] [OH]
KC = __________ = 0.1134
[H2O]2
KC [H2O]2 = [H3O+] [OH-]
[H2O] = 55.5 M constant
at 25 C
KW [H3O+] [OH-] = 1x10-14
KW has same value in neutral acidic or basic solutions. It is an
equilibrium constant. It is called ionization constant or dissociation
constant or ionic product of water.
in pure water or neutral solution :
[H+] = [OH-] = 1X10-7 M
acid makes [H+] > 1 X 10-7 M
base makes [H+] < 1 X 10-7 or [OH-] > 1 X 10-7
96
in acidic solution [H+] is considered to come safely from acid
for two reasons :
1- Ionization of water is very small.
2- The presence of H+ from acid makes ionization still lower. n
basic solution. OH- is considered safely to come from base for
the same reasons in basic solutions.
Strong acids is considered to ionize completely.
Strong base is considered to ionize completely.
H2O
H+ + OH-
* E.x. Calculate [H+] and [OH-] for 0.1 M HCl (aq).
HCl(aq) is a strong acid.
H2O
HCl(aq)
H3O+ + Cl-
o.1 M
0.1 M
0.1 M
[HCl] = [H+] = 0.1 M
Kw
1x10-14
[OH-] =_____ = _______ = 1 X 10-13
[H+]
0.1
97
* E.x. Calculate [OH-] and [H+] for 0.1 M Ba (OH)2
which is a strong base.
Ba+2 + 2OH0.1 M 0.2 M
Ba(OH)2
[OH-] = 0.2 M
Kw
1x10-14
[H+] =_____ = _______ = 5 X 10-14
[OH-]
0.2
* PH :1
PH = -log [H3O ] = Log ________
[H3O+]
+
1
POH = -log [OH ] = Log ________
[OH-]
-
KW = 1 X 10-14 = [H+] [OH-]
PKW = -log 1x10-14 = 14 = - log [H+] + (-log[OH-]
= PH + POH
For solution neutral, basic or acidic
PH + POH = 14
98
* Ex : Calculate PH and POH for 0.1 M HCl (strong acid)
[H+]= 0.1 M
PH = log [H+] – log 1x10-1 = 1
POH = 14 – 1 = 13
* Ex : Calculate POH and PH for 0.1 M Ba (OH)2
[OH-]= 0.2 M
POH = log 2xb-1 = 1-1092 = 0.7
PH = 14 – 0.7 = 13.3
- Note :
Neutral
solution
Acidic
solution
Basic
solution
PH
POH
7
7
<7
>7
>7
<7
* If PH changes from 7 to 5
10-5
_____ = 102
10-7
‫ مرة‬100 ‫يتغري‬
* If PH changes from 5.5 to 7
- 5.5 shift log
3.16X6-6
=_____ _______ = ________31.6
-7 shift log
1X10-7
99
‫يتغري‬
- Dissociation of weak acids and bases :A- + H3O+
HA+H2O
[H3O+] [A-]
Keq = ___________
[HA] [H2O]
[H3O+] [A-]
Keq [H2O] =___________ = Ka
[HA]
conc. At start
change
at eq
M+ + AO
O
+X
+X
X
X
HA
C
-X
C-X=C
X2
[H+]2
Ka = ____ = ________
C
C
Ka. C = [H+]2
[H+] =
Ka.C
100
* Ex : Calculate PH and POH for 0.1 M HC2 H3 O2
(actitic acid) Ka = 1.8 X 10-5
[H+] =
Ka. C =
1.8 x 10-5 X 0.1 = 1.34 X 10-3
PH = -log 1.34 X 10-3 = 2.87
POH = 14-287 = 11.13
- for weak bases like NH3 or organic amines,
NH3 + H2O
NH+4 + OHin the same way :
[OH-] =
Kb.C
* Ex : Calculate POH and PH for 0.1 M NH3, Kb=1.8 X 10-5
[OH-] =
Kb.C
= 1.34 X 10-3
POH = 2.87
PH = 14 – 2.87 = 1.13
101
- Buffers :
It is a solution which undergoes slight change in PH on
addition of small quantity of strong acid or strong base.
acidic buffer PH < 7
weak acid + salt of same acid + strong base
HC2 H3 O2 / Na C2 H3 O2
Basic buffer PH > 7
weak base + salt of same base + strong base
NH3 / NH4+
* How buffer works : for acidic buffers :
a) addition of acid
H+ + C2 H3 O2HC2 H3 O2
from salt
H+ consumed
b) addition of strong base :
OH- + HC2 H3 O2
H2O + C2 H3 O2* if you have basic buffer :
a) H+ + NH3
NH4+
b) OH- + NH4+
OH- consumed
H+ is consumed
H2O+ NH3
102
OH- is consumed
In acidic buffer :
H+ + C2 H3 O2- common lon
Na+ + C2 H3 O2-
HC2 H3 O2
salt ionize Na+ C2H3O2
completely
The presence of acetate from salt causes decrease in he
ionization of the acid, because of common ion effect. The
concentration of acetate can be considered safety to come from salt.
[H+] [C2H3O2-]
Ka =________________
[HC2H3O2]
[acid]
[H ] = Ka _______
[salt]
+
In the same way for basic buffer :
[base]
[OH ] = Kb _______
[salt]
-
Log [H+] = log ka + log [acid] – log [salt]
[salt]
pH = pKa +log ________
[acid]
‫ابلضرب يف سالب‬
pKa ‫يصبح احلمض ضعيف كلما زادت قيمة‬
This is Herdersn Hasselbakh equation
[salt]
pOH = pKb + log _______
[base]
103
(anti) in calculator = shift then log
* Ex : for a buffer solution 0.1 M acetic acid and 0.1 M sodium
acetate, calculate [H+] and pH. Ka = 1.8X10-5
[acid]
0.1 M
-5
[H ]= Ka. _______ = 1.8X10 x _______ = 1.8x10-5
[salt]
0.1 M
+
pH = log 1.8X10-5 = 4.74
____________________
Note [acid] = [salt]
pH = pKa
* Ex : What ratio of acetic acid / sodium acetate concentration is
needed to form a buffer whose
pH = 5 , Ka = 1.8x10-5
[H+] = anti log – pH = 1x10-5
1X10
-5
[acid]
= Ka _______
[salt]
1X10
-5
[acid]
= 1.8x10 X _______
[salt]
-5
1
[acid]
55
____=______ or ____
18 [salt]
100
as long as ratio of
conc. is 55%, the pH =5
104
Dilution of the buffer has no effect on PH of the buffer.
To have effective buffer :
1) Use concentrated solution.
2) Ph
3) pH around pKa, [acid] = [salt].
105
Calculation on Equilibrium
13.54 At a certain temperature Kc = 7.5 for the reaction
2NO2
N2O4
If 2.0 mol of NO2 are placed in 2.0 dm3 container and permitted to reach, what will be
the concentrations of NO2 and N2O4 at equilibrium ? what will be the equilibrium
concentrations if the size of the container is doubled ? Does this conform to what you
would expect from the chatelier's principle ?
2NO2
conc. at start
2.0
moldn-3
2
change in conc. -X
N2O4
O
X
2
[N2O4]
X/2
7.5 = _______ = ______
[NO2]2
(1-X)2
X
15 = _______
1-2X+X2
15-30X+15X2 = X
15-31X+15X2 = O , X = -b+ b2 – 4ac
2a
X= 31+ 961 – 900 = 31 +78=1.29 not acceptable X
30
30
X = 31- 961 -900
30
= 31-7.8 = 0.773
30
conc. of NO2 = 1M – 0.773 M = 0.227 M
conc. of N2O4 = 0.773 M = 0.387 M
2
_________________________________
Volume becomes 4dm3
concentration 0.5 X
X
eq.
2
7.5 = X/2
, 15 =
X
.
2
2
(0.5-X)
(0.25-X-X )
2
3.75 – 15 X + 15 X = X, 3-75-16X+15 X2 = O
X = 16 - 256 – 225 = 16-5.57 = 0.348 = 0.35M
30
30
conc. of N2O4 = 0.175 M
conc. of NO2 = 0.15 M
Total number moles of gas at V=2nd m3 = 0.46+0.78=1.24 mol
Total number moles of gas at V=4dm3 = 0.60+0.68=1.28 mol
Increasing the volume increase the total number of moles gas.
106
Draw Born-Haber Cycle for the formation of CaCl2 Given the
following data, calculate the lattice energy of CaCl2 in kilojoules per
mole. Energy needed to vapourize 1 mol of Ca(s) = 192 KJ; first
ionization energy of Ca=590KJ mol; second ionization energy of Ca
= 1146 KJ mol; electron affinity of Cl =-350 KJ mol-1, bond energy
of Cl2 = 238 KJ mol-1 of Cl-Cl bonds, energy change for the reaction,
Ca(s)+ Cl2(g)
Call2(s), - 795 KJ mol-1 of CaCl2(s)
formed
HF
Ca(s) +
Cl2(g)
CaCl2 (s)
(2)
(5)
(1)
Ca(g)
2Cl(g)
(4)
(3)
2Cl(g)
Ca2+(g)
Energy change for the above reaction is the sum of energy changes of
step (1) through step (5) as shown in the following :
Hf = H1 + H2 + H3 + H4 + H5
-795KJ = 192 KJ + 238 KJ + (590 KJ + 1146 KJ) + (2x(-350KJ)+ Hg
Hg = -2261 KJ mol-1
We have multiplied electron affinity by 2 because we have two chloride
cons.
If we have CaO, we must multiply bond energy by 1/2 because we need
only one mol of oxygen atoms. In addition we have to use 1 st electron
affinity an d 2nd electron affinity in order to get O2-.
In case of KB, we use also 1/2 bond energy of Br2, one ionization energy,
and one electron affinity.
Using fi2O, We have to multiply step (7) by 2 and step (3) by 2.
107
100 C the equilibrium constant, KC, for the reaction.
CO(g) + Cl2(g)
COCl2(g)
has a value of 4.6X109 if O.2 mol of COll2 is placed into a 100 dm3 Hask
at 100 C, what will be the concentration of all species at
equilibrium?
CO (g) + Cl2 (g)
Constant
O :
O
COCl2 (g)
(0.02-X)M
Constant
(0.02-X)M
X
:
X
From 0.2 mol
10dm3
[COCl2]
Kc = _________
[CO] (Cl2]
4-6X10
9
O.O2 - X
= __________
x.x
since the eq. constant has very high value we neglect X from
numerator, 4.6 X 109 = 0.02
x2
X2 = 0.00435 x 10-8 M2
X
= 0.0208x10-4 M
[CO] = 2.1X10-6M
[Cl2] = 2.1X10-6M
[Clc2]= O.O2 M
13.49 Sodium bicarbonate (baking soela) has many useful protection.
Among them in the ability to serve a fire extinguisher because of
thermal decomposition to produce CO2, which some there's the fire.
2NaHCOS(s)
Na2 CoS(s) + CO2 (g) + H2O (g)
It 125 C the value of Kp is 2.6X103 Kpa2.
What are the partial pressures of CO2(g) and H2O (g)…
This suplem at equilibrium ? Car yen explain why Na HCO 3 used in
banking ? This is a heterogeneous equilibrium.
Kp = CO2(g)
H2O(g)
26 X103 KP2c = P2CO2
PCO2(g) = 5.1 X 10 kPa
PHLO (g) = 5.1 X 10 kPa
108