Stoichiometry
Chemical Arithmetic
Two aspects of
chemistry
Qualitative
‫كمي‬
.
Quantitative
‫وصفي أو نوعي‬
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)
H 2O
Hydrogen + Oxygen
One mole hydrogen + half mole oxygen
H2 + 1/2 O2
Water (Qualitatively)
one mole water
(quantitatively).
Basic SI units:
Physical Quantity
Unit
Symbol
Mass
Kilogram
kg
Length
Meter
m
Time
Second
s
Temperature
Kelvin
K
Amount of substance
mole
mol
The Mole:
Is formally defined as the amount of substance of a system that
contains as many elementary 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
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
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
Ex.:
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
Molar mass = atomic weight or molecular weight in grams.
Percentage Composition:
Percentage by mass contributed by each element present in the
sample.
Ex.: 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.
E.F.
M.F.
CH2O
C 2 H4 O2
Acetic acid
H2O
H2O
Water
CH2O
C6H12O6
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 H 1.5
0.25
0.25
O 0.5
0.25
CH6 O2
PO2.5
C1.33 H2
P2 O5
C 4 H6
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
O0.131
0.0437
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
SO3
1.25
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 precede formulas. Start with elements that are
less frequent.
e.g.
C4 H10 + O2
CO2 + H2O
C4 H10 + O2
4CO2 + 5H2O
C4 H10 + 13/2 O2
4CO2 + 5H2O
2C4 H10 + 13O2
8CO2 + 10H2O
E.X:
NH3 + O2
NO + H2O
2NH3 + O2
2NO + 3H2O
2NH3 + 5/2O2
4NH3 + 5O2
E.g:
2NO + 3H2O
4NO + 6H2O
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
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 + O2
2H2O
The amount of product is determined by the reactant
which disappears first. The reactant which is consumed
first is called limiting reactant
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.
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
0.974 mol CCl4 = 0.325
2
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
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
E.g:
E.g:
% Yield = actual yield x 100
Theoretical Yield
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
Calculate the actual yield if the percentage yield of the previous reaction is
80%
actual yield = 101.5g x 80 = 81g
100
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
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 Na OH = 4 g x 1 mol = 0.1 mol
40g
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
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
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.
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 33+
3- Sum of oxidation number of all atoms in a molecule = Zero and
for polyatomic ion = charge on the ion.
PO4-3
S2O3-2
Mn O4-
y + (4 x -2) = -3
2y + (3 x -2) = -2
y + (4 x -2) = -1
Y – 8 = -3
2y – 6 = -2
y – 8 = -1
y = +5
2y = 4
y = +7
y = +2
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
+1-1
CF4
+4-1
MF2
+2-1
PF6+5-1
MF3
+3-1
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.
7- Oxygen in its compounds almost always has -2 O. N.
Exceptions:
Peroxides Na2O2 ,
H2O2 , BaO2
-1
-1
-1
Super Oxide
KO2
+1 -1/2
Fluorine Oxide
F2O
+2
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
Halogens in binary halides (Metal + halogen) have -1 O.N.
Fe Cl3
-1
For familiar ions the oxidation number can be considered the charge of the ions.
PO4-3
Na3PO4
Ca3(PO4)2
CO3-2 NO3SO4-2
NH4+
-2
-1
-2
+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
Zn+2 + Cu it is Redox reaction
+2
O
is
is
is
is
Oxidized
Reduced
Reducing agent
Oxidizing agent
E.g:
Which of the following are redox reaction:
H2 + ½
O O
H+ + OH+1
-1
H+ + H2O
+1 +1-2
O2
+1 -2
H2O
H2O
+1 -2
H3O+
H -2
Ca CO3 heat Ca O + CO2
+2 +1 -2
+2 -2
+4 -2
H2 + F2
O
O
Ag+ + Cl+1
-1
Cu+2 + 4Cl-
Fe+3 + 6 CN-
x
neutralization
x
acid-base reaction
x
decomposition
x
precipitation
2HF
+1 -1
Ag Cl
+1 -1
Cu Cl4
+2 -1
Fe (CN)6-3
+3
-1
x
x
complex formation
(Lewis acid – base Reaction)
complex formation
(Lewis acid – base Reaction)
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.
Balance in Acid medium
Cl2
Cl- + Cl O-3
Divide then carry out balancing
Cl2
Cl2
ClCl O-3
I- + I O-3 + Cl-
I Cl
I- + Cl-
I Cl
IO-3 + ClThen carry on as usual
I Cl
Balance in acid medium
CN- + As O-34
As O-2 + CN O-
CN-
CNO-
CN- + H2O
CNO- + 2H+ + 2e-
,
As O-34
As O-2
As O-34 + 4H+ + 2e-
As O-2 + 2H2O
CN- + As O-34 + 2H+
E.g.:
Balance in acid medium:
Zn + NO-3
NO-3
CNO- + As O-2 + H2O
Zn+2 + NH+4
NH4+
NO-3 + 10H+
NH4+ + 3H2O
NO-3 + 10H+ + 8e-
NH4++ 3H2O ……….. (2) x 1
4Zn + NO-3 + 10H+ + 8e-
4Zn+2 + 8e- + NH4+ + 3H2O
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 . (Nm-2) unit of pressure. It is called Pascal.
area
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.
E.g.:
0.8 atm in torrs
0.8
?
1
760
0.8 atm in Pascal
1.0
101325
0.8
?
608 torrs
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
,
V = constant
P
P
PV = constant
P1 V1 = P2 V2 = P3 V3
Graphically
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.
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.
P = constant
PαT
T
P1 = P2 = …… P = constant X T or
T1 T2
Combined gas law:
Pi Vi = Pf Vf
i = initial
Ti
Tf
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
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
Tf
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
:
2HCl (g)
1
:
2
The ideal gas law:
Vα 1
P
VαT
Vαn
V α nT
P
V = R nT
P
PV = nRT
R: universal gas constant
Ideal gas law or equation
of state for an ideal gas.
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
E.g:
Calculate the volume of 4g methane, CH4, at 25°C and 80 kPa.
,
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
P1 = n1 = X1
mole fraction
Pt
nt
P1 = Pt x X1
C = 12
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.
Inert gases have
49
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 نحسب معه ضغط الماء‬
PO2 only dry oxygen (gas): ‫أما‬
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
Inert gases have
nHe = 6g x 1 = 1.5 mo.
monatomic molecules
49
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
‫ اذا ذكر السؤال نحسب معه ضغط الماء‬Wet oxygen (Gas) Pt = Po2 + PH2o
‫ أما‬dry oxygen (gas): PO2 only
Density of a gas:
PV = n RT
n= m
M
mass of gas
molar mass
P = n RT
V
P = m RT
m =d
density
M V
V
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
Real Gases:
Gas laws apply perfectly to ideal gases, but real gases
deviate from ideal behavior.
Example: PV = n RT 
‫للغاز الحقيقي‬
‫خاصة عند درجات منخفضة وضغط عالي‬
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.
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
for n mole
V2
(P + a ) (V – b) = RT
for 1 mole
V2
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
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
R
0.0521 x 303
Mass of O2 = 4.5 x 10-3 x 32 = 0.144g
= 4.5 x 10-3 mol
b) No of moles of KClO3 consumed = 4.5 x 10-3 x 2 = 3 x 10-3 mol
3
Molar mass of KClO3 =122.5g mol-1
Mass of KClO3 consumed = 3 x 10-3 x 122.5 = 0.369g
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
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:
CO(g) ;ΔH2 = -110.5kJ
C(graphite) + 1/2O2(g)
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 )
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)
N2O5(g) + H2O(l)
2H2O(l)
ΔH° = -571.5kJ
2HNO3(l)
ΔH° = 76.6kJ
½ N(g) + 3/2 O2(g)
HNO3(l), ΔH° = -174kJ
Calculate ΔH for the reaction:
2N2(g) + 5O2(g)
2N2O5 (g)
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)
4HNO3(l), ΔH° = -696KJ
2N2O5 (g) + 2H2O(l) ,
2H2O(l)
Add
2H2(g) + O2(g)
ΔH° = +153.2 KJ
ΔH° = +571.5 kJ
2N2(g) + 5O2 (g)
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)
H2SO4(l)
S(s) + H2(g) + 2O2(g)
H2SO4(l)
S(g) + H2(g) + 2O2(g)
H2SO4(l) because S(g) is not most stable
H2(s) + Br2(l)
½ H2(g)
2HBr(g)
H(g)
ΔH°f
Example:
H2(g) + ½ O2(g)
C(graphite) + 2H2 (g)
H2O(l)
ΔH°f (1)
CH4 (g)
ΔH°(2)
1
C(graphite) + ½ O2
CO(g)
ΔH°f (3)
Reactions like (2) and (3) are difficult to measure. So, we use calculations to know ΔH°f
Ex.:
CO2 (g) ΔH° = -393.5kJ (1)
C(graphite) + O2(g)
H2(g) + ½ O2(g)
CH4(g) + 2O2(g)
Calculate ΔH°f for CH4.
ΔH° = -285.9kJ (2)
H2O(l)
CO2(g)+2H2O(l)ΔH° = -890.4kJ (3)
C(graphite) + 2H2(g)
CO2(g)
Change direction of (2), multiply (2) by 2 and add to (1)
C(graphite) + O2(g)
2H2(g) + O2(g)
CO2(g) + 2H2O(l)
CO2(g) ΔH° = -393.5kJ
2H2O(l)
ΔH° = -571.8kJ
CH4(g)+2O2(g)ΔH°= +890.4kJ
C(graphite) + 2H2(g)
ΔH° = -74.9kJ mol
CH4(g)
C (graphite) + 2O2(g) + 2H2(g)
ΔH1
CO2(g) + 2H2O(l)
ΔHf
ΔH2
ΔH1 = ΔHf + ΔH2
CH4 + 2O2
ΔH depends on status of reactants and products
Ex:
Calculate ΔH vap
H2(g) + ½ O2
H2O(l)
H2(g) + ½ O2
H2O(g)
H2O(l)
ΔH° = -286kJ (1)
ΔH° = -242 kJ (2)
H2O(g)
ΔH vap
Change direction of (1) and add to (2)
H2O(l)
H2 + ½ O2
H2 + ½ O2
ΔH° = 286 kJ
H2O(g)
ΔH° = - 242 kJ
H2O(g)
H2(g) +
H2O(g)
1/2 O2(g) ΔH1
ΔH° vap = 44 kJ
H2O(g)
ΔH2
ΔH vap
ΔH1 = ΔH2 + ΔHvap
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) + 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
ΔH = ?
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
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
Heat capacity:
Heat required or released to raise the temperature of the system 1ºC.
= heat compacity x Δt Heat released
= heat capcity x (t2 – t1)
Then calculate for one mole, ΔE.
heat at constant volume when ΔV = 0
ΔE = qv
Y: specific heat, M: mass
For pure substances = M x Y x Δt
Heat change
The relation ship between ΔH and ΔE:
P ΔV = Δn RT
H = E = PV
Δn = number of moles of gases produced
ΔH = ΔE+ PΔV
- number of moles of gases consumed
ΔH = ΔE + nRT
Neglect Δn change in solid and liquids.
E.x: Relation ship between ΔH and ΔE:
1. C(s) +
O2(g)
ΔH = ΔE
CO2(g)
because Δn( gas) = O
2. CO(g) +
1/2O2(g)
CO2(g)
ΔH = ΔE – 1/2 RT because Δn( gas) =½
3. Zn(s) +
2HCl(aq)
ΔH = ΔE + RT
Zn Cl2(aq) + H2(g)
because Δn(gas) = O
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:
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
9F
1s2 2s2 2p6 3s1
1s2
2s2
2p5
11Na
9F
.
.
:
[Na+] + [: F- : ]
Na +: F:
:
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 1s2 2s2 2p6 3s2 3p6 4s2 3d6
26Fe+2 [Ar] 366
24Cr
30Zn
[Ar]
[Ar] 4s1
3d5
[Ar] 4s2
3d10
24Cr+3
30Zn+2
[Ar] 3d3
[Ar] 3d10 pseudo noble gas
structure
The ratio between is not always 1 : 1
[Ca-]+2 + 2 [Cl]-
Ca + 2 Cl
2 Li + O
2 [Li]+ + [O]-2
ratio
E.x: deducing formula from charges:
Ca+2
(PO4)-3
2 :
2
1
Ca3 (PO4)2
Al+3 SO4-2
Mg+2 O-2
ratio 1 :
Al2 (SO4)3
Mg2 O2
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) +
1/2 F2(g)
Li+F-(s)
1 mole
1/2 mole
1mole
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)
F(g) (4) + 1e-
Li(g)
(3) - 1e-
F- (g)
Li+ (g)
H5H4 + H3 + H2 + H1 + Hf = 
Li(g); H=155 kJ
Step (1) : Vaporization or sublimation. Li(s)
endothermic
Step (2) : dissociation 1/2 F2 (g)
Step (3) : ionization Li(g)
endothermic
Step (4) : electron affinity
H = -330 kJ exothermic 
H= 79 kJ endothermic
F(g) ;
Li+ (g) + 1e-
F(g) + 1e-
Step (5) :Lattice energy: Li+(g)+ F -(g)
highly exothermic
;
H = 520 kJ
F-(g) ;
Li+F-(s) ;
H =- 1016kJ
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 (15P – 9F)
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:
NO3(9N, 8 O)
1- (1x5) + (3 x 6) = 24eo
2- o:N:o
o
3- o:N:o
4- X :o:
5- :o: :N:
:o:
____________________________________
CO2
4 + (2 x 6) = 16 eO:C:O
o:c:o
X
O:C:O
-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
Increase in bond energy
H- C-C- H
H
Bond order
between carbon
atoms
C=C
-C=C-
H
1
2
3
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.
between
State of mater and Intermolecular forces :
Intra bonding(intra means inside) is
stronger than intermolecular forces.
a) Dipole - dipole Interaction :
dipole
it occurs between polar molecules. It amounts to 1% strength of the covalent
bond.
HS+ - 98In solids > liquids > gases
98- - H8+
b) hydrogen bonding :
Occurs between molecules when hydrogen is covalently bonded to 1 of the
three small most electronegative atoms which are of F, O, N. example : H2O
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
Hydrogen bonding strength between H…F>H…O>H…N because of electro
negativity.
H2O
HF
NH3
Because water make
o
o
o
morehydrogenbonds
B.P
100C
19C
-35C
F
O
H
H2O
o
B.P 100C
F
H
H
F
H
CH3O-H
o
68 C
N
H
o
H
120
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 ,…..
B
non polar CO2 O = C = O
BF3 F
F
compound
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
Definitions of acids
and bases
* Arrhenius definition in modern concepts :
An acid is a substance that increases concentration of H3O+ in aqueous solution
H3O+ + Cl-
Hcl + H2O
CO2 + H2O
H2CO3
H2CO3+H2O
H3O+ + HCO3- 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
NH3 + H2O
- metals oxides : CaO, Na2O , BaO, K2O, M9O
are basic anhydrides
CaO + H2 O
Ca (OH)2
Neutralization : H3O+ + OHor simply :
H+ + OH-
2H2O
H2O
NH3+ OH-
* Bronsted – Lowry Definition :
An acid is a proton donor, a base is a proton acceptor
Hcl + H2O
acid
base
acid
H3O+ + Clbase
There are two acids and bases. An acid and abase that differ by only one proton are called acidbase conjugate pairs.
* E.x :- H2S , S-2 x
H2O , OH- 
-3
H2S , H5 
H3PO4, PO4 x
+
H3O, OH-= x
+
H3 O, H2O  +
NH3
+ H2O
NH4 + OH
base
acid
acid
base
conjugated pairs
* water behaves as an acid and a base, it is amphitricha or amphiprotic
NH3, CH3 COOH
+
H2O + H2O
H3O + OH
+
NH3 + NH3
NH4 + NH2
+
CH3COOH + CH3COOH
Al3+ (H2O) 6 hydrated iron is acidic
-
CH3 COOH2 + CH3 COO
+
Al+3 (H2O)6 + H2O
H3O+ [A13+ (H2O)5OH-]+2
* Bronsted – Lowry definition is not restricted to water.
HCl + NH3
NH4+ + ClRelative strength of acids and bases :
Hd + tho
H3O+ + d-
10%
H3O+ + Fonly 3%
common
Base
Hd is more able to donate its proton than HF to the same base..: Hd is stronger acid than HF.
F- is able to accept 97% protons from the same acid.
: F- is stronger base than Cl-.
as strong acid has a week conjugate base.
as weak acid has a strong conjugate base.
NH2 + H2O
NH3+ OH100%
+
NH3 + H2O
NH4 OH
0.4%
with the same reference acid NH2 is much stronger base than NH3.
with the same reference base, NH4+ is much stronger acid than NH3.
HF + H2O
strength
of acid
decreases
Acids
Hdo4
Hd
bases
dO4d-
Strength
of base
decreases
H2O
- Relative strength oxy acids :
CdO4 > HdO3 > HdO2 HdO
H2SO4 > H2SO3
HNO3 > HNO2
HdO4 > H2SO4 >H3PO4
P
If you increases no of oxygen
atom, acid strength increases.
increases of acid strength, increases,
with electro negativity.
S Cl
Electro negativity
HdO4 > 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.
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.
form covalent bond.
H
1
O–H
+
H + O–H
Acid base
+
+
H + H-N-H
1
H
H + O-H
1
H
H
H- N- H
1
H
1
: O-H
1
Acidity and basicity is associated with a particular electronic configuration.
‫ترتيب الكتروني معين‬
An acid has a complete valence shell
A base has electronic pairs to donate.
Acidity and basicity is not restricted to any element protn in an acid.
+2
NH3
2+
Complex ion
CU
Lewis
Acid
+ 4 NH3
H3N
C
Lewis
base
NH3
NH3
complex compound
H2O
OH2
OH2
Al
Al3+ + 6H2O
Lewis
Lewis
Acid
base
H2O
OH2
OH2
O
H2O
Lewis
base
+ SO3
Lewis
acid
HO - S - O
O
O
2O +
S–O
O- S–O
O
with rearrangement
HO +
O
O
11
C
C
HO
O
with rearrangement
bi coronate HCO3Hd + H2O
H3O+ + ClH2O
stronger base than d- and displace it.
Lewis base is also a Bronsted base.
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
- Concentration units :
na
male fraction Xa = ____________
na+nb+nc+…
no of moles of one component over total number of moles of all components. sum of
more fractional.
mole percent = mole fraction X 100
mass of one component
male fraction = ___________________
total mass of solution
mass % = mass fraction X 100
- 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 gmola) mass of 1L solution = 1000 ml X 1.107 gml-3 = 107g
121
mass percent =____ X 100 = 10.93%
1107
b) mass of water = 1107 – l2l = 9869
no. mde of Zn (NO3)2 = l2l 9 x 1md = -.639 mol
189.49
no. mde of H2O = 9869 x 1mol = 54.78 mol
189
mole fraction = 0.639
. = 0.0115
0.639+54.78
c) molarity = 0.639 mol = 0.639 M or (mdL-1)
IL
D) molality = 0.639 mol = 0.648 molkg-1 or m
0.986 Kg
- Conversions amongst concentration units :
to convert mass percent to or from mal fraction or to molality we need to know molecular
weight. To convert these into molarity we must know also density of solution.
* E.X : An aqueous solution of 1 M H2SO4 has density of 1.05g L-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 mdes 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
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 :
Solvent molecules
Solute particles
Solute – solvent attraction forces
CC/4
+
O
relatively
relatively
weak London
weak London
force
force
These two liquids are miscible. They dissolve in each other (similar attraction forces).
CC/4
+
H2O
relatively
relatively
weak London
strong hydrogen
force
bonding
they are immiscible. They form two layers.
C2H5OH
+ H2O
relatively
relatively
strong hydrogen
strong hydrogen
force
bonding
Between miscibility and immiscibility, these is partial miscibility i.e.: if you increase length
of the chain, solubility decreases.
* Solidsin 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 CC/4
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
- Heat of solution, H sdn :Hsdn = Hsolution – Hcomponents
If H sdn is – Ve, energy is released (exothermic)
If H sdn 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) solvent-
Head solution
CCl4 + benzene
Hsdn = O
solute
together
Energy required
(1)
To expand solvent
Solvent + solute
solvent – solvent,
solute – solute =
solvent – solute
attractive forces
A-A, B-B=A-B
2nd case
acetone / H2O
H-Ve (exothermic)
solute-solute, solvent-solvent <
volume of solution < volume of
solvent + volume of solute
(2)
(1)
(3)
Hsolution
3rd case
hexanet ethanol
H-Ve (endothermic)
solvent – solvent, solute-solute >
Solute-solvent attractive force
volume of solution > volume of
(3) solvent + volume of solute
(2)
(1)
Hsolution
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- 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), Hsdn
+
K+I-(S) + XH2O
Kaq + Iaq ; Hsdn
If lattice energy > hydration energy
Hsdn + Ve (endothermic)
If lattice energy < hydration energy
Hsdn – Ve (exothermic)
Energy diagram
K+(g)
+ I-(a)
Lattice
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Energy
hydration energy
+
Kaq + Ia3
KI(s)
Hsdn
endothermic
for lid, it is exothermic
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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 withy 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 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.
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gas (solute) + Solvent
solution
Two applications :carbonated beverages are canned or bottled under high pressure of CO2.
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 Pg
Partial
Conc. of gas
Pressure
of the gas
constant
characteristic
for the gas and
liquid
Henry's law applies for relatively low concentration and pressures, and for
gases that do not interact extensively with solvent.
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- 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
PA = XA. PA
V.P of pure solvent
in solution
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
Xa = 0.875
nA
0.875 = __________
nsolute+nA
5.25 mol
0.875 = __________
nsolute+5.25
nsolute = 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
O
PB
O
1
XA
XO
1
0
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Volume of solution
<volume of A+
Volume of B
O
- Ve deviation A-B>A-A+B-B
PA
exothermic H - Ve
rise temp.
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O
PB
O
1
XA
XO
1
0
O
PA
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O
PB
+ Ve deviation endothermic
H + Ve decreased in temp
volume of solution > volume of
A + volume of B
A-B< A-A+B-B
O
XA
1
1
XO
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
Xheptane Pleptans +Xoctane
Poctane
Xhep =
25/100
, Xoctane =
35/114
25/100+35/114
35/114+25/100
Psolution = 791 X xhep + 352x Xoctane
Psolution = 355.16 torr + 193.9 torr = 549.06 torr
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‫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
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1atm
ice
vapor
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tb
Pressure
0C
100C
boiling point
tf at freezing
t
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.
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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
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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 on.
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
h
hydrostatic
pressure =
osmotic pressure
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II & C. T
osmotic
pressure
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dilute conc.
malarity
II=RCT
Universal gas constant
IIV = nRT
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absolute temp.
equation
II = CRT
II = n/v RT
volume
Van't Hoff
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This phenomenon is used for determination molecular
masses particularly high molecular mass like portions or
plastics (polymers).
* E.X : The osmotic pressure found for solution of 59 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 b-5 mol
59
7.4 X 10-5 = _____________
molecular mass
59
Molar mass =___________ = 68000 g mol-1
7.4X10-5md
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
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 remains constant at equilibrium.
Rate
forward
time
There is relationship between concentration of reactants and products at
equilibrium called equilibrium law or law of mass action.
[E] . [F]
[ ] = molarity
Kc = _______
[C] . [D]
Kc : equilibrium constant. It is constant at constant temperature. In general for :-
cC + dD
eE + fF
[E]e . [F]f
Kc = _________
[C]c . [D]d
for gases we can use also Kp.
PEe . PFf
P : Partial pressure
Kp = _______
PCc . PDd
If you have large value Kc,
H2 + Cl2
2HCl
Kc = 5 X 1032
The reaction goes to large extant to the right.
For small value of Kc
2H2O + Cl2
2H2+ + O2 Kc = 1.2 o 15-82
The equilibrium lies for to the left.
This is dynamic equilibrium. The reaction does not stop but the forward rate equals
backward rate.
Some important relationship :
H2g + 3H2(g)
2NH3 (g)
Kc
2NH3 (g)
N2 (g) + 3H2 (g) Kc
1
Kc = _____
Kc
1/2 N2 (g) + 3/2 H2 (g)
NH3 (g)
Kc
Kc = Kc = 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).
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.
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.
2N2(g)
N2O4(g)
_____________________
- Reaction Quotient :
cC + dD
Ee + Ff
[E]e . [F]f
Q = _________
at any time before after or at eq
[C]c . [D]d
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
nD
PD = ____ RT = [ D ] R T
V
PDd = [ D ]d (RT)d
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. therare : 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
concentration.
[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
change in no, of moles+X
no. of moles at eq.
0.1+X
conc.
0.1+X
10
2HI
0.1
+X
0.1+X
0.1+X
10
0.74+0.5
-2X
1.24-2X
1.24-2X
10
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heterogeneous Equilibrium
2NaHCO3(g)
Na2CO3(g) H2O(g) + CO2(g)
[Na2CO3(S)] [H2O] [CO2(g)]
KC = _________________________________
[NaH CO3(S)]2
concentration of pure solids and liquids constant and does not
chang
[NaHCO3(S)]2
KC = ___________ = Kc [H2ocy] [LCO]
[Na2 CO3(S)]
Kp = PH2O – PCO2
KP = Kc(RT)org= Kc (RT)2
H2O(L) = H2O (g)
[H2O] = cent
Kp = PH2O (g)
= 3.17 kpu
Kc = [H2Ocg]
il P pvo at 25
Kp = 3.17 Kpa
-3
Kc = Kp = 3.17 hp
= 1.28 X 10-3 M
RT 8.31hpalmx243
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[HI]2
(1.24 – 2x)2
KC = ________ = 54.76 = _____________
[H2][I2]
(0.1+X) (0.1+X)
1.24 – 2X 10
10
54.76 = 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)
2HI(g)
at 27 C if Kc = 7.5
Kp = Kc. (RT)n
n = 0
Kp = Kc = 7.5
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* Acid – Base Equilibrium in aqueous solution :
Ionic equilibrium for acids and bases.
Ionization of water : H2O + H2O
H3O+ + OH[H3O+] [OH]
KC = __________ = 0.1134
[H2O]2
KC [H2O]2 = [H3O+] [OH-]
[H2O] = 55.5 M constant
KW [H3O+] [OH-] = 1x10-14
at 25 C
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 of 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
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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 in basis
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+ + Clo.1 M
0.1 M 0.1 M
[HCl] = [H+] = 0.1 M
Kw
1x10-14
[OH-] =_____ = _______ = 1 X 10-13
[H+]
0.1
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* E.x. Calculate [OH-] and [H+] for 0.1 M Ba (OH)2
which is a strong base.
Ba(OH)2
Ba+2 + 2OH0.1 M 0.2 M
[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-]
1
PKW = -log 1x10-14 = 14 = log [H+] + (-log[OH-3)]
= PH + POH
For solution neutral, basic or acidic
PH + POH = 14
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* 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 :
* If PH changes from 7 to 5
10-5
_____ = 102
10-7
* If PH changes from 5.5 to 7
5.5
shift log
3.16X6-6
=_____ = _______ = ________31.6
-7
shift log
1X10-7
‫ مرة‬100 ‫يتغير‬
‫يتغير‬

- Dissociation of weak acids and bases :-

HA+H2O

[H3O+] [A-]
Keq = ___________
[HA] [H2O]
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A- + H3O+
[H3O+] [A-]
Keq [H2O] =___________ = Ka
[HA]
conc. At start HA
M+ + Achange
C
O
at eq
-X
+X
C-X=C
X
X2
[H+]2
Ka = ____ = ________
C
C
Ka. C = [H+]2
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[H+] =
Ka.C
O
+X
X
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* 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 amins,
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 105
[OH-] = Kb.C = 1.34 X 10-3
POH = 2.87
PH = 14 – 2.87 = 1.13
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- Buffer :
It is a solution which under goes 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+ + H2 H3 O2HC2 H3 O2 H+ consumed
from salt
b) addition of strong base :
OH- + HC2 H3 O2
H2O + C2 H3 O2- OH- consumed
* if you have basic buffer :
a) H+ + NH3
NH4+
H+ is consumed
b) OH- + NH4+
H2O+ NH3 OH- is consumed
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In acidic buffer :
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HC2 H3 O2
H+ + C2 H3 O2common lon
ionize Na+ C2H3O2
Na+ + C2 H3 O2completely
The presence of acetate from salt causes decrease in he ionization of the acid, became 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]
‫بالضرب في سالب‬
[salt]
PH = Pka +
log ________
‫كلما زادت قيمة‬ka ‫يصبح الحمض ضعيف‬
[acid]
Herdersn Hasselbakh equation
[salt]
POH = PKb + log _______
[base]
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(anti) in calculator = shift + 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
[H+]= Ka. _______ = 1.8X10-5 X _______ = 1.8x15-5
[salt]
0.1 M
PH = log 1.8X10-5 = 4.74
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Note [acid] = [salt] PH = PKa
* Ex : What ratio of acetic acid / sodium acetate concentration is needed to
form a buffer whose
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PH = 5 , Ka = 1.8b-5
[H+] = anti log – PH = 1x10-5
[acid]
1X10-5 = Ka _______
[salt]
[acid]
1X10-5 = 1.8X10-5 X _______
[salt]
1
[acid]
55
____=______ or ____
18
[salt]
100
as long as ratio of
conc. is 55%, the PH=5
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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].
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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
N2O4
conc. at start
2.0
O
moldn-3
2
change in conc.
-X
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
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X = 31- 961 -900
= 31-7.8 = 0.773
30
30
conc. of NO2 = 1M – 0.773 M = 0.227 M
conc.
of N2O4 = 0.773 M = 0.387 M
2
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Volume becomes 4dm3
concentration 0.5 X
X
eq.
2
7.5 = X/2
, 15 =
X
.
(0.5-X)2
(0.25-X-X2)
3.75 – 15 X + 15 X2 = 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.
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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)
2Cl(g)
(4)
2Cl(g)
Ca(g)
(3)
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 1st 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.
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)
COCl2 (g)
Constant
O :
O
(0.02-X)M From 0.2 mol
10dm3
Constant
X
:
X
(0.02-X)M
[COCl2]
Kc = _________
[CO] (Cl2]
O.O2 - X
4-6X109 = __________
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 becaus
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 HCO3 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