Stoichiometry Chemical Arithmetic Two aspects of chemistry Quantitative وصفي أو نوعي كمي Qualitative Stoichiometry in the quantitative aspect of chemical.

Stoichiometry Chemical

Arithmetic

Two aspects of chemistry

Quantitative وصفي أو نوعي

كمي

Qualitative

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 H2O

Hydrogen + Oxygen Water (Qualitatively)

One mole hydrogen + half mole oxygen one mole water (quantitatively) .

Basic SI units: Physical QuantityUnitSymbol

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 4Contains 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 = 1MW = ( 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 CNumber 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.

CH2OC2H4O2Acetic acid

H2OH2O Water

CH2OC6H12O6Glucose 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 O 0.5 CH6 O2

0.25 0.25 0.25

PO2.5 P2 O5

C1.33 H2 C4 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 S1O3 SO3

0.0437 .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

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 14Molecular 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 + 5/2O2 2NO + 3H2O

4NH3 + 5O2 4NO + 6H2O

2NH3 + O2 2NO + 3H2O

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 1 molecule

12 g + 32 g 44 g

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 gmolar mass of SbF3 = (1 x 122) + (3 x 19) = 179g

no. of moles of CCl4 = 150 gCCl4 x 1mol CCl4 = 0.974mol SbF3

154g CCl4no. of moles of SbF3 = 100g SbF3 x 1mol SbF3 = 0.559mol SbF3

179g SbF3no. 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 CCl2F2Mass of CCl2F2= 0.839mol CCl2F2 x 121g Freon=101.5g Freon

1mol Freon121g is molar mass of CCl2F2b) 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 % Yield = actual yield x 100

Theoretical YieldE.g: Calculate the percentage yield of the previous

reaction (in the previous example) if the actual yield is 76g.

% yield = actual yield x 100Theoretical 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

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 YieldE.g.: Calculate the percentage yield of the previous reaction (in the previous example) if the actual

yield is 76g. % yield = actual yield x 100Theoretical 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

MolarityTo 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

0.4Y = 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 0.2Volume = 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 Na, Cu, K, Fe H2, F2, N2, O2, Cl2 P4 S8

2- Oxidation Number of atoms in simple ions (monatomic ions) equals charge of the ion.

+2 3 -3 +

M2+, X3-, M3 +

3- Sum of oxidation number of all atoms in a molecule = Zero and for polyatomic ion = charge

on the ion. PO4

-3S2O3-2Mn O4

-

y + (4 x -2) = -32y + (3 x -2) = -2y + (4 x -2) = -1

Y – 8 = -32y – 6 = -2y – 8 = -1

y = +52y = 4y = +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 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.

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

+2Hydrogen 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 -1Halogens 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 NO3- SO4-2 NH4+ -2 -1 -2 +1Oxidation: 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 Zn+2 + Cu it is Redox reaction O +2 +2 O

Zn is Oxidized Cu+2 is Reduced Zn is Reducing agent Cu+2 is Oxidizing agent

E.g: Which of the following are redox reaction:

H2 + ½ O2 H2O

O O +1 -2

H+ + OH- H2O x neutralization +1 -1 +1 -2

H+ + H2O H3O+ x acid-base reaction +1 +1-2 H -2

Ca CO3 heat Ca O + CO2 x decomposition +2 +1 -2 +2 -2 +4 -2

H2 + F2 2HF O O +1 -1

Ag+ + Cl- Ag Cl x precipitation +1 -1 +1 -1

Cu+2 + 4Cl- Cu Cl4 x complex formation +2 -1 (Lewis acid – base Reaction)

Fe+3 + 6 CN- Fe (CN)6-3 x complex formation +3 -1 (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 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- CNO- , As O-34 As O-

2

CN- + H2O CNO- + 2H+ + 2e-

As O-34 + 4H+ + 2e- As O-

2 + 2H2O

CN- + As O-34 + 2H+ CNO- + As O-

2 + H2O

E.g.: Balance in acid medium:

Zn + NO-3 Zn+2 + NH+

4

NO-3 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.

areaAtmospheric 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.

608 torrs

E.g.: 0.8 atm in torrs

0.8 ?

1.0 101325

0.8 ? 81060kPa

1 760 0.8 atm in Pascal

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. V α 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.

V α 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 α T P = constant T

P = constant X T or P1 = P2 = …… 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 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

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) 2HCl (g)

1 : 1 : 2

The ideal gas law: V α 1 V α nT P P V α T V α n V = R nT R: universal gas

constant P

PV = nRT 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. 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 m3J = 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

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

HeP total = PHe + Po2 = 3.7 + 3.06 = 6.73 atm

Or Pt = nt RT

V Pt = Po2 + PH2o Wet oxygen (Gas) معه نحسب السؤال ذكر اذا

الماء ضغطPO2 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

nHe = 6g x 1 = 1.5 mo. Inert gases have 49 monatomic 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) معه نحسب السؤال ذكر اذاالماء ضغط

PO2 only dry oxygen (gas) :أما

Density of a gas: PV = n RT n = m mass of gas

M molar mass P = n RT VP = 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-1or 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 الحقيقي للغاز

منخفضة درجات عند خاصةعالي وضغط

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 = 4.5 x 10-3 molMass 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

3Molar 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

ΔH1 = ΔH2 + ΔH3 We can calculate ΔH of a reaction from known heats of other reactions

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) CO2(g) , ΔH3 = -283KJ

Add the two equation

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 )

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) 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), ΔH° = -696KJ

4HNO3(l) 2N2O5 (g) + 2H2O(l) , ΔH° = +153.2 KJ

2H2O(l) 2H2(g) + O2(g) ΔH° = +571.5 kJ

Add

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) 2HBr(g)

½ H2(g) H(g) ΔH°f

Example: H2(g) + ½ O2(g) H2O(l) ΔH°f (1)

C(graphite) + 2H2 (g) 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.: C(graphite) + O2(g) CO2 (g) ΔH° = -393.5kJ (1)

H2(g) + ½ O2(g) H2O(l) ΔH° = -285.9kJ (2) CH4(g) + 2O2(g) CO2(g)+2H2O(l)ΔH° = -

890.4kJ (3) Calculate ΔH°f for CH4.

C(graphite) + 2H2(g) CO2(g)

Change direction of (2), multiply (2) by 2 and add to (1)

C(graphite) + O2(g) CO2(g) ΔH° = -393.5kJ

2H2(g) + O2(g) 2H2O(l) ΔH° = -571.8kJ

CO2(g) + 2H2O(l) CH4(g)+2O2(g)ΔH°= +890.4kJ

C(graphite) + 2H2(g) CH4(g) ΔH° = -74.9kJ

mol

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: H2(g) + ½ O2 H2O(l) ΔH° = -286kJ (1)

H2(g) + ½ O2 H2O(g) ΔH° = -242 kJ (2)

Calculate ΔH vap

H2O(l) H2O(g) ΔH vap

Change direction of (1) and add to (2)

H2O(l) H2 + ½ O2 ΔH° = 286 kJ

H2 + ½ O2 H2O(g) ΔH° = - 242 kJ

H2O(g) H2O(g) ΔH° vap =

44 kJH2(g) + 1/2 O2(g) ΔH1

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) ΔH = ?

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

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

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) CO2(g)

ΔH = ΔE – 1/2 RT because Δn( gas) =½

3. Zn(s) + 2HCl(aq) Zn Cl2(aq) + H2(g)

ΔH = ΔE + RT because Δn(gas) = O

Heat capacity :Heat required or released to raise the temperature of the

system 1ºC .Heat released = heat compacity x Δt =heat capcity x (t2 – t1)

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 P ΔV = Δn RT

ΔH = ΔE+ PΔV Δn = number of moles of gases produced ΔH = ΔE + nRT - number of moles of gases consumed

Neglect Δn change in solid and liquids .

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 8ALi 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

11Na 1s2 2s2 2p6 3s19F 1s2 2s2 2p5

. . :

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

[Ar] 24Cr [Ar] 4s1 3d5 24Cr+3 [Ar] 3d3 30Zn [Ar] 4s2 3d10 30Zn+2 [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 E.x: deducing formula from charges:

Ca+2 (PO4)-3 Ca3 (PO4)2

Al+3 SO4-2 Al2 (SO4)3

Mg+2 O-2 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 1moleWe can analyze the factors that contribute to the energy change of this reaction:- Born Haper cycle : Δ

HfLi(s) + 1/2 F2 (g)

Li+F- (s)

F(g) (4) + 1e- F- (g)

Li(g) (3) - 1e- Li+ (g)

Hf = H1 + H2 + H3 + H4 + H5

Step (1) : Vaporization or sublimation. Li(s) Li(g); H=155 kJ

endothermic

Step (2) : dissociation 1/2 F2 (g) F(g) ; H= 79 kJ endothermic

Step (3) : ionization Li(g) Li+ (g) + 1e- ; H = 520 kJ endothermicStep (4) : electron affinity

F(g) + 1e- F-(g) ; H = -330 kJ exothermic

Step (5) :Lattice energy: Li+(g)+ F -(g) Li+F-(s) ; H =- 1016kJ

highly exothermic

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 2s51- (1x5) + (3x7) = 26 e-2- F:P:F

F 4- F:P:F: F

3- :F:P:F: :F:

NO3- (9N, 8 O)1- (1x5) + (3 x 6) = 24e-

o2- o:N:o o

3- o:N:o4- X :o:

5- :o: :N: :o:

____________________________________ CO2

4 + (2 x 6) = 16 e-O:C:Oo:c:o

XO: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

H - C - C - H C=C -C=C-

H HBond order

between carbon 1 2 3atoms

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.

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+ - 98-

In 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 morehydrogenbondsB.P 100C 19C -35C

O F F N H H H H H H

HH2O CH3O-H CH3 CH2 O-H

o o oB.P 100C 68 C 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 o 120

Bnon 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 acidsand bases* Arrhenius definition in modern concepts : An acid is a substance that increases concentration of H3O+ in aqueous solution

Hcl + H2O H3O+ + Cl-

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 OH- in aqueous solution

NaOH(s) + H2O Na+aq + OH-aq aq=

aqueous

NH3 + H2O NH3+ OH-

- metals oxides : CaO, Na2O , BaO, K2O, M9O

are basic anhydrides CaO + H2 O Ca (OH)2

Neutralization : H3O+ + OH- 2H2O

or simply : H+ + OH- H2O

* Bronsted – Lowry Definition : An acid is a proton donor, a base is a proton acceptor

Hcl + H2O H3O+ + Cl-

acid base 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 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 CH3 COOH2 + CH3 COO

Al3+ (H2O) 6 hydrated iron is acidic +

Al+3 (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 :

Hd + tho H3O+ + d- 10%

HF + H2O H3O+ + F- only 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+ OH- 100%

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.

Acids basesstrength Hdo4 dO4

- Strength

of acid Hd d- of base decreases decreases H2O

- Relative strength oxy acids :CdO4 > HdO

3 > HdO2 HdO

H2SO4 > H2SO3 If you increases no of oxygen

HNO3 > HNO2 atom, acid strength increases.

HdO4 > H2SO4 >H3PO4 increases of acid strength, increases,

with electro negativity.P S Cl

Electro negativity HdO4 > HBrO4 >HIO4 increases of electro negativity,

Increases in acid strength.-Binary acid :

HI > HBr > Hd > HFAlthough 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 FCl decreases BrI

- 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 H + O–H O – HAcid base

H + H-N-H H- N- H 1 1

H H

+ + H

H + O-H 1 1 : O-H

H 1

Acidity and basicity is associated with a particular electronic configuration.

معين الكتروني ترتيبAn 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 + 4 NH3 H3N C NH3 complex compound

Lewis Lewis Acid base

NH3

H2O OH2 OH2

Al

Al3+ + 6H2O H2O OH2

Lewis LewisAcid base OH2

O

H2O + SO3 HO - S - OLewis Lewis base acid O

O 2-O + S – O O - S – O with rearrangement

O

- O O

11 C -HO + C H O O with rearrangement

bi coronate HCO3-Hd + H2O H3O+ + Cl-H2O stronger base than d- and displace it.Lewis base is also a Bronsted base.

SolutionsHomogeneous 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 :

namale 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=1molar mass of Zn (NO3)2 = 65.4 + (2X14) + (16X16) = 189.4 gmol-a) mass of 1L solution = 1000 ml X 1.107 gml-3 = 107g 121 mass percent =____ X 100 = 10.93% 1107b) 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 = 32mass of 1L = 1000 ml X 1.05 gml-1 = 1050 gmolar mass of H2SO4 = 98 g mol-1mass of water = 1050 – 98 = 952 gmolality = 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 = 1Assume we have 1 mol of solution no. of mdes of alcohol = 0.25 molno. of moles of water = 1-0.25 = 0.75 molmass of alcohol = 0.25 mol X 60g = 15 g

1molmass of water = 0.75 mol X 18g = 13.5 g

1molmass 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 + Orelatively relativelyweak London weak Londonforce forceThese two liquids are miscible. They dissolve in each other (similar

attraction forces).CC/4 + H2O relatively relativelyweak London strong hydrogenforce bonding they are immiscible. They form two layers.

C2H5OH + H2O relatively relativelystrong hydrogen strong hydrogenforce 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/4ionic 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 Head solution

(2) CCl4 + benzene to expand solute (3) solvent- Hsdn = O

solute together

Energy required solvent – solvent,

(1) solute – solute = To expand solvent solvent – solute

Solvent + solute attractive forces

A-A, B-B=A-B

2nd case acetone / H2OH-Ve (exothermic)

(2) solute-solute, solvent-solvent <

volume of solution < volume of

solvent + volume of solute (1) (3)

Hsolution

3rd case hexanet ethanolH-Ve (endothermic)

(2) solvent – solvent, solute-solute >

Solute-solvent attractive force

volume of solution > volume of

(1) (3) solvent + volume of solute

Hsolution

- Energetic of solution of ionic compounds in water :-- Energetic of solution of ionic compounds in water :- Two processes Two processes K+I-(S)K+I-(S) K+(g) + I-(g) K+(g) + I-(g) endothermic equals lattice endothermic equals lattice

energyenergy + -+ - K+(g) + I-(g) + XH2O Kaq + Iaq hydration energy K+(g) + I-(g) + XH2O Kaq + Iaq hydration energy

(exothermic)(exothermic) K+I-(S)+K+(g) + I-(g) + XH2O K+(g)+I-(g)+K+aq+I-K+I-(S)+K+(g) + I-(g) + XH2O K+(g)+I-(g)+K+aq+I-

(aq), (aq), HsdnHsdn + -+ - K+I-(S) + XH2O Kaq + Iaq ; K+I-(S) + XH2O Kaq + Iaq ; HsdnHsdn If lattice energy > hydration energy If lattice energy > hydration energy Hsdn + Ve (endothermic)Hsdn + Ve (endothermic) If lattice energy < hydration energy If lattice energy < hydration energy Hsdn Hsdn –– Ve (exothermic) Ve (exothermic) Energy diagram Energy diagram K+(g) + I-(a)K+(g) + I-(a)

Lattice Lattice hydration energy hydration energy + -+ - Energy Energy Kaq + Ia3 Kaq + Ia3

KI(s)KI(s) Hsdn Hsdn endothermic endothermic for lid, it is exothermic for lid, it is exothermic

increasing charge of the ion, increase both lattice energy and hydration increasing charge of the ion, increase both lattice energy and hydration energy.energy.

Decreasing the size of the ion, increases both lattice energy and Decreasing the size of the ion, increases both lattice energy and hydration energy.hydration energy.

التنبؤالتنبؤ It is difficult to predict in advance which phenomenon predominates.It is difficult to predict in advance which phenomenon predominates. - Solubility and Temperature :-- Solubility and Temperature :- If the process is exothermic, solubility decreases withy rising temp.If the process is exothermic, solubility decreases withy rising temp. If the process is endothermic, solubility increases with increasing If the process is endothermic, solubility increases with increasing

tempt.tempt. Solvent + gas solution always exothermicSolvent + gas solution always exothermic Solubility of a gas decreases with increasing temperature.Solubility of a gas decreases with increasing temperature. That is why gaseous beverages and preferred to drunk while cold.That is why gaseous beverages and preferred to drunk while cold. - The effect of pressure on solubility :-- The effect of pressure on solubility :- Changing pressure has no effect on solubility of liquid in liquid or solid Changing pressure has no effect on solubility of liquid in liquid or solid

in liquid.in liquid. But increasing pressure increases solubility of gas in liquid.But increasing pressure increases solubility of gas in liquid.

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.

- - Vapor pressure of solutions :Vapor pressure of solutions : When a solute dissolve in a solvent, the vapor pressure of the solvent in 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, 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.the vapor pressure of the solution (also the solvent) is given by Rauole's law.

OO V.P. of solvent PA = XA. PA V.P of pure solvent V.P. of solvent PA = XA. PA V.P of pure solvent in solution in solution mole fraction mole fraction of solvent of solvent ** OO PAPA PAPA

0,00,0 XA XA 1 1 * * E.XE.X:- A solution prepared from 96.0g of non volatile, non dissociating :- A solution prepared from 96.0g of non volatile, non dissociating

solute in 5.25 mol toluene O CH3 has a vapor solute in 5.25 mol toluene O CH3 has a vapor pressure of 16.31 Kpa at 60pressure of 16.31 Kpa at 60C. what is the molecular mass of the solute. The C. what is the molecular mass of the solute. The

V.P. of pure toluene at 60V.P. of pure toluene at 60 C is 18.63 KPa. C is 18.63 KPa. oo PA = XA. PA PA = XA. PA 16.31 KPa = XA. 18.63 KPa16.31 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.

oPA = XA. P A applying Rauol's law

oPB = XB. P B

o o

Ptotal= PA. P B = XAPA+XB PBGraphically V.P. of solution

O PA

OPB

O XA 1 1 XO 0

Volume of solution Volume of solution <volume of A+<volume of A+ Volume of BVolume of B OO - Ve deviation A-B>A-- Ve deviation A-B>A-

A+B-BA+B-B PAPA exothermic exothermic H - VeH - Ve OO rise temp. rise temp. PB PB

OO XA XA 1 1 11 XO XO 0 0

OO + Ve deviation endothermic + Ve deviation endothermic PAPA H + Ve decreased in temp H + Ve decreased in temp OO volume of solution > volume of volume of solution > volume of PB PB A + volume of B A + volume of B A-B< A-A+B-BA-B< A-A+B-B

O XAO XA 1 1 1 XO1 XO 0 0 * * E.XE.X Heptance, C7H16, has a vapor pressure of 791 torr at Heptance, C7H16, has a vapor pressure of 791 torr at

100100c. At this temp., octane C8H18 has a vapor pressure of 352 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 torr. What will vapor pressure in torr of a mixture of 25 g heptane and 35 g octane assume ideal solution behavior.heptane and 35 g octane assume ideal solution behavior.

Psolution = Pheptane + PoctanePsolution = Pheptane + Poctane Xheptane Pleptans +Xoctane Xheptane Pleptans +Xoctane

Poctane Poctane Xhep = Xhep = 25/100 25/100 , Xoctane = , Xoctane = 35/114 35/114 25/100+35/11425/100+35/114 35/114+25/100 35/114+25/100 Psolution = 791 X xhep + 352x Xoctane Psolution = 791 X xhep + 352x Xoctane Psolution = 355.16 torr + 193.9 torr = Psolution = 355.16 torr + 193.9 torr = 549.06 torr549.06 torr

التجمعية التجمعية الصفات الصفات - Colligative Properties :- Colligative Properties : These properties depend on the number of solute particle in solution and not on its nature. These properties depend on the number of solute particle in solution and not on its nature.

These are :These are : 1- Lowering of vapor pressure.1- Lowering of vapor pressure. 2- Elevation of boiling point and depression (lowering) of freezing point.2- Elevation of boiling point and depression (lowering) of freezing point. 3- Osmotic pressure.3- Osmotic pressure. Phase Diagram of WaterPhase Diagram of Water Water liquidWater liquid for solution for solution

1atm1atm ice ice vapor vapor Pressure Pressure tbtb

00C 100C 100CC tf at freezing tf at freezing t boiling point t boiling point point point ______________________________________________________________________________________ Normal boiling point : the temperature at which vapor pressure of liquids = 1 atm.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 Normal freezing point : the temperature at which 1 atmosphere line cross solid liquid

equilibrium line.equilibrium line. As a result of solution, vapor pressure decreases and causes rise in boiling point and loweing of As a result of solution, vapor pressure decreases and causes rise in boiling point and loweing of

freezing point.freezing point.

These properties are used for determination of molecular masses.These properties are used for determination of molecular masses. tb = Kb molality tb = Kb molality

Boiling point constant Boiling point constant Characteristic for each liquid Characteristic for each liquid tf = Kf molality tf = Kf molality

Freezing point constant Freezing point constant Characteristic for each liquid Characteristic for each liquid * For water : Bp = 100 + * For water : Bp = 100 + tbtb fp = 0 - fp = 0 - tftf * * E.XE.X : What is the molecular mass and molecular formula of a non- : What is the molecular mass and molecular formula of a non-

dissociating compound whose empirical formula is C4H2N if 3.84 g of the dissociating compound whose empirical formula is C4H2N if 3.84 g of the compound in 500 g benzene, C6H6, gives a freezing point dep. 0.307compound in 500 g benzene, C6H6, gives a freezing point dep. 0.307C.C.

Kf (benzene) = 5.12Kf (benzene) = 5.12Cm-1.Cm-1. tf = Kf. molality tf = Kf. molality 0.3070.307C = 5.12 C = 5.12 Cm-1 X molality Cm-1 X molality Molality = 0.06m or mol Kg-1Molality = 0.06m or mol Kg-1 0.06 mol Kg-1 0.06 mol Kg-1 nsolutensolute 0.5 Kg0.5 Kg nsolute = 0.03 mol nsolute = 0.03 mol

3.8453.845 molar mass = ________ = 128 g mol-1molar mass = ________ = 128 g mol-1 0.03 mol0.03 mol E.F. Wt = (4X12) + (2X1) + (1X14) = 64E.F. Wt = (4X12) + (2X1) + (1X14) = 64 128128 no. of time E.F. in M.F = _____ = 2no. of time E.F. in M.F = _____ = 2 64 64 Molecular formula Molecular formula = C8 H4 N2= C8 H4 N2 3- osmotic pressure :3- osmotic pressure : Osmosis is a process whereby solvent molecules pass through a semi permeable membrane from dilute Osmosis is a process whereby solvent molecules pass through a semi permeable membrane from dilute

solution to a more concentrated on.solution to a more concentrated on. Dialysis : occurs at cell wall permits water and small particles but restricts large molecules.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. There is a net transfer of solvent from dilute to conc. solutions. 11 h hydrostatic 11 h hydrostatic pressure = pressure = osmotic osmotic

pressure pressure

II & C. T absolute temp.II & C. T absolute temp. osmotic osmotic dilute conc. dilute conc. pressure malarity pressure malarity II=RCTII=RCT Universal gas constant Universal gas constant IIV = nRTIIV = nRT

II = CRTII = CRT Van't HoffVan't Hoff II = n/v RTII = n/v RT equation equation

volume volume

This phenomenon is used for determination This phenomenon is used for determination molecular masses particularly high molecular mass molecular masses particularly high molecular mass like portions or plastics (polymers).like portions or plastics (polymers).

* * E.XE.X : The osmotic pressure found for solution of : The osmotic pressure found for solution of 59 horse hemoglobin in IL solution is 1.8X10-3 atm. 59 horse hemoglobin in IL solution is 1.8X10-3 atm. At 25At 25C, what is the molar mass of hemoglobin.C, what is the molar mass of hemoglobin.

IIV = nRTIIV = nRT 1.8X10-3 X 1 = n X 0.0821 X 298 K1.8X10-3 X 1 = n X 0.0821 X 298 K n = 7.4 X b-5 moln = 7.4 X b-5 mol 5959 7.4 X 10-5 = _____________7.4 X 10-5 = _____________ molecular massmolecular mass 5959 Molar mass =___________ = 68000 g mol-1Molar mass =___________ = 68000 g mol-1 7.4X10-5md7.4X10-5md

- 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

timeThere 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]dfor gases we can use also Kp. PEe . PFf P : Partial pressure Kp = _______ PCc . PDdIf you have large value Kc, H2 + Cl2 2HCl Kc = 5 X 1032The reaction goes to large extant to the right.For small value of Kc2H2O + Cl2 2H2+ + O2 Kc = 1.2 o 15-82The 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) Kc2NH3 (g) N2 (g) + 3H2 (g) Kc 1Kc = _____ Kc

1/2 N2 (g) + 3/2 H2 (g) NH3 (g) Kc

Kc = Kc = Kc 1/22NO(g) + O2 (g) 2NO2 (g) K1N2 (g) + O2 (g) 2NO (g) K2___________________________________N2 (g) + 2O2 (g) 2NO2 K3K3 = K1 x K2 [NO2]2 [NO]2 [NO2]2 __________ X __________ = __________ [NO]2 [O2] [N]2 [O2] [N2] [O2] 2K 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 – VeWe 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]dIf 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 nDPD = ____ RT = [ D ] R T VPDd = [ D ]d (RT)d

PEe . PFf [E]e [RT]e. [F]f [RT]fKp = _________ = ___________________ 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 20.74 mol HI

If we insert 05 mol of HI after equilibrium has been reached calculate the new equilibrium concentration. [HI]2 (0.74)2KC = _________ = _____________ = 54.76 [H2] [I2] (0.1) X (0.1) 10 10 H2(g) + I2(g) 2HI

H2(g) + I2(g) 2HIno. of moles at start 0.1 0.10.74+0.5change in no, of moles+X +X -2Xno. of moles at eq. 0.1+X 0.1+X 1.24-2Xconc. 0.1+X 0.1+X 1.24-2X

10 10 10

heterogeneous Equilibriumheterogeneous Equilibrium 2NaHCO3(g) 2NaHCO3(g) Na2CO3(g) H2O(g) + CO2(g)Na2CO3(g) H2O(g) + CO2(g) [Na2CO3(S)] [H2O] [CO2(g)] [Na2CO3(S)] [H2O] [CO2(g)] KC = _________________________________ KC = _________________________________ [NaH CO3(S)]2 [NaH CO3(S)]2 concentration of pure solids and liquids constant and does not concentration of pure solids and liquids constant and does not

chang chang [NaHCO3(S)]2 [NaHCO3(S)]2 KC = ___________ = Kc [H2ocy] [LCO]KC = ___________ = Kc [H2ocy] [LCO] [Na2 CO3(S)] [Na2 CO3(S)] Kp = PH2O Kp = PH2O –– PCO2 PCO2 KP = KP = Kc(RT)org= Kc (RT)2Kc(RT)org= Kc (RT)2 H2O(L) = H2O (g)H2O(L) = H2O (g) [H2O] = cent[H2O] = cent Kp = PH2O (g)Kp = PH2O (g) = 3.17 kpu= 3.17 kpu Kc = [H2Ocg]Kc = [H2Ocg] il P pvo at 25il P pvo at 25 Kp = 3.17 KpaKp = 3.17 Kpa -3 -3 Kc = Kc = KpKp = = 3.17 hp3.17 hp = 1.28 X 10-3 M = 1.28 X 10-3 M RT 8.31hpalmx243RT 8.31hpalmx243

[HI]2 (1.24 [HI]2 (1.24 –– 2x)2 2x)2 KC = ________ = 54.76 = _____________KC = ________ = 54.76 = _____________ [H2][I2] [H2][I2] (0.1+X)(0.1+X) (0.1+X)(0.1+X) 1.24 1.24 –– 2X 10 10 2X 10 10 54.76 = 7.4 = _____________54.76 = 7.4 = _____________ 0.1 + X0.1 + X X = 0.053 molX = 0.053 mol 0.1 + 0.053 0.1530.1 + 0.053 0.153 [H2] = _________ = _______ = 0.0153[H2] = _________ = _______ = 0.0153 10 1010 10 [I2] = 0.0153[I2] = 0.0153 1.24 + 0.106 1.24 + 0.106 [HI] = __________ = 0.1134[HI] = __________ = 0.1134 10 10 * * E.xE.x. Calculate Kp for the reaction :. Calculate Kp for the reaction : H2 (g) + I2(g) 2HI(g)H2 (g) + I2(g) 2HI(g) at 27 at 27 C if Kc = 7.5C if Kc = 7.5 Kp = Kc. (RT)Kp = Kc. (RT)nn n = 0n = 0 Kp = Kc = 7.5Kp = Kc = 7.5

* * Acid Acid –– Base Equilibrium in aqueous solution : Base Equilibrium in aqueous solution : Ionic equilibrium for acids and bases.Ionic equilibrium for acids and bases. Ionization of water : H2O + H2O H3O+ + OH-Ionization of water : H2O + H2O H3O+ + OH- [H3O+] [OH] [H3O+] [OH] KC = __________ = 0.1134KC = __________ = 0.1134 [H2O]2 [H2O]2 KC [H2O]2 = [H3O+] [OH-]KC [H2O]2 = [H3O+] [OH-] [H2O] = [H2O] = 55.5 M55.5 M constant constant KW [H3O+] [OH-] = 1x10-14 at 25 KW [H3O+] [OH-] = 1x10-14 at 25 CC KW has same value in neutral acidic or basic solutions. It is KW has same value in neutral acidic or basic solutions. It is

an equilibrium constant. It is called ionization constant an equilibrium constant. It is called ionization constant oror dissociation constant dissociation constant oror ionic product of water. ionic product of water.

in pure water of neutral solution :in pure water of neutral solution : [H+] = [OH-] = 1X10-7 M[H+] = [OH-] = 1X10-7 M acid makes [H+] > 1 X 10-7 Macid makes [H+] > 1 X 10-7 M base makes [H+] < 1 X 10-7 or [OH-] > 1 X 10-7base makes [H+] < 1 X 10-7 or [OH-] > 1 X 10-7

in acidic solution [H+] is considered to come safely from acid for in acidic solution [H+] is considered to come safely from acid for two reasons :two reasons :

1- Ionization of water is very small.1- Ionization of water is very small. 2- The presence of H+ from acid makes ionization still lower in 2- The presence of H+ from acid makes ionization still lower in

basis solution. OH- is considered safely to come from base for the basis solution. OH- is considered safely to come from base for the same reasons in basic solutions.same reasons in basic solutions.

Strong acids is considered to ionize completely.Strong acids is considered to ionize completely. Strong base is considered to ionize completely.Strong base is considered to ionize completely.

H2O H+ + OH-H2O H+ + OH- * * E.xE.x. Calculate [H+] and [OH-] for 0.1 M HCl (aq).. Calculate [H+] and [OH-] for 0.1 M HCl (aq). HCl(aq) is a strong acid.HCl(aq) is a strong acid. H2OH2O HCl(aq) H3O+ + Cl-HCl(aq) H3O+ + Cl- o.1 Mo.1 M 0.1 M 0.1 M 0.1 M 0.1 M [HCl] = [H+] = 0.1 M[HCl] = [H+] = 0.1 M Kw 1x10-14Kw 1x10-14 [OH-] =_____ = _______ = 1 X 10-13[OH-] =_____ = _______ = 1 X 10-13 [H+] 0.1[H+] 0.1

* * E.xE.x. Calculate [OH-] and [H+] for 0.1 M Ba (OH)2. Calculate [OH-] and [H+] for 0.1 M Ba (OH)2 which is a strong base.which is a strong base. Ba(OH)2 Ba+2 + 2OH-Ba(OH)2 Ba+2 + 2OH- 0.1 M 0.2 M0.1 M 0.2 M [OH-] = 0.2 M[OH-] = 0.2 M Kw 1x10-14Kw 1x10-14 [H+] =_____ = _______ = 5 X 10-14[H+] =_____ = _______ = 5 X 10-14 [OH-] 0.2[OH-] 0.2 * PH :-* PH :- 11 PH = -log [H3O+] = Log ________PH = -log [H3O+] = Log ________ [H3O+][H3O+] 11 POH = -log [OH-] = Log ________POH = -log [OH-] = Log ________ [OH-][OH-] KW = 1 X 10-14 = [H+] [OH-]KW = 1 X 10-14 = [H+] [OH-] 11 PKW = -log 1x10-14 = 14 = log [H+] + (-log[OH-3)]PKW = -log 1x10-14 = 14 = log [H+] + (-log[OH-3)] = PH + POH= PH + POH For solution neutral, basic or acidic For solution neutral, basic or acidic PH + POH = 14PH + POH = 14

* Ex* Ex : Calculate PH and POH for 0.1 M Hcl (strong acid) : Calculate PH and POH for 0.1 M Hcl (strong acid) [H+]= 0.1 M[H+]= 0.1 M PH = log [H+] PH = log [H+] –– log 1x10-1 = 1 log 1x10-1 = 1 POH = 14 POH = 14 –– 1 = 13 1 = 13 * Ex* Ex : Calculate POH and PH for 0.1 M Ba (OH)2 : Calculate POH and PH for 0.1 M Ba (OH)2 [OH-]= 0.2 M[OH-]= 0.2 M POH = log 2xb-1 = 1-1092 = 0.7POH = log 2xb-1 = 1-1092 = 0.7 PH = 14 PH = 14 –– 0.7 = 13.3 0.7 = 13.3 Note :Note :

* If PH changes from 7 to 5* If PH changes from 7 to 5 10-5 10-5 _____ = 102_____ = 102 مرة مرة 100100يتغير يتغير 10-7 10-7 * If PH changes from 5.5 to 7* If PH changes from 5.5 to 7 5.5 shift log 3.16X6-65.5 shift log 3.16X6-6 يتغيريتغير 31.6________ = _______ = _____= 31.6________ = _______ = _____= -7 shift log 1X10-7-7 shift log 1X10-7

- Dissociation of weak acids and bases :-- Dissociation of weak acids and bases :-

HA+H2O A- + H3O+HA+H2O A- + H3O+

[H3O+] [A‑][H3O+] [A‑] Keq = ___________Keq = ___________ [HA] [H2O][HA] [H2O]

[H3O+] [A‑][H3O+] [A‑] Keq [H2O] =___________ = KaKeq [H2O] =___________ = Ka [HA][HA]

conc. At startconc. At start HA M+ + A-HA M+ + A- change change C C O O O O at eqat eq -X -X +X+X +X +X C-X=C X XC-X=C X X X2X2 [H+]2 [H+]2 Ka = ____ = ________Ka = ____ = ________ C C C C Ka. C = [H+]2Ka. C = [H+]2 [H+] = Ka.C[H+] = Ka.C

* Ex* Ex : Calculate PH and POH for 0.1 M HC2 H3 O2 : Calculate PH and POH for 0.1 M HC2 H3 O2 (actitic acid) Ka = 1.8 X 10-5(actitic acid) Ka = 1.8 X 10-5

[H+] = Ka. C = 1.8 x 10-5 X 0.1 = 1.34 X 10-3[H+] = Ka. C = 1.8 x 10-5 X 0.1 = 1.34 X 10-3 PH = -log 1.34 X 10-3 = 2.87PH = -log 1.34 X 10-3 = 2.87 POH = 14-POH = 14-287 = 11.13287 = 11.13 - fOR weak bases like NH3 or organic amins, - fOR weak bases like NH3 or organic amins, NH3 + H2O NH+4 + OH-NH3 + H2O NH+4 + OH- in the same way :in the same way : [OH-] = Kb.C[OH-] = Kb.C

* Ex* Ex : Calculate POH and PH for 0.1 M NH3, Kb=1.8 X 10- : Calculate POH and PH for 0.1 M NH3, Kb=1.8 X 10-55

[OH-] = Kb.C = 1.34 X 10-3[OH-] = Kb.C = 1.34 X 10-3 POH = 2.87POH = 2.87 PH = 14 PH = 14 –– 2.87 = 1.13 2.87 = 1.13

- Buffer : - Buffer : It is a solution which under goes slight change in PH on addition of It is a solution which under goes slight change in PH on addition of

small quantity of strong acid or strong base.small quantity of strong acid or strong base. acidic buffer PH < 7acidic buffer PH < 7 weak acid + salt of same acid + strong base weak acid + salt of same acid + strong base HC2 H3 O2 / Na C2 H3 O2HC2 H3 O2 / Na C2 H3 O2 Basic buffer PH > 7Basic buffer PH > 7 weak base + salt of same base + strong base weak base + salt of same base + strong base NH3 / NH4+NH3 / NH4+ * How buffer works : for acidic buffers :* How buffer works : for acidic buffers : a) addition of acid a) addition of acid H+ + H2 H3 O2- HC2 H3 O2 H+ consumed H+ + H2 H3 O2- HC2 H3 O2 H+ consumed from salt from salt b) addition of strong base :b) addition of strong base : OH- + HC2 H3 O2 H2O + C2 H3 O2- OH- consumedOH- + HC2 H3 O2 H2O + C2 H3 O2- OH- consumed * if you have basic buffer :* if you have basic buffer : a) H+ + NH3 NH4+ H+ is consumed a) H+ + NH3 NH4+ H+ is consumed b) OH- + NH4+ H2O+ NH3 OH- is consumed b) OH- + NH4+ H2O+ NH3 OH- is consumed

In acidic buffer :In acidic buffer : HC2 H3 O2 HC2 H3 O2 H+ + C2 H3 O2- common lon H+ + C2 H3 O2- common lon ionizeionize Na+ C2H3O2Na+ C2H3O2 Na+ + C2 H3 O2-Na+ + C2 H3 O2- completely completely The presence of acetate from salt causes decrease in he ionization of the acid, became 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 of common ion effect. The concentration of acetate can be considered safety to come from salt.from salt.

[H+] [C2H3O2-][H+] [C2H3O2-] Ka =________________ Ka =________________ [HC2H3O2][HC2H3O2] [acid][acid] [H+] = Ka _______ [H+] = Ka _______ [salt][salt] In the same way for basic buffer :In the same way for basic buffer : [base][base] [OH-] = Kb _______ [OH-] = Kb _______ [salt][salt] Log [H+] = log ka + log [acid] Log [H+] = log ka + log [acid] –– log [salt] log [salt] سالب في سالب بالضرب في بالضرب [salt][salt] PH =PH = Pka +Pka + log ________ log ________ قيمة زادت قيمة كلما زادت ضعيف kaka كلما الحمض ضعيف يصبح الحمض يصبح [acid][acid] Herdersn Hasselbakh equation Herdersn Hasselbakh equation [salt][salt] POH = PKb + log _______ POH = PKb + log _______ [base][base]

(anti) in calculator = shift + log(anti) in calculator = shift + log * Ex* Ex : for a buffer solution 0.1 M acetic acid and 0.1 M sodium Acetate, calculate : for a buffer solution 0.1 M acetic acid and 0.1 M sodium Acetate, calculate

[H+] and PH. Ka = 1.8X10-5[H+] and PH. Ka = 1.8X10-5 [acid][acid] 0.1 M 0.1 M [H+]= Ka. _______ = 1.8X10-5 X _______ = [H+]= Ka. _______ = 1.8X10-5 X _______ = 1.8x151.8x15-5-5 [salt][salt] 0.1 M 0.1 M PH = log 1.8X10-5 = PH = log 1.8X10-5 = 4.744.74 ________________________________________ Note [acid] = [salt] PH = PKaNote [acid] = [salt] PH = PKa * Ex* Ex : What ratio of acetic acid / sodium acetate concentration is needed to form : What ratio of acetic acid / sodium acetate concentration is needed to form

a buffer whose a buffer whose PH = 5 , Ka = 1.8b-5PH = 5 , Ka = 1.8b-5 [H+] = anti log [H+] = anti log –– PH = 1x10-5 PH = 1x10-5 [acid][acid] 1X10-5 = Ka _______ 1X10-5 = Ka _______ [salt][salt] [acid][acid] 1X10-5 = 1.8X10-5 X _______ 1X10-5 = 1.8X10-5 X _______ [salt][salt] 1 [acid] 551 [acid] 55 ____=______ ____=______ oror ____ ____ as long as ratio of as long as ratio of 18 [salt] 10018 [salt] 100 conc. is 55%, the PH=5conc. is 55%, the PH=5

Dilution of the buffer has no effect Dilution of the buffer has no effect on PH of the buffer.on PH of the buffer.

To have effective buffer :To have effective buffer : 1- Use concentrated solution.1- Use concentrated solution. 2- Ph2- Ph 3- PH around PKa, [acid] = [salt].3- PH around PKa, [acid] = [salt].

Calculation on EquilibriumCalculation on Equilibrium 13.54 At a certain temperature Kc = 7.5 for the reaction 13.54 At a certain temperature Kc = 7.5 for the reaction 2NO22NO2 N2O4 N2O4 If 2.0 mol of NO2 are placed in 2.0 dm3 container and permitted to reach, what will be the concentrations of 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 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 ?doubled ? Does this conform to what you would expect from the chatelier's principle ?

2NO22NO2 N2O4 N2O4 conc. at start conc. at start 2.02.0 O O moldn-3moldn-3 2 2 change in conc.change in conc. -X -X XX 22 [N2O4] X/2[N2O4] X/2 7.5 = _______ = ______7.5 = _______ = ______ [NO2]2 (1-X)2[NO2]2 (1-X)2 XX 15 = _______15 = _______ 1-2X+X21-2X+X2 15-30X+15X2 = X15-30X+15X2 = X 15-31X+15X2 = O , X = -b+ 15-31X+15X2 = O , X = -b+ b2 b2 –– 4ac 4ac 2a2a X= X= 31+ 961 31+ 961 –– 900 900 = = 31 +7831 +78=1.29 not acceptable X=1.29 not acceptable X 30 3030 30

X = X = 31- 961 -90031- 961 -900 = = 31-7.831-7.8 = 0.773 = 0.773 3030 30 30 conc. of NO2 = 1M conc. of NO2 = 1M –– 0.773 M = 0.227 M 0.773 M = 0.227 M conc.conc. of N2O4 = of N2O4 = 0.7730.773 M = 0.387 M M = 0.387 M 22 __________________________________________________________________

Volume becomes 4dm3Volume becomes 4dm3 concentration 0.5 Xconcentration 0.5 X XX eq.eq. 22 7.5 =7.5 = X/2 X/2 , 15 = , 15 = X . X . (0.5-X)2(0.5-X)2 (0.25-X-X2) (0.25-X-X2) 3.75 3.75 –– 15 X + 15 X2 = X, 3-75-16X+15 X2 = O 15 X + 15 X2 = X, 3-75-16X+15 X2 = O

X = X = 16 - 256 16 - 256 –– 225 225 = = 16-5.5716-5.57 = 0.348 = 0.35M = 0.348 = 0.35M 3030 30 30 conc. of N2O4 = 0.175 Mconc. of N2O4 = 0.175 M conc. of NO2 = 0.15 Mconc. of NO2 = 0.15 M Total number moles of gas at V=2nd m3 = 0.46+0.78=1.24 Total number moles of gas at V=2nd m3 = 0.46+0.78=1.24

molmol Total number moles of gas at V=4dm3 = 0.60+0.68=1.28 molTotal number moles of gas at V=4dm3 = 0.60+0.68=1.28 mol Increasing the volume increase the total number of moles gas.Increasing the volume increase the total number of moles gas.

Draw Born-Haber Cycle for the formation of CaCl2 Given the following data, calculate the 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) = 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 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 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, bonds, energy change for the reaction,

Ca(s)+ Cl2(g) Ca(s)+ Cl2(g) Call2(s), - 795 KJ mol-1 of CaCl2(s)Call2(s), - 795 KJ mol-1 of CaCl2(s) formed formed HFHF Ca(s)Ca(s) + Cl2(g) CaCl2 (s) + Cl2(g) CaCl2 (s) (2)(2) (5) (5) (1)(1) 2Cl(g) 2Cl(g) (4)(4) 2Cl(g) 2Cl(g) Ca(g)Ca(g) (3) (3) Ca2+(g) Ca2+(g) Energy change for the above reaction is the sum of energy changes of step (1) through step Energy change for the above reaction is the sum of energy changes of step (1) through step

(5) as shown in the following :(5) as shown in the following : Hf = Hf = H1 + H1 + H2 + H2 + H3 + H3 + H4 + H4 + H5H5

-795KJ = 192 KJ + 238 KJ + (590 KJ + 1146 KJ) + (2x(-350KJ)+ -795KJ = 192 KJ + 238 KJ + (590 KJ + 1146 KJ) + (2x(-350KJ)+ HgHg Hg = -2261 KJ mol-1 Hg = -2261 KJ mol-1 We have multiplied electron affinity by 2 because we have two chloride cons.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 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 atoms. In addition we have to use 1st electron affinity an d 2nd electron affinity in order to get O2-.get O2-.

In case of KB, we use also 1/2 bond energy of Br2, one ionization energy, and one electron In case of KB, we use also 1/2 bond energy of Br2, one ionization energy, and one electron affinity.affinity.

Using fi2O, We have to multiply step (7) by 2 and step (3) by 2.Using fi2O, We have to multiply step (7) by 2 and step (3) by 2.

100 100 C the equilibrium constant, KC, for the reaction.C the equilibrium constant, KC, for the reaction. CO(g) + Cl2(g) CO(g) + Cl2(g) COCl2(g)COCl2(g) has a value of 4.6X109 if O.2 mol of COll2 is placed into a 100 dm3 Hask at 100 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?C, what will be the concentration of all species at equilibrium? CO (g) + Cl2 (g) COCl2 (g)CO (g) + Cl2 (g) COCl2 (g) Constant O : OConstant O : O (0.02-X)M (0.02-X)M From 0.2 From 0.2 molmol 10dm310dm3 Constant X :Constant X : X X (0.02-X)M (0.02-X)M [COCl2][COCl2] Kc = _________Kc = _________ [CO] (Cl2][CO] (Cl2] O.O2 - XO.O2 - X 4-6X109 = __________4-6X109 = __________ x.xx.x since the eq. constant has very high value we neglect X from numerator, 4.6 X 109 = since the eq. constant has very high value we neglect X from numerator, 4.6 X 109 = 0.020.02 x2 x2 X2 = 0.00435 x 10-8 M2X2 = 0.00435 x 10-8 M2 X = 0.0208x10-4 MX = 0.0208x10-4 M [CO] = 2.1X10-6M[CO] = 2.1X10-6M [Cl2] = 2.1X10-6M[Cl2] = 2.1X10-6M [Clc2]= O.O2 M[Clc2]= O.O2 M 13.49 13.49 Sodium bicarbonate (baking soela) has many useful protection. Among them in the ability to serve a fire extinguisher because of thermal 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.decomposition to produce CO2, which some there's the fire. 2NaHCOS(s) 2NaHCOS(s) Na2 CoS(s) + CO2 (g) + H2O (g) Na2 CoS(s) + CO2 (g) + H2O (g) It 125 It 125 C the value of Kp is 2.6X103 Kpa2.C the value of Kp is 2.6X103 Kpa2. What are the partial pressures of CO2(g) and H2O (g)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.This suplem at equilibrium ? Car yen explain why Na HCO3 used in banking ? This is a heterogeneous equilibrium. Kp = CO2(g) H2O(g)Kp = CO2(g) H2O(g) 26 X103 KP2c = P2CO226 X103 KP2c = P2CO2 PCO2(g) = 5.1 X 10 KPaPCO2(g) = 5.1 X 10 KPa PHLO (g) = 5.1 X 10 KPaPHLO (g) = 5.1 X 10 KPa