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CHAPTER 5

THERMOCHEMISTRY

Prof. Dr. Nizam M. El-Ashgar

THERMOCHEMISTRY

Thermochemistry: is the study of the

relationship between ENERGY and

chemical reactions.

What is energy?

Energy: is the ability to do work ortransfer heat.

Work: Energy used to cause an object thathas mass to move .

Heat: Energy used to cause thetemperature of an object to rise.

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POTENTIAL ENERGY

Potential energy: is energy of an object

possesses by virtue of its position or

chemical composition.

chemical energy :

stored within structural units of chemical substances.

POTENTIAL ENERGY UNDER THE

MICROSCOPE

Gravitational forces play little to no role in interactions

between atoms and molecules. Predict what plays a

much more important role.

Electrostatic potential energy, Eel : One of the

most important form of P.E.

This energy is proportional to electrical charges on two

interacting objects, Q1 and Q2, and inversely

proportional to the distance, d. Predict an equation.

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ELECTROSTATIC POTENTIAL ENERGY (EEL)

�κ is a constant of proportionality =

8.99 x 109 J.m/C2

�How can we make electrostatic potential

energy absolute zero?

Answer: if d is ∞ (infinite)

At finite

separation

distances for two

charged particles:

Eel is positive for

like charges and

negative for

opposite charges.

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KINETIC ENERGY

Kinetic energy is energy an object possesses

by virtue of its motion (depends on mass and

speed).

1

2KE = mv 2

�Thermal energy is kinetic energy in the

form of random motion of particles in any

sample of matter. As temperature rises,

thermal energy increases.

�Heat is energy that causes a change in the

thermal energy of a sample. Addition of heat

to a sample increases its temperature.

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UNITS OF ENERGY

� The SI unit of energy is the joule (J).

A mass of 2 kg moving at a speed of 1m/s possesses a

kinetic energy of 1 J:

� An older, non-SI unit is still in widespread use: the

calorie (cal).

Originally defined as: amount of energy require to raise the

temperature of 1 g of water from 14.5 C to 15.5 C.

1 cal = 4.184 J exactly

In nutrition is the nutritional Calorie (note the capital

C): 1 Cal = 1000 cal = 1 kcal

1 J = 1 kg m2

s2

DEFINITIONS: SYSTEM AND SURROUNDINGS

�The system includes the

molecules we want to

study (here, the hydrogen

and oxygen molecules).

�The surroundings are

everything else (here, the

cylinder and piston).

The system: The portion of universe under study.

The surroundings: Everything else is called (remain of universe)

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TYPES OF SYSTEMS

Open system: matter and energy can be exchanged

with surroundings (ie: boiling water)

Closed System: can exchange energy but not matter with

surroundings (ie: soft drink bottle)

Isolated System: no exchange of matter or energy. (ie: coffee

thermos).

Open Closed Isolated

DEFINITIONS: FORCE , WORK & HEAT

�Force F: is any push or pull exertedon an object.

�Work w : Energy used to movean object over some distance iswork.

w = F × d

where w is work, F is the force, and d isthe distance over which the force isexerted.

� Heat q: Energy transferred (as heat)from warmer objects to cooler objects.

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EXAMPLE PROBLEM 5.1

� A bowler lifts a 5.4 kg (12 lb) bowling ball from ground

level to a height of 1.6 m (5.2 ft) and then drops it.

a) What happens to the potential energy of the ball as

it is raised?

Because the bowling ball is raised to a greater

height above the ground, its potential energy

increases.

b) What is the quantity of work, in J, used to raise the

ball? (Note: The force due to gravity is F = m x g,

where m is the mass of the object and g is the

gravitational constant; g = 9.8 m/s2.)

c) After the ball is dropped, it gains kinetic energy. If all

the work done in part b has been converted to K.E at

the point of impact with the ground, what is the speed

of the ball at the point of impact? (5.6 m/s)

PE converted to KE:

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THE FIRST LAW OF THERMODYNAMICS

� Energy can neither be created nor destroyed

� Any energy lost by a system must be gained by the

surroundings (vice versa)

� In other words, the total energy of the universe is a

constant

(Energy is conserved)

� Energy can be converted from one type to another.

Potential energy is converted to kinetic energy.

INTERNAL ENERGY: IE

� The internal energy of a system is the sum of

all kinetic (molecular motion) and Potential

energies (attractive/repulsive

interactions) of all components of the

system; we call it E.

Note: Numerical values of a systems IE are

not known

In thermodynamics, we are mainly concerned

with the changes in E of the system (∆E).

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INTERNAL ENERGY

By definition, the change in internal energy,

∆E, is the final energy of the system minus the

initial energy of the system:

∆E = Efinal – Einitial

Efinal & Einitial (can’t be determined exactly) for a practical

system we need only ∆E to apply the law

To deal with thermodynamic values (Ex. ∆E ) we need

three things:

1- number 2- unit 3- sign

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CHANGES IN INTERNAL ENERGY

� If ∆E > 0, Efinal > Einitial

� Therefore, the system

absorbed energy from

the surroundings.

� This energy change is

called endothermic

� ∆E is +ve

2 H2(g) + O2(g) � 2 H2O(l)

CHANGES IN INTERNAL ENERGY

� If ∆E < 0, Efinal < Einitial

� Therefore, the system

released energy to the

surroundings.

� This energy change is

called exothermic

� ∆E is -ve

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CHANGES IN INTERNAL ENERGY

�When energy is

exchanged between

the system and the

surroundings, it is

exchanged as either

heat (q) or work (w).

We algebraically represent a

change in internal energy by:

∆E = q + w

∆E, Q, W, AND THEIR SIGNS

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SAMPLE EXERCISE 5.2

� Two gases, A(g) and B(g), are confined in a cylinder-and-piston

arrangement like that in the Figure below. Substances A and B react to

form a solid product: A(g) + B(g) � C(s). As the reaction occurs, the

system loses 1150 J of heat to the surroundings. The piston moves

downward as the gases react to form a solid. As the volume of the gas

decreases under the constant pressure of the atmosphere, the

surroundings do 480 J of work on the system. What is the change in the

internal energy of the system?

∆E = q + w

q = -1150 J (heat lost from sys, to surr.)

W = + 480 J (work done by sys. On surr.)

So, ∆E = -1150 J + 480 J = -670 J

The system transfers heat to surrounding

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EXCHANGE OF HEAT BETWEEN SYSTEM AND SURROUNDINGS

� When heat is absorbed by the system from the

surroundings, the process is endothermic.

� When heat is released by the system into the

surroundings, the process is exothermic.

Ice melting, Evaporation of acetone Combustion of Gasoline

STATE FUNCTIONS

� A state function is a property of a system that is

determined only by the present state of the system,

not the path that system took.

50 oC 150 oC - 90 oC 30 oC -50 oC 57 oC 39 oC 120 oC

Ti Tf

�∆T = Tf – Ti = 120 – 50 = 70 oC

Examples of state functions: ∆V, ∆E, ∆P, ∆H, ∆n

�

=

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STATE FUNCTIONS

� The internal energy of a system is independent of the

path by which the system achieved that state.

� In the system below, the water could have reached room

temperature from either direction.

And so, ∆E depends only on Einitial and Efinal. , (state

function) ∆E = Efinal ─ Einitial

STATE FUNCTIONS�However, q and w are

not state functions.

�Whether the battery isshorted out or isdischarged by runningthe fan, its ∆E is thesame.� But q and w are

different in the twocases.

� Path a: ∆E1 = q1+w1

� Path b: ∆E2 = q2+w2

� ∆E1 = ∆E2

� q1 ≠ q1 and w1 ≠ w1

q max w max

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WORK

In an open container:

The only work done is

by a gas pushing on the

surroundings (or by the

surroundings pushing

on the gas).

WORK

We can measure the work done by the gas if

the reaction is done in a vessel that has

been fitted with a piston.

w = -P∆V

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

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ENTHALPY

� If a process takes place at constant pressure (as the

majority of processes we study do), ie open systems.

� The only work done is this pressure-volume work.

� Enthalpy is the internal energy plus the product of

pressure and volume:

H = E + PV

PRESSURE – VOLUME WORK

� Work involved in expansion or compression of gas

� When pressure is constant:

w = -P∆V

∆V is +ve (expansion) and –ve (compression)

� Q: If a system does not change its volume during the

course of a process, does it do pressure-volume work?

No work done ie w = 0

because ∆V = 0

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ENTHALPY

� When the system changes at constant pressure, the

change in enthalpy, ∆H, is derived as follows:

∆H = ∆(E + PV)

� This can be written

∆H = ∆E + P∆V------------------(1)

� Since ∆E = q + w and w = -P∆V, we can substitute

these into the enthalpy expression:

∆E = qP -P∆V

qp = ∆E + P∆V---------------- (2)

From (1) and (2) : ∆H = qp

� So, at constant pressure, the change in enthalpy is the

heat gained or lost.

H = E + PV

ENDOTHERMICITY AND EXOTHERMICITY

�A process is

endothermic when

∆H is positive.

�A process is

exothermic when

∆H is negative.

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ENTHALPY OF REACTION ,∆H

∆H is a state function

In reactions:

The change in enthalpy,

∆H, is the enthalpy of

the products minus the

enthalpy of the

reactants:

∆H = Hproducts - Hreactants

This quantity, ∆H, is called the enthalpy of reaction, or the heat of reaction.

SAMPLE EXCERCISE 5.3

� Indicate the sign of the enthalpy change, ∆H, in

these processes carried out under atmospheric

pressure and indicate whether each process is

endothermic or exothermic:

�An ice cube melts: Endothermic

�1 g of butane is combusted in sufficient oxygen to

give complete combustion to CO2 and H2O:

Exothermic

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THERMOCHEMICAL REACTIONS

� A thermochemical equation is a balanced chemical

equation that shows the associated enthalpy change

but does not specify amount of chemical involved.

� Consider the thermochemical equation

� The thermochemical equivalencies are

� 1 mol C3H8(g) react to give off 2.22×103 kJ

� 5 mol O2(g) react to give off 2.22×103 kJ

� 3 mol CO2(g) are formed and give off 2.22×103 kJ

� 4 mol H2O(l) are formed and give off 2.22×103 kJ

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H = -2.22×103 kJ

Mass of

CO2

Molar mass

of CO2

Moles of

CO2

Thermochemical

equivalent

Enthalpy

change

THE TRUTH ABOUT ENTHALPY

1- Enthalpy is an extensive property.

Directly proportional to the amount of material

Ex:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = -890 kJ

1 mol CH4(g) → -890 kJ

2 mol CH4(g) → -1780 kJ

2- ∆H for a reaction in the forward direction is equal in

size, but opposite in sign, to ∆H for the reverse

reaction.

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆H = +890 kJ

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3- ∆H for a reaction depends on the state of the

products and the state of the reactants.

� CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = -890 kJ

� CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = -802 kJ

As: 2H2O(l) → 2H2O(g) ∆H = +88 KJ

4- Multiplying or dividing the equation by a factor (n)

the same thing should be for the value of ∆H

2H2O(l) → 2H2O(g) ∆H = +88 KJ

H2O(l) → H2O(g ∆H = +44 KJ

SAMPLE EXERCISE 5.4

� How much heat is released when 4.50 g of methane gas is burned in a constant pressure system?

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆H = -890 kJ

4.50 g ?? Heat

1mol CH4(g) ↔ = -890 kJ

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Example: Given the following equation:

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆H = − 2803 kJ/mol

calculate the mass in grams of glucose burned in

oxygen for releasing 700.0 KJ of energy.

PRACTICE EXERCISE

Hydrogen peroxide can decompose to water and oxygen by the following reaction:

2 H2O2(l) → 2 H2O(l) + O2(g) ∆H = –196 kJ

Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.Answer: –14.4 kJ

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�We cannot know the exact

enthalpy of the reactants and

products. We can measure ∆H

through calorimetry.

�Calorimetry is the experimental

measurement of heat released or

absorbed by a chemical reaction.

�A calorimeter is the device used

to measure heat.

�The quantity of heat transferred

by the reaction causes a change

of temperature inside the

calorimeter.

CALORIMETRY

A simple calorimeter can

be made from styrofoam

coffee cups.

HEAT CAPACITY AND SPECIFIC HEAT

Heat capacity C: The amount of energy required to raise

the temperature of a substance by 1 K (1°C).

Specific heat CS: The amount of energy required to raise

the temperature of 1 g of a substance by 1 K (1°C).

Molar heat capacity Cm: The amount of energy required

to raise the temperature of 1 mole a substance by 1 K

(1°C).

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HEAT CAPACITY AND SPECIFIC HEAT

Specific heat, then, is

Specific heat =heat transferred

mass × temperature change

cs =q

m × ∆T

Units : Heat capacity, C : J/°C or J/K

Specific Heat Cs: J/g °C or J/g Κ

Molar Heat Capacity Cm: J/mol °C or J/mol Κ

where : q is heat, m is mass, C is heat capacity and

∆T = change in temp (∆T = Tfinal − Tinitial)

SAMPLE EXERCISE 5.5

� (a) How much heat is needed to warm 250 g of water (about1 cup) from 22 °C (about room temperature) to near itsboiling point, 98 °C? The specific heat of water is 4.18 J/g-K.

� (b) What is the molar heat capacity of water?

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PRACTICE EXERCISE

(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume thatthe specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by50.0 kg of rocks if their temperature increases by 12.0 °C.

(b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?

Answers: (a) 4.9 × 105 J, (b) 11 K decrease = 11 °C decrease.

CONSTANT PRESSURE CALORIMETRY

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CONSTANT PRESSURE CALORIMETRY

�The calorimeter and its contents are the

surroundings (qsurr).

�The chemical reaction is the system (qsys).

The law of conservation of energy requires that

qsys = − qsurr

qsurr = m × cs × ∆T

�All dilute, aqueous solutions have the same

density (1.00 g/mL) and specifi c heat capacity

(4.18 J/(g·°C) as water .

�Note that: qsys means qrxn

10

/18

/20

15

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SAMPLE EXERCISE 5.6

When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 °C to

27.5 °C. Calculate the enthalpy change for the reaction in kJ/mol HCl, assuming thatthe calorimeter loses only a negligible quantity of heat, that the total volume of thesolution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

50 mL, 1.0M 50 mL, 1.0M

Total m:

SAMPLE EXERCISE 5.6

When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in aconstant-pressure calorimeter, the temperature of the mixture increases from 22.30°C to 23.11 °C. The temperature increase is caused by the following reaction:.

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

Calculate ∆H for this reaction in kJ/mol AgNO3, assuming that the combined solutionhas a mass of 100.0 g and a specific heat of 4.18 J/g °C.

Answer: –68,000 J/mol = –68 kJ/mol

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BOMB CALORIMETRY

•Constant-Volume Calorimetry

•Studies Combustion rxn’s

• May only use after a

standardization measurement

•Difference between ∆E and ∆H

is very small

q rxn = −q cal

q cal = qwater + q walls

Once you have heat capacity of

calorimeter, you may use:

qrxn = -Ccal X ∆T

Typical procedure used in a bombcalorimeter

�Known amount of sample placed in steelcontainer and then filled with oxygengas

�Steel chamber submerged in knownamount of water

�Sample ignited electrically

�Temperature increase of water isdetermined

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�Because the volume in the bomb calorimeteris constant, what is measured is really thechange in internal energy, ∆E, not ∆H.

∆H = ∆E + P∆V

As ∆V = 0

So ∆H = ∆E

�For most reactions, the difference is verysmall.

SAMPLE EXERCISE 5.7

The combustion of methylhydrazine with oxygen produces N2(g), CO2(g), and H2O(l):

2 CH6N2(l) + 5 O2(g) → 2 N2(g) + 2 CO2(g) + 6 H2O(l)

When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, thetemperature of the calorimeter increases from 25.00 °C to 39.50 °C. In aseparate experiment the heat capacity of the calorimeter is measured tobe 7.794 kJ/°C. Calculate the heat of reaction for the combustion of amole of CH6N2.

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PRACTICE EXERCISE

A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heatcapacity is 4.812 kJ/°C. The temperature increases from 23.10 °C to 24.95 °C.Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole.

Answers: (a) –15.2 kJ/g, (b) –1370 kJ/mol.

HESS’S LAW

Hess’s law states

that: “If a reaction is

carried out in a series

of steps, ∆H for the

overall reaction will

be equal to the sum of

the enthalpy changes

for the individual

steps.”

Because ∆H is a state function, the total enthalpy

change depends only on the initial state of the reactants

and the final state of the products.

Used for calculating enthalpy for a reaction that cannot

be determined directly

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SAMPLE EXERCISE 5.8

The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C,and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol CO:

Solution:

PRACTICE EXERCISE

Carbon occurs in two forms, graphite and diamond. The enthalpy of thecombustion of graphite is –393.5 kJ/mol and that of diamond is –395.4kJ/mol:

Calculate for the conversion of graphite to diamond:

Answer: ∆H3 = +1.9 kJ

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SAMPLE EXERCISE 5.9

� Calculate ∆H for the reaction:

2 C(s) + H2(g) → C2H2(g)given the following chemical equations and their respective

enthalpy changes

Solution:

PRACTICE EXERCISE

Calculate ∆H for the reaction

NO(g) + O(g) → NO2(g)given the following information:

Answer: –304.1 kJ

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ENTHALPIES OF FORMATION

Enthalpy of formation, ∆Hf : is defined as the enthalpy

change for the reaction in which a compound is made

from its constituent elements in their elemental forms.

∆Hf depends on: T, P and State (G, L and S,

crystalline) of reactants and products.

STANDARD ENTHALPIES OF FORMATION

Standard enthalpies of formation, ∆Hf°, The enthalpy

change that results when 1 mole of a compound is

formed from its elements in their standard states (25

°C and 1.00 atm pressure).

∆Hf° for an element in its standard state is defined

as zero.

∆Hf°: O2 (g) = 0, O3 (g ) ≠ 0, O ≠ 0

∆Hf° : I2 (s)= 0, I2(g) ≠ 0, I2(l) ≠ 0

∆Hf° : C(Graph)= 0, C(diam) ≠ 0, C(60) ≠ 0

∆Hf° : Cl2(g) = 0.0 ∆Hf

° : I2(s) = 0.0

� Values of ∆Hf° for substances are found in reference

tables

� Used to calculate the ∆H°rxn

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SAMPLE EXERCISE 5.10

For which of the following reactions at 25°C would the enthalpy changerepresent a standard enthalpy of formation? For each that does not, whatchanges are needed to make it an equation whose ∆H is an enthalpy offormation?

Solution:

a) The eq. Represents ∆Hf° of Na2O(s).

b) Does not represent ∆Hf° of KCl(s)

Correction: K(s) + 1/2Cl2 (g) → KCl(s)

c) Does not represent ∆Hf° of C6H12O6(s)

Correction:

6 C(graphite) + 6 H2(g) + 3 O2(g) →C6H12O6(s)

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PRACTICE EXERCISE

Write the equation corresponding to the standard enthalpy offormation of liquid carbon tetrachloride (CCl4).Answer: C(graphite) + 2 Cl2(g) →CCl4(l)

CALCULATION OF ∆H

We can use Hess’s law in this way:

∆Horxn = Σ n ∆Hf°products – Σ m ∆Hf° reactants

where n and m are the stoichiometric coefficients.

∆Horxn = [3(-393.5 kJ) + 4(-285.8 kJ)] – [1(-103.85 kJ) + 5(0 kJ)]

∆Horxn = -2220 kJ

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (l)

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SAMPLE EXERCISE 5.11

(a) Calculate the standard enthalpy change for the combustion of 1 mol ofbenzene, C6H6(l), to form CO2(g) and H2O(l).

(b) (b) Compare the quantity of heat produced by combustion of 1.00 gpropane to that produced by 1.00 g benzene.

PRACTICE EXERCISE

Using the standard enthalpies of formation listed in Table 5.3, calculate the enthalpy change for the combustion of 1 mol of ethanol:

Answer: –1367 kJ.

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SAMPLE EXERCISE 5.12

The standard enthalpy change for the reaction:

is 178.1 kJ. From the values for the standard enthalpies of

formation of CaO(s) and CO2(g) given in Table 5.3, calculate the

standard enthalpy of formation of CaCO3(s).

PRACTICE EXERCISE

Given the following standard enthalpy change, use the standard

enthalpies of formation in Table 5.3 to calculate the standard

enthalpy of formation of CuO(s):

Answer: –156.1 kJ/mol

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ENERGY IN FOODS

Most of the fuel in the food we eat comes from

carbohydrates and fats.

SAMPLE EXERCISE 5.14

a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milkprovides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values ofthese kinds of substances, estimate the energy value (caloric content) of this serving.

This corresponds to 160 kcal:

(b) A person of average weight uses about 100 Cal/mi when running or jogging.How many servings of this cereal provide the energy value requirements forrunning 3 mi?

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ENERGY IN FUELS

The vast majority of the energy consumed in

this country comes from fossil fuels.

UNIT 3 Section 5.4

Alternative Renewable Energy SourcesChapter 5: Energy Changes

Renewable energy sources in Ontario:

• account for about 25% of energy production

• are projected to increase to as high as 40% by 2025

Include:

• Hydroelectric power: (major source).

• Wind energy: (currently ~ 1% and projected to 15% in 2025),

• Solar energy: (currently low but may be as high as 5% in2025).

Much lower contributors:

• Biomass, wave power, and geothermal energy.

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UNIT 3 Section 5.4

What Is a “Clean” Fuel?

Chapter 5: Energy Changes

Different fuels have differing impacts on the environment.One way this impact is measured is through emissions.

For example, CO2(g) emissions per kJ of energy produced.

Fuel kg CO2/ kJ energy

Anthracite coal 108.83

Oil 78.48

Natural gas 56.03

Nuclear 0.00

Renewables 0.00

HW PROBLEMS

3, 5, 8, 11, 14, 18, 20, 24, 26,

30, 36, 39, 52, 58, 61, 60, 65, 68,

73, 76, 82, 85, 87, 91, 94, 97, 106

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�vkRh gf و�Vnوا واiqh، ا�ب tl gf �lt� gvhور bf إS{ وbnm�dlن زو[bن bahن�kal �yW وbcن :أ�Vى Vfة اb�Sب �Vخ �aLة !!sSbc|� سPq ٢٥ b� KL VmXب VQ�lف �vxL ا��Vاج

"،Khأ Vان� }Sإ PcVWSا bfو byvL gf ،تbان`v� Vم..أن�`v�Sا Vvd� ef رbsRSا".. Vmواس� �an� gv]و�Sا gf �lt� بb�Sة اVf ىV�.أ

و�Vخ bndSbhدة و[iy إ�f£ اNSي اb�Sب، bd�� }kX tl¡¢ اbmSء و¡Vsات ا bsfر، ھs`ل thأ ث�."أbl Kh ان�tl Vي، �mS واbmSء �Vsm، إنby أKh" : أ�Vى Vfة

KLه وNھ P�¥kSا �S es�dl نb]و�Sت ا`xdSوس�  ا �]VSز ا`anSذا" :اbmS   م`R� رةbl�h �vWsSا إKqh أن ��v اb¡ gf }|��dmSدf`ن إنbq ":اanS`ز اb¡ �]VSل ھqh� ” bq¨؟ X\ج kX{ واQ¥S`ل

t¡ ©Wاً أ�VvQh ة  ولVf KL i�bv�“

"اVn� �c ®­bR¥Sف ��{ اªk«�d� ¬­b�qS  " :دائماً تذكر