unit 4 1st sec chemistry 2018

1. Chemistry 1st
sec. by Mr. Nabil
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Unit 4 Thermochemistry
Energy is : the ability to do work
 There are many forms of energy such as: light – sound – kinetic –potential
–heat energy …………………………………………………………………………… etc.
Law of conservation of energy:
Energy can be neither created nor destroyed, but can change from one form to another.
Thermodynamics:
the science deals with studying heat energy and its transferring
Thermochemistry:
One of the branches of thermodynamics which studies thermal changes which associate
with chemical and physical transformations
System is a part of the universe in which chemical or physical changes occur
Or : The certain part of matter which is studied.
Surrounding: is the part outside the
system and exchanges energy with it in the
form of heat or work.
Isolated system:
 It doesn't allow the exchange of matter
or energy between the system and
surrounding.
 Example: Water in an isolated container.
GR: The energy + work in isolated system is constant value
……………………………………………………………………………………………………………………………………………………………………………
……………………………………………………………………………………………………………………………………………………………………………
 Opened system:
It allows the exchange of both energy and matter between the system
and surrounding.
 Example: Water in an open sea
GR: Water in an open sea is example of open system
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2. Chemistry 1st
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Closed system:
It allows the exchange of energy only between the system and surrounding in the form of heat
or work Example: Thermometer
Thermometer is example of closed system
……………………………………………………………………………………………………………………………………………………………………………………………………………………
………….……………………………………………………………………………………………………………………………….………………………………………………
system Isolated Closed open
Matter exchange with surrounding ……………………… …………………… ………………………
energy exchange with surrounding ……………………… ……………………… ………………………
First law of thermodynamics:
The total energy of isolated system remains constant, even if the state
(solid , liquid , gas ) of the system changed.
Any change occurs to the energy of the system is associated with an
equivalent change in that of the surrounding, but with different sign
to keep the total energy constant.
 ΔE system = - ΔE surrounding
Temperature: The measurement of the average kinetic energy of
the molecules of substance, which determines if it is hot or cold.
System:
 The group of molecules which react with each other.
 When the average kinetic energy of the molecules of a substance
increases, the temperature increases.
Heat:
 A form of energy which is transferred between two objects of different
temperatures
 The internal energy of a system is directly proportional to the kinetic energy
of its particle.

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 Measuring units of heat:
 Calorie (cal.): The amount of heat required to raise the temperature of 1g of
water 1C
 Joule (J): The amount of heat required the raise the temperature of 1g of
water
184
1
.
C (N.B: calorie = 4.184 joule)
a) 10 joules = ………………………………………………. Calories b) 10 Calories = ………………………………. joules
Specific heat Cs: The amount of heat required to raise the temperature of
1g of matter 1C
 Measuring unit of specific heat = J/gC or J/gK
What is meant by: specific heat of water = 4.18 J / g °C
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 Specific heats depends on the kind of substance
 As the specific heat increases, this requires great amounts of heat to
raise their temperatures, and they take much time to lose heat and vice
versa.
 The change in temperature will be difficult "resists the change "
 Human body resists the change in temperature of surrounding
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 Water causes a moderate climate in coastal areas in both winter and
summer
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(N.B: water has the greatest specific heat on the surface of the earth)
Substance Al Carbon Copper Iron Liquid water Water vapour
Specific heat 0.9 0.711 0.385 0.444 4.184 2.01
To calculate the amount of heat absorbed or lost "q":
q = m x c x ΔT Where:
q: amount of measured heat m = mass c = specific heat
ΔT = the difference in temperature ΔT = T2 – T1

4. Chemistry 1st
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Amount of heat is measured by a device called "Calorimeter"
When using a calorimeter, 0.28g of propanol fuel was burned, which
raised the temperature of water 21.5C (the mass of water in calorimeter
equals 100g) Find the amount of energy resulted from burning the fuel.
Solution:-
The mass of water (m) = 100 g
The change of temperature ( ΔT) = 21.5C
The specific heat of water (c) = 4.184 j/gC
The amount of heat = m x c x ΔT = 100 x 4.184 x 21.5 = 9030 Joules
When dissolving sodium nitrates in a quantity of water, and this
quantity was raised to 100ml, the temperature of the solution decreased
from 25C to 17C. Calculate the quantity of absorbed energy
Mass of water (m) = 100g
Specific heat (c) = 4.18
The change of temperature (T) = 25-17 = 8C
 The amount of absorbed energy (q) = m x c x T
= 100x 4.18 x 8 = 3334 J/mol
The "coffee-cup" calorimeter:
 It provides us with an isolated system, to calculate
temperature.
 This enables us to use a certain amount of
substance with which thermal exchange occurs.
 Then, we calculate the change of temperature by
the following law ΔT = T2 – T1
 ΔT = Change of temperature.

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 T1 = Initial temperature (temperature before heating)
 T2 = Final temperature (temperature after heating)
 Components of calorimeter:
1- Isolated vessel 2-Thermometer 3-Stirrer
4- liquid (usually water)
Combustion calorimeter (bomb)
 Function: Measure the heat of combustion of some
substances.
 Idea of operation: We burn certain amounts of the
substance (by using an electric wire) in oxygen gas
at constant pressure in a vessel called"
Decomposition vessel".
 Decomposition vessel is surrounded by a certain
amount of water.
Water is used usually in calorimeter as heat exchange liquid
 Because it has high specific heat which allows it gain and lose great
amount of energy.
HW1
Write the scientific term:
1) One of the branches of thermodynamics which studies thermal changes
associated with chemical and physical transformations.
2) A part of the universe in which chemical or physical changes occur
Or the certain part of matter which we study
3) The region surrounding the system which exchanges energy with it in the
form of heat or work.

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4) Energy can be neither created nor destroyed, but can change from one form
to another.
5) The total energy of any isolated systems remains constant, even if the form
of the system changed.
6) The measurement of the average kinetic energy of the molecules of
substance, which determines if it is hot or cold.
7) Form of energy which is transferred between two objects of different
temperatures
8) The amount of heat required to raise the temperature of 1g of matter 1C
9) The amount of heat required to raise the temperature of 1g of water1C
10) The amount of heat required the raise the temperature of 1g of water
184.4
1 C
11) The amount of heat required to raise the temperature of an object 1C
Choose the correct answer:
1) The science deals with studying heat energy and its transferring is :
a) Thermodynamic
b) thermochemistry
c) nanochemistry
d) nanotechnology
2) The certain part of matter which is studied:
a) Molecule
b) chemical formula
c) system
d) surrounding
3) System doesn't allow the exchange of matter or energy between the
system and surrounding
a) Isolated

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b) Opened
c) Closed
d) surrounding
4) The total energy of isolated system remains constant, even if the state of
the system changed…………………
a) Thermodynamic 1st
law
b) mass conservation
c) Avogadro's law
d) Hess's law
5) When the energy of system increases , the energy of universe :
a) Increases b) decreases c) be constant d) no answer
6) The measurement of the average kinetic energy of the molecules of
substance :
a) energy
b) heat
c) temperature
d) heat content
7) A form of energy which is transferred between two objects of different
temperatures:
a) light b) internal c) heat d) magnetic
8) The amount of heat required to raise the temperature of 1g of water 1C
a) joule b) calorie c) watt d) volt
9) 10 joules =……………… calorie
a) 41.8 b)20.9 c) 13.5 d) 2.39
10) 10 calorie =……………… joules
a) 41.8 b)20.9 c) 13.5 d) 2.39

8. Chemistry 1st
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11) Measuring unit of specific heat is ………………..
a) C .J/g b) J/gC c) gK/J d) J. g / K
Give Reason for:
1. The energy + work in isolated system is constant value
2. Water in an open sea is example of open system
3. Thermometer is example of closed system
4. Human body resists the change in temperature of surrounding
5. Water causes a moderate climate in coastal areas in both winter and
summer
6. Water is used usually in calorimeter as heat exchange liquid
Problems
1) When dissolving sodium hydroxide in a quantity of water, and this
quantity was raised to 200 ml, the temperature of the solution decreased
from 18C to 38 C. Calculate the quantity of released amount of heat
(C of water =4.18 J/gC)
2) What does it mean?
a. Specific heat of water = 4.18 J / g.°C
Answer the following questions
1. When a chocolate stuffed cake is taken out of an oven of 200°C, does
the temperature of the cake and filling have the same temperature?
Are they different? Explain your answer?
2. In a medical thermometer, is the system opened or closed? And how
did this system transform into an isolated system?
3. When is the value of change of heat content of the reaction and heat
of combustion are equal?
4. Farmers in cold areas spray fruit trees with water.

9. Chemistry 1st
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Heat content
Every substance has energy stored in it , this energy is known as "Internal
energy"
Chemical energy stored in the atom:
 The energy of electrons in their levels, which is the sum of the
potential and kinetic energies of the electron in its level.
Chemical energy stored in molecule:
 It exists in the chemical bonds between its atoms (covalent, ionic
bonds)
Molecules binding forces:
 The attraction force between molecules is known as "Van-der waal
force" which is a potential energy.
 There are many other forces between molecules such as Hydrogen
bonds,
 These forces depend on the nature and polarity of molecules.
 We conclude from the previous remarks that each substance has a great
amount of stored energy, which is known as "Heat content" or "Molar
enthalpy"
Heat content (molar enthalpy):
The sum of energies stored in one mole of matter
 Substances have different heat contents due to :
 The difference of the number of atoms, their kinds and the bonds between
them.
 We can't measure the heat content of a substance practically, but we can
calculate the change in heat content which occurs due of the different
transformations of the substance

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Heat content differs from substance to another
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Change in heat content (ΔH): The difference between the total heat
content of products and that of reactants in a chemical reaction.
ΔH = H products - H reactants
 The change in standard heat content (ΔH)
Change in heat contents is calculated in standard conditions which are:-
- Pressure = 1 atomic pressure (1atm)- Temperature = 25C
- Concentration of solution = 1 mole
- The heat content of an element = zeroC
We calculate the change in heat content by the following law:
ΔH =
n
qp
ΔH: change in heat content qp : Heat amount n :no. of moles
calculate ∆H of the reaction: 2 C2H2(g)+5O2(g)  4CO2(g)+2H2O(l)
Knowing that heat contents (C2H2=226, CO2 = -393.5 , H2O=-285.85)
Kj/mole
Solution:
ΔH = H products - H reactants
= (4 x-393.5 + 2x – 285 ) - (2x226 + zero of O2)=- 2599.2
KJ/mole
∆H of the reaction: 2 C2H2(g)+5O2(g)  4CO2(g)+2H2O(g) larger or less
 We can divide chemical reactions which associate change in temperature
into:-

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First: Exothermic reactions: ∆H=-ve
They are reactions which give off energy as a
product to the surrounding, which increases its
temperature.
Example: The reaction of one mole of
hydrogen gas with half mole of oxygen gas
forms one mole of water and heat of
285.5kJ/mol
H2 (g) + 2
1
O2 (g) H2O (l) + 285.5kJ/mol
 From the previous reaction, we conclude that:-
1) Heat transfers from the system to the surrounding, The reaction gives
energy
2) The total heat content of products is less than that of reactants.
3) (ΔH) = -ve value = - 285 Kj/mole
second: Endothermic reactions ∆H=+ve
 Reactions in which energy is absorbed from the surrounding, which
decreases its temperature.
 When one mole of magnesium carbonates
decomposes into carbon dioxide gas and magnesium
oxide, it needs to absorb energy of 117.3kJ/mol.
MgCO3 (s) + 117.3 kJ/mol MgO (s) + CO2
(g)
From the previous reaction, we conclude that:-
 Heat transfers from the surrounding to the system,
The reaction gains energy
 The total heat content of products is greater than
that of reactants.
 (ΔH) = +ve value = +117.3 Kj/mole

12. Chemistry 1st
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 During chemical reactions, the bonds of reactants break up to form new
bonds in the products.
 When a bond breaks up, it absorbs an amount of energy from the
surrounding.
 When a new bond is formed, it gives away an amount of energy to the
surrounding (which increases its temperature)
 Bond Energy:
 Energy must be absorbed to break the bond or energy releases when
the bond is formed in one mole of the substance.
Bond Bond energy Kj/mol Bond Bond energy kj/mol
H-H 432 C-C 346
C-O 358 C=C 610
C=O 745 C≡C 835
O-H 467 C-H 413
O=O 498 Si-H 318
 In exothermic reactions:
 (-ΔH) the energy released from forming new bonds in product
molecules is greater than that formed from breaking up the bonds in
reactant molecules.

13. Chemistry 1st
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 In endothermic reactions
1.(+ΔH) the energy absorbed by reactant molecules to break up their
bonds is greater that the energy released from forming new bonds in
the products.
Example: Calculate the heat of the following
reactions, and find its type:
CH4 (g) + 2O2(g) CO2(g) + 2H2O(g)
(O=O) 498, (C-H) 413, (O-H) 467,
(C = O) 745
Solution:
 The energy required to
break up the reactant
molecules
= [4 x(C–H)] + [2x(O=O)]
= [4x413]+[2x498] = 2648 kJ
The energy released from forming the product molecules
=[2x(C=O)] + [4 (O- H)]
=[2x745]+ [4 x 467] = 3358 kJ
ΔH = breaking energy+ formation energy
= +2648 +(–3358)= - 710 kJ / mol.
The change in heat content is negative, therefore the reaction is exothermic.
Q1: Calculate the ΔH of the reaction: 2H2(g)+O2(g)2H2O(l)
If (O-H = 463kJ, H-H=435kJ,O=O = 494kJ)
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Q2: Calculate the change in heat content for the following reaction
CH4 + I2 → CH3I +HI, Given the bond energy in
CH3-H=104Kj.I-I=36KJ, CH3-I=56KJ. H-I=71KJ.
What is the type of reaction! Draw the energy graph.
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Q: Calculate the change in heat content at the reaction of the hydrogen
molecule with a molecule of chlorine to form 2 molecules of hydrogen
chloride. (H2 = 435 K. Joule, Cl2 = 240 K. Joule, HCl = 430 K joules)
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Thermochemical equation:
A balanced chemical equation which includes the thermal (enthalpy)
change associated with the reaction as one of the reaction products or reactants.
Conditions of thermochemical reactions
1- It should be balanced
2- In thermochemical reactions, we can write these numbers in the form of
fractions. Such as
H2(g) + 1/2 O2(g) H2O(l) + 285.5 kJ / mol.

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3- The physical state of the reactants and products should be mentioned (s),
(l),(g), (aq) Because the heat content change by the change of physical
state.
Example:-
 H2 (g) + 1/2 O2 (g) H2O(l) + 285.5 kJ/mol.
 (water in liquid state)
 H2 (g) + 1/2 O2 (g) H2O(g) + 242 kJ/mol.
 (water in gaseous state)
4- The value and sign of heat content is mentioned to determine whether the
reaction is endothermic or exothermic.
 H2O (s) H2O (l) ΔH = + 6 kJ/mol (endothermic reaction)
5- When multiplying the sides of thermochemical equations, we shouldn't
forget to multiply the amount of heat.
 H2O (s) H2O(l) ΔH = + 6 kJ / mol
 2H2O(s) 2H2O(l) ΔH = 2 x 6 = 12 kJ / mol.
HW2
1. Write the scientific term:
1)The amount of heat stored inside the matter
2)The sum of energies stored in one mole of substance
3)The difference between sum of heat contents of products and sum of the heat
contents of reactants
4)The chemical equation that includes the heat change accompanying the reaction.
5)The reaction in which the heat contents of products is greater than that of reactants
6)The reaction in which the heat contents of products is less than that of reactants
7)The reaction in which the heat transfers from system to surrounding
8)The reaction in which the heat transfers from surrounding to system
9)The energy required to break the bonds between molecules in one mole of matter.

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10) A balanced chemical equation which includes the thermal (enthalpy) change
associated with the reaction (as one of the reaction products or reactants)
2. Give reason for :
1.Heat content differs from substance to another
………………………………………………………………………………………………………..………………………………………………………………………………………………………..…
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2.The physical state must be written in thermochemical equation
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3.Some reactions are exothermic and some are endothermic.
………………………………………………………………………………………………………..………………………………………………………………………………………………………..…
…………………………………………………………………………..……………………………………………………..……………………………………………………………………………….
4.∆H of endothermic reaction =+ve value
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5.∆H of exothermic reaction =-ve value
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6.Bond formation is exothermic
………………………………………………………………………………………………………..………………………………………………………………………………………………………..…
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3. What is meant by:
1. specific heat of aluminum = 0.888 J / g °C
……………………………………………………………………..…………………………………………………………………………………………………………..…………………………
1. Find the amount of energy which raised the temperature of water from 20C
to 50C (the mass of water in calorimeter equals 400gm )
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2. When dissolving sodium hydroxide in a quantity of water, and this quantity
was raised to 100ml, the temperature of the solution increased from 25C to
40C. Calculate the quantity of released heat energy
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17. Chemistry 1st
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1. calculate ∆H of the reactions , and draw the graph :
a) 2 CH4(g)+3O2(g)  2CO2(g)+2H2O(l)
Knowing that heat contents
(CH4=-78.85, CO2 = -393.5 , H2O=-285.85) Kj/mole
b) H2(g) +Br2(g)  2HBr(g)
Knowing that bond energies are
(H-H=435, Br-Br = 193 , HBr =366) Kj/mole
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Lesson 2 :Heat change with physical changes
Calculating heat content associated with the combustion of different types of
fuel is very important because:-
1) It helps firefighters determine amount of heat accompanies with combustion.
2) It helps them also choose the best methods to put out fire
Dissolution – DilutionExamples of physical changes:
(ΔHs ): The amount of heat released orStandard heat of solution
absorbed when dissolving one mole of solute in a certain quantity of
solvent to get a saturated solution in standard conditions
Endothermic solution process:-
When dissolving one mole of ammonium nitrate (NH4NO3)
in water, the temperature of the solution decreases, which
is known as endothermic solution
NONHNONH -
)aq(3)aq(4
OH
(s)34
2
 
ΔHs = +25.7 kJ / mol
Exothermic solution process:-
When dissolving one mole of NaOH in water, the temperature
of the solution increases, which is known as exothermic
dissolution.
OHNaNaOH -
)aq()aq(
OH
(s)
2
 
,
ΔHs = -51 kJ /
mol
Explanation of standard heat of dissolution
1- Separating between the molecules of solvent:
 An endothermic process which absorbs energy to
overcome the attraction forces between the solvent
molecules, denoted by ΔH1

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2- Separating between the molecules of solute:
 An endothermic process which absorbs energy to
overcome the attraction forces between the particles of
solvent, denoted by ΔH2
3- Dissolution process:
 An exothermic process which gives out energy because solute molecules
bind with solvent molecules, denoted ΔH3
 If ΔH1+ ΔH2 > ΔH3 ,
o solution will be endothermic ∆H= +ve
 If ΔH1+ ΔH2 < ΔH3 ,
o solution will be exothermic ∆H= -ve
Solution process is associated with thermal change.
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To calculate the standard heat of solution, we use the following law:-
q = m. c. ΔT
 In diluted solutions, their masses are equal to their volumes
o because the density of water equals 1g/mol
 We use the specific heat of water (4.18 J/gC).
Molar heat of solution: Thermal change resulted from the dissolution
of one mole of solute in solvent forming one liter of solution
Molar heat of solution=
soluteofmolesofnumber
soluteofdissolvingwhenabsorbedorreleasedheatofamountThe
1) When one mole of sulphuric acid is dissolved to form 1000 ml of its
solution the temperature increased by 17C find the amount of heat release

20. Chemistry 1st
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Solution:
q = m. c. ∆T = 1000 x 4.18 x 17 = 71060 joules
2) When 160 gram of NaOH is dissolved to form 1000 ml of its solution the
temperature increased by 24C find
a) the amount of heat release b) molar heat of solution
Solution:
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Standard heat of dilution "ΔH dil.: The amount of heat absorbed or
released per every mole of solute when decreasing the
concentration of solution in standard conditions
 In high-concentrated solutions, the ions of the solvent are very close to each
other.
 When we decrease their concentrations by adding more quantity of solvent,
ions pull away from each other, which absorb energy.
 When the no. of solvent molecules increases, ions bind to more molecules,
which releases energy.
Dilution process is accompanied with release of energy
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Heat Changes Accompanying Chemical Changes
Heat of combustion: Combustion is a sequence of exothermic chemical
reactions between matter and oxygen.
 Complete combustion of substances release a great quantity of energy
in the form of heat or light. This released energy is known as "heat of
combustion ∆Hc"
Standard heat of combustion ∆Hc:

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The amount of energy released from the complete combustion of one
mole of matter in standard conditions.

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Examples of combustion reactions:-
1- Combustion of natural gas (mixture of
Butane C4H10 and propane C3H8) in
oxygen of air releases great quantity of
energy, which is used in cooking at
home
C3H8 (g) +5O2(g) 3CO2(g)+ 4H2O (g) +2323.7 kJ/ mol
2- Combustion of Glucose (C6H12O6) with oxygen inside living
organisms, which supplies living organisms with energy required to
carry out vital processes.
C6H12O6 (s) + 6O2 (g) 6CO2 (g) +6H2O, ΔHc = -2808 kJ/mol
3) The combustion always exothermic reaction
Standard heat of formation ∆HF :
The quantity of heat absorbed of released when forming one mole
from its elements, its elements should be in standard conditions
Heat of formation and the stability of compounds
 Compounds which have negative heat content are more stable at the
ordinary temperature of the room, and don't tend to decompose
because their heat contents are low and need more energy to
dissociate
 Compounds which have positive heat contents are less stable at the
ordinary temperature of room, and tend to decompose spontaneously
into their elements because the excess amount of energy they
contain.

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Comparison between exothermic and endothermic compounds
exothermic endothermic
Thermally stable Thermally unstable
Negative heat of formation positive heat of formation
its heat content < its elements Its heat content > its elements
Need energy to be dissociate lose energy and easily dissociate
Exothermic reactions form thermally stable products
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endothermic reactions form thermally unstable products
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(N.B: Most chemical reactions occur in the way to make stable compounds)
Using calculating standard heat of formation in change in heat content
 Formation heat of an element equals zero
 in the following conditions Pressure of 1 atmospheric -Temperature of
25C
ΔH = Sum ∆HF product – Sum of ∆HF of reactants
Example: If the heat of formation of methane is -74.6 and carbon dioxide
is -393.5 and water is -241.8 kJ/mol, calculate the change in the
heat content of the reaction shown in the following equation:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
Solution: (ΔH°f) = the sum of Hf of products – the sum of Hf of
reactants
F = (CO2 + 2H2O) − (CH4+2O2)
= [(−393.5)+(2 ✕−241.8)] − [(−74.6)+(2✕0)] = 802.5 kJ/mol

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If the heat of formation of aluminum oxide and sodium chloride are
- 1390.8 and -410.8 k J / mole respectively Calculate the change in
heat content of the reaction AlCl3 + 6Na  6NaCl+ 2Al
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 The following equation represents the burning of acetylene.
C2H2+ 5/2 O2 → 2CO2+H2O, ∆H=-1300KJ.Calculate the heat of
formation of acetylene given that the heat of formation of water and
carbon dioxide are -285.85KJ,and-393.7KJ respectively
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Hess's law:
o Scientists use indirect methods to calculate the heat of a reaction
because:-
1. Products and reactants may be mixed with other substances
2. Some reactions take long time to occur (Ex. Rusting of iron)
3. It may be dangerous to measure the heat of reaction practically
4. It is hard to measure the heat of some reactions in standard conditions for
heat and pressure

25. Chemistry 1st
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o Scientists use "Hess's law" to measure the heat of such reactions
Hess's law: Heat of reaction has a constant value which doesn't change
in standard conditions, even if the reaction took place in one or more steps.
Mathematical formula of Hess's law: ΔH = ΔH1 + ΔH2 + ΔH3 +…
Importance of Hess's law: We can use it to measure the change in heat
content of reactions indirectly, using other reactions with known heat
contents.
Example: By using Hess's law, we can measure the heat content of the
reaction as the following:
1. C (s, graphite) + O2 (g) CO2 (g) ∆H1= -393.5 kJ/mol.
2. C (s, diamond) + O2 (g) CO2 (g) ∆H2 = -395.4 kJ/mol
 (N.B: When reversing equations, do not forget to change the sign
of heat content) By reversing equation (1),we get equation (3)
3. CO2 (g) C (s, graphite) + O2(g) ∆H1=+393.5 kJ/mol.
 By adding equation 2 and 3 and simplifying them
C (s, diamond) C (s, graphite)
∆H =-395.4+393.5=-1.9kJ/mol
From the following data,
 CH4 + 2O2  CO2 + 2H2O ∆Ho
= -890 kJ/mol
H2O(l)  H2O(g) ∆Ho
= 44 kJ/mol at 298 K
 Calculate the enthalpy of the reaction
 CH4 + 2 O2(`g)  CO2(g) + 2 H2O(g)
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HW4
1) Write the scientific term
1) Rapid reaction of the substance with oxygen which releases amount of heat
2) The amount of heat released or absorbed when dissolving one mole of solute
in a certain quantity of solvent to get a saturated solution in standard
conditions.
3) The linkage of disassociated ions with water molecules.
4) The amount of heat absorbed or released per every mole of solute when
decreasing the concentration of solution (in case it is in its standard conditions.
5) The amount of heat absorbed of released when forming one mole from its
elements.
6) Heat of reaction has a constant value which doesn't change in standard conditions,
even if the reaction took place in one or more steps.
2) Problems:
1) Hydrogen it burns in oxygen, it releases a huge quantity of energy, its combustion reaction
is :- 2H2 (g) + O2 (g)  2H2O (g) ∆H = -484 kJ/mol . Find the amount of energy
released from burning 1g of hydrogen gas.
2) Calculate the standard change in heat content of the following reaction:-
H2 S (g) + 4F2 (g)  2HF(g) + SF6 (g)
H2S = - 21 kJ/mol, HF = -273 kJ/mol, SF6 = -1220 kJ/mol
3) Calculate the heat of formation of sodium chloride by using the following reactions:
2Na(s) + 2HCl (g)  2NaCl(s) + H2 (g) , ΔH= - 637.4 kJ
H2 (g) +Cl2 (g)  2HCl (g) , ΔH = -184.6Kj
4) Calculate ∆H of dissolving 160g of sodium nitrate in water to form 1 liter of solution, the
initial temp was 20 C and decreased to 8 C after the reaction.
5) If nitric oxide (NO) gas burned forming nitric dioxide gas (NO2), as in the following:
NO + 2
1
O2 (g)  NO2 (g)

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Calculate the change in heat content of the previous equation using the following two
equations:-
(1) 2
1
N2 (g) + 2
1
O2 (g)  NO(g) ∆H = +90.29 kJ/mol
(2) 2
1
N2 (g) + O2 (g)  NO2 (g) ∆H = +33.2 kJ/mol
3) Re-write the following statements after correcting the
underlined:
1) Heat is considered a measurement for the average kinetic energy of the molecules
that form the substance or the system.
2) The joule is known as the amount of required heat to raise the temperature of 1 g
of water one degree Celsius (from 15°C to 16°C).
3) The specific heat measuring unit is the J.
4) Chemical energy derives in the molecule from level energy which is the sum of
the electron's kinetic energy in addition to its potential energy.
5) The change in the heat content is the sum of the stored energy in one mole of the
substance.
6) In the open system there is no transportation of either the energy of the substance
between the system and the surrounding medium.
7) The thermometer is used as an isolated system to measure the absorbed or
released heat in the chemical reaction.