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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
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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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Vmhور و�Vn اNLbqSة iltl gf ا�Vج .اNLbqSة bahن� bc �kalن اNSي اb�Sب و[kX i{ واS|}`ل bv�bm�f ef P�VLً اanS`ز ا�dW�L �]VS "!!وراءنVvd� b ا��baر [evm ان�V أKh" و�Vخ اyS`اء
iqhإ.
�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