Thermodynamics Chapter 5 Written by JoAnne Swanson University of Central Florida Topics Types of energy and units of energy Exothermic vs.
Download ReportTranscript Thermodynamics Chapter 5 Written by JoAnne Swanson University of Central Florida Topics Types of energy and units of energy Exothermic vs.
Slide 1
Thermodynamics
Chapter 5
Written by
JoAnne Swanson
University of Central Florida
Slide 2
Topics
Types
of energy and units of energy
Exothermic vs. endothermic reactions
Heat capacity and specific heat
Energy transfers in changes of state
Calorimetry
Enthalpy and entropy
Hess’s Law
Slide 3
Energy is defined as the
capacity to do work.
There
are two categories of energy:
Kinetic energy.
Potential energy
Kinetic Energy is the energy of motion
Potential energy is energy of position,
Slide 4
Chemical energy is a form of
potential energy stored in the
structure of a chemical substance
due to its composition.
Slide 5
Consider
the potential energy of a
rock up on a cliff. If one rock is up on
a 1000 ft. cliff and another is up on a
10 ft. hill, the rock that is 1000ft. up
will hit the ground with much more
energy than the other if it falls.
Slide 6
Chemical
energy has similar
differences in that the structure
of one substance may contain
much more potential energy
than another substance.
Slide 7
Kinetic Energy of an object depends on
its mass and velocity:
Ek = ½ (mv2)
or mv2/2
mass is in Kg,
and Kg.m2/s2 = 1 Joule, the unit used to
designate energy.
From this equation you should see that
Ek increases with mass and with velocity
Slide 8
Potential Energy depends on gravity,
mass, and position:
Ep = mgh = (mass, gravity, height)
(gravitational constant = 9.8 m/s2)
But Ep of submicroscopic substances
(molecules, ions, etc.)
electrical charges between particles.
Slide 9
Electrical charge is designated by the
symbol ‘Q’ and has the value of an
electron charge = 1.60 x 10-19 C
The unit for electrical charge is the
coulomb ( C).
Eel = kQ1Q2 / d
where k=8.99x109Jm
C2
This equation shows the electrostatic
forces of attraction between two particles,
1 and 2, are inversely proportional to the
distance between the particles.
Slide 10
Another energy unit we will use is the
calorie. 1 c = 1 cal = 4.184 J
This differs from the nutritional calorie
which is 1C = 1000c = 1Kcal
Slide 11
Thermochemistry is the study
of heat transfers occurring in
chemical and physical changes
of substances.
Thermal
energy
random motion of atoms and
molecules.
Slide 12
The Universe is made up of
the system and the
surroundings
The system is the part of the universe
that is of interest to us.
The surroundings is everything outside
of the system.
Slide 13
Energy is transferred as heat and work.
Work = Force x Distance
= pressure x volume
It usually pertains to either the
electrical work (for instance in a battery)
or mechanical work (like the pressure
exerted on or by a gas in a reaction).
Slide 14
The change in Internal Energy of a
system = DE = q + w
DE is the change in internal energy
q = heat
w = work
Heat can flow into the system or out of
the system.
•
sign on q
Work can be done on the system or by
the system.
•
sign on w
Slide 15
Exothermic vs. Endothermic
Processes
Any
process that gives off
heat to the surroundings is an
Exothermic process.
bonds are made
Slide 16
When
a process absorbs
heat from the surroundings
it is an Endothermic
process. When ice melts it
absorbs heat from the
surroundings, thus it is an
endothermic process.
Slide 17
To measure the change in internal energy
of a chemical reaction, the reaction is
often carried out in a “bomb calorimeter”.
There is no change in volume in this
sealed container, therefore there is no
work involved and
DE = qreaction (at constant volume)
Slide 18
The symbol DH is used to
represent the change in heat
into or out of the system. It is
defined as the change in
enthalpy.
Slide 19
Enthalpy and Internal Energy are nearly
equal. Enthalpy takes into account
reactions at constant pressure where an
expanding gas does work on the
surroundings.
Remember w = PV, but if w is done by the
system, w = -PV
H = E + PV
w = - DPV
and DH = DE + DPV
so, DH = (q + w) – w
DH = q (at constant
pressure)
Slide 20
•1 mole gas at 25oC and 1atm has a
volume of 24.5L =( 24.5 L.atm / mol )
•1 L.atm = 0.1013 kJ
• DPV makes little difference in the DH
value, and even less difference for liquids
or solids.
why?
20
Slide 21
If 1 L.atm = 0.1013 kJ how many kJ are in
the molar volume of a gas at 25oC?
24.5 L.atm x 0.1013 kJ = 2.5 kJ
1 L.atm
see next example
Slide 22
Examples: given the values for PV,
calculate the change in enthalpy for the
combustion of methane where DE = -885 kJ
PV at 25oC
CH4(g)
2.5 KJ/ mol
O2(g)
2.5 KJ/ mol
CO2(g)
2.5 KJ/ mol
H20(l)
0.0018 KJ/ mol
DH = DE + DPV
(DPV = PV products – PV reactants)
continued………
Slide 23
CH4(g) + 2 O2 (g) CO2 (g) + 2 H2O (l)
DPV = PV CO2 + 2PV H2O –[PV CH4 + 2PV O2
DPV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ]
DPV = -5.0 KJ
DH = -885 KJ - 5.0 KJ = -890 KJ
Carried out at constant pressure,
DH = -885 KJ
Not a big difference when DPV is added
Slide 24
EXAMPLE 2:
a. Calculate the kinetic energy in joules,
of a 45 g golf ball moving at 61 m/s.
b. Convert this energy to calories.
c. What happens to the energy when the
ball strikes a tree?
Since 1J = 1kg m2/s2
change g to kg.
45 g = 0.045 kg
Slide 25
Ek = mv2 / 2
= 0.045 kg x (61m/s)2
2
= 84 kg m2 = 84 J
s2
b. 84 J x 1cal
= 20 cal
4.184 J
c. When the ball hits the tree, velocity = 0
and so does Kinetic energy. Kinetic
Energy is transferred as potential energy
to the tree and the deformed golf ball.
Slide 26
Example 4:
Calculate DE when a balloon is inflated
completely by heating it. The volume changes
from 4.00 x 106 L to 4.50 x 106 L, with the addition
of 1.3 x 108 J of heat. The balloon expands
against an internal pressure of 1.0 atm. DE = q+w
w = - DPV because ________________
q is positive because _________________
Slide 27
DV = 4.50 x 106L - 4.00 x 106L = 5.00 x 105 L
-PDV = 5.00 x 105 L x (-1.0 atm) =5.00 x 105 Latm
= -5.07 x 107 J
DE = 1.3 x 108J – 5.07 x 107J = 7.9 x 107 J
27
Slide 28
Example 3:
Calculate DE when a gas is compressed
from 6.0 x 104L to 5.3 x 104L under an
external pressure of 1.3 atm. The amount
of heat transferred to the surroundings was
4.3 x 102 J.
DE = w + q, where w = +DPV why?
DV = 5.3 x 104L - 6.0 x 104L = - 7.0 x 103L
DPV = (1.3 atm)(- 7.0 x 103L)
= -9.1 x 103Latm
Slide 29
-9.1 x 103 Latm x 0.1013 kJ = -9.2 x 102 KJ
Latm
In this case since the gas was
compressed, w = + and since heat was
transferred to the surroundings, q = - .
DE = w – q, = -9.2 x 105 J – 4.3 x 102 J =
= - 9.2043 x 105 J
= - 9.2 x 105 J
Slide 30
The enthalpy of reaction is the
difference between the enthalpies
of the products and reactants.
DH(rxn) = S DHf(products) – S DHf(reactants)
Slide 31
energy expelled
energy added
Ea
Ea
reactants
Energy
Energy
products
products
reactants
reaction progression
reaction progression
Endothermic
Exothermic
Slide 32
The diagram illustrates energy given
off when bonds are made between
H2 and O2 , (a). Energy is absorbed
when bonds are broken in HgO , (b).
Slide 33
The units of energy are
Kilojoules and Calories.
Enthalpy of reaction will be
expressed as Kilojoules (Kj).
Slide 34
It is important to note the following
when calculating the enthalpy of
reaction:
The enthalpy of a substance in its
standard state is equal to zero.
Enthalpy is dependent on the
quantity of the substance therefore the
enthalpy of a substance must be
multiplied by its coefficient in the
chemical equation.
Slide 35
When a rxn is reversed, the sign on the
enthalpy of rxn is changed.
If a rxn is divided through or multiplied
through by a number, so is the enthalpy
of rxn.
the state of matter is important in
enthalpy calculations.
35
Slide 36
Enthalpy of formation values of
compounds are found in tables of
thermodynamic data.
The enthalpy of formation is defined as the
energy associated with the formation of
1 mole of a substance from its elements in
their standard states.
Na(s) + 1/2 Cl2 (g) NaCl
The formation equation for sodium chloride
Slide 37
Ex. 1
Determine the enthalpy for the following
chemical reaction:
CO2 (g) + 2 H2O (l) -----> 2 O2 (g) + CH4 (g)
Slide 38
Use appendix C pg. 1041 in your text to
find the individual enthalpy values.
CO2 (g) = -393.509, H2O (l) = -285.83
CH4 (g) = -74.81 ,
O2 (g)= 0
CO2 + 2 H2O 2 O2 + CH4
DH =S DHf(products) – S DHf(reactants)
DH = (0 + -74.81)-(-393.509 + 2(- 285.83))
= -74.81 + 965.17 = 890.36 Kj
Slide 39
HESS’S LAW
ENTHALPY IS A STATE FUNCTION.
ENTHALPY OF A REACTION WILL
BE THE SAME NO MATTER WHAT
PATH IS TAKEN TO ARRIVE AT THE
PRODUCTS.
Slide 40
This means that if we need to
calculate the enthalpy of a reaction
for which we do not know the
enthalpies of formation, we can
algebraically manipulate other
reactions to arrive at the enthalpy for
the desired reaction.
Slide 41
Ex. Determine the DHrxn for
Sn(s) + 2 Cl2 (g) SnCl4(l)
Given the following:
Sn(s) + Cl2 (g) SnCl2(s)
DH1= -349.8 kJ
SnCl2(s) + Cl2 (g) SnCl4(l) DH2= -195.4 kJ
Sn (s) + 2 Cl2 (g) SnCl4(l) DH3= ?
DH1 + DH2 = DH3 = -545.2 kJ
Slide 42
Ex. Calculate the DHrxn for
S (s) + O2 (g) SO2(g)
Given
S (s) + 3/2 O2 (g) SO3(g) DH= -395.2 kJ
SO2 (g) + O2 (g) 2 SO3(g) DH= -198.2 kJ
NOTICE THAT WE NEED SO2 (g) TO BE
ON THE PRODUCT SIDE. THEREFORE
WE MUST REVERSE THE SECOND RXN
CONTINUED
Slide 43
S (s) + O2 (g) SO2(g)
S (s) + 3/2 O2 (g) SO3(g) DH= -395.2 kJ
2 SO3 (g) O2 (g) + 2 SO2(g) DH= +198.2 kJ
IT IS ALSO NECESSARY TO DIVIDE
EQUATION 2 BY 2, SO THAT WE HAVE
1 SO2(g) IN THE PRODUCT. WE MUST
ALSO DIVIDE THE DH
+198.2 kJ / 2 = 99.10 kJ
Slide 44
Now add the reactions and the enthalpies.
S (s) + 3/2 O2 (g) SO3(g)
DH= -395.2 kJ
SO3 (g) 1/2 O2 (g) + SO2(g) DH= +99.1 kJ
S (s) + O2 (g) SO2(g)
DH= -296.1 kJ
Slide 45
The enthalpy of formation of a
substance in a chemical reaction can
also be calculated if the enthalpy of
reaction is known and the enthalpies of
the other substances in the reaction are
known. Simply call the unknown
enthalpy ‘X’ and plug in all known values
to solve for ‘X’.
example on transparency
Slide 46
Specific heat and heat Capacity
Heat capacity is the amount of heat
required to raise the temperature of a
given quantity of a substance by 1o C.
Specific Heat is the amount of heat
required to raise the temperature of 1
gram of a substance by 1o C. J/goC
Molar heat capacity is J / mol oC
Slide 47
•a substance’s ability to absorb heat and
to store heat.
For example, a metal _______________
________________________________
A liquid like water (which contains strong
intermolecular forces, hydrogen bonds),
_________________________________
_________________________________
Slide 48
The metal has a ______________ and
the water has a ________________.
The specific heat capacity ( C ) of
water is an important and easy number
to remember ; C = 1 cal / g oC ,
(or 4.184 J / goC )
Since oC and Kelvin have the same size
increments, they are interchangeable in
these equations.
Slide 49
Calorimetry is a technique used
in the lab to measure the change
in enthalpy of a reaction. One
apparatus used is the Bomb
Calorimeter. It is a Heavy
walled, steel container which has
a known specific heat.
Slide 50
Types of Calorimetry problems we will cover:
1. simple change in water temperature
2. change in state of water
3. a rxn in a coffee cup calorimeter
4. a rxn in a bomb calorimeter
5. a mixture of hot substance with cold
substance in a calorimeter.
Slide 51
Equations needed:
1. q = C x m x DT
2. q = qice + DHf + qwater + DHv + qsteam
3. qrxn = DHrxn = -qwater
4. DHrxn = -[qwater + qbomb], or = -(Ccal x DT)
5. qcold = -qhot
Slide 52
Some needed values:
specific heat of water(l) = 4.184 J/gK
specific heat of water(s) = 2.05 J/gK
specific heat of water(g) = 2.01 J/gK
DHf = enthalpy of fusion = energy
involved in a change of state from liquid
to solid or vice versa = 333 J/g
DHv = enthalpy of vaporization = energy
involved in a change of state from liquid
to vapor or vice versa = 2444 J/g
Slide 53
The heat of the reaction (q) is
equal to the negative of the heat
absorbed by the (bomb plus the
water surrounding the bomb).
qrxn = DHrxn = - (qbomb + q water )
Slide 54
q is used to represent the
quantity and direction of heat
transferred, using a calorimeter.
Slide 55
Necessary Equations:
q = C x m x DT
q = (specific heat)(mass)(change in Temp.)
q = (J/gK)(g)(DT)
q(rxn) = - (q of the water + q of the bomb)
Slide 56
Subliminal message.....
Wake up !!!
Slide 57
1.
Calorimetry
A 466g sample of water is heated from
8.50oC to 74.60oC. Calculate the amount
of heat absorbed by the water.
q = (sp. heat)(mass)(DT)
q = (4.184 J/goC)(466g)(74.60-8.50oC)
q = 1.29 x 105 J = 129 KJ
Slide 58
change of state of water example on
transparency.
coffee cup calorimeter on transparency
58
Slide 59
Ex. 3
1.435 g of Naphthalene (molar mass
=128.2) was burned in a bomb
calorimeter. The temp. rose from
20.17oC to 25.84 oC. The mass of
the water surrounding the
calorimeter was 2000.g and the heat
capacity of the bomb was 1.80
KJ/oC. Calculate the heat of
combustion of Naphthalene.
Slide 60
q(rxn) = -(q water + q bomb)
q(water) = (2000g)(4.184 J/goC)(5.67oC)
= 4.74 x 104
q(bomb) = (1.80 x 103J / oC)(5.67oC)
= 1.02 x 104 J
q(rxn)
= -(4.74 x 104 J + 1.02 x 104 J)
= -5.76 x 104 J
Slide 61
MORE EXAMPLES WILL BE GIVEN IN
CLASS.
SINCE I HAD TO RUSH TO GET
THESE NOTES ON LINE, THEY
WERE NOT COMPLETELY EDITED.
THERE IS SOME REPETITION, and
they may need some corrections.
61
Thermodynamics
Chapter 5
Written by
JoAnne Swanson
University of Central Florida
Slide 2
Topics
Types
of energy and units of energy
Exothermic vs. endothermic reactions
Heat capacity and specific heat
Energy transfers in changes of state
Calorimetry
Enthalpy and entropy
Hess’s Law
Slide 3
Energy is defined as the
capacity to do work.
There
are two categories of energy:
Kinetic energy.
Potential energy
Kinetic Energy is the energy of motion
Potential energy is energy of position,
Slide 4
Chemical energy is a form of
potential energy stored in the
structure of a chemical substance
due to its composition.
Slide 5
Consider
the potential energy of a
rock up on a cliff. If one rock is up on
a 1000 ft. cliff and another is up on a
10 ft. hill, the rock that is 1000ft. up
will hit the ground with much more
energy than the other if it falls.
Slide 6
Chemical
energy has similar
differences in that the structure
of one substance may contain
much more potential energy
than another substance.
Slide 7
Kinetic Energy of an object depends on
its mass and velocity:
Ek = ½ (mv2)
or mv2/2
mass is in Kg,
and Kg.m2/s2 = 1 Joule, the unit used to
designate energy.
From this equation you should see that
Ek increases with mass and with velocity
Slide 8
Potential Energy depends on gravity,
mass, and position:
Ep = mgh = (mass, gravity, height)
(gravitational constant = 9.8 m/s2)
But Ep of submicroscopic substances
(molecules, ions, etc.)
electrical charges between particles.
Slide 9
Electrical charge is designated by the
symbol ‘Q’ and has the value of an
electron charge = 1.60 x 10-19 C
The unit for electrical charge is the
coulomb ( C).
Eel = kQ1Q2 / d
where k=8.99x109Jm
C2
This equation shows the electrostatic
forces of attraction between two particles,
1 and 2, are inversely proportional to the
distance between the particles.
Slide 10
Another energy unit we will use is the
calorie. 1 c = 1 cal = 4.184 J
This differs from the nutritional calorie
which is 1C = 1000c = 1Kcal
Slide 11
Thermochemistry is the study
of heat transfers occurring in
chemical and physical changes
of substances.
Thermal
energy
random motion of atoms and
molecules.
Slide 12
The Universe is made up of
the system and the
surroundings
The system is the part of the universe
that is of interest to us.
The surroundings is everything outside
of the system.
Slide 13
Energy is transferred as heat and work.
Work = Force x Distance
= pressure x volume
It usually pertains to either the
electrical work (for instance in a battery)
or mechanical work (like the pressure
exerted on or by a gas in a reaction).
Slide 14
The change in Internal Energy of a
system = DE = q + w
DE is the change in internal energy
q = heat
w = work
Heat can flow into the system or out of
the system.
•
sign on q
Work can be done on the system or by
the system.
•
sign on w
Slide 15
Exothermic vs. Endothermic
Processes
Any
process that gives off
heat to the surroundings is an
Exothermic process.
bonds are made
Slide 16
When
a process absorbs
heat from the surroundings
it is an Endothermic
process. When ice melts it
absorbs heat from the
surroundings, thus it is an
endothermic process.
Slide 17
To measure the change in internal energy
of a chemical reaction, the reaction is
often carried out in a “bomb calorimeter”.
There is no change in volume in this
sealed container, therefore there is no
work involved and
DE = qreaction (at constant volume)
Slide 18
The symbol DH is used to
represent the change in heat
into or out of the system. It is
defined as the change in
enthalpy.
Slide 19
Enthalpy and Internal Energy are nearly
equal. Enthalpy takes into account
reactions at constant pressure where an
expanding gas does work on the
surroundings.
Remember w = PV, but if w is done by the
system, w = -PV
H = E + PV
w = - DPV
and DH = DE + DPV
so, DH = (q + w) – w
DH = q (at constant
pressure)
Slide 20
•1 mole gas at 25oC and 1atm has a
volume of 24.5L =( 24.5 L.atm / mol )
•1 L.atm = 0.1013 kJ
• DPV makes little difference in the DH
value, and even less difference for liquids
or solids.
why?
20
Slide 21
If 1 L.atm = 0.1013 kJ how many kJ are in
the molar volume of a gas at 25oC?
24.5 L.atm x 0.1013 kJ = 2.5 kJ
1 L.atm
see next example
Slide 22
Examples: given the values for PV,
calculate the change in enthalpy for the
combustion of methane where DE = -885 kJ
PV at 25oC
CH4(g)
2.5 KJ/ mol
O2(g)
2.5 KJ/ mol
CO2(g)
2.5 KJ/ mol
H20(l)
0.0018 KJ/ mol
DH = DE + DPV
(DPV = PV products – PV reactants)
continued………
Slide 23
CH4(g) + 2 O2 (g) CO2 (g) + 2 H2O (l)
DPV = PV CO2 + 2PV H2O –[PV CH4 + 2PV O2
DPV = [2.5 KJ + 2(0.0018 KJ)] – [2.5 KJ + 2(2.5 KJ) ]
DPV = -5.0 KJ
DH = -885 KJ - 5.0 KJ = -890 KJ
Carried out at constant pressure,
DH = -885 KJ
Not a big difference when DPV is added
Slide 24
EXAMPLE 2:
a. Calculate the kinetic energy in joules,
of a 45 g golf ball moving at 61 m/s.
b. Convert this energy to calories.
c. What happens to the energy when the
ball strikes a tree?
Since 1J = 1kg m2/s2
change g to kg.
45 g = 0.045 kg
Slide 25
Ek = mv2 / 2
= 0.045 kg x (61m/s)2
2
= 84 kg m2 = 84 J
s2
b. 84 J x 1cal
= 20 cal
4.184 J
c. When the ball hits the tree, velocity = 0
and so does Kinetic energy. Kinetic
Energy is transferred as potential energy
to the tree and the deformed golf ball.
Slide 26
Example 4:
Calculate DE when a balloon is inflated
completely by heating it. The volume changes
from 4.00 x 106 L to 4.50 x 106 L, with the addition
of 1.3 x 108 J of heat. The balloon expands
against an internal pressure of 1.0 atm. DE = q+w
w = - DPV because ________________
q is positive because _________________
Slide 27
DV = 4.50 x 106L - 4.00 x 106L = 5.00 x 105 L
-PDV = 5.00 x 105 L x (-1.0 atm) =5.00 x 105 Latm
= -5.07 x 107 J
DE = 1.3 x 108J – 5.07 x 107J = 7.9 x 107 J
27
Slide 28
Example 3:
Calculate DE when a gas is compressed
from 6.0 x 104L to 5.3 x 104L under an
external pressure of 1.3 atm. The amount
of heat transferred to the surroundings was
4.3 x 102 J.
DE = w + q, where w = +DPV why?
DV = 5.3 x 104L - 6.0 x 104L = - 7.0 x 103L
DPV = (1.3 atm)(- 7.0 x 103L)
= -9.1 x 103Latm
Slide 29
-9.1 x 103 Latm x 0.1013 kJ = -9.2 x 102 KJ
Latm
In this case since the gas was
compressed, w = + and since heat was
transferred to the surroundings, q = - .
DE = w – q, = -9.2 x 105 J – 4.3 x 102 J =
= - 9.2043 x 105 J
= - 9.2 x 105 J
Slide 30
The enthalpy of reaction is the
difference between the enthalpies
of the products and reactants.
DH(rxn) = S DHf(products) – S DHf(reactants)
Slide 31
energy expelled
energy added
Ea
Ea
reactants
Energy
Energy
products
products
reactants
reaction progression
reaction progression
Endothermic
Exothermic
Slide 32
The diagram illustrates energy given
off when bonds are made between
H2 and O2 , (a). Energy is absorbed
when bonds are broken in HgO , (b).
Slide 33
The units of energy are
Kilojoules and Calories.
Enthalpy of reaction will be
expressed as Kilojoules (Kj).
Slide 34
It is important to note the following
when calculating the enthalpy of
reaction:
The enthalpy of a substance in its
standard state is equal to zero.
Enthalpy is dependent on the
quantity of the substance therefore the
enthalpy of a substance must be
multiplied by its coefficient in the
chemical equation.
Slide 35
When a rxn is reversed, the sign on the
enthalpy of rxn is changed.
If a rxn is divided through or multiplied
through by a number, so is the enthalpy
of rxn.
the state of matter is important in
enthalpy calculations.
35
Slide 36
Enthalpy of formation values of
compounds are found in tables of
thermodynamic data.
The enthalpy of formation is defined as the
energy associated with the formation of
1 mole of a substance from its elements in
their standard states.
Na(s) + 1/2 Cl2 (g) NaCl
The formation equation for sodium chloride
Slide 37
Ex. 1
Determine the enthalpy for the following
chemical reaction:
CO2 (g) + 2 H2O (l) -----> 2 O2 (g) + CH4 (g)
Slide 38
Use appendix C pg. 1041 in your text to
find the individual enthalpy values.
CO2 (g) = -393.509, H2O (l) = -285.83
CH4 (g) = -74.81 ,
O2 (g)= 0
CO2 + 2 H2O 2 O2 + CH4
DH =S DHf(products) – S DHf(reactants)
DH = (0 + -74.81)-(-393.509 + 2(- 285.83))
= -74.81 + 965.17 = 890.36 Kj
Slide 39
HESS’S LAW
ENTHALPY IS A STATE FUNCTION.
ENTHALPY OF A REACTION WILL
BE THE SAME NO MATTER WHAT
PATH IS TAKEN TO ARRIVE AT THE
PRODUCTS.
Slide 40
This means that if we need to
calculate the enthalpy of a reaction
for which we do not know the
enthalpies of formation, we can
algebraically manipulate other
reactions to arrive at the enthalpy for
the desired reaction.
Slide 41
Ex. Determine the DHrxn for
Sn(s) + 2 Cl2 (g) SnCl4(l)
Given the following:
Sn(s) + Cl2 (g) SnCl2(s)
DH1= -349.8 kJ
SnCl2(s) + Cl2 (g) SnCl4(l) DH2= -195.4 kJ
Sn (s) + 2 Cl2 (g) SnCl4(l) DH3= ?
DH1 + DH2 = DH3 = -545.2 kJ
Slide 42
Ex. Calculate the DHrxn for
S (s) + O2 (g) SO2(g)
Given
S (s) + 3/2 O2 (g) SO3(g) DH= -395.2 kJ
SO2 (g) + O2 (g) 2 SO3(g) DH= -198.2 kJ
NOTICE THAT WE NEED SO2 (g) TO BE
ON THE PRODUCT SIDE. THEREFORE
WE MUST REVERSE THE SECOND RXN
CONTINUED
Slide 43
S (s) + O2 (g) SO2(g)
S (s) + 3/2 O2 (g) SO3(g) DH= -395.2 kJ
2 SO3 (g) O2 (g) + 2 SO2(g) DH= +198.2 kJ
IT IS ALSO NECESSARY TO DIVIDE
EQUATION 2 BY 2, SO THAT WE HAVE
1 SO2(g) IN THE PRODUCT. WE MUST
ALSO DIVIDE THE DH
+198.2 kJ / 2 = 99.10 kJ
Slide 44
Now add the reactions and the enthalpies.
S (s) + 3/2 O2 (g) SO3(g)
DH= -395.2 kJ
SO3 (g) 1/2 O2 (g) + SO2(g) DH= +99.1 kJ
S (s) + O2 (g) SO2(g)
DH= -296.1 kJ
Slide 45
The enthalpy of formation of a
substance in a chemical reaction can
also be calculated if the enthalpy of
reaction is known and the enthalpies of
the other substances in the reaction are
known. Simply call the unknown
enthalpy ‘X’ and plug in all known values
to solve for ‘X’.
example on transparency
Slide 46
Specific heat and heat Capacity
Heat capacity is the amount of heat
required to raise the temperature of a
given quantity of a substance by 1o C.
Specific Heat is the amount of heat
required to raise the temperature of 1
gram of a substance by 1o C. J/goC
Molar heat capacity is J / mol oC
Slide 47
•a substance’s ability to absorb heat and
to store heat.
For example, a metal _______________
________________________________
A liquid like water (which contains strong
intermolecular forces, hydrogen bonds),
_________________________________
_________________________________
Slide 48
The metal has a ______________ and
the water has a ________________.
The specific heat capacity ( C ) of
water is an important and easy number
to remember ; C = 1 cal / g oC ,
(or 4.184 J / goC )
Since oC and Kelvin have the same size
increments, they are interchangeable in
these equations.
Slide 49
Calorimetry is a technique used
in the lab to measure the change
in enthalpy of a reaction. One
apparatus used is the Bomb
Calorimeter. It is a Heavy
walled, steel container which has
a known specific heat.
Slide 50
Types of Calorimetry problems we will cover:
1. simple change in water temperature
2. change in state of water
3. a rxn in a coffee cup calorimeter
4. a rxn in a bomb calorimeter
5. a mixture of hot substance with cold
substance in a calorimeter.
Slide 51
Equations needed:
1. q = C x m x DT
2. q = qice + DHf + qwater + DHv + qsteam
3. qrxn = DHrxn = -qwater
4. DHrxn = -[qwater + qbomb], or = -(Ccal x DT)
5. qcold = -qhot
Slide 52
Some needed values:
specific heat of water(l) = 4.184 J/gK
specific heat of water(s) = 2.05 J/gK
specific heat of water(g) = 2.01 J/gK
DHf = enthalpy of fusion = energy
involved in a change of state from liquid
to solid or vice versa = 333 J/g
DHv = enthalpy of vaporization = energy
involved in a change of state from liquid
to vapor or vice versa = 2444 J/g
Slide 53
The heat of the reaction (q) is
equal to the negative of the heat
absorbed by the (bomb plus the
water surrounding the bomb).
qrxn = DHrxn = - (qbomb + q water )
Slide 54
q is used to represent the
quantity and direction of heat
transferred, using a calorimeter.
Slide 55
Necessary Equations:
q = C x m x DT
q = (specific heat)(mass)(change in Temp.)
q = (J/gK)(g)(DT)
q(rxn) = - (q of the water + q of the bomb)
Slide 56
Subliminal message.....
Wake up !!!
Slide 57
1.
Calorimetry
A 466g sample of water is heated from
8.50oC to 74.60oC. Calculate the amount
of heat absorbed by the water.
q = (sp. heat)(mass)(DT)
q = (4.184 J/goC)(466g)(74.60-8.50oC)
q = 1.29 x 105 J = 129 KJ
Slide 58
change of state of water example on
transparency.
coffee cup calorimeter on transparency
58
Slide 59
Ex. 3
1.435 g of Naphthalene (molar mass
=128.2) was burned in a bomb
calorimeter. The temp. rose from
20.17oC to 25.84 oC. The mass of
the water surrounding the
calorimeter was 2000.g and the heat
capacity of the bomb was 1.80
KJ/oC. Calculate the heat of
combustion of Naphthalene.
Slide 60
q(rxn) = -(q water + q bomb)
q(water) = (2000g)(4.184 J/goC)(5.67oC)
= 4.74 x 104
q(bomb) = (1.80 x 103J / oC)(5.67oC)
= 1.02 x 104 J
q(rxn)
= -(4.74 x 104 J + 1.02 x 104 J)
= -5.76 x 104 J
Slide 61
MORE EXAMPLES WILL BE GIVEN IN
CLASS.
SINCE I HAD TO RUSH TO GET
THESE NOTES ON LINE, THEY
WERE NOT COMPLETELY EDITED.
THERE IS SOME REPETITION, and
they may need some corrections.
61