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Chapter 6 Thermochemistry

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Người gửi: Trương Minh Chiến (trang riêng)
Ngày gửi: 02h:03' 12-01-2011
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Mr. Truong Minh Chien ; losedtales@yahoo.com
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Chemistry, Julia Burdge, 2st Ed.
McGraw Hill.
Chapter 6
Thermochemistry
2011, NKMB Co., Ltd.
2
Heating Your Home
most homes burn fossil fuels to generate heat
the amount the temperature of your home increases depends on several factors
how much fuel is burned
the volume of the house
the amount of heat loss
the efficiency of the burning process
can you think of any others?
Chemistry, Julia Burdge, 2nd e., McGraw Hill.
3
Nature of Energy
even though Chemistry is the study of matter, energy effects matter
energy is anything that has the capacity to do work
work is a force acting over a distance
Energy = Work = Force x Distance
energy can be exchanged between objects through contact
collisions
Chemistry, Julia Burdge, 2nd e., McGraw Hill.
Tro, Chemistry: A Molecular Approach
4
Classification of
Energy
Kinetic energy is energy of motion or energy that is being transferred
thermal energy is kinetic
Tro, Chemistry: A Molecular Approach
5
Classification of Energy
Potential energy is energy that is stored in an object, or energy associated with the composition and position of the object
energy stored in the structure of a compound is potential
Tro, Chemistry: A Molecular Approach
6
Law of Conservation of Energy
energy cannot be created or destroyed
First Law of Thermodynamics
energy can be transferred between objects
energy can be transformed from one form to another
heat → light → sound
Tro, Chemistry: A Molecular Approach
7
Some Forms of Energy
Electrical
kinetic energy associated with the flow of electrical charge
Heat or Thermal Energy
kinetic energy associated with molecular motion
Light or Radiant Energy
kinetic energy associated with energy transitions in an atom
Nuclear
potential energy in the nucleus of atoms
Chemical
potential energy in the attachment of atoms or because of their position
8
Units of Energy
the amount of kinetic energy an object has is directly proportional to its mass and velocity
KE = ½mv2
Tro, Chemistry: A Molecular Approach
9
Units of Energy
joule (J) is the amount of energy needed to move a 1 kg mass a distance of 1 meter
1 J = 1 N∙m = 1 kg∙m2/s2
calorie (cal) is the amount of energy needed to raise one gram of water by 1°C
kcal = energy needed to raise 1000 g of water 1°C
food Calories = kcals
Tro, Chemistry: A Molecular Approach
10
Energy Use
11
Energy Flow and
Conservation of Energy
we define the system as the material or process we are studying the energy changes within
we define the surroundings as everything else in the universe
Conservation of Energy requires that the total energy change in the system and the surrounding must be zero
DEnergyuniverse = 0 = DEnergysystem + DEnergysurroundings
D is the symbol that is used to mean change
final amount – initial amount
Tro, Chemistry: A Molecular Approach
12
Internal Energy
the internal energy is the total amount of kinetic and potential energy a system possesses
the change in the internal energy of a system only depends on the amount of energy in the system at the beginning and end
a state function is a mathematical function whose result only depends on the initial and final conditions, not on the process used
DE = Efinal – Einitial
DEreaction = Eproducts - Ereactants
Tro, Chemistry: A Molecular Approach
13
State Function

Tro, Chemistry: A Molecular Approach
14
Energy Diagrams
energy diagrams are a “graphical” way of showing the direction of energy flow during a process
if the final condition has a
larger amount of internal
energy than the initial
condition, the change in the
internal energy will be +
if the final condition has a
smaller amount of internal
energy than the initial
condition, the change in the
internal energy will be ─
Tro, Chemistry: A Molecular Approach
15
Energy Flow
when energy flows out of a system, it must all flow into the surroundings
when energy flows out of a system, DEsystem is ─
when energy flows into the surroundings, DEsurroundings is +
therefore:
─ DEsystem= DEsurroundings

Tro, Chemistry: A Molecular Approach
16
Energy Flow
when energy flows into a system, it must all come from the surroundings
when energy flows into a system, DEsystem is +
when energy flows out of the surroundings, DEsurroundings is ─
therefore:
DEsystem= ─ DEsurroundings

Tro, Chemistry: A Molecular Approach
17
How Is Energy Exchanged?
energy is exchanged between the system and surroundings through heat and work
q = heat (thermal) energy
w = work energy
q and w are NOT state functions, their value depends on the process
DE = q + w
Tro, Chemistry: A Molecular Approach
18
Energy Exchange
energy is exchanged between the system and surroundings through either heat exchange or work being done
Tro, Chemistry: A Molecular Approach
19
Heat & Work
on a smooth table, most of the kinetic energy is transferred from the first ball to the second – with a small amount lost through friction
Tro, Chemistry: A Molecular Approach
20
Heat & Work
on a rough table, most of the kinetic energy of the first ball is lost through friction – less than half is transferred to the second
Tro, Chemistry: A Molecular Approach
21
Heat Exchange
heat is the exchange of thermal energy between the system and surroundings
occurs when system and surroundings have a difference in temperature
heat flows from matter with high temperature to matter with low temperature until both objects reach the same temperature
thermal equilibrium
Tro, Chemistry: A Molecular Approach
22
Quantity of Heat Energy Absorbed
Heat Capacity
when a system absorbs heat, its temperature increases
the increase in temperature is directly proportional to the amount of heat absorbed
the proportionality constant is called the heat capacity, C
units of C are J/°C or J/K
q = C x DT
the heat capacity of an object depends on its mass
200 g of water requires twice as much heat to raise its temperature by 1°C than 100 g of water
the heat capacity of an object depends on the type of material
1000 J of heat energy will raise the temperature of 100 g of sand 12°C, but only raise the temperature of 100 g of water by 2.4°C
Tro, Chemistry: A Molecular Approach
23
Specific Heat Capacity
measure of a substance’s intrinsic ability to absorb heat
the specific heat capacity is the amount of heat energy required to raise the temperature of one gram of a substance 1°C
Cs
units are J/(g∙°C)
the molar heat capacity is the amount of heat energy required to raise the temperature of one mole of a substance 1°C
the rather high specific heat of water allows it to absorb a lot of heat energy without large increases in temperature
keeping ocean shore communities and beaches cool in the summer
allows it to be used as an effective coolant to absorb heat
Tro, Chemistry: A Molecular Approach
24
Quantifying Heat Energy
the heat capacity of an object is proportional to its mass and the specific heat of the material
so we can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat, and the temperature change of the object
Heat = (mass) x (specific heat capacity) x (temp. change)
q = (m) x (Cs) x (DT)
Example 6.2 – How much heat is absorbed by a copper penny with mass 3.10 g whose temperature rises from
-8.0°C to 37.0°C?



q = m ∙ Cs ∙ DT
Cs = 0.385 J/g (Table 6.4)
the unit and sign are correct
T1= -8.0°C, T2= 37.0°C, m=3.10 g
q, J
Check:
Check
Solution:
Follow the Concept Plan to Solve the problem
Concept Plan:



Relationships:
Strategize
Given:

Find:
Sort Information
Tro, Chemistry: A Molecular Approach
26
Pressure -Volume Work
PV work is work that is the result of a volume change against an external pressure
when gases expand, DV is +, but the system is doing work on the surroundings so w is ─
as long as the external pressure is kept constant
─Work = External Pressure x Change in Volume
w = ─PDV
to convert the units to joules use 101.3 J = 1 atm∙L
Example 6.3 – If a balloon is inflated from 0.100 L to 1.85 L against an external pressure of 1.00 atm, how much work is done?



the unit and sign are correct
V1=0.100 L, V2=1.85 L, P=1.00 atm
w, J
Check:
Solution:
Concept Plan:



Relationships:
Given:

Find:
101.3 J = 1 atm L
Tro, Chemistry: A Molecular Approach
28
Exchanging Energy Between
System and Surroundings
exchange of heat energy
q = mass x specific heat x DTemperature
exchange of work
w = −Pressure x DVolume
Tro, Chemistry: A Molecular Approach
29
Measuring DE,
Calorimetry at Constant Volume
since DE = q + w, we can determine DE by measuring q and w
in practice, it is easiest to do a process in such a way that there is no change in volume, w = 0
at constant volume, DEsystem = qsystem
in practice, it is not possible to observe the temperature changes of the individual chemicals involved in a reaction – so instead, we use an insulated, controlled surroundings and measure the temperature change in it
the surroundings is called a bomb calorimeter and is usually made of a sealed, insulated container filled with water
qsurroundings = qcalorimeter = ─qsystem
─DEreaction = qcal = Ccal x DT
Tro, Chemistry: A Molecular Approach
30
Bomb Calorimeter
used to measure DE because it is a constant volume system
31
Example 6.4 – When 1.010 g of sugar is burned in a bomb calorimeter, the temperature rises from 24.92°C to 28.33°C. If Ccal = 4.90 kJ/°C, find DE for burning 1 mole
qcal = Ccal x DT = -qrxn
MM C12H22O11 = 342.3 g/mol
the units and sign are correct
1.010 g C12H22O11, T1 = 24.92°C, T2 = 28.33°C, Ccal = 4.90 kJ/°C
DErxn, kJ/mol
Check:
Solution:
Concept Plan:



Relationships:
Given:

Find:
Tro, Chemistry: A Molecular Approach
32
Enthalpy
the enthalpy, H, of a system is the sum of the internal energy of the system and the product of pressure and volume
H is a state function
H = E + PV
the enthalpy change, DH, of a reaction is the heat evolved in a reaction at constant pressure
DHreaction = qreaction at constant pressure
usually DH and DE are similar in value, the difference is largest for reactions that produce or use large quantities of gas
33
Endothermic and Exothermic Reactions
when DH is ─, heat is being released by the system
reactions that release heat are called exothermic reactions
when DH is +, heat is being absorbed by the system
reactions that absorb heat are called endothermic reactions
chemical heat packs contain iron filings that are oxidized in an exothermic reaction ─ your hands get warm because the released heat of the reaction is absorbed by your hands
chemical cold packs contain NH4NO3 that dissolves in water in an endothermic process ─ your hands get cold because they are giving away your heat to the reaction
34
Molecular View of
Exothermic Reactions
in an exothermic reaction, the temperature rises due to release of thermal energy
this extra thermal energy comes from the conversion of some of the chemical potential energy in the reactants into kinetic energy in the form of heat
during the course of a reaction, old bonds are broken and new bonds made
the products of the reaction have less chemical potential energy than the reactants
the difference in energy is released as heat
Tro, Chemistry: A Molecular Approach
35
Molecular View of
Endothermic Reactions
in an endothermic reaction, the temperature drops due to absorption of thermal energy
the required thermal energy comes from the surroundings
during the course of a reaction, old bonds are broken and new bonds made
the products of the reaction have more chemical potential energy than the reactants
to acquire this extra energy, some of the thermal energy of the surroundings is converted into chemical potential energy stored in the products
Tro, Chemistry: A Molecular Approach
36
Enthalpy of Reaction
the enthalpy change in a chemical reaction is an extensive property
the more reactants you use, the larger the enthalpy change
by convention, we calculate the enthalpy change for the number of moles of reactants in the reaction as written
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g) DH = -2044 kJ

37
Example 6.6 – How much heat is evolved in the complete combustion of 13.2 kg of C3H8(g)?




1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol
the sign is correct and the value is reasonable
13.2 kg C3H8,
q, kJ/mol
Check:
Solution:
Concept Plan:



Relationships:
Given:

Find:
Calorimetry animation
Tro, Chemistry: A Molecular Approach
38
Tro, Chemistry: A Molecular Approach
39
Measuring DH
Calorimetry at Constant Pressure
reactions done in aqueous solution are at constant pressure
open to the atmosphere
the calorimeter is often nested foam cups containing the solution
qreaction = ─ qsolution = ─(masssolution x Cs, solution x DT)
DHreaction = qconstant pressure = qreaction
to get DHreaction per mol, divide by the number of moles
40
Example 6.7 – What is DHrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
100.0 mL of solution changes the temperature from 25.6°C to 32.8°C?




1 kg = 1000 g, 1 mol C3H8 = -2044 kJ, Molar Mass = 44.09 g/mol
the sign is correct and the value is reasonable
0.158 g Mg, 100.0 mL,
q, kJ/mol
Check:
Solution:
Concept Plan:



Relationships:
Given:

Find:
41
Example 6.7 – What is DHrxn/mol Mg for the reaction
Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) if 0.158 g Mg reacts in
100.0 mL of solution to change the temperature from 25.6°C to 32.8°C?



qsoln = m x Cs x DT = -qrxn
the units and sign are correct
0.158 g Mg, 100.0 mL sol’n, T1 = 25.6°C, T2 = 32.8°C, Cs = 4.18 J/°C, dsoln = 1.00 g/mL
DHrxn, J/mol Mg
Check:
Solution:
Concept Plan:


Relationships:
Given:

Find:
Tro, Chemistry: A Molecular Approach
42
Relationships Involving DHrxn
when reaction is multiplied by a factor, DHrxn is multiplied by that factor
because DHrxn is extensive
C(s) + O2(g) → CO2(g) DH = -393.5 kJ
2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2(-393.5 kJ) = -787.0 kJ
if a reaction is reversed, then the sign of DH is reversed
CO2(g) → C(s) + O2(g) DH = +393.5 kJ
Hess’s Law animation
Tro, Chemistry: A Molecular Approach
43
Tro, Chemistry: A Molecular Approach
44
Relationships Involving DHrxn
Hess’s Law
if a reaction can be expressed as a series of steps, then the DHrxn for the overall reaction is the sum of the heats of reaction for each step
Tro, Chemistry: A Molecular Approach
45
Sample – Hess’s Law
Given the following information:
2 NO(g) + O2(g)  2 NO2(g) DH° = -173 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = -255 kJ
N2(g) + O2(g)  2 NO(g) DH° = +181 kJ
Calculate the DH° for the reaction below:
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = ?
[2 NO2(g)  2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+173 kJ)
[2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)] x 0.5 DH° = 0.5(-255 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
[3 NO2(g)  3 NO(g) + 1.5 O2(g)] DH° = (+259.5 kJ)
[1 N2(g) + 2.5 O2(g) + 1 H2O(l)  2 HNO3(aq)] DH° = (-128 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = - 49 kJ
Tro, Chemistry: A Molecular Approach
46
Standard Conditions
the standard state is the state of a material at a defined set of conditions
pure gas at exactly 1 atm pressure
pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
usually 25°C
substance in a solution with concentration 1 M
the standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states
the standard enthalpy of formation, DHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements
the elements must be in their standard states
the DHf° for a pure element in its standard state = 0 kJ/mol
by definition
Tro, Chemistry: A Molecular Approach
47
Formation Reactions
reactions of elements in their standard state to form 1 mole of a pure compound
if you are not sure what the standard state of an element is, find the form in Appendix IIB that has a DHf° = 0
since the definition requires 1 mole of compound be made, the coefficients of the reactants may be fractions
48
Writing Formation Reactions
Write the formation reaction for CO(g)
the formation reaction is the reaction between the elements in the compound, which are C and O
C + O → CO(g)
the elements must be in their standard state
there are several forms of solid C, but the one with DHf° = 0 is graphite
oxygen’s standard state is the diatomic gas
C(s, graphite) + O2(g) → CO(g)
the equation must be balanced, but the coefficient of the product compound must be 1
use whatever coefficient in front of the reactants is necessary to make the atoms on both sides equal without changing the product coefficient
C(s, graphite) + ½ O2(g) → CO(g)
Tro, Chemistry: A Molecular Approach
49
Calculating Standard Enthalpy Change for a Reaction
any reaction can be written as the sum of formation reactions (or the reverse of formation reactions) for the reactants and products
the DH° for the reaction is then the sum of the DHf° for the component reactions
DH°reaction = S n DHf°(products) - S n DHf°(reactants)
S means sum
n is the coefficient of the reaction
Tro, Chemistry: A Molecular Approach
50
The Combustion of CH4

Tro, Chemistry: A Molecular Approach
51
Sample - Calculate the Enthalpy Change in the Reaction
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound and
determine the DHf° for each
2 C(s, gr) + H2(g) ® C2H2(g) DHf° = +227.4 kJ/mol
C(s, gr) + O2(g) ® CO2(g) DHf° = -393.5 kJ/mol
H2(g) + ½ O2(g) ® H2O(l) DHf° = -285.8 kJ/mol
Tro, Chemistry: A Molecular Approach
52
Sample - Calculate the Enthalpy Change in the Reaction
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
2 C2H2(g) ® 4 C(s) + 2 H2(g) DH° = 2(-227.4) kJ
4 C(s) + 4 O2(g) ® 4CO2(g) DH° = 4(-393.5) kJ
2 H2(g) + O2(g) ® 2 H2O(l) DH° = 2(-285.8) kJ
2. Arrange equations so they add up to desired reaction
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DH = -2600.4 kJ
Tro, Chemistry: A Molecular Approach
53
Sample - Calculate the Enthalpy Change in the Reaction
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
DH°reaction = S n DHf°(products) - S n DHf°(reactants)

DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]

DHrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]

DHrxn = -2600.4 kJ
54
Example 6.11 – How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?




MMoctane = 114.2 g/mol, 1 kg = 1000 g
the units and sign are correct
the large value is expected
1.0 x 1011 kJ
mass octane, kg
Check:
Solution:
Concept Plan:





Relationships:
Given:
Find:
Write the balanced equation per mole of octane
from
above
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Look up the DHf°
for each material
in Appendix IIB
Tro, Chemistry: A Molecular Approach
55
Energy Use and the Environment
in the U.S., each person uses over 105 kWh of energy per year
most comes from the combustion of fossil fuels
combustible materials that originate from ancient life
C(s) + O2(g) → CO2(g) DH°rxn = -393.5 kJ
CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) DH°rxn = -802.3 kJ
C8H18(g) +12.5 O2(g) → 8 CO2(g) + 9 H2O(g) DH°rxn = -5074.1 kJ
fossil fuels cannot be replenished
at current rates of consumption, oil and natural gas supplies will be depleted in 50 – 100 yrs.
Tro, Chemistry: A Molecular Approach
56
Energy Consumption
the distribution of energy consumption in the US
the increase in energy consumption in the US
Tro, Chemistry: A Molecular Approach
57
The Effect of Combustion Products
on Our Environment
because of additives and impurities in the fossil fuel, incomplete combustion and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy
therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming
Tro, Chemistry: A Molecular Approach
58
Global Warming
CO2 is a greenhouse gas
it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the earth to escape into outer space
it acts like a blanket
CO2 levels in the atmosphere have been steadily increasing
current observations suggest that the average global air temperature has risen 0.6°C in the past 100 yrs.
atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed
some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats
Tro, Chemistry: A Molecular Approach
59
CO2 Levels

Tro, Chemistry: A Molecular Approach
60
Renewable Energy
our greatest unlimited supply of energy is the sun
new technologies are being developed to capture the energy of sunlight
parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity
solar energy used to decompose water into H2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity
H2(g) + ½ O2(g) → H2O(l) DH°rxn = -285.8 kJ
hydroelectric power
wind power

 
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