CHM 110 CHAPTER 6 THERMOCHEMISTRY: Energy Flow and Chemical Change - PowerPoint PPT Presentation

1 / 40
About This Presentation
Title:

CHM 110 CHAPTER 6 THERMOCHEMISTRY: Energy Flow and Chemical Change

Description:

this is because enthalpy is an extensive. property ... Another way to state this: the overall enthalpy ... CO the enthalpy change for the formation of 1 ... – PowerPoint PPT presentation

Number of Views:239
Avg rating:3.0/5.0
Slides: 41
Provided by: chem211
Category:

less

Transcript and Presenter's Notes

Title: CHM 110 CHAPTER 6 THERMOCHEMISTRY: Energy Flow and Chemical Change


1
CHM 110CHAPTER 6THERMOCHEMISTRY Energy Flow
and Chemical Change
  • Dr. Floyd Beckford
  • Lyon College

2
(No Transcript)
3
ENERGY
  • Energy the capacity to do work or supply heat
  • ENERGY Work heat
  • Energy is classified into two types
  • 1. Kinetic energy, EK energy of motion
  • EK ½ mv2
  • where m mass of object
  • v velocity of object

4
  • 2. Potential energy, EP energy a system
  • possesses by virtue of its composition or
  • position
  • The SI unit of energy is the Joule, J
  • 1 J 1 kgm2/s2
  • An older unit, the Calorie, is still used
  • particularly in nutrition
  • 1 cal 4.184 J

5
ENERGY CHANGES
  • Energy changes are dictated by the Law of
  • Conservation of Energy - energy cannot be
  • created or destroyed it can only be converted
    from
  • one form into another
  • HEAT the energy transferred from one object
  • to another as a result of as temperature
    difference
  • between them

6
INTERNAL ENERGY
  • The substances involved in the process under
  • investigation makes up the system
  • Everything in the systems environment
  • constitutes its surroundings
  • The internal energy, E, of the system represents
  • all the energy contained within the system the
  • sum of all types of energies present in the system

7
  • The First Law of Thermodynamics states the
  • total energy of an isolated system is constant
  • It is important to be able to measure energy
  • changes which occur during reactions
  • With respect to internal energy
  • ?E Efinal Einitial
  • Energy changes are always measured with
  • respect to the system

8
  • Energy flow from the system ? ?E is negative
  • - heat is lost and Efinal lt Einitial
  • When Efinal gt Einitial heat flows into the
    system
  • from the surroundings ?E is positive

9
  • The thermodynamic state of a system is defined
  • by a set of conditions that completely specifies
    all
  • the properties of the system
  • These properties are called state functions
  • The value of a state function depends ONLY on
  • the state of the system and NOT on the way in
  • which the system came to be in that state
  • e.g. altitude

10
  • A change in a state function describes a
  • difference between two states it is independent
  • of the process or path way by which the change
  • takes place
  • It is the changes relating to state functions
  • which usually interests us
  • For any state function, X
  • ?X Xfinal - Xinitial

11
WORK
  • All energy transfer to and from the system
  • occur as heat or work
  • In chemical systems, the most common type of
  • work observed is pressure-volume or PV work
  • - work done as a result of volume changes
  • Volume changes may be expansion or
  • contraction

12
w -P ?V
13
  • During expansion, work is done BY the system
  • ON the surroundings
  • - sign of work is negative
  • V2 gt V1 so ?V V2 V1 gt 0
  • So w -P ?V lt 0
  • This work can be due to the number of moles

of gas CO2(s) ? CO2(g)
14
  • The reverse process is called contraction work
  • is done ON the system BY the surroundings
  • V1 gt V2 so ?V V2 V1 lt 0
  • And
  • w -P ?V positive
  • This change can be due to a decrease in the
  • number of moles of gas
  • When ?V 0 no PV work is done

15
Calculate the work (in kilojoules) done
during the synthesis of ammonia in which the
volume contracts from 8.6 L to 4.3 L at a
constant external pressure of 44 atm. In which
direction does the work energy flow? What is the
sign of the energy change? 1 Latm 101 J
16
ENERGY AND ENTHALPY
  • The change in internal energy is ?E
  • ?E q w q heat transferred
  • System gains heat q is positive
  • System loses heat q is negative
  • ?E q - P ?V and so
  • q ?E P ?V
  • At constant volume qv ?E

17
  • If the reaction is carried out at constant
    pressure
  • the heat transferred is called the heat of
    reaction
  • or the enthalpy change, ?H of the reaction
  • qp ?E P ?V ?H
  • Enthalpy is a state function. So
  • ?H Hproducts Hreactants
  • For ordinary reactions ?E and ?H are roughly
  • the same value

18
THERMODYNAMIC STANDARD STATE
  • Consider the following equation
  • This equation is referred to as a thermochemical
  • equation
  • Note that the physical states of the substances
  • MUST be specified when enthalpy changes are
  • reported

19
  • C3H8(g) 5O2(g) ? 3CO2(g) 3H2O(g)
  • ?H -2043 kJ
  • C3H8(g) 5O2(g) ? 3CO2(g) 3H2O(l)
  • ?H -2219 kJ
  • It is also necessary to specify the conditions
  • under which the reaction was run
  • The thermodynamic standard state for a system
  • must be defined

20
  • 1. Most stable form of a substance at 1 atm
  • pressure and at a specified temperature, usually
  • 25 C
  • 2. 1 M concentration for all substances in
    solution
  • Measurements done under standard conditions
  • are are indicated with a superscript
  • Reactants ? Products ?Hrxn
  • ?Hrxn standard enthalpy change for reaction

21
  • H2O(s) ? H2O(g) ?Hsubl
  • H2O(s) ? H2O(l) ?Hfusion
  • H2O(l) ? H2O(g) ?Hvap
  • For the above processes the enthalpy changes
  • are referred to as enthalpy of sublimation,
    fusion,
  • and vaporization respectively
  • Note that during a phase change the temperature
  • remains constant

22
  • 1. Ba(OH)28H2O(s) 2NH4Cl(s) ? BaCl2(aq)
  • 2NH3(aq) 10H2O(l) ?H 80.3 kJ
  • 2. 2H2(g) O2(g) ? 2H2O(g) ?H -484 kJ
  • Reaction 1 absorbs heat from the surroundings
  • the reaction is said to be endothermic
  • When heat flows OUT of the system the reaction
  • is exothermic reaction 2

23
  • Points to note about ?H
  • 1. The value of ?H is for the thermochemical
  • equation
  • - this is because enthalpy is an extensive
  • property
  • 2. ?H values refer to the reaction proceeding in
  • the direction as written
  • - the sign of ?H must be changed for the
  • reverse reaction

24
How much heat (in kilojoules) is evolved or
absorbed in the burning 15.5 g of propane
as given in the following equation C3H8(g)
5O2(g) ? 3CO2(g) 3H2O(l) ?H -2219 kJ
25
CALORIMETRY
  • The amount of heat transferred during a
  • reaction can be measured by calorimetry
  • Based on observing the temperature change
  • when a system absorbs or releases heat
  • The reaction is carried out in a calorimeter
  • The calorimeter measures ?H or ?E depending
  • on the setup

26
(No Transcript)
27
  • Various heat transfer processes occur during a
  • calorimetry experiment
  • The heat released warms both the surrounding
  • water as well as heat the calorimeter
  • The amount of heat absorb by the calorimeter
  • depends on its heat capacity
  • - the amount of heat required to raise the
  • temperature of an object by a certain amount

28
  • A more useful quantity is the specific heat
    the
  • amount of heat necessary to raise the temperature
  • of exactly 1 g of a substance by exactly 1 K
  • In general
  • Heat transferred heat gained by water heat
  • gained by calorimeter
  • For any substance
  • q mc?T

29
Assuming that Coca Cola has the same
specific heat as water (4.18 J/gC), calculate
the amount of heat (in kilojoules) transferred
when one can (about 350 g) is cooled from 25 C
to 3 C. What is the specific heat of lead if it
takes 96 J to raise the temperature of a 75 g
body by 10.0 C?
30
HESSS LAW
  • Hesss Law the enthalpy change for a reaction
  • is the same whether it occurs by one step or by
  • any series of steps
  • Another way to state this the overall enthalpy
  • change for a reaction is equal to the sum of the
  • enthalpy changes for the individual steps in the
  • reaction

31
  • ?Hrxn ?H1 ?H2 ?H3
  • 1, 2, 3, refers to balanced thermochemical
  • equations
  • Consider the following equation
  • C(graphite) ½ O2(g) ? CO(g) ?Hrxn ?
  • The following reactions are known
  • C(graphite) O2(g) ? CO2(g) ?Hrxn -393.5
  • CO(g) ½ O2(g) ? CO2(g) ?Hrxn -283.0

32
  • C(graphite) O2(g) ? CO2(g) ?Hrxn -393.5
    (1)
  • CO2(g) ? CO(g) ½ O2(g) ?Hrxn 283.0 (2)
  • _________________________________________
  • C(graphite) ½ O2(g) ? CO(g) ?Hrxn ?
  • From Hesss Law
  • ?Hrxn ?Hrxn (1) ?Hrxn (2)
  • - 393.5 283.0
  • - 110.5 kJ/mol

33
Determine the heat of formation of liquid
hydrogen peroxide at 25 C from the following
thermochemical data. H2(g) ½ O2(g) ? H2O(g)
?H -241.82 kJ/mol 2H(g) O(g) ? H2O(g)
?H -926.92 kJ/mol 2H(g) 2O(g) ? H2O2(g) ?H
-1070.60 kJ/mol 2O(g) ? O2(g) ?H
-498.34 kJ/mol H2O2(l) ? H2O2(g) ?H
51.46 kJ/mol
34
(No Transcript)
35
HEAT OF FORMATION
  • Consider again the final reaction we just saw
  • The 110.5 kJ is the amount of energy liberated
  • when CO is formed from carbon and oxygen
  • This is the standard heat of formation, ?Hf,
    for
  • CO the enthalpy change for the formation of 1
  • mole of a substance in its standard state from
    its
  • constituents in their standard states

36
  • In using this definition the ?Hf for any
    element
  • in its standard state is zero
  • ?H ?Hf(products) - ?Hf(reactants)
  • For any reaction
  • aA bB .. ? cC dD
  • ?H c ?Hf (C) d ?Hf (D)
  • a ?Hf (A) b ?Hf (B)

37
e.g. SiH4(g) 2O2(g) ? SiO2(s) 2H2O(l) ?Hrxn
?Hf (SiO2) 2 ?Hf (H2O) ?Hf
(SiH4) 2 ?Hf (O2) But ?Hf (O2) 0 So
?Hrxn ?Hf (SiO2) 2 ?Hf (H2O)
?Hf (SiH4)
38
Acetic acid (CH3COOH) is made by the reaction of
C2H5OH with oxygen C2H5OH(l) O2(g) ?
CH3COOH(l) H2O(l) Use the following data to
calculate ?H (in kJ) for the reaction ?Hf
(C2H5OH) -277.7 kJ/mol ?Hf (CH3COOH) -484.5
kJ/mol ?Hf (H2O) -284.8 kJ/mol
39
Calculate ?Hf (in kilojoules per mol) for
benzene, C6H6, from the following data 2C6H6(l)
15O2(g) ? 12CO2(g) 6H2O(l) ?H -6534
kJ ?Hf (CO2) - 393.5 kJ/mol ?Hf (H2O)
285.8 kJ/mol
40
BOND ENERGIES
  • Bond energy, D the amount of energy needed
  • to break one mole of bonds to form products
  • X-Y ? X Y ?H D
  • Hesss law can be applied to bond energies
  • ?H D(bonds broken) D(bonds formed)
  • e.g. CH4(g) 3Cl2(g) ? CHCl3(g) 3HCl(g)
  • ?H (3DCl-Cl 3DC-H) (3DC-Cl 3DH-Cl)
Write a Comment
User Comments (0)
About PowerShow.com