Chapter 7 Thermodynamics:

1

• Chapter 7 Thermodynamics
• System
• Energy
• The First Law
• Work/Heat
• Internal energy (E)
• Heat capacity
• State changes (s,l,g)
• HE PV
• Reaction Enthalpies

2

• Thermodynamics is the science of heat (energy)
  transfer.

Heat energy is associated with intramolecular and
extramolecular motions.
Heat transfers until thermal equilibrium is
established.
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UNITS OF ENERGY

• 1 calorie heat required to raise temp. of 1.00
  g of H2O by 1.0 oC.
• 1000 cal 1 kilocalorie 1 kcal
• 1 kcal 1 Calorie (a food calorie)
• But we use the unit called the JOULE
• 1 cal 4.184 joules

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A system Ex. System is a reaction
mixture Surroundings atmosphere around the
reaction mixture
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Directionality of Heat Transfer

• Heat always transfer from hotter object to cooler
  one.
• EXOthermic heat transfers from SYSTEM to
  SURROUNDINGS.

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Combustion of octane

• 2C8H18(l) 25O2(g)? 18H2O (l) 16CO2 (g)
• DHrxn -10.9 103 kJ -10.9 MJ
• 5.45 103 kJ of energy released per mol of
  octane
• Combustion of 1 gallon of gas ?136 MJ energy

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Directionality of Heat Transfer

• Heat always transfer from hotter object to cooler
  one.
• ENDOthermic heat transfers from SURROUNDINGS to
  the SYSTEM.

The example in the figure illustrates the
melting of ice.Note that for state changes the
temperature is constant. All energy is consumed
by thestate change. Only the T(surr)goes down
in this case.
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Energy capacity to do work or to produce heat
1
ek
mv2

• Kinetic energy (KE) motion of object,
• Potential energy (PE) due to position or
  composition
• (bonds, height of a mass)
• Ex. Combustion of methane (CH4) potential
  energy stored in bonds in converted to kinetic
  energy (thermal energy) via heat.
• Heat is the form of energy that flows between 2
  objects because of their difference in
  temperature, always transfer from hotter object
  to cooler one.
• Electrical energy, Light energy

2
kg m2
J
s2
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Energy Chemistry

• All of thermodynamics depends on the law of
• CONSERVATION OF ENERGY.
• The total energy is unchanged in a chemical
  reaction.
• If PE of products is less than reactants, the
  difference must be released as KE.

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First law of thermodynamics

• The total energy of the universe is constant.
• Also known as Law of Conservation of Energy
  energy is neither created nor destroyed in
  ordinary chemical reactions
• Energy in Joules

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Internal Energy changeHeat Work

• The internal energy of a system E is the sum of
  kinetic and potential energies of all particles.
• The internal energy is state function E(p,v,T)
• The internal energy is the capacity of a system
  to do work its total store of energy.
• DE Efinal Einitial q w
• q is heat w is work
• The heat and the work are not state functions.

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Work

• Work change in energy when an object is moved
  through a distance
• Example of work
• compression or expansion of a gas
• W -Pex D V

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A system doing work (i.e. expansion of a gas).
Work is proportional to the external pressure
and change in volume.
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• 3.1 HW. A gas in a cylinder of an engine expands
  by 500.0 mL against a pressure of 1.20 atm. What
  is the magnitude of the work involved? Is work
  done by or on the gas?

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Internal Energy, E or U

• Total energy (potential and kinetic) in a system.

• Translational kinetic energy.
• Molecular rotation.
• Bond vibration.
• Intermolecular attractions.
• Chemical bonds.
• Electrons.

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Maxwell distributions One gas (50g/mol) at
three different temperatures
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State Functions

• Any property that has a unique value for a
  specified state of a system is said to be a
  State Function.
• Water at 293.15 K and 1.00 atm is in a specified
  state.
• d 0.99820 g/mL
• This density is a unique function of the state.
• It does not matter how the state was established.

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Functions of State

• U is a function of state.
• Not easily measured.
• ?U has a unique value between two states.
• Is easily measured.

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7-6 Heats of Reaction ?U and ?H
Reactants ? Products
Ui Uf
?U Uf - Ui
?U qrxn w
In a system at constant volume
?U qrxn 0 qrxn qv
But we live in a constant pressure world! How
does qp relate to qv?
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Heats of Reaction
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Heats of Reaction
qV qP w We know that qV ?U, w -P?V and
therefore ?U qP - P?V qP ?U P?V These
are all state functions, so define a new
function. Let H U PV Then ?H Hf Hi ?U
?PV If we work at constant pressure and
temperature ?H ?U P?V qP
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Comparing Heats of Reaction
qP -566 kJ/mol ?H P?V P(Vf Vi)
RT(nf ni) -2.5 kJ ?U ?H - P?V
-563.5 kJ/mol qV
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Changes of State of Matter
Molar enthalpy of vaporization
H2O (l) ? H2O(g) ?H 44.0 kJ at 298 K
Molar enthalpy of fusion
H2O (s) ? H2O(l) ?H 6.01 kJ at 273.15 K
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Example 7-3
Enthalpy Changes Accompanying Changes in States
of Matter. Calculate ?H for the process in which
50.0 g of water is converted from liquid at
10.0C to vapor at 25.0C.
Break the problem into two steps Raise the
temperature of the liquid first then completely
vaporize it. The total enthalpy change is the
sum of the changes in each step.
3.14 kJ 122 kJ 125 kJ
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Standard States and Standard Enthalpy Changes

• Define a particular state as a standard state.
• Standard enthalpy of reaction, ?H
• The enthalpy change of a reaction in which all
  reactants and products are in their standard
  states.
• Standard State
• The pure element or compound at a pressure of 1
  bar and at the temperature of interest.

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Enthalpy Diagrams
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7-7 Indirect Determination of ?HHesss Law

• ?H is an extensive property.
• Enthalpy change is directly proportional to the
  amount of substance in a system.

N2(g) O2(g) ? 2 NO(g) ?H 180.50 kJ
½N2(g) ½O2(g) ? NO(g) ?H 90.25 kJ

• ?H changes sign when a process is reversed

NO(g) ? ½N2(g) ½O2(g) ?H -90.25 kJ
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Hesss Law

• Hesss law of constant heat summation
• If a process occurs in stages or steps (even
  hypothetically), the enthalpy change for the
  overall process is the sum of the enthalpy
  changes for the individual steps.

½N2(g) O2(g) ? NO2(g) ?H 33.18 kJ
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Hesss Law Schematically
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7-8 Standard Enthalpies of Formation
?Hf

• The enthalpy change that occurs in the formation
  of one mole of a substance in the standard state
  from the reference forms of the elements in their
  standard states.
• The standard enthalpy of formation of a pure
  element in its reference state is 0.

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Standard Enthalpies of Formation
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Standard Enthalpies of Formation
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Standard Enthalpies of Reaction
?Hoverall -2?HfNaHCO3 ?HfNa2CO3 ?HfCO2
?HfH2O
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Enthalpy of Reaction
?Hrxn ??Hfproducts- ??Hfreactants
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7-9 Fuels as Sources of Energy

• Fossil fuels.
• Combustion is exothermic.
• Non-renewable resource.
• Environmental impact.

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The work is not state function

• If the pressure is changing during the volume
  change a Pex(V) function can be constructed
• Then the work is an integral, and not a simple
  state function
• Consequently the heat is not a state function

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The work at constant pressure

• If Pex is constant, then the integral is
  simplified to
• W -Pex (Vfinal Vinitial) -Pex DV
• In this special case the work will depend only on
  the Vfinal and Vinitial, and it will behave like
  a state function
• As the heat (q) is DE w DE Pex DV, it will
  also behave like a state function

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Enthalpy, state function

• The Enthalpy change is equal to the transferred
  heat at constant pressure
• DH qp DE Pex DV
• Hfinal Hinitial Efinal Einitial Pex
  (Vfinal Vinitial)
• DH gt 0 Endothermic
• DH lt 0 Exothermic
• The enthalpy is a state function H(P,V,T)
  E(P,V,T) P V or simpler
• H E P V

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HEAT CAPACITY

• The heat required to raise an objects T by 1 C.

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Specific Heat Capacity

• How much energy is transferred due to T
  difference?
• The heat (q) lost or gained is related to
• a) sample mass
• b) change in T and
• c) specific heat capacity

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Specific Heat Capacity

• Substance Spec. heat (Jg-1K-1)
• H2O 4.184
• Ethylene glycol 2.39
• glass 0.84
• Aluminum 0.897

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per gram of substance per mol
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Specific Heat Capacity

• If 25.0 g of Al cool from 310 oC to 37 oC, how
  many joules of heat energy are lost by the Al?

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Determination of Specific Heat Capacity
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Example 7-2
Determining Specific Heat from Experimental
Data. Use the data presented on the last slide to
calculate the specific heat of lead.
qlead -qwater
qwater mc?T (50.0 g)(4.184 J/g C)(28.8 -
22.0)C
qwater 1.4x103 J
qlead -1.4x103 J mc?T (150.0 g)(c)(28.8 -
100.0)C
clead 0.13 Jg-1C-1
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Specific Heat Capacity

• If 25.0 g of Al cool from 310 oC to 37 oC, how
  many joules of heat energy are lost by the Al?

where ?T Tfinal - Tinitial q (0.897
J/gK)(25.0 g)(37 - 310)K q - 6120 J
Notice that the negative sign on q signals heat
lost by or transferred OUT of Al.
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Heat TransferNo Change in State

• q transferred (sp. ht.)(mass)(?T)

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Heat Transfer with Change of State

• Changes of state involve energy (at constant T)
• Ice 333 J/g (heat of fusion) ? Liquid water
• q (heat of fusion)(mass)

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Heat Transfer and Changes of State
Liquid ? Vapor

• Requires energy (heat).
• This is the reason
• a) you cool down after swimming
• b) you use water to put out a fire.

energy
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Heating/Cooling Curve for Water
Evaporate water
Heat water
Note that T is constant as ice melts
Melt ice
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Heat Changes of State

• What quantity of heat is required to melt 500. g
  of ice and heat the water to steam at 100 oC?

Heat of fusion of ice 333 J/g Specific heat of
water 4.2 J/gK Heat of vaporization 2260 J/g
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Heat Changes of State

• What quantity of heat is required to melt 500. g
  of ice and heat the water to steam at 100 oC?
• 1. To melt ice
• q (500. g)(333 J/g) 1.67 x 105 J
• 2. To raise water from 0 oC to 100 oC
• q (500. g)(4.2 J/gK)(100 - 0)K 2.1 x 105
  J
• 3. To evaporate water at 100 oC
• q (500. g)(2260 J/g) 1.13 x 106 J
• 4. Total heat energy 1.51 x 106 J 1510 kJ

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Enthalpies of Phase Change

• Condensation (exothermic)
• Vaporization (endothermic)
• Freezing (exothermic)
• Melting (endothermic)
• Solidification (exothermic)
• Sublimation (endothermic)

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• 3.2. HW How much heat is needed to vaporize
  150. g of benzene?
• The boiling point of benzene is 80o C.
• The DHvapo (benzene) is 30.8 kJ/mol.

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How do molcules store energy?
Bond energy each C-H bond 400 kJ Total 1600
kJ
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How do molcules store energy?

Translational 3.7 kJ/mol Rotational 3.7
kJ/mol Vibrational 11.1 kJ/molBonds 1600
kJ/mol (part of the electronic energycalculated
from quantum mechanics) E Etrans Erot
Evib Eelec
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How can we extract energy?
Translational 3.7 kJ Rotational 3.7
kJ Vibrational 134 kJ Bonds 1600 kJ

CH4 2O2 ? CO2 2H2O DH -890
kJ Burning in this chemical reaction the bonds
of methane and oxygen are broken and new bonds in
CO2 and H2O are formed. These latter bonds are
more stable. The reaction is exothermic because
of the weak bond of oxygen, compared to the
strong bonds of CO2 and H2O.
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• Calorimeter To determine the heat associated
• with a chemical reaction
• temperature change of a reaction or substance is
  measured

fancy one!
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CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
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Example 7-3

• Using Bomb Calorimetry Data to Determine a Heat
  of Reaction.
• The combustion of 1.010 g sucrose, in a bomb
  calorimeter, causes the temperature to rise from
  24.92 to 28.33C. The heat capacity of the
  calorimeter assembly is 4.90 kJ/C.
• What is the heat of combustion of sucrose,
  expressed in kJ/mol C12H22O11
• Verify the claim of sugar producers that one
  teaspoon of sugar (about 4.8 g) contains only 19
  calories.

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Example 7-3
Calculate qcalorimeter
qcal C?T (4.90 kJ/C)(28.33-24.92)C
(4.90)(3.41) kJ 16.7 kJ
per 1.010 g
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Example 7-3
Calculate qrxn in the required units
-16.7 kJ
qrxn -qcal
-16.5 kJ/g
1.010 g
Calculate qrxn for one teaspoon
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3.3. HWNaOH HCl ? H2O NaCl

• 50.0 mL of 1.0 M NaOH is mixed with 50.0 mL of
  1.0 M HCl at 25.0 oC. The temperature of the
  solution increases to 31.9 oC. What is the total
  heat released? How much energy is released per
  mol of reactant? What is the enthalpy (DH) for
  the reaction?

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3.4. HWSO2(g) O2(g) ? SO3(g)

• 2.00 mols SO2 are reacted with 1.00 mol O2 at 25o
  C and 1.00 atm. 198 kJ of energy is released as
  heat. Calculate DH and DE.

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P DV Dn R T
SO2(g) O2(g) ? SO3(g) at 25o C and 1.0 atm
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C2H5OH(l) 3O2(g) - 1367 kJ ? 2CO2(g) 3H2O(l)
per mol reaction means per total number mols
of each substance in balanced equation
DHrxn - 1367 kJ
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3.5. HW4Al(s) 3O2(g) ? 2Al2O3(s) DHrxn
-3352 kJ/mol rxn

• How much heat is released upon reaction of 24.2 g
  Al at 25o C and 1 atm?

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3.6. HWCuSO4(aq) 2NaOH(aq) ? Cu(OH)2(s)
Na2SO4(aq)

• 50.00 mL of 0.400 M CuSO4 reacts with 50.00
  mL of 0.600 M NaOH at 1 atm. The total heat
  evolved is 1.49 kJ. What is DHrxn?

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Thermochemical Standard States

• DHorxn standard heat of reaction
• (1 atm, 298 K, 1 M)
• DHfo standard molar enthalpy of formation
• change in DH when one mol of a substance is
  formed from its constituent elements
• H2(g) Br2(l) ? 2HBr(g) DHorxn -72.8 kJ
• ½H2(g) ½Br2(l) ? HBr(g) DHof (HBr)
  -36.4 kJ/mol

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2C(graphite) 3H2(g) ½O2(g) ? C2H5OH(l)
DHfo(C2H5OH) -277.7 kJ/mol DHorxn -277.7
kJ/mol rxn

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Note Scale and Sign

• CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
• DHr -890 kJ
• 2CH4(g) 4O2(g) ? 2CO2(g) 4H2O(l)
• DHr -1780 kJ
• CO2(g) 2H2O(l) ? CH4(g) 2O2(g)
• DHr 890 kJ

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Note Forms of Reactants and Products are
Important

• CH4(g) 2O2(g) ? CO2(g) 2H2O(g)
• DHr -802. kJ
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
• DHr -890. kJ

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Hesss Law Law of SummationThe enthalpy of a
reaction can be estimated by summing the
enthalpies of sub-steps of the reaction.
DHrxno DHao DHbo DHco DHdo .
(sub-steps of reaction) DHrxn is an extensive
property, it is proportional to the amounts of
substances (stoichiometry). Reversing a
chemical equation changes the sign of a function
of state (e.g. DHrxno) related to that equation.
By definition, ?Hfo 0 for elements in their
standard states.
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Hesss Law Energy Level Diagrams
Forming H2O can occur in a single step or in a
two steps. ?Htotal is the same no matter which
path is followed.
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Hesss Law Energy Level Diagrams

• Forming CO2 can occur in a single step or in a
  two steps. ?Htotal is the same no matter which
  path is followed.

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Using Standard Enthalpy Values

• Use ?Hs to calculate enthalpy change for
• H2O(g) C(graphite) ? H2(g) CO(g)

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Using Standard Enthalpy Values

• H2O(g) C(graphite) ? H2(g) CO(g)
• From reference books we find
• H2(g) 1/2 O2(g) ? H2O(g)
• ?Hf of H2O vapor - 242 kJ/mol
• C(s) 1/2 O2(g) ? CO(g)
• ?Hf of CO - 111 kJ/mol

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Using Standard Enthalpy Values

• H2O(g) ? H2(g) 1/2 O2(g) ?Ho 242 kJ
• C(s) 1/2 O2(g) ? CO(g) ?Ho -111 kJ
• --------------------------------------------------
  ------------------------------

H2O(g) C(graphite) ? H2(g) CO(g)
?Honet 131 kJ
To convert 1 mol of water to 1 mol each of H2 and
CO requires 131 kJ of energy. The water gas
reaction is ENDOthermic.
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Hesss Law Law of Summation
C(gr) O2(g) ? CO2(g) What is the DHrxno
? C(gr) ½O2(g) ? CO(g) DHo -110.5
kJ CO(g) ½O2(g) ? CO2(g) DHo -282.0
kJ C(gr) O2(g) ? CO2(g) By definition, ?Hfo
0 for elements in their standard states here
C(gr) and O2(g)
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Relationship between DHf and DHrxn
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Using Standard Enthalpy Values
CH3OH(g) 3/2 O2(g) ? CO2(g) 2 H2O(g) ?Horxn
? ?Hfo (prod) - ? ?Hfo (react)

• ?Horxn ?Hfo (CO2) 2 ?Hfo (H2O)
• (-393.5 kJ) 2 (-241.8 kJ)
• - 3/2 ?Hfo (O2) ?Hfo (CH3OH)
• - 0 (-201.5 kJ)
• ?Horxn -675.6 kJ per mol of methanol

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3.7. HWHesss Law another example

• Find DHrxno for
• C(gr) 2H2(g) ? CH4(g)
• - - - - - - - - - - - - - - - - - - - - - - - - -
  - - -
• CH4(g) 2O2(g) ? CO2(g) 2H2O(l) -890.3 kJ
• Standard enthalpies of formations
• C(gr) O2(g) ? CO2(g)
  -393.5 kJ
• H2(g) ½O2(g) ? H2O(l)
  -285.8 kJ

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DHrxno estimated based on DHfo of products and
reactants.DHrxno SnDHfo(products)
-nDHfo(reactants)

• Example C2H4(g) H2O(l) ? C2H5OH(l)
• What is DHrxno ?
• DHfo(C2H4 (g)) 52.3 kJ/mol
• DHfo(H2O (l)) -285.8 kJ/mol
• DHfo(C2H5OH (l)) -277.7 kJ/mol

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• 3.8. HW2 NH2CH2COOH (s) 3O2(g) ?
• H2NCONH2(s) 3CO2 (g) 3H2O(l)
• What is the standard enthalpy of reaction?
• What is the enthalpy per mol of NH2CH2COOH
  (glycine reactant)?

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Combustion

• DHco standard enthalpy of combustion
• Combustion reaction of substance with O2 to
  produce CO2 and H2O
• change in enthalpy per mol of substance burned
• Ex. CH4(g) 2O2(g) ? CO2(g) 2H2O(l)
• 
  DHrxn -890.kJ
• 1 mol CH4 combustion 890 kJ or DHco
  890. kJ/mol

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Bond Enthalpy

• Strength of a chemical bond DHb
• Use bond enthalpies to estimate DHrxno

What is the strength of a C-H bond?
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Average Single Bond Enthalpies (kJ/mol)
H2(g) ? H(g) H(g) DH BE 436
kJ/mol HCl(g) ? H(g) Cl(g) DH BE
432 kJ/mol
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Single and Multiple Bond Enthalpies (kJ/mol)
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Bond enthalpies are reported as average values.
410 kJ/mol 393 kJ/mol
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Using bond enthalpies to estimate DHrxnoWhat is
the DHrxn?

• DHrxn S BE (reactants) - S BE (products)
• DHrxn S BE (bonds broken) - S BE (bonds formed)
  

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• How much energy is released in this burning
  reaction approximatively?
• C3H8(g) 5O2(g) ? 3CO2(g) 4H2O(g)
• 2 C-C (345) 8 C-H (415) 5 OO (495) ? 6 CO (800)
  8 O-H (460)
• Broken bonds Formed bonds
• 6485 kJ 8480 kJ
• DE 6485-8480 -1995 kJ (release of energy)
• Notice the small OO and the large CO energies
• The correct value is -2200 kJ

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3.9. HWUsing bond enthalpies to estimate
DHrxnoWhat is the DHrxn?

• Br2(g) 3F2(g) ? 2BrF3(g)
• DHrxn S BE (reactants) - S BE (products)
• BE (Br2) 192 kJ
• BE (F2) 159 kJ
• BE(BrF) 197 kJ