Chapter 6 Thermochemistry

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Chapter 6Thermochemistry
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Preview

• Introduction to thermodynamics
• Chemical energy.
• Internal energy, work and heat.
• Exothermic vs. endothermic reactions.
• PV work.
• Enthalpy and calorimetry.
• Constant P and constant V calorimetries.
• Hesss law.
• Standard enthalpies of formation.

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The Nature of Energy
Chapter 6 Section 1

• Energy, including all different
  classes, is very essential for our lives.
• Macroscopic scale fossil fuels, water and winds,
  solar, heat (thermal), electricity, etc.
• Microscopic scale chemical reactions in living
  cells, nuclear energy, etc.
• Energy can be converted form one form to another.
• Energy is the capacity to do work or to
  produce heat.

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The Nature of Energy
Chapter 6 Section 1

• The law of conservation of energy.
• Energy can be converted from one form to another
  but can neither be created nor destroyed.
• EUniverse constant
• First law of thermodynamics

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Potential and Kinetic Energies
Chapter 6 Section 1

• Potential due to position or composition can be
  converted to work.
• Water behind a dam.
• Gasoline before it burns (energy stored in the
  chemical bonds).
• Kinetic due to motion of an object.
• Water when it falls down from the dam (work).
• Gasoline when it burns up (heat) and gives power
  to the engine (work).
• KE ½ (mv2)

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Potential and Kinetic Energies
Chapter 6 Section 1

• (a) Initial positions ball A has a higher
  potential energy than ball B.
• (b) Final positions the potential energy lost by
  A has been
• transferred to the surface of the hill as heat
  (friction).
• transferred to B, causing an increase in its
  potential energy (work).

Effect of the pathway used by A to travel (Is
the surface rough or soft??)
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State Function or State Property
Chapter 6 Section 1

• The change in the state function (property) is
  only dependent on the initial and final states,
  no matter which pathway is taken between these
  two states.
• Energy is state function.
• Work and heat are not.

Elevation is always the same (state
property). Distances traveled by the two hikers
differ (path or non-state property)
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Chemical Energy
Chapter 6 Section 1

• Combustion of methane
• CH4(g) 2O2(g) CO2(g) 2H2O(g)
  energy (heat)

System
Surroundings
Reactants and products
Reaction container, room, everything else
heat
Energy lost from the system Energy gained by
the surroundings
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Exothermic and Endothermic Reactions
Chapter 6 Section 1

• In methane combustion the energy (heat) flows out
  of the system.
• It is an exothermic reaction.
• H2O(l) H2O(s) energy (heat)
• Nitric oxide formation is an endothermic
  reaction.
• N2(g) O2(g) energy (heat)
  2NO(g)
• Here the heat flows from the surroundings into
  the system.
• H2O(s) energy (heat) H2O(l)

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Exothermic Reactions
Chapter 6 Section 1
In Exothermic reactions, some PE stored in the
chemical bonds is converted to thermal energy via
heat.
CH4(g) 2O2(g)
?(PE)
CO2(g) 2H2O(g)
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Endothermic Reactions
Chapter 6 Section 1
In endothermic reactions, products have higher PE
and weaker bonds on average.
2NO(g)
?(PE)
N2(g) O2(g)
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Internal Energy (E)
Chapter 6 Section 1

• E can be changed by a flaw of work, heat, or
  both
• ?E q w

Exothermic
Endothermic
Heat is going into the system (q is ) Work is
done on the system (w is ) E is increasing (?E gt
0)
Heat is flowing out of the system (q is -) Work
is done by the system (w is -) E is decreasing
(?E lt 0)
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Internal Energy (E)
Chapter 6 Section 1

• Sample Exercise 6.1
• What is the ?E for a system undergoing a process
  in which 15.6 kJ of heat flows into it and 1.4 kJ
  of work is done on it?
• It is an endothermic process where
• q 15.6 kJ and w 1.4 kJ.
• ?E (15.6 1.4) kJ 17.0 kJ
• Joule is the fundamental SI unit for energy.
• 1 kilojoule (kJ) 1000 J

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The PV Work
Chapter 6 Section 1

• work force distance
• pressure force / area
• Thus
• work pressure area distance
• w P A ?h
• w P ?V
• w - P ?V
• This gives how much work is being done by the
  system to expand a gas ?V against a constant
  pressure P.

Work done by or on gas
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?E, h, and w Problem
Chapter 6 Section 1

• Sample Exercise 6.3
• Reading the problem on page 234, we have the
  following
• Heat q 1.3 108 J
• P 1.0 atm
• ?V Vfinal Vinitial 0.5 106 L
• We need to calculate ?E q w
• w - P?V - 0.5 106 L atm
• - 0.5 106 L atm - 5.1
  107 J
• ?E 1.3 108 J - 5.1 107 J 8 107 J

101.3 J
1 L atm
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Enthalpy (H)
Chapter 6 Section 2

• Enthalpy of a system is defined as
• H E PV
• Is H a state function? and way?
• Yes it is, because E, P and V are all state
  functions.
• For a system at constant P and its w is only PV
  work
• ?E qP w qP P?V
• Then qP ?E P?V ?H (_at_ constant P)
• At constant P where only PV work is
  allowed
• ?H qP

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Enthalpy Change (?H) for Chemical Reactions
Chapter 6 Section 2

• For a chemical reaction studied at constant P,
  the heat flow to or from the reaction is a
  measure of the change in enthalpy (?H) for that
  chemical reaction.
• Heat of reaction and change in enthalpy are
  equivalent terms when P is constant.
• ?Hrxn Hproducts Hreactants
• For Hproducts gt Hreactants
• ?Hrxn ve gt Endothermic reaction.
• For Hproducts lt Hreactants
• ?Hrxn -ve gt Endothermic reaction.

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Sample Exercise 6.4
Chapter 6 Section 2

• When 1 mole of CH4 is burned at constant P, 890
  kJ of energy is released as heat. Calculate ?H
  for a process in which a 5.8 g sample of CH4 is
  burned at constant P.

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Exercise 37 on Page 267
Chapter 6 Section 2
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Calorimetry
Chapter 6 Section 2

• Calorimetry is the science of measuring heat by
  observing the change in T when a body absorbs or
  releases energy as heat.
• Some materials need too much energy to raise
  their T 1oC. Some other materials need much less
  energy to raise their T 1oC.
• Heat Capacity, C, is a measure of that property.
• C

heat absorbed
increase in T
J
J
oC
K
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Specific and Molar Heat Capacities
Chapter 6 Section 2

• It would be more useful when C is given per unit
  mass, such as grams. (J / oC g) which is
  specific heat capacity.
• C can be also expressed in terms of moles (J / oC
  mol) which is molar heat capacity.

• Specific Heat Capacity of a substance is the
  amount for energy (normally heat) required to
  raise T of 1 gram of that substance by 1oC.
• Molar Heat Capacity of a substance is the
  amount for energy (normally heat) required to
  raise T of 1 mole of that substance by 1oC.

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Specific Heat Capacity
Chapter 6 Section 2

• Specific heat capacities for some common
  substances

It takes much less energy to raise the energy of
1g of metals than for 1g for water
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Constant Pressure Calorimeter
Chapter 6 Section 2

• P which is the atmospheric pressure remains
  constant.
• No heat goes in or out from the calorimeter.
• Surrounding medium is water.
• Example

HCl 50.0 mL 1.0 M
NaOH 50.0 mL 1.0 M

Well-insulated walls
T went up from 25.0oC to 31.9oC.
A simple Styrofoam-cup calorimeter
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Constant Pressure Calorimeter
Chapter 6 Section 2

• Net ionic equation is
• H(aq) OH-(aq) H2O(l)
• Heat required to raise T of water for 1oC is 4.18
  J/oCg.
• E released by rxn E absorbed by water
• specific heat capacity
  for water (s)
• mass of solution
  (m) increase in T (?T)
• (4.18 )
  (100.0 mL)(100 g/ml)(6.9oC)
• 2.9103 J
• ? Find ?H in kJ/mol for the above neutralization
  reaction.

J
oC g
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Constant Volume Calorimeter
Chapter 6 Section 2

• Because V is constant, P?V 0 and there will be
  no work done (w 0).
• At constant V calorimeter ?E qV
• It is also know as bomb calorimeter, where the
  sample is ignited.

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Constant Volume Calorimeter
Chapter 6 Section 2

• A combustion of 0.5269g of octane (C8H18) in a
  bomb calorimeter produces an increase of 2.25oC
  in T. Calculate ?E for the reaction if Ccal.
  11.3 kJ/mol.
• The reaction is
• C8H18(s) 25/2 O2(g) 8CO2(g)
  9H2O(g)
• E released by rxn ?T Ccal.
  (2.25oC)(11.3kJ/oC) 25.4kJ
• To get ?E (in kJ/mol) for the reaction
• 0.5269 g octane
  4.61410-3 mol octane
• For 1 mole of octane 25.4 kJ / 4.61410-3 mol
  octane
• 
  5.50103 kJ/mol
• Thus q ?E - 5.50103 kJ/mol

Energy required to raise of water and other parts
of the bomb calorimeter by 1oC
excess
1 mol octane
114.2 g octane
-
(Exothermic)
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Calorimetry
Chapter 6 Section 2
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Hesss Law
Chapter 6 Section 3

• For the reaction
• A B C D
• ?H is the same whether
• The reaction takes place in one step.
• The reaction takes place in a series of steps
  (two or more).
• Also notice that
• When A B C D has ?H x
  ,
• then C D A B has ?H
  - x
• and 2A 2B 2C 2D has ?H
  2x

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Oxidation of N2 to Produce NO2
Chapter 6 Section 3
Step 2
Step 1
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Hesss Law
Chapter 6 Section 3

• Calculate ?H for conversion from graphite to
  diamond.
• Cgraph(s) Cdiam(s)
• given that
• Cgraph(s) O2(g) CO2(g)
  ?H -394 kJ
• Cdiam(s) O2(g) CO2(g)
  ?H -396 kJ
• Cgraph(s) O2(g) CO2(g)
  ?H -394 kJ
• CO2(g) Cdiam(s) O2(g)
  ?H 396 kJ
• Cgraph(s) Cdiam(s)
  ?H 2 kJ

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Graphite vs. Diamond
Chapter 6 Section 3

• Is it easy to get diamond from graphite?
• Needs very high P and very high T which can be
  approached almost only under the earth crust.

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Standard Enthalpy of Formation (?Hfo)
Chapter 6 Section 4

• It is defined as the change in H that accompanies
  the formation of 1 mole of a compound from its
  elements with all substances in their standard
  states.
• N2(g) 2O2(g) 2NO2(g) ?H 68
  kJ
• ½ N2(g) O2(g) NO2(g) ?H
  34 kJ
• There is no method to determine the absolute H
  value. We can only calculate the change in H (?H)
  from thermodynamics experiments.
• ?Hfo of an elements or a compound in its standard
  state is zero

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Standard Enthalpy of Formation (?Hfo)
Chapter 6 Section 4
Remember that ?Hfo 0 for standard states

• What does Standard State mean?
• For a compound
• For gases P 1atm.
• For solids and liquids The pure form of solid or
  liquid.
• For solutions M 1M
• For an element
• The standard state is the form in which the
  element exists at 1 atm and 25oC.
• Standard state of oxygen is O2(g) at 1 atm
• Standard state of sodium is Na(s).
• The standard state is different from the STP
  condition.

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Example Combustion of CH4
Chapter 6 Section 4

• CH4(g) 2O2(g) CO2(g) H2O(l)
  ?Hfo ?
• The detailed procedure is to apply Hesss law by
  breaking up the reactant components into
  substances in their standard states, which would
  lead to the products.

CH4 only is not ignited
CH4 with O2 makes the reaction start
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CH4(g) 2O2(g) CO2(g) 2H2O(l)
Chapter 6 Section 4

• Rxn (a) C(s) 2H2(g) CH4(g)
  ?Hfo (a) -75 kJ
• Rxn (b) O2(g) standard state
  ?Hfo (b) 0 kJ
• Rxn (c) C(s) O2(g) CO2(g)
  ?Hfo (c) -394 kJ
• Rxn (d) H2(g) ½ O2(g) H2O(l)
  ?Hfo (d) -286 kJ

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CH4(g) 2O2(g) CO2(g) 2H2O(l)
Chapter 6 Section 4
Rxn (a) C(s) 2H2(g) CH4(g)
?Hfo (a) -75 kJ Rxn (b) O2(g)
standard state ?Hfo (b) 0 kJ
Rxn (c) C(s) O2(g) CO2(g)
?Hfo (c) -394 kJ Rxn (d) H2(g) ½
O2(g) H2O(l) ?Hfo (d) -286 kJ

• ?Horeaction ?Hfo (a) ?Hfo (b) ?Hfo
  (c) 2?Hfo (d)
• 75 kJ 0 kJ 394
  kJ 2286 kJ
• -891 kJ
• In general
• ?Horeaction ?nproducts ?Hfo (products)
  ?nreactants ?Hfo (reactants)

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Standard Enthalpy of Formation (?Hfo)
Chapter 6 Section 4
?Horeaction ?nproducts ?Hfo (products)
?nreactants ?Hfo (reactants)
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