Thermochemistry:

1
Chapter 6
Thermochemistry Energy Flow and Chemical Change
2
Thermochemistry Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy Heats of Reaction and Chemical
Change
6.3 Calorimetry Laboratory Measurement of
Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hesss Law of Heat Summation
6.6 Standard Heats of Reaction (DH0rxn)
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Thermochemistry

• thermodynamics - heat and its transformations
• thermochemistry - energy changes in chemical
  reactions
• system - stuff we are concerned with, a chemical
  reaction
• surroundings - other stuff , the container, the
  air, etc
• closed system - one in which the matter is all
  contained, energy transfer is still possible

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Internal Energy

• Internal energy - all the kinetic and potential
  energy stored in a system.
• The internal energy of a system can change when
  it loses/gains energy to/from the surroundings.
• ?E Ef - Ei
• a positive ?E indicates Ef gt Ei, the system has
  gained energy
• a negative ?E indicates Ef lt Ei, the system has
  lost energy
• ex 2H2 O2 ? 2H2O E
• the reaction produces energy and therefore lost
  energy so ?E is negative

5
Energy, heat and work

• Energy can be transferred from one object to
  another in the form of heat or work.
• work (w)- moving objects against a force
  (gravity, friction,..) w f x d
• heat (q)- energy transfer from hotter objects to
  colder ones
• energy - the capacity to do work or transfer heat
• energy of a system changes as heat is
  added/removed or work is done on/by
• ?E q w
• q, system has gained energy from surroundings
• - q, system has lost energy to surroundings
• w, work done on the system by the surroundings
• -w, work done by the system on the surroundings
• endothermic - process in which the system absorbs
  heat (q)
• exothermic - process in which the system produces
  heat (-q)

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DE q w
Figure 6.3
A system transferring energy as heat only.
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Figure 6.4
A system losing energy as work only.
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Table 6.1 The Sign Conventions for q, w and DE


DE
q
w




-
depends on sizes of q and w
-

depends on sizes of q and w
-
-
-
For q means system gains heat - means
system loses heat.
For w means word done on system - means
work done by system.
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Sample Problem 6.1
Determining the Change in Internal Energy of a
System
PLAN
Define system and surrounds, assign signs to q
and w and calculate DE. The answer should be
converted from J to kJ and then to kcal.
SOLUTION
q - 325J
w - 451J
DE q w
-325J (-451J)
-776J
0.776kJ
0.776kcal
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• 1st law of thermodynamics
• energy cannot be created or destroyed
• The total energy of the universe is constant
• ?Euniverse ?Esystem ?Esurroundings 0
• Units of Energy
• joule kg m2/s2
• energy of reaction usually expressed in kJ
• 1 cal 4.184 J
• 1 Cal 1kcal 1000 cal
• British thermal units
• 1Btu 1055J

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State Functions

• state function - a property that depends only on
  the current condition of a system, not how the
  condition was arrived at.
• ?E is a state function
• It does not depend on how the change occurs, only
  the initial and final values.
• work done is not a state function
• It does depend on how the change occurs

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Two different paths for the energy change of a
system.
Figure 6.6
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Figure 6.7
Pressure-volume work.
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Enthalpy

• Enthalpy (?H) heat transferred at constant
  pressure. qp
• ?H q in any reaction that occurs open to the
  atmosphere
• ?H much easier to measure than ?E
• sign is same as for energy, - loss, gain by
  system
• ?H ? ?E in
• Reactions that do not involve gases
• Reactions in which the number of moles of gas
  doesnt change
• Reactions in which the number of moles of gas
  does change, but q gtgtgt P?V

15
Figure 6.8
Enthalpy diagrams for exothermic and endothermic
processes.
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Sample Problem 6.2
Drawing Enthalpy Diagrams and Determining the
Sign of DH
PLAN
Determine whether heat is a reactant or a
product. As a reactant, the products are at a
higher energy and the reaction is endothermic.
The opposite is true for an exothermic reaction
SOLUTION
(a) The reaction is exothermic.
(b) The reaction is endothermic.
EXOTHERMIC
DH -285.8kJ
DH 40.7kJ
ENDOTHERMIC
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Some Important Types of Enthalpy Change
heat of combustion (DHcomb)
heat of formation (DHf)
heat of fusion (DHfus)
heat of vaporization (DHvap)
Notice all are tabulated per mole of substance
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Calorimetry

• How much energy does it take to heat up a
  battleship, a B.B., a glass of water?
• What factors are important?
• Mass
• Temperature change, ?T Tf -Ti
• I.D. of substance, steel, water, (s.h.)
• Heat gained or lost, q (s.h.)(mass)(?T)

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Specific Heat Capacity

• heat required to raise 1g of a substance up by
  1C
• different substances require different amounts of
  heat
• units J/gC or cal/gC or J/g K or cal/g K
• Note the magnitude of 1 C is equal to that of 1
  K
• substances with small specific heats absorb
  little energy when warming and give off little
  energy when cooled.
• Similar concepts
• heat capacity - amount of heat energy required to
  raise an object 1C or K
• molar heat capacity - heat capacity of a mole of
  substance

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Calorimetry

• measurement of heat flow, used to measure
  enthalpy changes experimentally.
• q (s.h.)(mass)(?T) can be rearranged to find q
  or s.h., m or T 
• specific heat of water defined as 1 cal/gC
  4.184J/g C
• can be used to measure specific heats of other
  substances and ?H of reactions

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constant pressure calorimetry

• because the change in energy of a solution can be
  calculated by q (s.h.)(mass)(?T). qrxn (and at
  constant pressure ?Hrxn) can be calculated for a
  reaction
• the energy produced or absorbed in a chemical
  reaction can be measured experimentally if one
  considers the 1st law of thermo
• in a perfectly insulated container qlost -
  qgained or qrxn - qsol
• in an imperfect container qrxn - (qsol qcal)
• qcal can be determined experimentally
• So qrxn - (s.h.sol)(mass sol)(?T sol)

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Figure 6.10
Coffee-cup calorimeter
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Sample Problem 6.3
Calculating the Quantity of Heat from the
Specific Heat Capacity
PLAN
Given the mass, specific heat capacity and change
in temperature, we can use q c x mass x DT to
find the answer. DT in 0C is the same as for K.
SOLUTION
1.33x104J
q
125g
(300-25)0C
x
x
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Sample Problem 6.4
Determining the Specific Heat Capacity of a Solid
PLAN
It is helpful to use a table to summarize the
data given. Then work the problem realizing that
heat lost by the system must be equal to that
gained by the surroundings.
SOLUTION
? x
25.64g x
-71.51K

-
csolid
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Table 6.4 Specific Heat Capacities of Some
Elements, Compounds, and Materials
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Constant V or bomb calorimetry

• combustible compounds are burned in a sealed
  container, energy given off is measured and used
  to calculate ?E not ?H because pressure not
  constant.
• the heat capacity of the calorimeter must be
  previously calculated using a standard
• hc is in energy/degree
• ex 50kJ produced in standard reaction raise
  calorimeter by 10 C
• qrxn - hccal ?T

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Figure 6.11
A bomb calorimeter
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Sample Problem 6.5
Calculating the Heat of Combustion
PLAN
- q sample qcalorimeter
SOLUTION
qcalorimeter
heat capacity x DT
8.151kJ/K x 4.937K
40.24kJ
9.62kcal or Calories
The manufacturers claim is true.
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Heat or Enthalpy of Reaction

• ?Hrxn
• The energy absorbed or released in a chemical
  change is due to differences between the
  strengths of reactant bonds and product bonds.
• Reactants absorb energy to break bonds, energy is
  released as the products form new bonds
• Exothermic rx product bonds stronger
• Endothermic rx product bonds weaker
• Reactive bonds are less stable, weaker, and
  higher in energy than stronger, more stable bonds.

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Enthalpy of Reaction

• Since ?X Xf - Xi,
• it follows that for a reaction, ?H H(products)
  - H(reactants)
• this is called enthalpy of Rx or ?HRx
• thermochemical equations show a rx with the
  associated ?H
• 2H2(g) O2(g) ? H2O(g) ?H -483.6 kJ
• rx is exothermic (it lost)

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Enthalpy of Reaction

• 2H2(g) O2(g) ? 2H2O(g) ?Hrxn -483.6 kJ
• rx is exothermic (it lost)
• guidelines
• 1. magnitude of ?H is proportional to amount of
  reactant used/product produced
• Energy to form 4 moles water 2x(-483.6 kJ)
• 2. reversing a reaction, change in enthalpy
  magnitude is the same but sign is opposite
• 2H2O(g) ? 2H2(g) O2(g) ?Hrxn 483.6 kJ
• 3. enthalpy change depends on states of products
• 2H2(g) O2(g) ? 2H2O(l) ?Hrxn -572 kJ
• 2H2O(g) ? 2H2O(l) ?Hrxn ?

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Sample Problem 6.6
Using the Heat of Reaction (? Hrxn) to Find
Amounts
SOLUTION
PLAN
heat(kJ)
1.000x103kJ x
1676kJ2mol Al
mol of Al
32.20g Al
x M
g of Al
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Hesss Law

• A reaction enthalpy is the sum of the enthalpies
  of any sequence of reactions into which the
  overall reaction can be divided.
• (because enthalpy is a state function)
• ex. H2O(g) ? H2O(l) ?Hrxn ?
• 2 H2(g) O2(g) ? 2 H2O(g) ?Hrxn - 483.6 kJ
• 2 H2(g) O2(g) ? 2 H2O(l) ?Hrxn - 572 kJ
• So
• ½( 2 H2(g) O2(g) ? 2 H2O(l)) ?Hrxn
  - 286 kJ
• ½( 2 H2O(g) ? 2 H2(g) O2(g)) ?Hrxn
  241.8 kJ
• H2O(g) ? H2O(l) ?Hrxn
  - 44.2 kJ

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Sample Problem 6.7
Using Hesss Law to Calculate an Unknown DH
PLAN
Equations A and B have to be manipulated by
reversal and/or multiplication by factors in
order to sum to the first, or target, equation.
SOLUTION
Multiply Equation B by 1/2 and reverse it.
DHB -90.6kJ
DHrxn -373.6kJ
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Standard Enthalpies of Formation

• ?Hf - the enthalpy change of forming a compound
  from it's elements
• since magnitude of ?H depends upon T, P, and
  state, a standard must be defined.
• standard state - the form of the substance at 1
  atm. and 298K
• standard enthalpy - ?H - enthalpy of a reaction
  when everything is in standard state
• ?Hf - is ?H for the reaction that forms 1 mole
  of the cmpd from its elements with all substances
  in their standard states
• most stable forms of elements used in formation
  reaction ex O2 not O3
• The standard enthalpy of formation of the most
  stable form of the elements is zero.
• ex 2C(graphite) 3H2(g) 1/2O2(g) ? C2H5OH(l)
  ?Hf -277.7kJ

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Sample Problem 6.8
Writing Formation Equations
(a) Silver chloride, AgCl, a solid at standard
conditions.
(b) Calcium carbonate, CaCO3, a solid at standard
conditions.
(c) Hydrogen cyanide, HCN, a gas at standard
conditions.
PLAN
Use the table of heats of formation for values.
SOLUTION
DH0f -127.0kJ
DH0f -1206.9kJ
DH0f 125kJ
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Standard Enthalpies of Formation

• ?Hf can be used to find ?Hrx
•  
• ?Hrxn ?n?Hf(products) - ?m?Hf(reactants)
•  

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Table 6.5 Selected Standard Heats of Formation
at 250C(298K)
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Sample Problem 6.9
Calculating the Heat of Reaction from Heats of
Formation
PLAN
Look up the DH0f values and use Hesss Law to
find DHrxn.
SOLUTION
DHrxn S mDH0f (products) - S nDH0f (reactants)
DHrxn 4(DH0f NO(g) 6(DH0f H2O(g)
- 4(DH0f NH3(g) 5(DH0f O2(g)
(4mol)(90.3kJ/mol) (6mol)(-241.8kJ/mol) -
(4mol)(-45.9kJ/mol) (5mol)(0kJ/mol)
DHrxn -906kJ
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Figure B6.1
The trapping of heat by the atmosphere
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Figure B6.2
The accumulating evidence for the greenhouse
effect.
A Atmospheric CO2 concentrations
B Average global temperatures
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Table 6.2 Heats of Combustion(DHcomb) of Some
Carbon Compounds
Two-Carbon Compounds
One-Carbon Compounds
Ethane (C2H6)
Ethanol (C2H6O)
Methane (CH4)
Methanol (CH4O)
Structural Formulas
Sum of C-C and C-H Bonds
7
6
4
3
Sum of C-O and O-H Bonds
0
2
0
2
DHcomb(kJ/mol)
-1560
-1367
-890
-727
DHcomb(kJ/g)
-51.88
-29.67
-55.5
-22.7
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Table 6.3 Heats of Combustion of Some Fats and
Carbohydrates
Fats
Carbohydrates