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Chapter 6Chapter 6
EnergyEnergy
ThermodynamicsThermodynamics
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Energy is...Energy is... ……the ability to do work or produce heat.the ability to do work or produce heat. ……conserved.conserved. ……defined as kinetic or potential.defined as kinetic or potential. ……a state function (independent of the a state function (independent of the
path, or how you get from point A to path, or how you get from point A to B...depends only on its B...depends only on its present statepresent state).).
Work is a force acting over a distance.Work is a force acting over a distance. Heat is energy Heat is energy transferred transferred between between
objects because of temperature objects because of temperature difference.difference.
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The universe…The universe… ……is divided into two halves.is divided into two halves. ……the the systemsystem and the and the
surroundingssurroundings.. The system is the part you are The system is the part you are
concerned with (as a chemist).concerned with (as a chemist). The surroundings are the rest.The surroundings are the rest. Exothermic reactions Exothermic reactions releaserelease energy energy
to the surroundings.to the surroundings. Endothermic reactions Endothermic reactions absorbabsorb
energy from the surroundings.energy from the surroundings.
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CH + 2O CO + 2H O + Heat4 2 2 2
CH + 2O 4 2
CO + 2 H O 2 2
Pote
nti
al en
erg
y
Heat
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N + O2 2
Pote
nti
al en
erg
y
Heat
2NO
N + O 2NO2 2 + heat
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DirectionDirection Every energy measurement has Every energy measurement has
three parts:three parts:
1.1. A unit ( Joules, kJ or Calories).A unit ( Joules, kJ or Calories).
2.2. A number.A number.
3.3. A sign to tell direction of energy A sign to tell direction of energy flow.flow.
negative - negative - exothermicexothermic Positive - Positive - endothermicendothermic
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System
Surroundings
Energy
E <0Energy is leavingthe system!
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System
Surroundings
Energy
E >0Energy is enteringthe system!
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Same rules for heat and workSame rules for heat and work Heat given off is negative.Heat given off is negative. Heat absorbed is positive.Heat absorbed is positive. Work done by system on Work done by system on
surroundings is negative.surroundings is negative. Work done on system by Work done on system by
surroundings is positive.surroundings is positive. Thermodynamics- The study of Thermodynamics- The study of
energy and the changes it energy and the changes it undergoes.undergoes.
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First Law of ThermodynamicsFirst Law of Thermodynamics The energy of the universe is constant.The energy of the universe is constant. Law of conservation of energy.Law of conservation of energy. qq = heat and = heat and ww = work = work E = q + w (Internal Energy = E = q + w (Internal Energy = E)E) E is the sum of the KE and PE of all E is the sum of the KE and PE of all
particles in a system…which can be particles in a system…which can be changed by a flow of work and/or heat!changed by a flow of work and/or heat!
Take the systems point of view to Take the systems point of view to decide signs.decide signs.
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What is work?What is work? Work is a force acting over a Work is a force acting over a
distance.distance. w= F x w= F x dd P = F/ areaP = F/ area
Typically, we will let the physicist Typically, we will let the physicist worry about the work!worry about the work!
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Work needs a sign…Work needs a sign… If the volume of a gas increases, the If the volume of a gas increases, the
system has done work on the system has done work on the surroundings (it pushed against the surroundings (it pushed against the surroundings).surroundings).
work is negative (or lost to the surr.)work is negative (or lost to the surr.) w = - Pw = - PVV Expanding work is negative.Expanding work is negative. Contracting, surroundings do work on Contracting, surroundings do work on
the system…w is positive.the system…w is positive.
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Enthalpy (for nerds)Enthalpy (for nerds) abbreviated Habbreviated H H = E + PV (that’s the definition)H = E + PV (that’s the definition) Calculated at a constant pressure.Calculated at a constant pressure. H = H = E + PE + PVV
the heat at constant pressure qthe heat at constant pressure qpp can can
be calculated frombe calculated from
E = qE = qpp + w = q + w = qpp - P - PVV
qqpp = = E + P E + P V = V = HH
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Enthalpy for ChemistsEnthalpy for Chemists Since qSince qpp = = E + P E + P V = V = H…then we H…then we
can say qcan say qpp = = HH
In a “nutshell”: at constant pressure, In a “nutshell”: at constant pressure, the change in enthalpy of the system the change in enthalpy of the system is equal to the energy flow as heat.is equal to the energy flow as heat.
Practical applications: the change in Practical applications: the change in enthalpy is given by the equation enthalpy is given by the equation
H = H H = H productsproducts - H - H reactantsreactants
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CalorimetryCalorimetry Science of measuring heat.Science of measuring heat. Use a calorimeter: Two types…Use a calorimeter: Two types… #1: #1: Constant pressure calorimeter Constant pressure calorimeter
(called a coffee cup calorimeter)(called a coffee cup calorimeter) heat capacity for a material, C is heat capacity for a material, C is
calculated calculated C= heat absorbed/C= heat absorbed/T = T = H/H/TT specific heat capacity = C/mass specific heat capacity = C/mass
(units J/ºC•g or J/K•g)(units J/ºC•g or J/K•g)
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CalorimetryCalorimetry molar heat capacity = C/molesmolar heat capacity = C/moles heat = specific heat x m x heat = specific heat x m x TT heat = molar heat x moles x heat = molar heat x moles x TT Make the units work and you’ve done Make the units work and you’ve done
the problem right.the problem right. A coffee cup calorimeter measures A coffee cup calorimeter measures H.H. An insulated cup, full of water. An insulated cup, full of water. The specific heat of water is 1 cal/gºCThe specific heat of water is 1 cal/gºC Heat of reaction= Heat of reaction= H = Cp x mass x H = Cp x mass x
TT
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ExamplesExamples The specific heat of graphite is 0.71 The specific heat of graphite is 0.71
J/gºC. Calculate the energy needed to J/gºC. Calculate the energy needed to raise the temperature of 75 kg of raise the temperature of 75 kg of graphite from 294 K to 348 K.graphite from 294 K to 348 K.
A 46.2 g sample of copper is heated to A 46.2 g sample of copper is heated to 95.4ºC and then placed in a 95.4ºC and then placed in a calorimeter containing 75.0 g of water calorimeter containing 75.0 g of water at 19.6ºC. The final temperature of at 19.6ºC. The final temperature of both the water and the copper is both the water and the copper is 21.8ºC. What is the specific heat of 21.8ºC. What is the specific heat of copper?copper?
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CalorimetryCalorimetry #2 #2 Constant volume calorimeter Constant volume calorimeter is is
called a bomb calorimeter.called a bomb calorimeter. Material is put in a container with Material is put in a container with
pure oxygen. Wires are used to pure oxygen. Wires are used to start the combustion. The container start the combustion. The container is put into a container of water.is put into a container of water.
The heat capacity of the The heat capacity of the calorimeter is known and tested.calorimeter is known and tested.
Since Since V = 0, PV = 0, PV = 0, V = 0, E = q E = q
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Bomb CalorimeterBomb Calorimeter thermometerthermometer
stirrerstirrer
full of waterfull of water
ignition wireignition wire
Steel bombSteel bomb
samplesample
2020
PropertiesProperties Intensive property - Intensive property - not related not related
to the to the amountamount of substance. of substance. density, specific heat, density, specific heat,
temperature.temperature. Extensive property Extensive property - does - does
depend on the amount of depend on the amount of substance.substance.
Heat capacity, mass, heat from a Heat capacity, mass, heat from a reaction.reaction.
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Hess’s LawHess’s Law Enthalpy is a state function Enthalpy is a state function (independent of (independent of
the path).the path).
We can “add equations” to come up We can “add equations” to come up with the desired final product, and with the desired final product, and then add the then add the H values to arrive at the H values to arrive at the final enthalpy change!final enthalpy change!
Two rulesTwo rules
#1 - If the reaction is reversed the sign #1 - If the reaction is reversed the sign of of H is changedH is changed
#2 - If the reaction is multiplied, so is #2 - If the reaction is multiplied, so is HH
2222
State FunctionState Function One two-step pathway…One two-step pathway…
NN22 + O + O22 2NO 2NOΔΔH = 180 kJH = 180 kJ
2NO + O2NO + O22 2NO2NO22 Δ ΔH = -112 kJH = -112 kJ
A different single-step pathway…A different single-step pathway…
NN22 + 2O + 2O22 2NO2NO22 Δ ΔH = 68 kJH = 68 kJ
Regardless of the pathway…68 kJ are Regardless of the pathway…68 kJ are absorbed in this endothermic reaction!absorbed in this endothermic reaction!
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N2 +2O2
O2+2NO
68 kJ
2NO2180 kJ
-112 kJ
H (
kJ)
N2 +2O2
Two-step
One-step
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Standard EnthalpyStandard Enthalpy The enthalpy change for a reaction The enthalpy change for a reaction
under standard conditions (25ºC, 1 under standard conditions (25ºC, 1 atm, 1 M solutions)atm, 1 M solutions)
Symbol Symbol HºHº When using Hess’s Law, work by When using Hess’s Law, work by
adding the equations up to obtain adding the equations up to obtain the the ““target equation”target equation”. .
The other species will cancel out…The other species will cancel out…and you will sum the enthalpy and you will sum the enthalpy values!values!
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Procedure for solving…Procedure for solving… Goal: Establish a pathway to the Goal: Establish a pathway to the
Target Equation by rearranging the Target Equation by rearranging the “given equations”“given equations”
Hints:Hints:– Work backwards from the target equation– Look for unique formulas (which only appear
once)– Avoid formulas which appear multiple times
until you get started
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H (g) + 1
2O (g) H (l) 2 2 2O
C(s) + O (g) CO (g) 2 2Hº= -394 kJ
Hº= -286 kJ
C H (g) + 5
2O (g) 2CO (g) + H O( ) 2 2 2 2 2 l
Hess’s Law ExampleHess’s Law Example GivenGiven
calculate calculate Hº for this reaction Hº for this reaction (target (target equation)equation)
Hº= -1300. kJ
2C(s) + H (g) C H (g) 2 2 2
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Standard Enthalpies of FormationStandard Enthalpies of Formation Hess’s Law is useful when you are Hess’s Law is useful when you are
given multiple sets of reactions….given multiple sets of reactions…. However, the standard enthalpy of However, the standard enthalpy of
formation (formation (HHffº) for a compound is º) for a compound is defined as defined as – the change in enthalpy that accompanies the
formation of one mole of a compound from its elements in their standard states.
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Standard Enthalpies of FormationStandard Enthalpies of Formation Standard states are 1 atm, 1M and Standard states are 1 atm, 1M and
25ºC25ºC See p. 246 of our text.See p. 246 of our text. For an element it is 0 (zero)!For an element it is 0 (zero)!
– Why? It comes to us already formed! There is a table of values listed in There is a table of values listed in
Appendix 4 (pg A19-22).Appendix 4 (pg A19-22).
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Standard Enthalpies of FormationStandard Enthalpies of Formation First, you need to be able to write the First, you need to be able to write the
balanced equation from the elements.balanced equation from the elements. What is the equation for the formation What is the equation for the formation
of NOof NO22 ? ?
½N½N2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) (g)
Have to make Have to make one mole one mole to meet the to meet the definition.definition.
Write the equation for the formation of Write the equation for the formation of methanol CHmethanol CH33OH.OH.
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Since we can manipulate the Since we can manipulate the equations…equations…
We can use heats of formation to We can use heats of formation to figure out the heat of reaction.figure out the heat of reaction.
Lets do it with this equation.Lets do it with this equation. CC22HH55OH +3OOH +3O22(g) (g) 2CO 2CO22 + 3H + 3H22OO
which leads us to this rule:which leads us to this rule:
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Since we can manipulate the Since we can manipulate the equations…equations…
We can use heats of formation to We can use heats of formation to figure out the heat of reaction.figure out the heat of reaction.
Lets do it with this equation.Lets do it with this equation. CC22HH55OH +3OOH +3O22(g) (g) 2CO 2CO22 + 3H + 3H22OO
which leads us to this rule.which leads us to this rule.
reactants)H(-products)H(H of
of
o
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Solve the problem:Solve the problem:CC22HH55OH(l) +3OOH(l) +3O22(g) (g) 2CO 2CO22(g) + 3H(g) + 3H22O(l)O(l)
ΔΔHf:Hf: -278 -278 0 0 -393.5 -393.5 -286-286(kJ/mol)(kJ/mol)
ΔΔH = [(2•-393.5kJ)+(3•-286kJ)]-[-278kJ + H = [(2•-393.5kJ)+(3•-286kJ)]-[-278kJ + 3•0kJ]3•0kJ]
ΔΔH = -1367kJ of energy (released) per H = -1367kJ of energy (released) per mole of ethanol burned.mole of ethanol burned.
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Chapter 6 problems:Chapter 6 problems: 59. 60. 62. 66. 69. 71. 77. 8459. 60. 62. 66. 69. 71. 77. 84