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33
Chemical ReactionsChemical Reactions Chemical reactions involve not just
the conversion of reactants into products, but also involve an energy change in the form of heat—heat released as the result of a reaction, or heat absorbed as a reaction proceeds.
Energy changes accompany all chemical reactions and are due to rearranging of chemical bonding.
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Making BondsMaking Bonds Addition of energy is always a
requirement for the breaking of bonds but the breaking of bonds in and of itself does not release energy.
Energy release occurs when new bonds are formed.
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Bond EnergyBond Energy If more energy is released when new
bonds form than was required to break existing bonds, then the difference will result in an overall release of energy.
If, on the other hand, more energy is required to break existing bonds than is released when new bonds form, the difference will result overall in energy being absorbed.
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Overall Reaction EnergyOverall Reaction Energy Whether or not an overall reaction
releases or requires energy depends upon the final balance between the breaking and forming of chemical bonds.
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Energy is...Energy is... the ability to do work or produce the ability to do work or produce
heat.heat. conserved.conserved. made of heat and work.made of heat and work. a state function. a state function. ((Energy is a property that
is determined by specifying the condition or “state” (e.g., temperature, pressure, etc.) of a system or substance.)
independent of the path, or how you independent of the path, or how you get from point A to B.get from point A to B.
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EnergyEnergy While the total internal energy of a
system (E) cannot be determined, changes in internal energy (E) can be determined.
The change in internal energy will be the amount of energy exchanged between a system and its surroundings during a physical or chemical change.
Δ E = E final - E initial
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DefinitionsDefinitions
Work is a force acting over a Work is a force acting over a distance.distance.
Heat is energy transferred Heat is energy transferred between objects because of between objects because of temperature difference. temperature difference. ((Heat is not a property of a system or substance and is not a state function. Heat is a process—the transfer of energy from a warm to a cold object.)
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System vs SurroundingsSystem vs Surroundings The Universe is divided into two The Universe is divided into two
halves.halves.• system and the surroundings.
In a chemistry setting, a system includes all substances undergoing a physical or chemical change.
The surroundings would include everything else that is not part of the system.
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HeatHeat
Most commonly, energy is exchanged between a system and its surroundings in the form of heat.
Heat will be transferred between objects at different temperatures.
Thermochemistry is the study of thermal energy changes.
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Exo vs EndoExo vs Endo
Exothermic reactions release Exothermic reactions release energy to the surroundings.energy to the surroundings.
Endothermic reactions absorb Endothermic reactions absorb energy from the surroundings.energy from the surroundings.
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Three PartsThree PartsEvery energy measurement has Every energy measurement has three parts:three parts:
1.1. A unit ( Joules of calories).A unit ( Joules of calories).
2.2. A number.A number.
3.3. a sign to tell direction.a sign to tell direction.negative - exothermic
positive- endothermicpositive- endothermic
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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 Work done by the systemby the system on the on the surroundings is surroundings is negativenegative..
Work done Work done on the systemon the system by the by the surroundings is surroundings is positivepositive..
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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. It is also called the: It is also called the: Law of conservation of energy.Law of conservation of energy.
q = heatq = heat w = workw = work In a chemical system, the energy exchanged
between a system and its surroundings can beaccounted for by heat (q) and work (ww).
E = q + wE = q + w Take the system’s point of viewTake the system’s point of view to decide to decide
signs.signs.
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Conservation of EnergyConservation of Energy
Energy exchanged between a system and its surroundings can be considered to off set one another.
The same amount of energy leaving a system will enter the surroundings (or vice versa), so the total amount of energy remains constant.
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Heat and WorkHeat and Work E = q + wE = q + w - q is exothermic - q is exothermic -q = -∆H-q = -∆H +q is endothermic+q is endothermic -w is done “by” the system-w is done “by” the system +w is done “on” the system+w is done “on” the system
Note: Note: ∆∆H stands for H stands for enthalpyenthalpy which is the which is the
heat of reactionheat of reaction
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Practice ProblemPractice Problem A gas absorbs 28.5 J of heat and
then performs 15.2 J of work. The change in internal
energy of the gas is:(a) 13.3 J(b) - 13.3 J(c) 43.7 J(d) - 43.7 J(e) none of the above
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Practice ProblemPractice ProblemWhich of the following statements correctly describes the signs of q and w for the following exothermic process at 1 atmosphere pressure and 370 Kelvin?
H2O(g) → H2O(l)
(a) q and w are both negative(b) q is positive and w is negative(c) q is negative and w is positive(d) q and w are both positive(e) q and w are both zero
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AnswerAnswer
(c). An exothermic indicates q is negative and the gas is condensing to a liquid so it is exerting less pressure on its surroundings indicating w is positive.
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What is work?What is work? Work is a force acting over a distance.Work is a force acting over a distance.
w= F x w= F x ddP = F/ areaP = F/ aread = V/aread = V/areaw= (P x area) x w= (P x area) x (V/area)= P (V/area)= PVV
Work can be calculated by multiplying Work can be calculated by multiplying pressure by the change in volume at pressure by the change in volume at constant pressure.constant pressure.
Use units of liter•atm or L•atmUse units of liter•atm or L•atm
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Pressure and Volume WorkPressure and Volume Work
Work refers to a force that moves an object over a distance.
Only pressure/volume work (i.e., the expansion/contraction of a gas) is of significance in chemical systems and only when there is an increase or decrease in the amount of gas present.
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Work needs a signWork needs a sign If the volume of a gas increases, If the volume of a gas increases,
the system has done work on the the system has done work on the surroundings.surroundings.
work is negativework is negative w = - Pw = - PVV Expanding work is negative.Expanding work is negative. Contracting, surroundings do work Contracting, surroundings do work
on the system W is positive.on the system W is positive. 1 L•atm = 101.3 J1 L•atm = 101.3 J
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Example When, in a chemical reaction,
there are more moles of product gas compared toreactant gas, the system can be thought of as performing work on its surroundings (making w < 0) because it is “pushing back,” or moving back the atmosphere to make room for the expanding gas. When the reverse is true, w > 0.
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Compressing and Expanding GasesCompressing and Expanding Gases
Compressing gasCompressing gas– Work on the system is positive– Work is going into the system
Expanding gasExpanding gas– Work on the surroundings is negative– Work is leaving the system
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Clarification InfoClarification Info If the reaction is performed in a rigid
container, there may be a change in pressure,but if there is no change in volume, the atmosphere outside the container didn’t “move” and without movement, no work is done by or on the system.
If there is no change in volume (V= 0), then no work is done by or on the system (w= 0) and the change in internal energy will be entirely be due to the heat involved ( ΔE = q).
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ExamplesExamples What amount of work is done when What amount of work is done when
15 L of gas is expanded to 25 L at 15 L of gas is expanded to 25 L at 2.4 atm pressure?2.4 atm pressure?
If 2.36 J of heat are absorbed by the If 2.36 J of heat are absorbed by the gas above. what is the change in gas above. what is the change in energy?energy?
How much heat would it take to How much heat would it take to change the gas without changing change the gas without changing the internal energy of the gas? the internal energy of the gas?
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EnthalpyEnthalpy The symbol for Enthalpy is HThe symbol for Enthalpy is H H = E + PV (that’s the definition)H = E + PV (that’s the definition) at constant pressure.at 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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H = H = E + PE + PVV
Using Using H = H = E + PE + PVV
the heat at constant pressure qthe heat at constant pressure qpp can can
be calculated from:be calculated from:
E = qE = qpp + w (if w = - P + w (if w = - PV then…) V then…)
E = qE = qpp – P – PV (now rearrange)V (now rearrange)
qqpp = = E + P E + P V = V = HH
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Examples of Enthapy Changes
KOH(s) → K(aq) + OH-1 (aq) ΔHsolution = - 57.8 kJ mol1
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔHcombustion = -2221kJ/mol
H2O(s) → H2O(l) ΔHfusion = 6.0 kJ/mol
Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s) ΔHreaction - 852 kJ/mol
Ca(s) + O2(g) H2(g) → Ca(OH)2(s) ΔHformation - 986 kJ/mol1
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3 Methods3 Methods There are a variety of methods
for calculating overall enthalpy changes that you should be familiar with.
The three most common are the:
1. the use of Heats of Formation 2. Hess’s Law 3. the use of Bond Energies
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Heat of ReactionHeat of Reaction To compare heats of reaction for
different reactions, it is necessary to know the temperatures at which heats of reaction are measured and the physical states of the reactants and products.
Look in the Appendix of the textbook to find Standard Enthapy tables.
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Standard Enthalpy of FormationStandard Enthalpy of Formation Measurements have been made and tables
constructed of Standard Enthalpies of Formation with reactants in their “standard states”.
Use the symbol Use the symbol HºHºff
Standard state is the most stable physical state of reactants at:
1 atmosphere pressurespecified temperature—usually 25 °C 1 M solutions
For solids which exist in more than one allotropic form, a specific allotrope must be specified.
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ΔH°formation
It is important to recognize that the ΔH°formation (abbreviated as ΔH°f) is really just the heat of reaction for a chemical change involving the formation of a compound from its elements intheir standard states.
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Standard Enthalpies of FormationStandard Enthalpies of Formation The standard heat of formation is The standard heat of formation is
the amount of heat needed to form the amount of heat needed to form 1 mole of a compound from its 1 mole of a compound from its elements in their standard states.elements in their standard states.
See the table in the Appendix See the table in the Appendix Remember: For an element the Remember: For an element the
value is 0value is 0
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Equation PracticeEquation Practice You need to be able to write the You need to be able to write the
equation correctly before solving equation correctly before solving the problem.the problem.
Try…Try… What is the equation for the What is the equation for the
formation of NOformation of NO22 ? Try writing the ? Try writing the
equation.equation.
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Practice AnswerPractice Answer
½N½N2 2 (g) + O(g) + O22 (g) (g) NO NO22 (g) (g)
You must make You must make one mole one mole to meet to meet the definition.the definition.
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Since we can manipulate the equationsSince we can manipulate the 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.
( H products) - ( H reactants) = Hfo
fo o
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Hess’s LawHess’s Law Definition:Definition: When a reaction may be expressed as
the algebraic sum of other reactions, the enthalpy change of the reaction is the algebraic sum of the enthalpy changes for the combined reactions.
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Hess’ LawHess’ Law Enthalpy is a state function.Enthalpy is a state function. It is independent of the path.It is independent of 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 add the add the H values.H values.
Two rules to remember:Two rules to remember: If the reaction is reversed the sign If the reaction is reversed the sign
of of H is changedH is changed If the reaction is multiplied, so is If the reaction is multiplied, so is HH
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EnthalpyEnthalpy As enthalpy is an extensive property,
the magnitude of an enthalpy change for a chemical reaction depends upon the quantity of material that reacts.
This means: if the amount of reacting
material in an exothermic reaction is doubled, twice the quantity of heat energy will be released.
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For the oxidation of sulfur dioxide gas
SO2(g) + ½O2(g) → SO3(g) ΔH° = - 99 kJ/mol
Doubling the reaction results in: 2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
Notice that if you double the reaction, you must double the ΔH° value.
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Sign Change of Sign Change of ΔH 2SO2(g) + O2(g) → 2SO3(g)
ΔH° = - 198 kJ/mol
If the reaction is written as an endothermic reaction:
2SO3(g) → 2SO2(g) + O2(g)
ΔH° = + 198 kJ/mol
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Tips for Hess’s Law ProblemsTips for Hess’s Law Problems It is always a good idea to begin by looking
for species that appear as reactants and products in the overall reaction.
This will provide a clue as to whether a reaction needs to be reversed or not.
Second, consider the coefficients of species that appear in the overall reaction.
This will help determine whether a reaction needs to be multiplied before the overall summation.
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ExampleExample
C(s) + O2(g) → CO2(g) ΔH°f = - 394 kJ/mol
2H2(g) + O2(g) → 2H2O(l) ΔH°f = - 572 kJ/mol
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ΔH°f = + 891 kJ/mol
-----------------------------------------------------
C(s) + 2H2(g) → CH4(g) ΔH°f = - 75 kJ/mol
5252
Hess’s Law ExampleHess’s Law Example
O (g) + H (g) 2OH(g) 2 2 O (g) 2O(g)2 H (g) 2H(g)2
O(g) + H(g) OH(g)
Given
Calculate Hº for this reaction
Hº= +77.9kJHº= +495 kJ
Hº= +435.9kJ
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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
ExampleExample GivenGiven
calculate calculate Hº for this reactionHº for this reaction
Hº= -1300. kJ
2C(s) + H (g) C H (g) 2 2 2
5656
CalorimetryCalorimetry Calorimetry is the study of the heat
released or absorbed during physical and chemical reactions.
For a certain object, the amount of heat energy lost or gained is proportional to
the temperature change. The initial temperature and the final temperature in the calorimeter
are measured and the temperature difference is used to calculate the heat of reaction.
5757
Equipment: CalorimeterEquipment: Calorimeter There are two kinds of There are two kinds of
calorimeters: calorimeters: – constant pressure
– bomb
5858
CalorimetryCalorimetry Constant pressure calorimeter Constant pressure calorimeter
(called a coffee cup (called a coffee cup calorimeter)calorimeter)
A coffee cup calorimeter measures A coffee cup calorimeter measures H.H.
The calorimeter can be an The calorimeter can be an insulated cup, full of water. insulated cup, full of water.
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Definitions: Heat capacity is the amount of energy
required to raise the temperature of an object 1 kelvin or 1 °C.
Specific heat capacity is the heat capacity of 1 gram of a substance.
Molar heat capacity is the heat capacity of1 mole of a substance.
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Heat CapacityHeat Capacity Heat capacity is an extensive property,
meaning it depends on the amount present—a large amount of a substance would require more heat to raise the temperature 1 K than a small amount of the same substance.
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Specific Heat CapacitySpecific Heat Capacity
Definition: The specific heat capacity of each
substance is an intensive property which relates the heat capacity to the mass of the substance.
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Specific Heat Capacity “c”Specific Heat Capacity “c” q = mass x c x q = mass x c x TT Or written as Or written as
q = mcq = mcTT
This is the main equation for This is the main equation for calorimetry calculationscalorimetry calculations
mass will be in gramsmass will be in grams Units for “c” are J/g K or J/g °CUnits for “c” are J/g K or J/g °C The specific heat of water is 1 cal/g ºCThe specific heat of water is 1 cal/g ºC
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Molar Heat CapacityMolar Heat Capacity Molar Heat Capacity is the heat capacity Molar Heat Capacity is the heat capacity
of 1 mole of a substance.of 1 mole of a substance.
molar heat capacity = c/molesmolar heat capacity = c/moles
heat = molar heat x moles x heat = molar heat x moles x TT
Remember that heat is shown as ∆HRemember that heat is shown as ∆H Make the units work and you’ve done Make the units work and you’ve done
the problem right.the problem right.
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Sign of qSign of q• If a process results in the sample losing heat energy, the loss in heat is designated as q is negative.The temperature of the surroundings will increase during this exothermic process.
• If the sample gains heat during the process, then q is positive. The temperature of thesurroundings will decrease during an endothermic process.
The amount of heat that an object gains or loses is directly proportional to the change intemperature.
6565
ExamplesExamples The specific heat of graphite is The specific heat of graphite is
0.71 J/gºC. 0.71 J/gºC. Calculate the energy needed to 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.
6666
Extra ProblemExtra Problem A 46.2 g sample of copper is A 46.2 g sample of copper is
heated to 95.4ºC and then placed heated to 95.4ºC and then placed in a calorimeter containing 75.0 g in a calorimeter containing 75.0 g of water at 19.6ºC. The final of water at 19.6ºC. The final temperature of both the water and temperature of both the water and the copper is 21.8ºC. the copper is 21.8ºC.
What is the specific heat of What is the specific heat of copper?copper?
6767
CalorimetryCalorimetry Constant volume calorimeter is Constant volume calorimeter 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
6868
Bomb CalorimeterBomb Calorimeter thermometerthermometer
stirrerstirrer
full of waterfull of water
ignition wireignition wire
Steel bombSteel bomb
samplesample
6969
PropertiesProperties intensive properties are not related intensive properties are not related
to the amount of substance.to the amount of substance. Examples: density, specific heat, Examples: density, specific heat,
temperature.temperature.
Extensive property - does depend Extensive property - does depend on the amount of substance.on the amount of substance.
Examples: Heat capacity, mass, Examples: Heat capacity, mass, heat from a reaction.heat from a reaction.
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Collegeboard. (2007-2008). Professional Development workshop Collegeboard. (2007-2008). Professional Development workshop materials: Special focus thermochemistry. materials: Special focus thermochemistry. http://apcentral.collegeboard.com/apc/public/repository/5886-http://apcentral.collegeboard.com/apc/public/repository/5886-3_Chemistry_pp.ii-88.pdf3_Chemistry_pp.ii-88.pdf