Barnegat High SchoolBarnegat High SchoolAP Chemistry Chapter 5AP Chemistry Chapter 5

ThermodynamicsThermodynamics

SystemSystem SurroundingsSurroundings State property – depend only State property – depend only

on the state of the system, on the state of the system, not on the way the system not on the way the system reached the statereached the state

System and Surroundings

• The system includes the molecules we want to study (here, the hydrogen and oxygen molecules).

• The surroundings are everything else (here, the cylinder and piston).

volume – state propertyvolume – state property heat flow – not a state heat flow – not a state

property – value depends on property – value depends on how the system reached that how the system reached that state.state.

The universeThe universe

is divided into two halves.is divided into two halves. the system and the surroundings.the system and the surroundings. The system is the part you are concerned The system is the part you are concerned

with.with. The surroundings are the rest.The surroundings are the rest. Exothermic reactions release energy to the Exothermic reactions release energy to the

surroundings.surroundings. Endothermic reactions absorb energy from Endothermic reactions absorb energy from

the surroundings.the surroundings.

EnergyEnergy

The ability to do work or transfer The ability to do work or transfer heat.heat. Work: Energy used to cause an object Work: Energy used to cause an object

that has mass to move.that has mass to move. Heat: Energy used to cause the Heat: Energy used to cause the

temperature of an object to rise.temperature of an object to rise.

Potential EnergyPotential Energy

Energy an object possesses by virtue of Energy an object possesses by virtue of its position or chemical composition.its position or chemical composition.

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Initial PositionInitial Position

In the initial position, ball A has a In the initial position, ball A has a higher potential energy than ball B.higher potential energy than ball B.

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Final PositionFinal Position

After A has rolled down the hill, the potential After A has rolled down the hill, the potential energy lost by A has been converted to random energy lost by A has been converted to random motions of the components of the hill (frictional motions of the components of the hill (frictional heading) and to the increase in the potential heading) and to the increase in the potential energy of B.energy of B.

Kinetic Energy

Energy an object possesses by virtue of its motion.

12KE = mv2

Units of Energy

• The SI unit of energy is the joule (J).

• An older, non-SI unit is still in widespread use: The calorie (cal).

1 cal = 4.184 J

1 J = 1 kg m2

s2

Work

• Energy used to move an object over some distance.

• w = F d,where w is work, F is the force, and d is the distance over which the force is exerted.

Work needs a sign

• If the volume of a gas increases, the system has done work on the surroundings.

• work is negative

• Expanding work is negative.• Contracting surroundings do work on the

system w is positive.

Heat

• Energy can also be transferred as heat.

• Heat flows from warmer objects to cooler objects.

First Law of Thermodynamics• Energy is neither created nor destroyed.• In other words, the total energy of the universe is a

constant; if the system loses energy, it must be gained by the surroundings, and vice versa.

Use Fig. 5.5

Internal EnergyThe internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.

Use Fig. 5.5

Internal EnergyBy definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system:

E = Efinal − Einitial

Use Fig. 5.5

Changes in Internal Energy

• If E > 0, Efinal > Einitial

– Therefore, the system absorbed energy from the surroundings.

– This energy change is called endergonic.

Changes in Internal Energy

• If E < 0, Efinal < Einitial

– Therefore, the system released energy to the surroundings.

– This energy change is called exergonic.

Changes in Internal Energy

• When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).

• That is, E = q + w.

E, q, w, and Their Signs

• Text Pg. 171 Sample Exercise 5.2

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6–23

Work vs. Energy Flow

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

Exchange of Heat between System and Surroundings

• When heat is absorbed by the system from the surroundings, the process is endothermic.

• When heat is released by the system to the surroundings, the process is exothermic.

State Functions

Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem.

State Functions• However, we do know that the internal energy of a

system is independent of the path by which the system achieved that state.– In the system below, the water could have reached room

temperature from either direction.

State Functions• Therefore, internal energy is a state function.• It depends only on the present state of the system,

not on the path by which the system arrived at that state.

• And so, E depends only on Einitial and Efinal.

State Functions

• However, q and w are not state functions.

• Whether the battery is shorted out or is discharged by running the fan, its E is the same.– But q and w are different

in the two cases.

Work

When a process occurs in an open container, commonly the only work done is a change in volume of a gas pushing on the surroundings (or being pushed on by the surroundings).

WorkWe can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston.

w = −PV

Enthalpy• If a process takes place at constant pressure

(as the majority of processes we study do) and the only work done is this pressure-volume work, we can account for heat flow during the process by measuring the enthalpy of the system.

• Enthalpy is the internal energy plus the product of pressure and volume:

H = E + PV

Enthalpy

• When the system changes at constant pressure, the change in enthalpy, H, is

H = (E + PV)• This can be written

H = E + PV

Enthalpy

• Since E = q + w and w = −PV, we can substitute these into the enthalpy expression:

H = E + PVH = (q+w) − w H = q

• So, at constant pressure the change in enthalpy is the heat gained or lost.

Endothermicity and Exothermicity

• A process is endothermic, then, when H is positive.

Endothermicity and Exothermicity

• A process is endothermic when H is positive.

• A process is exothermic when H is negative.

Pg. 176 Sample and practice 5.3Pg. 176 Sample and practice 5.3

Enthalpies of Reaction

The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants:

H = Hproducts − Hreactants

Enthalpies of Reaction

This quantity, H, is called the enthalpy of reaction, or the heat of reaction.

The Truth about Enthalpy

Pg. 1781. Enthalpy is an extensive property.2. H for a reaction in the forward direction is

equal in size, but opposite in sign, to H for the reverse reaction.

3. H for a reaction depends on the state of the products and the state of the reactants.

• Pg. 179 Sample and practice (most important out of all the samples)

Calorimetry

Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow.

Heat Capacity and Specific Heat

• The amount of energy required to raise the temperature of a substance by 1 K (1C) is its heat capacity.

• We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

Heat Capacity and Specific Heat

Specific heat, then, is

Specific heat =heat transferred

mass temperature change

Cs =q

m T

Constant Pressure Calorimetry

By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter.

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A Coffee A Coffee Cup Cup CalorimetCalorimeter Made er Made of Two of Two StyrofoaStyrofoam Cupsm Cups

Constant Pressure Calorimetry

Because the specific heat for water is well known (4.184 J/mol-K), we can measure H for the reaction with this equation:q = m s T

Pg. 181 samplePg. 181 sample Pg. 182 samplePg. 182 sample

Bomb Calorimetry

Reactions can be carried out in a sealed “bomb,” such as this one, and measure the heat absorbed by the water.

Bomb Calorimetry

• Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.

• For most reactions, the difference is very small.

Bomb Calorimetry

• Pg. 184 sample

EnthalpyEnthalpy

HeatHeat Remember:Remember: qqreactionreaction at constant pressure = at constant pressure = ∆∆H = HH = Hproductproduct – H – Hreactantsreactants

-∆H exothermic-∆H exothermic +∆H endothermic+∆H endothermic Remember reaction diagramsRemember reaction diagrams

Thermochemical equationsThermochemical equations

Specify ∆H in kilojoules Specify ∆H in kilojoules Workbook Handout – Pg. 158 Read Workbook Handout – Pg. 158 Read

through and do sample problems - up through and do sample problems - up to pg. 163to pg. 163

Rules of Thermochemistry Masterton Rules of Thermochemistry Masterton Handout Pg. 203 -205Handout Pg. 203 -205

Hess’s LawHess’s Law Summarized on page 206Summarized on page 206

Rules of Thermochemistry

1. The magnitude of ∆H is directly proportional to the amount of reactant or product

2. ∆H for a reaction is equal in magnitude but opposite in sign to ∆H for the reverse reaction

3. The value of ∆H for a reaction is the same whether it occurs in one step or in a series of steps. Called Hess’s Law

∆∆H is directly proportional to amount H is directly proportional to amount of reactants or products. of reactants or products.

When one mole of ice melts, 6.00 kJ When one mole of ice melts, 6.00 kJ of heat is absorbed, ∆H = +6.00 kJof heat is absorbed, ∆H = +6.00 kJ

If one gram of ice melts,If one gram of ice melts, ∆∆H = 6.00kJ/18.02 = 0.333 kJ.H = 6.00kJ/18.02 = 0.333 kJ.

In general, ∆H can be related In general, ∆H can be related to amount by the conversion to amount by the conversion

factor approachfactor approach HH2(g)2(g) +Cl +Cl2(g)2(g) 2HCl 2HCl(g) (g) ∆H = -185kJ ∆H = -185kJ When 1.00 g of ClWhen 1.00 g of Cl22 reacts, reacts, ∆∆H = 1.00g ClH = 1.00g Cl22 x x 1 molCl1 molCl22 x x -185kJ-185kJ

==

70.90g Cl70.90g Cl22 1 mol Cl 1 mol Cl22

-2.61 kJ

∆∆H for a reaction is equal in H for a reaction is equal in magnitude but opposite in sign magnitude but opposite in sign to ∆H for the reverse reactionto ∆H for the reverse reaction

HH22OO(s)(s) H H22OO(l)(l) ∆H = +6.00kJ; ∆H = +6.00kJ; 6.00kJ 6.00kJ

absorbed absorbed

HH22OO(l)(l) HH22OO(s)(s) ∆H = -6.00kJ; 6.00kJ ∆H = -6.00kJ; 6.00kJ

evolvedevolved

Hess’s LawHess’s Law

If Equation 1 + Equation 2 = If Equation 1 + Equation 2 = Equation 3, then ∆HEquation 3, then ∆H33 = ∆H = ∆H11 + ∆H + ∆H22

Often used to calculate ∆H for one Often used to calculate ∆H for one step, knowing ∆H for all other steps step, knowing ∆H for all other steps and for the overall reactionand for the overall reaction

CC(s)(s) + ½ O + ½ O2(g)2(g) CO CO(g)(g) ∆H ∆H11 = ? = ?

COCO(g)(g) + 1/2 O + 1/2 O2(g)2(g) CO CO2(g)2(g) ∆H ∆H22 = -283.0kJ = -283.0kJ

CC(s)(s) + O + O2(g)2(g) CO CO2(g) 2(g) ∆H∆H33 = -393.5 = -393.5

∆∆HH11 = -110.5 kJ = -110.5 kJ

Examples Pg. 204 – 205 8.4, 8.5, 8.6Examples Pg. 204 – 205 8.4, 8.5, 8.6

Go over homework for this topicGo over homework for this topic

Heat of Formation, Heats Heat of Formation, Heats of Reactionof Reaction

The standard heat of formation The standard heat of formation ((ΔΔHHooff) is ) is

the change of enthalpy for the reaction that the change of enthalpy for the reaction that forms a compound from its pure elements forms a compound from its pure elements under standard conditions.under standard conditions.

The standard heat of formation The standard heat of formation ((ΔΔHHooff) )

for pure elements at standard conditions is for pure elements at standard conditions is zero.zero.

Standard heats of formation can be used to Standard heats of formation can be used to estimate the estimate the ΔΔHHoo

f f of any reaction.of any reaction.

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Standard StatesStandard States

CompoundCompound For a gas, pressure is exactly 1 For a gas, pressure is exactly 1

atmosphere.atmosphere. For a solution, concentration is exactly 1 For a solution, concentration is exactly 1

molar.molar. Pure substance (liquid or solid)Pure substance (liquid or solid)

ElementElement The form [NThe form [N22((gg), K(), K(ss)] in which it exists at 1 )] in which it exists at 1

atm and 25°C.atm and 25°C.

ΔΔHHoof f and heats of formationand heats of formation

ΔHof and heat of formation

ΔHof = ΣHo

f (products) - ΣHof (reactants)

Enthalpies of FormationEnthalpies of Formation

Standard molar enthalpy of formation Standard molar enthalpy of formation – equal to the enthalpy change when – equal to the enthalpy change when one mole of the compound is formed one mole of the compound is formed at a constant pressure of 1 atm and a at a constant pressure of 1 atm and a fixed temp (usually 25fixed temp (usually 25oo C C

∆∆HHofof Can calculate Standard Enthalpy Can calculate Standard Enthalpy

values ∆Hvalues ∆Hoo from these values from these values

The standard enthalpy change , ∆HThe standard enthalpy change , ∆Ho ,o , for a given thermochemical equation for a given thermochemical equation is equal to the sum of the standard is equal to the sum of the standard enthalpies of formation of the enthalpies of formation of the product compounds minus the sum product compounds minus the sum of the standard enthalpies of of the standard enthalpies of formation of the reactant compoundsformation of the reactant compounds

∆∆HHoo = ∑∆H = ∑∆Hooff products - ∑∆H products - ∑∆Hoo

ff reactantsreactants

Elements in their standard states can Elements in their standard states can be omitted because their heats of be omitted because their heats of formation are zeroformation are zero

Use Reference Table on Pg. 207Use Reference Table on Pg. 207

ΔΔHHoof f and heats of formationand heats of formation

• Example: Calculate the Example: Calculate the ΔΔHHooff for for

the reaction,the reaction,• PbOPbO22(s) + 2H(s) + 2H22SOSO44 (l) + Pb(s) (l) + Pb(s)

2PbSO 2PbSO44 (s) + 2H (s) + 2H22O (l)O (l)

• Given the following Given the following ΔΔHHooff

informationinformation

•Species ΔHof (kJ)

•PbO2 (s) -66.3

•2H2SO4 -194.5•PbSO4 -219.9•H2O -68.3

•ΔHof = ΣHo

f (products) - ΣHof

(reactants)

•ΔHof = [(2 x -219.9) + (2

x -68.3)] – [(66.3) + (2 x -194.5)]•ΔHo

f = -121.1 kJ

Pg. 207 ExamplePg. 207 Example Pg. 208 do examples 8.7 and 8.8Pg. 208 do examples 8.7 and 8.8

• Bond energies are the amount of energy given off when bonds are formed, or the amount of energy used when bonds are broken.

• Bond energies deal with reactants and products in their gaseous state under standard conditions.

• Breaking bonds is an exothermic process.• Making bonds is an endothermic process.• Heat of reaction can be estimated by finding the

difference between the bonds made and the bond energies of the bonds broken.

• Bond energies used in this way to find heats of reaction is an example of Hess’s Law.

ΔHo and Bond Energies

ΔHo reaction = ΣHo bond energies (bonds broken) - ΣHo bond energies (bonds

made)

Bond enthalpy always positive

Pg. 211 of Masterton handout – table of bond enthalpies – heat always absorbed when bonds are broken

The bonds in the reactants are stronger than those in the products

And There are more bonds in the reactants than

in the products

Bond entalpy is larger for a multiple bond than for a single bond between the same two atoms

Bottom of pg. 211

First Law of Thermodynamics◦ The first law of thermodynamics is commonly

referred to as the law of Conservation of Energy.◦ More specifically, the first law states that the

changes in internal energy is equal to the difference between the energy supplied to the system as heat and the energy removed from the system as work performed on the surroundings.

The energy of the universe is constant. Law of conservation of energy. q = heat w = work E = q + w Take the systems point of view to decide

signs. In any process, the total change in energy

of a system,E, is equal to the sum of the heat, q, and the work,w, transferred between the system and the surroundings

Work is a force acting over a distance. w= F x d P = F/ area d = V/area w= (P x area) x (V/area)= PV Work can be calculated by multiplying

pressure by the change in volume at constant pressure.

If the volume of a gas increases, the system has done work on the surroundings.

work is negative w = - PV Expanding work is negative. Contracting, surroundings do work on the

system w is positive.

Pg. 213 Example 8.9