prentice hall © 2003chapter 5 chapter 6 thermochemistry chemistry

Prentice Hall © 2003 Chapter 5

Chapter 6Chapter 6ThermochemistryThermochemistry

CHEMISTRY

Prentice Hall © 2003 Chapter 5

ThermodynamicsThermodynamics

- is the study of energy and the transformations it undergoes in chemical reactions.

Units: Joules (J)calorie = 4.184 J

calorie – the amount of E needed to raise the temperature of 1 gram of water 1 oC

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EnergyEnergy

• Examples of energy:

• Potential E• Kinetic E• Work – E needed to move an object against a force• Heat – E transferred from hot to cold objects

- capacity to do work or transfer heat

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Today’s TopicsToday’s Topics

• 4 Thermodynamic Functions• Definition of State Function• Internal Energy• Point of Views – System, Surroundings Universe• Sign Conventions

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4 Thermodynamic Functions4 Thermodynamic Functions

• E - Internal Energy• H - Enthalpy• S - Entropy• G - Gibb’s Free Energy

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What is a state function?What is a state function?

• A state function is a property that depends on the present condition and not on how the change occurs.

• Derived from calculating the change: = [Final value – initial value]

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4 State Functions4 State Functions

• Therefore:•

E = Ef -Ei

H = Hf - Hi

S = Sf - Si

G = Gf - Gi

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Internal Energy (E)Internal Energy (E)

• Internal Energy - is the sum of the kinetic and potential energy of a system

• Is the sum of heat (q) and work (w)

E = Efinal – Einitial

E = q + w

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Point of ViewsPoint of Views

• System - the reaction we are studying

• Surroundings – anything else besides the reactionFor example: the container, you, etc….

• Universe – system + surroundings

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q = heat and w = workq = heat and w = work

• If q is (+), system gaining heat from surroundings (endo)

• If q is (-), system giving up heat to the surroundings (exo)

• If w is (+), system is the recipient of work from the surroundings. (in short, surroundings is doing work on the system.

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Thermodynamic FunctionsThermodynamic Functions

• Have value, unit and magnitude.

• The sign of E depends on the magnitude of q and w.

• Knowing the value of E does not tell us which variable is larger, q or w.

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Problem 1Problem 1

• Calculate the change in internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings.

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Problem 2Problem 2

• Consider the reaction of hydrogen and oxygen gases to produce water. As the reaction occurs, the system loses 1150 J of heat to the surroundings. The expanding gas does 480 J of work on the surroundings as it pushes against the atmosphere. Calculate the change in the internal energy of the system?

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Problem 3Problem 3

• Calculate E and determine whether the process is endothermic or exothermic.

• 1.) q = 1.62 kJ and w = -874 J• 2.) The system releases 113 kJ of heat to the

surroundings and does 39 kJ of work.• 3.) The system absorbs 77.5 kJ of heat while doing

63.5 kJ of work on the surroundings.

Prentice Hall © 2003 Chapter 5

Thermodynamic FunctionsThermodynamic Functions

• Have value, unit and magnitude.

• The sign of E depends on the magnitude of q and w.

• Knowing the value of E does not tell us which variable is larger, q or w.

Prentice Hall © 2003 Chapter 5

EnthalpyEnthalpy

• Enthalpy – accounts for heat flow in chemical reactions that occur at constant P when nothing other than P-V work are performed

H = Hfinal - Hinitial

• If: H = E + PV• Then:H = E + PV

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Ways of Measuring Ways of Measuring HH

• Calorimetry

• Hess’s Law

• Heats of Formation (Hof)

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CalorimetryCalorimetry

• - Is the measurement of heat flow

• Specific Heat capacity (C) - the amount of heat needed to raise the temperature of 1 g of substance 1 oC.

• The greater the heat capacity, the greater the heat required to produce a rise in temp.

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Calorimetry EquationCalorimetry Equation

• q = mCT

• - qsystem = qsurroundings

• - qsystem = qwater + qcalorimeter

• - qsubstance = (mCTsolution + mCTcalorimeter)

• Simplifies to:

• - qsubstance = (mCTsolution + CTcalorimeter)

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Calorimetry ProblemCalorimetry Problem

• A 30.0 gram sample of water at 280 K is mixed with 50.0 grams of water at 330 K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The specific heat capacity of the solution is 4.18 J/g-oC.

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Calorimetry ProblemCalorimetry Problem

• A 46.2 gram sample of copper is heated to 95.4 oC and then placed in a calorimeter containing 75.0 gram of water at 19.6 oC. The final temperature of the metal and water is 21.8 oC. Calculate the specific heat capacity of copper, assuming that all the heat lost by the copper is gained by the water.

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Calorimetry ProblemCalorimetry Problem

• A 15.0 gram sample of nickel metal is heated to 100.0 oC and dropped into 55.0 grams of water, initially at 23 oC. Assuming that no heat is lost to the calorimeter, calculate the final temperature of the nickel and water. The specific heat of nickel is 0.444 J/g-oC. The specific heat of water is 4.18 J/g-oC.

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Problem 4Problem 4

• The specific heat of water is 4.18 J/g-K.• How much heat is needed to warm 250 g of water

from 22 oC to 98 oC?

• What is the molar heat capacity of water?

• Molar heat Capacity = C x molar mass

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Problem 5Problem 5

• Large beds of rocks are used in solar heated homes to store heat. Assume that the specific heat of rocks is 0.82 J/g-K.

• A. Calculate the amount of heat absorbed by 50 kg. of rocks if their temperature rose by 12.0 oC.

• B. What temperature change would these rocks undergo if they emitted 450 kJ of heat?

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• Work – E needed to move an object against a force

• When P is constant, P-V work is given byw = - PV

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EnthalpyEnthalpy

• Enthalpy of a reaction or heat of Reaction: H = Hproducts - Hreactants

• 1. sign of H depends on the amount of reactant consumed

• 2. H sign is opposite for backwards reaction• 3. Hrxn depends on the physical state of the

reactants and products.

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Problem 1Problem 1

• Given the reaction:

• 2H2 (g) + O2 (g) 2 H2O (g) H = -483 kJ

• Calculate the H value for:

• 2 H2O (g) 2H2 (g) + O2 (g)

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Problem 2Problem 2

• Given the reaction:

• 2H2 (g) + O2 (g) 2 H2O (g) H = -483 kJ

• How much heat is released when 10.5 grams of H2 is burned in a constant-pressure system?

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Ways of Measuring Ways of Measuring HH

• Calorimetry

• Hess’s Law

• Heats of Formation (Hof)

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Hess’ LawHess’ Law

• If a reaction is carried out in steps, H for the reaction will equal the sum of the enthalpy changes for the individual steps.

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Problem 1Problem 1

• Given:

• C (s) + O2 (g) CO2 (g) H = -393.5 kJ

• CO(g) + ½ O2 (g) CO2 (g) H = -283.5 kJ

• Calculate the enthalpy of combustion for:C(s) + ½ O2 (g) CO (g)

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Problem 2Problem 2

• Given:

• C(graphite) + O2 (g) CO2 (g) H = -393.5 kJ

• C(diamond) + O2 (g) CO2 (g) H = -395.4 kJ

• Calculate the enthalpy of combustion for:C(graphite) C (diamond)

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Problem 3Problem 3

• Given:• C2H2 (g) + 5/2 O2 (g) 2CO2 (g) + H2O (l) H = -1299.6 kJ

• C (s) + O2 (g) CO2 (g) H = -395.4 kJ

• H2 (g) + 1/2 O2 (g) H2O (l) H = -285.8 kJ

• Calculate the enthalpy of combustion for:2C (s) + H2 (g) C2H2 (g)

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Enthalpies of FormationEnthalpies of Formation

• - known as heat of formation • - gives the energy needed for a compound to form

• Standard enthalpy - is the enthalpy change (H) when the reactants and products are in their standard state, usually 1 atm and 25 oC - denoted by Ho ( ex. Ho

f)

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Standard Enthalpy of Standard Enthalpy of Formation, (Formation, (HHoo

ff))

• By definition:

• The standard enthalpy of formation ( ex. Hof) of the

most stable form of any element is ZERO because there is no formation reaction needed when the element is in its standard state.

• - important for diatomic molecules• - need knowledge of standard states of compounds

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• If there is more than one state for a substance under standard conditions, the more stable one is used.

• Standard enthalpy of formation of the most stable form of an element is zero.

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• We use Hess’ Law to calculate enthalpies of a reaction from enthalpies of formation.

Enthalpies of FormationEnthalpies of Formation

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Enthalpies of FormationEnthalpies of Formation

Prentice Hall © 2003 Chapter 5

Using Enthalpies of Formation of Calculate Enthalpies of Reaction

• For a reaction

Enthalpies of FormationEnthalpies of Formation

reactantsproductsrxn ff HmHnH

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Next TopicNext Topic

S = Entropy = disorder

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Thermodynamic QuestionThermodynamic Question

• Can a process occur?• Spontaneous or Non-spontaneous?

– Forward vs. Reverse reactions

– Example: Gas Expansion

• Reversible or irreversible?

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EntropyEntropy

• Entropy, S, is a measure of the disorder of a system.• Spontaneous reactions proceed to lower energy or

higher entropy.• In ice, the molecules are very well ordered because of

the H-bonds.• Therefore, ice has a low entropy.

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• As ice melts, the intermolecular forces are broken (requires energy), but the order is interrupted (so entropy increases).

• Water is more random than ice, so ice spontaneously melts at room temperature.

• Conclusion: The higher the entropy the more spontaneous the reaction.

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• Generally, when an increase in entropy in one process is associated with a decrease in entropy in another, the increase in entropy dominates.

• Entropy is a state function. • For a system, S = Sfinal - Sinitial.• If S > 0 the randomness increases, if S < 0 the

order increases.

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Entropy• Suppose a system changes reversibly between state 1

and state 2. Then, the change in entropy is given by

– at constant T where qrev is the amount of heat added reversibly to the system. (Example: a phase change occurs at constant T with the reversible addition of heat.)

Entropy and the Second Entropy and the Second Law of ThermodynamicsLaw of Thermodynamics

)(constant revsys T

Tq

S

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The Second Law of Thermodynamics• Spontaneous processes have a direction.• In any spontaneous process, the entropy of the

universe increases. Suniv = Ssys + Ssurr: the change in entropy of the

universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.

• Entropy is not conserved: Suniv is increasing.

Entropy and the Second Entropy and the Second Law of ThermodynamicsLaw of Thermodynamics

Prentice Hall © 2003 Chapter 5

The Second Law of Thermodynamics

• Reversible process: Suniv = 0.

• Spontaneous process (i.e. irreversible): Suniv > 0.

Ssys for a spontaneous process can be less than 0 as long as Ssurr > 0. This would make Suniv still (+).

• For an isolated system, Ssys = 0 for a reversible process and Ssys > 0 for a spontaneous process.

Entropy and the Second Entropy and the Second Law of ThermodynamicsLaw of Thermodynamics

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Third Law of Third Law of ThermodynamicsThermodynamics

• The entropy of a perfect crystal at 0 K is zero.

• Entropy changes dramatically at a phase change.

• As we heat a substance from absolute zero, the entropy must increase.

Prentice Hall © 2003 Chapter 5

Prentice Hall © 2003 Chapter 5

• Absolute entropy can be determined from complicated measurements.

• Standard molar entropy, S: entropy of a substance in its standard state. Similar in concept to H.

• Units: J/mol-K. Note units of H: kJ/mol.• Standard molar entropies of elements (S ) are not 0.• For a chemical reaction which produces n moles of

products from m moles of reactants:

Entropy Changes in Entropy Changes in Chemical ReactionsChemical Reactions

reactantsproducts mSnSS

Prentice Hall © 2003 Chapter 5

Prentice Hall © 2003 Chapter 5

For a spontaneous reaction the entropy of the universe must increase.

• Reactions with large negative H values are spontaneous.

• How do we correlate S and H to predict whether a reaction is spontaneous?

• Gibbs free energy, G, of a state is

• For a process occurring at constant temperature

Gibbs Free EnergyGibbs Free Energy

TSHG

STHG

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Gibb’s Free EnergyGibb’s Free Energy

• Is the capacity to do maximum useful work.

• Heat decreases the amount of useful work done.

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WORKWORK

• At constant P, wsys = - PV

• At constant temperature, wsys = - TS

• If G = H – TS, for maximum useful work, H must be = 0.

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Three important conditions:– If G < 0 then the forward reaction is spontaneous.

– If G = 0 then reaction is at equilibrium and no net reaction will occur.

– If G > 0 then the forward reaction is not spontaneous. If G > 0, work must be supplied from the surroundings to drive the reaction.

• For a reaction the free energy of the reactants decreases to a minimum (equilibrium) and then increases to the free energy of the products.

Gibbs Free EnergyGibbs Free Energy

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ProblemProblem

• Correlate S and H to predict whether a reaction is non-spontaneous or spontaneous. If spontaneous, determine whether the reaction will be spontaneous at all temperature, at high temperature, or at low temperature.

• A. If S is (-) and H is (+).• B. If S is (-) and H is (-).• C. If S is (+) and H is (+).• D. If S is (+) and H is (-).

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• Consider the formation of ammonia from N2 and H2.

• Initially ammonia will be produced spontaneously (Q < Keq).

• After some time, the ammonia will spontaneously react to form N2 and H2 (Q > Keq).

• At equilibrium, ∆G = 0 and Q = Keq.

Gibbs Free EnergyGibbs Free Energy

N2(g) + 3H2(g) 2NH3(g)

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Standard Free-Energy Changes Gf• Standard states are: pure solid, pure liquid, 1 atm (gas), 1 M concentration (solution), and G = 0 for elements.

G for a process is given by

• The quantity G for a reaction tells us whether a mixture of substances will spontaneously react to produce more reactants (G > 0) or products (G < 0).

Gibbs Free EnergyGibbs Free Energy

reactantsproducts ff GmGnG

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Free Energy and Free Energy and TemperatureTemperature

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• Recall that G and K (equilibrium constant) apply to standard conditions.

• Recall that G and Q (equilibrium quotient) apply to any conditions.

• It is useful to determine whether substances under any conditions will react:

Free Energy and The Free Energy and The Equilibrium ConstantEquilibrium Constant

QRTGG ln

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• At equilibrium, Q = K and G = 0, so

• From the above we can conclude:– If G < 0, then K > 1.

– If G = 0, then K = 1.

– If G > 0, then K < 1.

Free Energy and The Free Energy and The Equilibrium ConstantEquilibrium Constant

eq

eq

KRTG

KRTG

QRTGG

ln

ln0

ln

Prentice Hall © 2003 Chapter 5

Calculating Calculating H, H, S, S, GG

• Use Hess’ Law• Standard Heats of Formation (Ho, So, Go )• Equations