Prentice Hall © 2003Chapter 19 Chapter 19 Chemical Thermodynamics CHEMISTRY The Central Science 9th Edition David P. White

Prentice Hall © 2003 Chapter 19

Chapter 19Chapter 19Chemical ThermodynamicsChemical Thermodynamics

CHEMISTRY The Central Science

9th Edition

David P. White


• Thermodynamics is concerned with the question: can a reaction occur?

• First Law of Thermodynamics: energy is conserved.• Any process that occurs without outside intervention is spontaneous.• When two eggs are dropped they spontaneously break.• The reverse reaction is not spontaneous.• A process that is spontaneous in one direction is not spontaneous in

the opposite direction.

Spontaneous ProcessesSpontaneous Processes

Spontaneity & TemperatureSpontaneity & Temperature

The direction of a spontaneous process can depend on temperature: ice turning to water is spontaneous at T > 0C, water turning to ice is spontaneous at T < 0C.

Packet Example Page 3Packet Example Page 3

• Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or are in equilibrium.

• a) When a piece of metal heated to 150C is added to water at 40C, the water gets hotter.

• b) Water at room temperature decomposed into hydrogen and oxygen gases.

• c) Benzene vapor at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene.


Reversible ProcessesReversible Processes

• Reversible processes can go back and forth between states along the same path.

• Example: changes of state

• Path taken back to original state is exactly the reverse of the forward process.

• No net change in system or surroundings when cycle is completed.


– When 1 mol of water is frozen at 1 atm at 0C to form 1 mol of ice, q = Hvap of heat is removed.

– To reverse the process, q = Hvap must be added to the 1 mol of ice at 0C and 1 atm to form 1 mol of water at 0C.

– Converting between 1 mol of ice and 1 mol of water at 0C is a reversible process.

Reversible ProcessesReversible Processes

Irreversible ProcessesIrreversible Processes

An irreversible process cannot be reversed to restore the system and surroundings back to their original state.A different path with different values of q and w are needed. Surroundings are not returned to original conditions.


• Chemical systems in equilibrium are reversible.

• In any spontaneous process, the path between reactants and products is irreversible.

• Thermodynamics gives us the direction of a process but it cannot predict the speed at which the process will occur.

Reversible & Irreversible Reversible & Irreversible ProcessesProcesses

• Why does the gas expand?

• Why is the reverse process nonspontaneous?

Spontaneous Spontaneous Expansion of a Expansion of a

GasGas

• Consider 2 gas molecules in one flask (a).

• Once the stopcock is open, there is a higher probability that one molecule will be in each flask than both molecules being in the same flask (b).

Spontaneous Expansion of Spontaneous Expansion of a Gasa Gas


• Entropy, S, is a measure of the disorder of a system.

• The more disordered or random the system, the large value of S.

• Spontaneous reactions proceed to lower energy or higher entropy.

EntropyEntropy

•In ice, molecules are very well ordered because of the H-bonds.•Ice has a low entropy.•As ice melts, intermolecular forces are broken , order is interrupted.•Water is more random than ice, has higher entropy.• Ice spontaneously melts at room temperature.

•A balance between energy and entropy considerations. •When an ionic solid is placed in water two things happen:

–Hydrates form (entropy decreases)–Ions dissociate (entropy increases)


• 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.

Entropy, Cont.Entropy, Cont.

Packet Example Page 6Packet Example Page 6

• By considering the disorder in the reactants and products, predict whether ∆S is positive or negative for the following:

• a) H2O (l) → H2O (g)

• b) Ag+(aq) + Cl-(aq) → AgCl (s)

• c) 4Fe(s) + 3O2 (g) → 2Fe2O3 (s)



• Second Law of Thermodynamics: In any spontaneous process, the entropy of the universe increases.

Suniv = Ssys + Ssurr

• Entropy is not conserved: Suniv is increasing.

• To predict spontaneity, we must know the sign of ∆Suniv.

Second Law of Second Law of ThermodynamicsThermodynamics

• For a reversible process: Suniv = 0.

• For a spontaneous process (i.e. irreversible): Suniv > 0. Suniv< 0 means spontaneous is opposite direction.

• Second law states that the entropy of the universe must increase in a spontaneous process. Entropy of a system can decrease as long as the entropy of the surroundings increases.

• 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


Entropy

• Suppose a system changes reversibly between state 1 and state 2. At constant T where qrev is the amount of heat added reversibly to the system, the change in entropy is given by

• 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

Example Packet p 7Example Packet p 7

• The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9C and its molar enthalpy of fusion is ∆H = +2.29 kJ/mol. Calculate the entropy change of the system when 50.0 g liquid mercury freezes at the normal freezing point.

• Remember: fusion = melting

• Use: )(constant revsys T

Tq

S

Practice ExamplePractice Example

• The normal boiling point of ethanol (C2H5OH) is 78.3C and its ∆H vaporization= +38.56 kJ/mol. Calculate ∆S when 68.3 g of ethanol at 1 atm condense to liquid at the normal boiling point.



• Consider the reversible melting of 1 mol of ice in a large, isothermal water bath at 0C. The enthalpy of fusion is 6.01 kJ/mol. Calculate the entropy change in the system and in the surroundings and the overall change in entropy of the universe for this process.


Practice ExamplePractice Example

• ∆Hvap = 30.71 kJ/mol for liquid bromine. Calculate the change in entropy for the system, the surroundings, and the universe for the reversible vaporization of liquid bromine at its normal boiling point of 59C.


Effect of Temperature on Effect of Temperature on SpontaneitySpontaneity

• ∆Ssurr are determined primarily by flow of energy into and out of system as heat.

• An exothermic process in system increases entropy in surroundings.

• Exothermicity is a driving force for spontaneity BUT it depends on the temperature at which the process occurs!

• Impact of transfer of a given quantity of energy as heat to or from surroundings will be greater at lower temperatures.


Entropy Changes in the Entropy Changes in the SurroundingsSurroundings

• Sign of ∆Ssurr depends on direction of heat flow.

• Magnitude of ∆Ssurr depends on the temperature.

• ∆Ssurr depends directly on quantity of heat transferred and inversely on temperature.

• ∆Ssurr = -∆H / T

• (We need a minus sign here to change the point of view from the system to the surroundings.)


Examples from Zumdahl Examples from Zumdahl pp 782 & 783pp 782 & 783

• Calculate ∆Ssurr for the following reactions at 25C and

1 atm.

• Sb2S3(s) + 3Fe(s) → 2Sb(s) + 3FeS(s) ∆H = -125 kJ

• Sb4O6(s) + 6C(s) → 4Sb(s) + 6CO(s) ∆H = +778 kJ

• Use ∆Ssurr = -∆H / T

• +419 J/K• -2.61 x 103 J/K


Determining the Sign of Determining the Sign of ∆S∆Sunivuniv

Signs of Entropy Changes •∆Ssys ∆Ssurr ∆Suniv Process Spontaneous?

• + + + Yes• - - - No, rxn goes in reverse

• + - ? Yes, if ∆Ssys ∆Ssurr

• - + ? Yes, if ∆Ssurr ∆Ssys


• Movement of molecules is related to energy and entropy.• Three atomic modes of motion:

– translation (moving from one point in space to another),

– vibration (shortening and lengthening of bonds, including the change in bond angles),

– rotation (spinning around an axis).

Atomic Modes of MotionAtomic Modes of Motion


• Energy is required to get a molecule to translate, vibrate or rotate.

• The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy.

• In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order.

The Molecular The Molecular Interpretation of EntropyInterpretation of Entropy

• Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

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

• Entropy changes dramatically at a phase change.• Entropy increases when

--Liquids or solutions are formed from solids

--Gases are formed from solids or liquids

--The number of gas molecules increases.

Third Law of Third Law of ThermodynamicsThermodynamics

Heat and EntropyHeat and Entropy

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

Boiling corresponds to a much greater change in entropy than melting.


• Choose the sample of matter that has the greater entropy in each pair and explain your choice:

• a) 1 mol of solid NaCl or 1 mol of gaseous HCl at 25C• b) 2 mol of HCl(g) or 1 mol of HCl(g) at 25C• c) 1 mol of HCl(g) or 1 mol of Ar(g) at 25C• d) 1 mol of N2(s) at 24 K or 1 mol of N2 at 25C



• Predict whether the entropy change of the system in each of the following isothermal reactions is positive or negative.

• a) CaCO3(s) → CaO(s) + CO2(g)

• b) N2(g) + 3H2(g) → 2NH3(g)

• c) N2(g) + O2(g) → 2NO(g)


• 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 are not zero.

• 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


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

• Reactions with large negative H values are spontaneous.• How to we balance 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


• There are 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.





• Consider the formation of ammonia from nitrogen and hydrogen:

• 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.


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


Standard Free-Energy Changes

• We can tabulate standard free-energies of formation, Gf (c.f. standard enthalpies of formation).

• 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).


reactantsproducts ff GmGnG


• Focus on G = H - TS:– If H < 0 and S > 0, then G is always negative.– If H > 0 and S < 0, then G is always positive. (That is, the

reverse of 1.)– If H < 0 and S < 0, then G is negative at low temperatures.– If H > 0 and S > 0, then G is negative at high temperatures.

• Even though a reaction has a negative G it may occur too slowly to be observed.

Free Energy and Free Energy and TemperatureTemperature


Free Energy and Free Energy and TemperatureTemperature


• 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


• 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


End of Chapter 19End of Chapter 19Chemical ThermodynamicsChemical Thermodynamics

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Prentice Hall © 2003Chapter 19 Chapter 19 Chemical Thermodynamics CHEMISTRY The Central Science 9th Edition David P. White