Thermodynamics Entropy, Energy and equilibrium

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ThermodynamicsEntropy, Energy and

equilibrium

Chapter 19

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19.1 Thermodynamics

Thermo: heat Dynamics: power Study of energy flow and its

transformations (heat and energy flow) Determines direction of reactions

(spontaneous or nonspontaneous under given conditions)

State Functions: considers only initial and final states

Does not consider pathways or rate Organized into three laws:

1st Law (ΔU = qp + w = qp – PΔV ) 2nd Law 3rd Law

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19.1 Basic Definitions

System Surrounding Open system Closed system Isolated system State of system: defined by values of

composition, pressure, T, V. State Function: defined only by initial and

final condition of the system (Enthalpy, Entropy, Gibbs Free Energy).

Energy change signaled by: accomplishment of work and/or appearance or disappearance of heat.

system surroundings

universe

system surroundings

universe

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19.1 1st Law of Thermodynamics

1. A gas does 135 J of work while expanding, and at the same time it absorbs 156 J of heat. What is the change in internal energy?

(21 J)2. The internal energy of a fixed quantity of ideal

gas depends only on its temperature. If a sample of an ideal gas is allowed to expand against a constant pressure at a constant temperature,a) What is ΔU for the gas? b) Does the gas do work?c) If any heat exchanged with the surroundings?

a) 0 b) w = -P ΔV c)no, only work

done by gas is energy leaving the system. Internal energy should decrease, so the temp; but temp is const.; therefore the int energy does not change; gas has to absorb enough heat from surroundings to compensate for the work. Q = -w; ΔU = -w+w

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19.1 First Law of Thermodynamics 1st Law: Energy can be neither created nor

destroyed, but it can be converted from one form to another or transferred from a system to the surroundings or vice versa. Energy of the universe is constant

Important concepts from thermochemistry Enthalpy Hess’s law

Purpose of 1st Law Energy bookkeeping

How much energy? Exothermic or endothermic? What type of energy? Δu = q + w q = heat; w = work system does on the

surroundings (-PΔV)

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19.2 Spontaneous Processes: Expansion

Spontaneous processes are those that can proceed without any outside intervention. Product favored at equilibrium. Product-favored at

equilibrium May be fast or slow May be influenced by

temperature The gas in vessel B will

spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously separate

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19.1 Spontaneous Processes

Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

Examples: rusting, neutralization reaction, dissolution of sugar in water, heat flow, expansion of gas, spontaneous combustion (CH4 + O2), reaction of sodium with water

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19.1 Spontaneous Processes

Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.

Above 0C it is spontaneous for ice to melt. Below 0C the reverse process is

spontaneous.

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19.2 Spontaneous vs. Nonspontaneous

3. Determine if the following processes are spontaneous or not and if they are exothermic or endothermic.(a) Gases expand into larger volumes at constant temperature ________________(b) H2O(s) melts above 0C ____________

(c) H2O(l) freezes below 0C ____________

(d) NH4NO3 dissolves spontaneously in H2O ___________

(e) Steel (iron) rusts in presence of O2 and H2O ________

(f) Wood burns to form CO2 and H2O __________

(g) CH4 gas burns to form CO2 and H2O __________

Evolution of Heat (Exothermicity) : not enough to predict spontaneity

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19.2 Spontaneous vs Nonspontaneous

Nonspontaneous process Does not occur unless there is outside

assistance (energy?) Reactants-favored at equilibrium

All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction Spontaneous processes have a definite

direction Spontaneous processes are irreversible. Can

be reversed with considerable input of energy.

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19.2 Factors That Favor Spontaneity

Spontaneous Processes driven by Enthalpy, H (Joules)

Many, but not all, spontaneous processes tend to be exothermic.

Entropy, S (Joules/K) Measure of the disorder of a system Many, but not all, spontaneous

processes tend to increase disorder of the system

Exothermicity favors spontaneity, but does not guarantee it.

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19.2 Factors That Favor Spontaneity: Enthalpy

Example of spontaneous reaction that is not exothermic:

NH4NO3(s) → NH4+ (aq) + NO3

-(aq) ΔH = 25 kJ/mol

Expansion of gas: energy neutral Phase changes: endothermic processes

that occurs spontaneously (ice to water). Chemical system: H2(g) + I2(g) ↔ 2HI(g)

Equilibrium can be approached from both sides (spontaneous both ways) even though the forward reaction is endothermic and the reverse is exothermic.

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19.2 Spontaneity: Examples

4. Based on your experience, predict whether the following processes are spontaneous, are spontaneous in reverse direction, or are in equilibrium:

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

(b) Water at room temperature decomposes into hydrogen and oxygen gases

(c) Benzene vapor, C6H6(g), at a pressure of 1 atm condenses to liquid benzene at the normal boiling point of benzene, 80.1 ºC.

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19.2 Reversible Processes

In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process.Example: melting ice at its melting point

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19.2 Irreversible Processes

Irreversible processes cannot be undone by exactly reversing the change to the system. Different path has to be used.

Spontaneous processes are irreversible. Example: expansion a gas into vacuum.

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19.2 Entropy

Entropy (S) is a term coined by Rudolph Clausius in the 19th century.

Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, q

T

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19.2 Entropy Direct measure of the randomness or

disorder of the system. Related to probability

describes # of ways the particles in a system can be arranged in a given state (position and/or energy levels)

The most likely state – the most random More possible arrangements, the higher

disorder, higher entropy Ordered state – low probability of occurring Disordered state: high probability of

occurring

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19.2 Entropy on the Molecular Scale

Ludwig Boltzmann described the concept of entropy on the molecular level.

Temperature is a measure of the average kinetic energy of the molecules in a sample.

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Molecules exhibit several types of motion: Translational: Movement of the entire molecule

from one place to another. Vibrational: Periodic motion of atoms within a

molecule. Rotational: Rotation of the molecule on about an

axis or rotation about bonds.

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Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the

molecules. He referred to this sampling as a microstate of

the thermodynamic system.

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Each thermodynamic state has a specific number of microstates, W, associated with it.

Entropy isS = k lnW

where k is the Boltzmann constant, 1.38 1023 J/K;W: number of microstates

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The change in entropy for a process, then, is

S = k lnWfinal k lnWinitial

lnWfinal

lnWinitial

S = k ln

• Entropy increases with the number of microstates in the system.

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19.2 Spontaneous Processes: Dispersal of Matter

Isothermal (constant temperature) expansion of gas

After opening stopcock the molecules could be in any arrangement shown (4 arrangements)Probability for each arrangement = (1/2)2

S = k (ln(4)

25% probability

Two molecules present:

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Gas Container = two bulbed flask

Gas Molecules

Ordered State

19.2 Spontaneous Process: Isothermal Gas Expansion

Consider why gases tend to isothermally (constant temp.) expand into larger volumes.

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Gas Container

S = k ln (W) = k (ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K

For 3 particles, probability = (1/2)3

For N particles, probability = (1/2)N

Ordered State

19.2 Spontaneous Process: Isothermal Gas Expansion

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More probable that the gas molecules will disperse between two halves than remain on one side

Disordered States

27Driving force for expansion is entropy (probability); gas

molecules have a tendency to spread out

Disordered States

28S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K

Disordered States

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Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K

Total Arrangements

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The number of microstates and, therefore, the entropy tends to increase with increases in Temperature. Volume. The number of independently moving

molecules.

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19.2 Entropy and Temperature

(a) A substance at a higher temperature has greater molecular motion, more disorder, and greater entropy than (b) the same substance at a lower temperature.

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19.2 Entropy and Physical States

Entropy increases with the freedom of motion of molecules.

Therefore,S(g) > S(l) > S(s)

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19.2 Entropy Changes

In general, entropy increases when Gases are formed from

liquids and solids. Liquids or solutions are

formed from solids. The number of gas

molecules increases. The number of moles

increases.

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19.2 Entropy in Temperature

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Sta

ndard

entr

op

y,

S(

J/K

)

Temperature (K)0

20

10

30

40

50

10050 250 300150 200

What kind of changes are represented here?

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19.2 Solutions

Generally, when a solid is dissolved in a solvent, entropy increases.

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19.2 Patterns of Entropy Change

6. Describe in words the entropy of the system

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19.2 Entropy

Like total energy, E, and enthalpy, H, entropy is a state function.

Therefore, S = Sfinal Sinitial

S > 0 represents increased randomness or disorder

Note: The magnitude of change in entropy depends on temperature.

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19.2 Entropy

For a process occurring at constant temperature (an isothermal process), the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature:

S =qrev

T

Units: Joule/K

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19.2 Entropy: Example

5. ΔS = q/TThe element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 ºC, and its molar enthalpy of fusion is ΔHfusion = 2.331 kJ/mol. What is the entropy change when 50.0 g of Hg(l) freezes at the normal freezing point?

(-2.48 J/K)

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19.2 Entropy: Examples

7. Predict if ΔS increases, decreases or does not change

(a) Freezing liquid mercury(b) Condensing H2O(vapor)

(c) Precipitating AgCl(d) Heating H2(g) from 60.0 ºC to 80 ºC

(e) Subliming iodine crystals(f) Rusting iron nail

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19.2 Entropy - Examples

8. Predict which substance has the higher entropy:a) NO2(g) or N2O4(g)b) I2(g) or l2(s)

9. Predict whether each of the following leads to increase or decrease in entropy of a system If in doubt, explain why.a) The synthesis of ammonia:

N2(g) + 3H2(g) ↔ 2NH3(g)

b) C12H22O11(s) → C12H22O11(aq)

c) Evaporation to dryness of a solution of urea, CO(NH2)2 in vapor.

CO(NH2)2(aq) → CO(NH2)2(s)

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19.3 Second Law of Thermodynamics

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19.3 Second Law of Thermodynamics: System

10. Predict the sign of ΔS0 for each of the following reactions:a) Ca+2(aq) + 2OH-(aq) → Ca(OH)2(s)

b) MgCO3(s) → MgO(s) + CO2(g)

d) H2(g) + Br2(g) → 2HBr(g)

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Second Law of ThermodynamicsThe second law of thermodynamics states that theentropy of the universe increases for spontaneousprocesses, and the entropy of the universe doesnot change for reversible processes.

In other words:For reversible processes:

Suniv = Ssystem + Ssurroundings = 0

For irreversible processes:

Suniv = Ssystem + Ssurroundings > 0

For nonspontaneous Process:

ΔS univ= ΔS syst. + ΔS surr. <0

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Second Law of Thermodynamics

These last truths mean that as a result of all spontaneous processes the entropy of the universe increases.

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19.3 Second Law: Entropy Changes

EQUILIBRIUM PROCESSES (reversible)

♦ΔS universe= ΔS syst. + ΔS surr. =0

♦ΔS syst = ΔS surr

♦ΔS syst = ΔSº final - ΔS0 initial

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19.3 Entropy Changes in a System (Reactions)

Entropy changes in a system aA + bB → cC + dD Standard entropy change ΔSº (25 ºC, 1atm). Only changes in entropy can be measured. Each element has an entropy value (compare to

enthalpy). Absolute value for each substance can be

determined. For a chemical system:

S° = nS°(products) - mS°(reactants)

where n and m are the coefficients in the balanced chemical equation.

Standard Molar entropy, S0, is the entropy of one mole of a substance in its standard state (298 K)

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19.2 Standard Entropies

These are molar entropy values of substances in their standard states.

Standard entropies tend to increase with increasing molar mass.

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19.2 Standard Entropies

Larger and more complex molecules have greater entropies.

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11. Using standard molar entropies, calculate S°rxn for the following reaction at 25°C:

2SO2(g) + O2(g) → 2SO3(g)

S° = 248.1 205.1 256.6 (J · K-

1mol-1)

(Ans.: -187.9 J/K)

19.3 Second Law: Example

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19.3 Entropy of Reactions (System)

12. Using thermodynamic tables, calculate the standard entropy changes for the following reactions at 25 ºC

a) Evaporation of 1.00 mol of liquid ethanol to ethanol vapor.

b) The oxidation of one mole pf ethanol vapor (combustion reaction)c) Are the reactions spontaneous under the given conditions.

Answers:a) 122.0 J/K b) 96.09 J/K

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19.3 Entropy Changes in the System

In a reaction More gas molecules produced:

entropy increases Less gas molecules produced:

entropy decreases No net change of # of gas molecules

produced: entropy changes, but slightly

Liquid, solid products: hard to estimate-needs calculations

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19.3 Entropy of Surrounding

(a)When an exothermic reaction occurs in the system (ΔH < 0), the surroundings gain heat and their entropy increases (Δ Ssurr > 0).

(b) When an endothermic reaction occurs in the system (Δ H > 0), the surroundings lose heat and their entropy decreases (Δ Ssurr < 0).

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19.3 Entropy Changes in Surroundings

Heat that flows into or out of the system changes the entropy of the surroundings.

For an isothermal process:

Ssurr =qsys

T

• At constant pressure, qsys is simply H for the system. (negative sign needed to make entropy positive in an exothermic process)

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19.3 Second Law of Thermodynamics (System)

13. Examples: H2O(s) melts above 0C (endothermic). What

about entropy? Steel (iron) rusts in presence of O2 and H2O

(exothermic). What about entropy? 4Fe(s) + 3O2(g) 2Fe2O3(s)

CH4 gas burns to form CO2 and H2O (exothermic). What about entropy?

CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

Each process increases entropy of the universe.

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19.3 Second Law of Thermodynamics Consider transferring the same amount of heat to two

different systems, one at 298K and another at 500K.

More entropy (disorder) is created in the system at lower temp.

At high temperatures the system is already disordered

T

qS rev

298K 500KQ Q

K

q

K

q

500298

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19.3 Second Law of Thermodynamics To determine Suniv for a process, both Ssystem and

Ssurroundings need to be known:

Ssystem

related to matter dispersal in system Ssurroundings

determined by heat exchange between system and surroundings and T at which it occurs

Sign of Ssurr depends on whether process in system is endothermic (Ssurr< 0) or exothermic (Ssurr >0)

Magnitude of Ssurr depends on T

Ssurr = -Hsystem/T

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19.3 Second Law of Thermodynamics

14. Reaction: N2(g) + 3H2(g) → 2NH3(g)

ΔHº = -92.6kJ ΔSsys = -199 J/K at 25 ºC

ΔSsurr = -(-92.6 x1000)J/298K = 311 J/K

ΔSuniv = -199 J/K + 311 J/K = 112 J/K

Reaction spontaneous at 25 ºC

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ΔSsystem ΔS surr ΔSuniv Spontaneity

+ + + Yes

- - - No, reaction towards reactants

+ - ? Yes, if ΔSsys> ΔSsurroun

- + ? Yes, if ΔSsurr> ΔSsyst

19.3 Conditions for Spontaneity

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19.3 Entropy and Spontaneity

15. Consider the vaporization of liquid water to steam at a pressure of 1 atm. a) Is this process endothermic or exothermic, explain.b) In what temperature range is the process spontaneous?c) In what temperature range is it a nonspontaneous process?d) At what temperature are the two phases at equilibrium?

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19.3 Third Law of Thermodynamics

The entropy of a pure crystalline substance at absolute zero is 0.Perfect crystal: its internal arrangement is absolutely regular. Nothing is in motion (vibrations, rotations and translations)

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Third Law of Thermo

Gives us a starting point, S at 0K is equal to zero.

All others must be >0.

Standard Entropies Sº ( at 298 K and 1 atm) of substances are listed.

Products - reactants to find Sº (a state function)

More complex molecules higher Sº.

19.3 Entropy and Third Law

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19.4 Gibbs Free Energy

Heat that flows into or out of the system changes the entropy of the surroundings.

For an isothermal process:

Ssurr =qsys

T

• At constant pressure, qsys is simply H for the system.

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The universe is composed of the system and the surroundings.

Therefore,Suniverse = Ssystem + Ssurroundings

For spontaneous processes

Suniverse > 0

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19.4 Gibb’s Free Energy

This becomes:

Suniverse = Ssystem +

Multiplying both sides by T,

TSuniverse = Hsystem TSsystem

Hsystem

T

TΔS universe is defined as the Gibbs free energy, G.When Suniverse is positive, G is negative.Therefore, when G is negative, a process is spontaneous.

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19.4 Gibb's Free Energy at any Conditions

G=H-TS Never used this way. At constant temperature

G=Hsys –TSsys

G function eliminates the need to deal with entropy of the surroundings

If G is negative at constant T and P, the process is spontaneous.

We deal only with the SYSTEM.

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The Gibbs Free Energy is a measure of the maximum amount of work, at a given temperature and pressure, that can be done on the surroundings by a system.

Never really achieved because some of the free energy is changed to heat during a change, so it can’t be used to do work.

Wmax = G For a spontaneous process:

Maximum amount of energy released by the system that can do useful work on the surroundings

Energy available from spontaneous process that can be used to drive non-spontaneous process.

For a nonspontaneous process: Minimum amount of work that must be done to

force the process to occur

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Summary of Conditions for Spontaneity G < 0

reaction is spontaneous in the forward direction

(Suniv > 0) G > 0

reaction is nonspontaneous in the forward direction

(Suniv < 0) G = 0

SYSTEM IS AT EQUILIBRIUM (Suniv = 0)

Remember- Spontaneity tells us nothing about rate.

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19.4 Gibbs Standard Energy Free Energy

The standard free energies of formation, G°, are the free energy values for the formation of a substance under standard conditions.

The standard free energy of formation for any element in its standard state is zero. (Like enthalpy, but not entropy)

Compare: G = Hsys - TSsys (any conditions)

Gº = H0sys - TS0

sys (standard conditions)

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19.4 Convention for Standard States

State of Matter Standard State

Gas 1 atm pressure

Liquid Pure liquid

Solid Pure solid

Elements Gºf = 0

Solution 1 molar

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19.4 Free Energy in Reactions

Gº = standard free energy change. Free energy change that will occur if reactants in their

standard state turn to products in their standard state. Can’t be measured directly, can be calculated from

other measurements. The reaction:

aA + bB → cC + dD

Gºrxn = ΣnGº (products) - ΣmGº (reactants)

or Gº=Hº-TSº

Use adapted Hess’s Law with known reactions.

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19.4 Free Energy Changes

At temperatures other than 25°C,G ° = H TS How does G change with temperature? There are two parts to the free energy

equation: H — the enthalpy term TS — the entropy term

The temperature dependence of free energy, then comes from the entropy term.

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19.4 Gibbs free Energy: Example

16. For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0?

Look at the equation G=H-TS Spontaneity can be predicted from the

sign of H and S.

(ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C)

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ΔH ΔS -|TΔS| ΔG = ΔH-TΔS Reaction Characteristics

Example

- + -Always negative Spontaneous at

all temperatures2O3(g) →3O2(g)

+ - +Always positive

Nonspontaneous at all temperatures

3O2(g) →2O3(g)

- - +

Negative at low T; positive at high T

Spontaneous at low T; nonspontaneous at high T. Enthalpy driven

H2O(l) →H2O(s)+

+ + -

Positive at low T; negative at high T

Nonspontaneous at low T; becomes spontaneous at high T. Entropy driven.

H2O(s) → H2O(l)

19.4 Gibbs Free Energy and SpontaneityEffect of Temperature on the Spontaneity of

Reactions

1

2

3

4

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Free energy change as a function of temperature.

19.4 Gibbs Free Energy and Spontaneity

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19.4 Gibbs Free Energy and Temperature

Spontaneous Processes H2O(s) melts above 0C (endothermic)

H > 0 (endo process), S > 0 ENTROPY DRIVEN

Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) 4Fe(s) + 3O2(g) 2Fe2O3(s) H < 0 (Exo) , S < 0 (entropy decreases) ENTHALPY DRIVEN

G < 0 for each process T determines sign of G: G = H -TS

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19.4 Gibbs Free Energy: Example (p. 748)

17. Predict which of the four cases in Table (slide #78) you expect to apply to the following reactions:a) C6H12O6 (s) + 6O2(g) → 6CO2(g) + 6H2O(g) ΔH = -2540 kJ

b) Cl2(g) → 2Cl(g)

18. Calculate ΔG0 at 298 K for the reaction 4HCl(g) + O2(g) → 2Cl2(g) + 2H2O(g) ΔH0 = 114.4 kJ

a) using Gibbs free Energy equationb) from standard free energies of formation.

( -76.0 kJ)

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19.4 Gibbs free Energy: Example

18. For the reaction SO2(g) + 2H2S(g) →3S(s) + 2H2O(g)

Calculate the temperature at which ΔG0 = 0Values of ΔH0, kJ/mol ΔS0 (kJ/(K mol) SO2 -296.8 0.2481

H2S -20.6 0.2057S 0.0 0.0318H2O -241.8 +0.1887

Answer: 780K

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19.4 Temperature and Chemical Reactions – Problem Set

19. Calculate the temperature at which the reaction:

CaCO3(s) → CaO(s) + CO2(g) becomes spontaneous.

ΔS º = 160.5 J/K; ΔH º = 177.8 kJ

Answer: 835 ºC

20. At its normal boiling point, the enthalpy of vaporization of pentadecane, CH3(CH2)13CH3, is 49.45 kJ/mol. What should be its approximate normal boiling point temperature be? ΔS0 of vaporization is 87 J mol-1 K-1

(570 K)

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21. Consider the reaction C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ΔH0 = -2220

kJ(a) Without using data from the Thermodynamic

Tables, predict whether ΔG0 for this reaction is more negative, or less negative than ΔH0.

(b) Use data from the Tables to calculate ΔG0 at 298 K. Is your prediction correct?Values of ΔG0 (in kJ/mol): C3H8(g), -23.5; O2(g), 0; CO2(g) -394.4; H2O(l) -237.2

(a) less negative; (b) -2108 kJ

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19.5 Gibbs free Energy and Equilibrium

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19.5 Free Energy and Equilibrium

ΔG and ΔGº are not the same. ΔG = ΔH – T ΔS

ΔGº = ΔH º – T ΔS º ΔGº is only at standard conditions

(values obtained from Tables)ΔGº = Gº(products) - ΔGº

(reactants)

ΔG any conditions (no Tables of values available)

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19.5 Free Energy and Equilibrium

Predicament: we start a reaction with all reactants in standard state (1 atm, 25ºC, 1M solution). Is the standard state preserved as the reaction progresses?

It can be shown mathematically that at non-standard conditions:

]tan[

[products] quotient reaction ,

tsreacQ

ΔG = ΔGº +RTlnQ

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If ΔG is negative, the forward reaction is spontaneous.

If Δ G is 0, the system is at equilibrium.If G is positive, the reaction is spontaneous in the

reverse direction.

K>Q

K<Q

K=Q

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19.5 Free Energy and the Equilibrium Constant

At equilibrium:

eq

eq

KRTG

KRTG

ln

ln0

From the above we can conclude:If G < 0, then K > 1 Product favoredIf G = 0, then K = 1 EquilibriumIf G > 0, then K < 1 Reactants favored

K = e-(ΔGº /RT)

Kc used for solutions and molarities, Kp used for gases

Useful equation to determine small

Keq

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19.5 Temperature Dependence of K

Gº= -RTlnK = Hº - TSº

ln(K) = Hº/R(1/T)+ Sº/R

A straight line of lnK vs 1/T Slope?

Y-intercept?

19.5 Free Energy and Chemical Equilibrium

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19.5 Reaction Path and ΔGº

ΔGº = -RTln K

Value of ΔGº Sign of K Path of reaction

Negative, <0 K>0 Products are favored

Positive, >0 K<0 Reactants are favored

zero, =0 K = 1 Equilibrium

92

19.5 Keq: Problem Set – 19.6 (p. 755)

22. Determine the value of Keq at 25 ºC for the reaction:

2NO2(g) ↔ N2O4(g).

Note: calculate ΔG0 from tables Use ΔG0 = -RTln Kp

( 6.9)

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19.5 Keq: Problem Set 19.7 (p. 755)

23. Using the solubility product of of silver iodide at 25 ºC (8.5 x 10-17), calculate ΔGº for the process:

AgI(s) ↔ Ag+(aq) + I-(aq)Answer:24. Estimate the value of ΔS0

298 for the dissociation of copper (II) oxide.

CuO(s) ↔ 2Cu2O(s) + O2(g) ΔH0298 = 283 kJ

( 0.203 kJ K-1

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19.5 Gibbs and Equilibrium: Example

25. Calculate Grxn for the reaction below:

2A(aq) + B(aq) C(aq) + D(g) if G°rxn = 9.9 x 103 J/mol and 150 °C(a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05

M, and PD = 0.05 atm, and (b) (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5

M, and PD = 0.5 atm. (c) Is the reaction spontaneous under

these conditions?

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26. Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard

state conditions? (ans.: 27 kJ)

27. Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions? For the reaction below, K = 1.0 x 1014.

H+ + OH- H2O

(ans.: -80 kJ)


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28. Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored?

(ans.: a) 0.15; b) 216; 250C)


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19.5 Summary

First Law of Thermodynamics: ΔU = Δq + Δw

Second Law: ΔSuniv = ΔS sys + ΔSsurr

ΔS sys= ΔH/TΔG = ΔH – T ΔSΔG = ΔGº +RTlnK

ΔGº = -RTlnK

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Many biological reactions essential for life are nonspontaneous Spontaneous reactions used to “drive”

the nonspontaneous biological reactions Example: photosynthesis

6CO2 + 6H2O C6H12O6 + 6O2 G > 0

What spontaneous reactions drive photosynthesis?

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19.5 Entropy and Life Processes

If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible? growth of a complex life form

represents an increase in order (less randomness) lower entropy

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19.5 Entropy and Life Processes

CO2

H2

Oheat

Organisms “pay” for their increased order by increasing Ssurr.

Over lifetime, Suniv > 0.

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Thermodynamics Entropy, Energy and equilibrium