Chapter 18 Entropy, Free Energy, and Equilibrium Part I

Chapter 18

Entropy, Free Energy, and Equilibrium

Part I

1Dr. Al-Saadi

Spontaneous Processes Chemical processes can be classifies as :

o Spontaneous processes: processes that occur “naturally” under a specific set of conditions.

• Combustion of hydrocarbons:CH4(g) + 2O2(g) CO2(g) + 2H2O(l) ΔH= ‒ 890 kJ/mol

A spontaneous process involves a decrease in the energy of the system. it is usually “but not always” exothermic.

o Nonspontaneous processes: processes that do not occur “naturally” under a specific set of conditions.CO2(g) + 2H2O(l) CH4(g) + 2O2(g) ΔH= + 890 kJ/mol

A nonspontaneous process involves an increase in the energy of the system. it is usually “but not always” endothermic.

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18.1

Spontaneous Processes

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Reaction of sodium with water is so vigorous such that water catches fire!!

http://www.youtube.com/watch?v=uqDWbknpiVk&feature=related



Spontaneous Processes At room temperature :

o Freezing of water:H2O(l) H2O(s) ΔH= ‒ 6.01 kJ/mol

is a nonspontaneous process although it is exothermic.

o Melting of ice:H2O(s) H2O(l) ΔH= + 6.01 kJ/mol

is a spontaneous process although it is endothermic.

To predict spontaneity, it is not enough to consider only the reaction energy or enthalpy. Another important thermodynamic quantity called entropy helps in predicting the spontaneity of a given system.

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What is Entropy?

Entropy (S): is a thermodynamic quantity that measures the disorder of a system.In general, a greater disorder means a greater entropy, and a greater order means a smaller entropy.

Like enthalpy (H) and internal energy (U), entropy (S) is a state function.

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Probability

A probable event can happen in many different ways.

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Shuffling

Reshuffling

An improbable event can happen in only a small number of ways (may be only one way).

Very highly ordered Disordered

Microstates

What will happen if the stopcock between the two bulbs is opened?The molecules originally confined in the left-side bulb expand and evenly occupy both bulbs. “a spontaneous process”

Let us inspect this more carefully by considering the microscopic states “microstates” of a system starting with a very small number of molecules.

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The state of problem:A sample of gaseous molecules is confined in the left bulb, and the molecules are not

allowed to go to the other bulb

stopcock

Microstates

The number of molecules is The number of possible positions (left or right) isThe number of microstates is

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122

Probability of finding the molecule in the left side bulb is 1/2 .

Microstates

The number of molecules is The number of possible positions isThe number of microstates is

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Microstates

The number of molecules is The number of possible positions isThe number of microstates is

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Number of microstates = nx

x : number of molecules n : number of possible positions


Microstates and Most Probable Distribution

The number of molecules is The number of possible positions isThe number of microstates for this distribution

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421

Distribution I 4 0


Number of microstates = nx = 24 = 16



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Distribution II 0 4



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Distribution III 3 1



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Distribution IV 1 3



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Distribution V2 2

The most probable distribution

because more microstates are

involved

Does this make sense??Yes, the gas molecules are more probably going to split evenly between the two bulbs.

Microstates and Least Probable Distribution

It is very highly improbable for the gas molecules to stay in the left bulb and not distributed between the two bulbs (nonspontaneous process). This will almost never occur if you have 1 mole of molecules.

The gas will spontaneously expand so that the molecules will fill up both bulbs evenly. This satisfies a larger number of microstates for the gas molecules.

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k : Boltzmann constant (1.38 x 1023 J/K)

The larger the value of W for a system (more microstates), the greater its entropy.

The entropy change (ΔS) for a process is given by:ΔS = Sf – Si

ΔS = k ln Wf – k ln Wi

When the number of microstates increases during a process, the entropy of that process increases (ΔS > 0).

Entropy and Microstates In 1868, Boltzmann showed that the entropy of a system is

related to the natural log of the number of microstates (W)

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S = k ln W

Entropy and Macrostates

The previous equation can not be used to calculate the entropy for macroscopic systems.

Normal calculators can’t handle x > 500.

Entropy is normally determined using calorimetry experiments.

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W = nx

S = k ln W

Standard Entropy The standard entropy (S) is the absolute entropy of a

substance at 1 atm (typically at 25C). Its units is (J/K mol).∙

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General Trends for Entropy Entropy for the gas phase is greater than that of the liquid or

solid phase of the same substance. More complex structures have a greater entropy. For allotropes, more ordered forms have a lower entropy. Heavier monoatomic elements have a greater entropy than

lighter ones.

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Entropy Change in a System

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One can predict qualitatively whether the entropy of a system increases (ΔS > 0) or decreases (ΔS < 0).

(a) Melting:

solid liquid

Particles are confined to fixed positions. They are highly ordered and occupy a small number of microstates.

Particles can now occupy more possible positions. They become less ordered and occupy a greater number of microstates compared to the solid state.Ssolid Sliquid<

ΔS > 0


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(b) Vaporization:

liquid

gas

Sliquid Sgas<<

ΔS > 0

Particles in the gaseous phase occupy far more microstates than the liquid phase.


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(c) Dissolving:

solution

Spure substance Ssolution<

ΔS > 0

When the highly ordered particles of the pure substance and the ordered solvent molecules are mixed, the order in both are disrupted. The molecules in the solution, as a result, will have a grater number of microstates.

pure substance

Entropy Change in a System(d) Heating:

Heating always increases both the energy and the entropy of a system.

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Slower temp. Shigher temp.<

ΔS > 0


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(e) Chemical reactions:A reaction that results in an increase in the number of moles of gas always increases the entropy of the system.

ΔS > 0

Sfewer moles Smore moles<

Entropy Exercise:

Determine the sign of ΔS for the following processes: o Liquid nitrogen evaporates.

ΔS > 0o Two clear liquids are mixed and a solid yellow

precipitate forms.ΔS > 0 for mixing of the two liquids.ΔS < 0 for formation of precipitation.

o Liquid water is heated from 22.5 C to 55.8 C ΔS > 0

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The Universe, the System, and the Surroundings

The universe is made up of two parts: the system and the surroundings

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SystemReactants

and products

Surroundings

Reaction container,

room, everything else

ΔSuniv = ΔSsys + ΔSsurr

Second Law of Thermodynamics There are two types of processes:

o Spontaneous processes: They occur under a specific set of conditions.In a spontaneous process, the entropy of the universe increases.

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o Equilibrium processes: They do not tend to occur. However, They can be made to occur by the addition or removal of energy. (e.g. melting of ice at 0C)In an equilibrium process, the entropy of the universe remains unchanged.

ΔSuniv = ΔSsys + ΔSsurr > 0

ΔSuniv = ΔSsys + ΔSsurr = 0

Entropy Change in a System (ΔSrxn)

For a system represented by the following equation:a A + b B c C + d D

The change in standard entropy of this reaction (ΔSrxn) is:

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ΔSrxn = [c S(C) + d S(D)] ‒ [a S(A) + b S(B)]

In general, the change of entropy of a chemical reaction (a system) is given by:

ΔSrxn = ΣnS(products) ‒ ΣmS(reactants)where n and m are the stoichiometric coefficients of the reactants and products in a given equation.

Entropy Change in a System (ΔSrxn) Exercise:

Calculate the standard entropy change for the following reaction using the table of standard values. First, predict the sign for ΔSrxn qualitatively.

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

The entropy is expected to increase (2 mol 4 mol) ΔSrxn = ΣnS(products) ‒ ΣmS(reactants)

= [(1)(191.5 J/K·mol) + (3)(131.0 J/K·mol)] ‒ [(2)(193.0 J/K·mol)]

= 584.5 J/K·mol ‒ 386 J/K·mol

= 198.5 J/K·mol (Entropy increases)Dr. Al-Saadi 30

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The values of S can be found in Appendix 2

Entropy Change in a System (ΔSrxn)

The ΔSrxn values of the following reactions are calculated from the standard entropy values.

ΔSrxn (J/K·mol)

N2(g) + 3H2(g) 2NH3(g) ‒ 198.5

CaCO3(s) CaO(s) + CO2(g) 160.5

H2(g) + Cl2(g) 2HCl(g) 20.0

Recall that the reaction that results in an increase in the number of moles of gas always increases the entropy of the system. The opposite is also true. When there is no change in the number of moles of gas, the calculated ΔSrxn will be a relatively a small value with a positive or negative sign.

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Entropy Change in the Surroundings (ΔSsurr)

Change in entropy of the surroundings is directly proportional to the enthalpy of the system. At constant temperature:o an exothermic process corresponds to an increase of the

entropy of the surroundings. (ΔSsurr > 0)o an endothermic process corresponds to a decrease of the

entropy of the surroundings. (ΔSsurr < 0)

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ΔSsurr α ‒ ΔHsys ①

Just think about it. Any system naturally tends to lower its energy (lose energy). This is in consistent with the second law of thermodynamics which states that the entropy of the universe is always positive.


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The transfer of a given quantity of energy from the system to the surroundings as heat produces a greater disorder in the surrounding at a low temperature that it does at a high temperature. Thus, the magnitude of change in entropy of the surroundings is inversely proportional to the temperature.

ΔSsurr α 1/T ②

Combining ① and ② gives:

ΔSsurr = ‒ ΔHsys

T


Exercise:Consider the following two reactions at 25C:Sb2S3(s) + 3Fe(s) 2Sb(s) + 3FeS(s) ΔH = ‒ 125 kJ/mol

Sb4O6(s) + 6C(s) 4Sb(s) + 6CO(g) ΔH = 778 kJ/mol

Calculate ΔSsurr for each reaction at 1 atm and 25C.

We use:

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ΔSsurr = ‒ ΔHsys

T

For the first reaction: ΔSsurr = ‒ (‒ 125 KJ/mol)

298 K

For the second reaction:

= 419 J/K·mol

ΔSsurr = ‒ (778 KJ/mol)

298 K= ‒ 2610 J/K·mol

Calculating ΔSuniv

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Exercise:For the reaction: 2NH3(g) N2(g) + 3H2(g) , ΔSrxn and ΔHrxn are 198.5 J/K·mol and ‒92.6 kJ/mol , respectively. Calculate ΔSuniv associated with the reaction above.

18.3

Predicting Spontaneity of a Process

Here are the possible combinations of , and how to predict whether a chemical process is spontaneous or not.

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ΔSuniv = ΔSsys + ΔSsurr

Third Law of Thermodynamics For a perfect crystalline substance at absolute

zero (0 K), there is only one way to arrange its constituents. They are not moving at all (the number of microstates is just one).

S = k ln Wwith W = 1S = k ln (1) = 0

The third law of thermodynamic states that entropy of a perfect crystalline substance at absolute zero temperature is zero.

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Third Law of Thermodynamics As the temperature of a perfect crystal is raised, the random

vibrational motions increase, and disorder (and number of microstates) increases within the crystal.

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A perfect crystal of HCl at 0 K. (S = 0)

Increasing temperature

More vibrations, more disorder and a greater number of microstates. (S = +ve)

Third Law of Thermodynamics The absolute entropy of a substance at any given temperature

can be determined using the third law of thermodynamics.ΔS = Sf ‒ Si

ΔS = Sf

The measured change in entropy is equal to the absolute entropy at the new temperature.

Unlike enthalpy and internal energy, the true value of entropy of any substance can be determined. We usually refer to the standard entropies (S)

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Si is zero if the substance starts at 0 K.

Entropy Change as a Function of Temperature

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Spontaneity of a Process Considering Only the System

Starting from the 2nd law of thermodynamics for a spontaneous process:

ΔSuniv = ΔSsys + ΔSsurr > 0

Also, we have: ΔSsurr = ‒ ΔSsys / T

Substituting ② into ① gives:

ΔSuniv = ΔSsys ‒ > 0

T ΔSuniv = T ΔSsys ‒ ΔHsys > 0

In terms of only the system: ΔHsys ‒ T ΔSsys < 0

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①

②

ΔHsys

T

At constant T and P, the process is spontaneous only when the change of entropy and enthalpy of the system are such that ΔHsys ‒ T ΔSsys is less than zero.

③

Gibbs Free Energy Gibbs free energy (or simply free energy) is another

thermodynamic quantity and is defined as:G = H – TS

The change in Gibbs free energy for a system at constant temperature is:

ΔG = ΔH – T ΔS Getting back to equation ③, we have:

ΔG = ΔH – T ΔS < 0o ΔG < 0 ; the forward reaction is spontaneous.o ΔG > 0 ; the forward reaction is nonspontaneous.o ΔG = 0 ; the reaction is at equilibrium.

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Predicting the Sign of ΔG

The sign of ΔG can be predicted from knowing the sign of ΔH and ΔS for the reaction.

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ΔG = ΔH – T ΔS < 0

Standard Free-Energy Change

The standard free energy (ΔGrxn) of a system is the change in free energy when reactants in their standard states are converted to products in their standard states.

Standard states are:o Gases 1 atmo Liquids Pure liquid o Solids Pure solido Elements The most stable form at 1 atm and 25Co Solutions 1 mol concentration

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Standard Free-Energy Change

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Consider the following equation:a A + b B c C + d D

The change in standard free-energy (ΔGrxn) is:

ΔGrxn = [c ΔGf (C) + d ΔGf (D)] ‒ [a ΔGf (A) + b ΔGf (B)]

In general, the change of entropy of a chemical reaction (a system) is given by:

ΔGrxn = Σn ΔGf (products) ‒ Σm ΔGf (reactants)where n and m are the stoichiometric coefficients of the reactants and products in a given equation.

Standard Free Energy of Formation

The standard free energy of formation (ΔGf) of a compound is the change in free energy that occurs when 1 mole of that compound is formed from reactants in their standard states. o It is very similar to the concept of the standard enthalpy of

formation (ΔHf) that you learned in CHEM 101.

N2(g) + 2O2(g) 2NO2(g) ΔG = 103.6 kJ

½ N2(g) + O2(g) 1NO2(g) ΔG = 51.8 kJ = ΔGof (NO2)

C(s) + 2H2(g) + ½O2(g) CH3OH(l) ΔGof = –166 kJ/mol

2Fe(s) + ½O2(g) Fe2O3(s) ΔGof = – 741 kJ/mol

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Elements @ standard statesΔG

o

f = 0 ΔGo

f = 34 kJ/mol

Standard Free-Energy ChangeExercise:Calculate the standard free-energy change for the following reaction at 25C.

2KClO3(s) 2KCl(s) + 3O2(g)

ΔGrxn = Σn ΔGf (products) ‒ Σm ΔGf (reactants) = 2(ΔGf (KCl(s)) + 3(ΔGf (O2(g)) – 2(ΔGf (KClO3(s)) = 2(– 408.3 kJ/mol) + 3(0) – 2(– 289.9 kJ/mol)

= – 816.6 – (– 579.8) = – 236.8 kJ/mol The reaction is spontaneous. A negative ΔG corresponds to a larger equilibrium constant

(K), while a positive ΔG corresponds to a smaller equilibrium constant (K).

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Standard Free-Energy Change Exercise:

The reaction CaCO3(s) CaO(s) + CO2(g) has ΔH= 177.8 kJ/mol and ΔS= 160.5 J/K mol. Predict the spontaneity of ∙the reaction at room temperature and at 1000C.

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Standard Free-Energy Change The temperature affects the sign of ΔG of chemical reactions. Consider the melting of ice:

H2O(s) H2O(l) ΔH= 6.03 kJ/mol ; ΔS= 22.1 J/K mol∙

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The process is :o spontaneous at 10C. (Ice melts at this temperature)o nonspontaneous at ‒ 10C. (Ice does not melt at this temperature)o at equilibrium at 0C. (Ice and water are at equilibrium at this

temperature)

ΔG is ‒ve ΔG is +ve ΔG is 0

Relationship between ΔG and ΔG Most of the reactions take place in states other than their

standard states. Thus, in order to determine spontaneity of a reaction, we need to know how to calculate ΔG when the reaction is not occurring at standard states.

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ΔG = ΔG + RT lnQ

where:ΔG is the non-standard free energyΔG° is the standard free energy (from appendix 2)

R = 8.314 J/K mol∙T is in Kelvin Q is the reaction quotient.

Relationship between ΔG and ΔG Exercise:

Consider the reaction: H2(g) + Cl2(g) 2HCl(g)

How does the value of ΔG change when the pressures of the H2, Cl2 and HCl gases are changed to 0.25 atm, 0.45 atm and 0.30 atm, respectively at 25 C?

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ΔG = ΔG + RT lnQFirst of all calculate ΔG using appendix 2. ΔG° = [2(‒95.27 kJ/mol)] [0 + 0] = 190.54 kJ/mol Secondly, calculate Q.

80.0)45.0)(25.0(

)30.0())((

)( 2

ClH

2HCl

22

PP

PQP

Cont.

Relationship between ΔG and ΔG

Thirdly. Find the non-standard free energy.ΔG = ΔG + RT lnQΔG = 190540 J/mol + (8.314 J/K·mol)(298 K) ln (0.80)ΔG = 191.09 kJ/mol The reaction becomes more spontaneous since the free energy is more negative. Thus, more HCl will be formed with the given pressures.The reaction will continue producing more HCl and consuming more H2 and Cl2 until QP = KP .

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Relationship between ΔG and K

At equilibrium:Q = K and ΔG = 0

Thus, at equilibrium:ΔG = ΔG + RT lnQ 0 = ΔG + RT lnK

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ΔG = ‒ RT lnK

The larger K is, the more negative ΔG is, and the reaction proceeds more towards the product side.

Chapter 18 Entropy, Free Energy, and Equilibrium Part I

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Transcript of Chapter 18 Entropy, Free Energy, and Equilibrium Part I