Chapter 17 Free Energy and Thermodynamics Lesson 1.

Chapter 17

Free Energy and

ThermodynamicsLesson 1

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First Law of Thermodynamics

• First Law of Thermodynamics: Energy cannot be created or destroyedthe total energy of the universe cannot changeit can be transfered from one place to another

Euniverse = 0 = Esystem + surroundings

system = reactants & products

surroundings = everything else

(the transfer of energy from one to the other does not change the energy of the universe)

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First Law of ThermodynamicsFor an exothermic reaction, heat from the system

goes into the surroundings• two ways energy can be “lost” from a system,

converted to heat, qused to do work, w Energy conservation requires that the energy change in the

system = heat exchanged + work done on the system. E = q + w (E = internal energy change) E = H – PV (at const. P, qp = H, enthalpy

change)

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Enthalpy, H• related to (includes) the internal energy• H generally kJ/mol• stronger bonds = more stable molecules• if products more stable than reactants, energy

released; exothermic H = negative

• if reactants more stable than products, energy absorbed; endothermic H = positive

• The enthalpy is favorable for exothermic reactions and unfavorable for endothermic reactions.

• Hess’ Law: H°rxn = (Hf°prod) - (Hf°react)

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Thermodynamics and Spontaneity• thermodynamics predicts whether a process will

proceed (occur) under the given conditionsspontaneous process

nonspontaneous process does not occur under specific conditions.

• spontaneity is determined by comparing the free energy (G) of the system before the reaction with the free energy of the system after reaction. if the system after reaction has less free energy than

before the reaction, the reaction is thermodynamically favorable.

• spontaneity ≠ fast or slow (rate); this is kinetics

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Spontaneous Nonspontaneousice melts @ 25oC water freezes @ 25oC

2Na(s) + 2H2O(l) H2(g) + 2NaOH(aq) H2(g) + 2NaOH(aq) 2Na(s) + 2H2O(l)

ball rolls downhill ball rolls uphill

water freezes @ -10 oC ice melts @ -10oC

Copyright McGraw-Hill 2009

Entropy Changes in a System Qualitative

• Ssolid < Sliquid



• Sliquid < Svapor



• Spure < Saqueous



Slower temp < Shigher temp

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Diamond → Graphite

Graphite is thermodynamically more stable than diamond, so the conversion of diamond into

graphite is spontaneous – but it’s kinetically too slow (inert) it will never happen in many, many

generations.

kinetics

Spontaneity:direction & extent

kinetics:how fast

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Factors Affecting Whether a Reaction Is Spontaneous

• The two factors that determine the thermodynamic favorability are the enthalpy and the entropy.

• The enthalpy is a comparison of the bond energy of the reactants to the products. bond energy = amount needed to break a bond. H

• The entropy factors relate to the randomness/orderliness of a system S

• The enthalpy factor is generally more important than the entropy factor

Substance Hf° kJ/mol

Substance Hf° kJ/mol

Al(s) 0 Al2O3 -1669.8 Br2(l) 0 Br2(g) +30.71

C(diamond) +1.88 C(graphite) 0 CO(g) -110.5 CO2(g) -393.5 Ca(s) 0 CaO(s) -635.5 Cu(s) 0 CuO(s) -156.1 Fe(s) 0 Fe2O3(s) -822.16 H2(g) 0 H2O2(l) -187.8

H2O(g) -241.82 H2O(l) -285.83 HF(g) -268.61 HCl(g) -92.30 HBr(g) -36.23 HI(g) +25.94

I2(s) 0 I2(g) +62.25 N2(g) 0 NH3(g) -46.19 NO(g) +90.37 NO2(g) +33.84 Na(s) 0 O2(g) 0 S(s) 0 SO2(g) -296.9

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Entropy, S

• Entropy is a thermodynamic function that increases as the number of energetically equivalent ways of arranging the components increases. S generally in J/K (joules/K)

• S = k lnWk = Boltzmann Constant (R/NA) = 1.38 10-23 J/K

W is the number of energetically equivalent ways, (microstates). It is unitless.

Entropy is usually described as a measure of the randomness or disorder; the greater the disorder of a system, the greater its S.The greater the order the smaller its S.

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Entropy & Microstates, WEnergetically Equivalent States for the Expansion of a Gas (4 gas molecules) 1 microstate

1 microstate

6 microstates(most probable distribution)

S = k ln W

S = k ln Wf - k ln Wi

if Wf > Wi , S > 0 & entropy increases.

i

f

W

Wk lnS

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Changes in Entropy, S• entropy change is favorable when the result is a

more random system (State C: higher entropy). S is positive (S > 0)

Some changes that increase the entropy are:rxns where products are in a more disordered state.

(solid > liquid > gas) less order(solid< liquid < gas) larger S (disorder)

reactions which have larger numbers of product molecules than reactant molecules.

increase in temperature (more movement)solids dissociating into ions upon dissolving



• Sfewer moles < Smore moles

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Changes in Entropy in a System(melting)

Particles fixed in space

Particles can occupy many positions

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Changes in Entropy in a System(vaporization)

Particles occupy more space (larger volume)

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Changes in Entropy in a System (solution process)

Structure of solute and solvent disrupted (also more solute particles)

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17.1

Strategy

To determine the entropy change in each case, we examine whether the number of microstates of the system increases or decreases. The sign of ΔS will be positive if there is an increase in the number of microstates and negative if the number of microstates decreases.

Solution(a) Upon freezing, the ethanol molecules are held rigid in position.

This phase transition reduces the number of microstates and therefore the entropy decreases; that is, ΔS < 0.

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17.1

(b) Evaporating bromine increases the number of microstates because the Br2 molecules can occupy many more positions in nearly empty space. Therefore, ΔS > 0.

(c) Glucose is a nonelectrolyte. The solution process leads to a greater dispersal of matter due to the mixing of glucose and water molecules so we expect ΔS > 0.

(d) The cooling process decreases various molecular motions. This leads to a decrease in microstates and so ΔS < 0.

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The 2nd Law of Thermodynamics

• The entropy of the universe increases in a spontaneous process. Suniverse = Ssystem + Ssurroundings > 0

Suniverse = Ssystem + Ssurroundings = 0 (equilibrium)

If Ssystem >> 0, Ssurroundings < 0 for Suniverse > 0! If Ssystem < 0, Ssurroundings >> 0 for Suniverse > 0!

• the increase in Ssurroundings often comes from the heat released in an exothermic reaction, system < 0.

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Entropy Changes in the Surroundings (Ssurr)

Exothermic ProcessSsurr > 0

Endothermic ProcessSsurr < 0

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The 3rd Law of Thermodynamics

S = k ln W

S = k ln W = k ln 1 = 0

S = Sf – Si; where Si = 0 @ 0 K

the absolute entropy of a substance is always (+) positive at the new T

-allows determination of entropy of substances.(W = 1, there is only one way to arrange the particles to form a perfect crystal)

the 3rd Law states that for a perfect crystal at absolute zero, the absolute entropy = 0 J/mol∙K

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Third Law of Thermodynamics

The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

S = k ln W

W = 1

S = 0

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Standard Entropies

• S°

• entropies for 1 mole at 298 K for a particular state, a particular allotrope, particular molecular complexity, a particular molar mass, and a particular degree of dissolution

Values can be used to calculate the standard

entropy change for a reaction, Sorxn (= So

sys)

Substance S° J/mol-K

Substance S° J/mol-K

Al(s) 28.3 Al2O3(s) 51.00 Br2(l) 152.3 Br2(g) 245.3

C(diamond) 2.43 C(graphite) 5.69 CO(g) 197.9 CO2(g) 213.6 Ca(s) 41.4 CaO(s) 39.75 Cu(s) 33.30 CuO(s) 42.59 Fe(s) 27.15 Fe2O3(s) 89.96 H2(g) 130.58 H2O2(l) 109.6

H2O(g) 188.83 H2O(l) 69.91 HF(g) 173.51 HCl(g) 186.69

HBr(g) 198.49 HI(g) 206.3 I2(s) 116.73 I2(g) 260.57 N2(g) 191.50 NH3(g) 192.5 NO(g) 210.62 NO2(g) 240.45 Na(s) 51.45 O2(g) 205.0 S(s) 31.88 SO2(g) 248.5

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Trends: Standard EntropiesMolar Mass

• For monatomic species, the larger the molar mass, the larger the entropy

• available energy states more closely spaced, allowing more dispersal of energy through the states

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Trends: Standard EntropiesStates

• the standard entropy of a substance in the gas phase is greater than the standard entropy of the same substance in the solid or liquid phase at a particular temperature

SubstanceS°,

(J/mol∙K)

H2O (l) 70.0

H2O (g) 188.8

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Trends: Standard EntropiesAllotropes

• the more highly ordered form has the smaller entropy

-different forms of an element

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Trends: Standard EntropiesMolecular Complexity (inc # of atoms)

larger, more complex molecules generally have larger entropy

•more available energy states, allowing more dispersal of energy through the states

SubstanceMolar

MassS°,

(J/mol∙K)

Ar (g) 39.948 154.8

NO (g) 30.006 210.8

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Trends: Standard EntropiesDissolution

• dissolved solids generally have larger entropy

• distributing particles throughout the mixture

SubstanceS°,

(J/mol∙K)

KClO3(s) 143.1

KClO3(aq) 265.7

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Q. Arrange the following in order of increasing entropy @ 25oC! (lowest to highest)

Ne(g), SO2(g), Na(s), NaCl(s) and H2(g)

Na(s) < NaCl(s) < H2(g) < Ne(g) < SO2(g)

Calculate S for the reaction4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g)

S is +, as you would expect for a reaction with more gas product molecules than reactant molecules

standard entropies from Appendix IIB

S, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

SSNH3, SO2, SNO, SH2O,

reactantsproducts SSS rp nn

K

J

K

J

K

J

K

J

K

J

)(O)(NH)O(H)NO(

reactantsproducts

8.178

)]2.205(5)8.192(4[)]8.188(6) 8.210(4[

)]S(5)S(4[)]S(6)S(4[

SSS

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gggg

rp nn

Substance S, J/molK

NH3(g) 192.8

O2(g) 205.2

NO(g) 210.8

H2O(g) 188.8

Consider 2 HConsider 2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25O(liq) @ 25ooCC

∆∆SSoo = 2 S = 2 Soo (H (H22O) - [2 SO) - [2 Soo (H (H22) + S) + Soo (O (O22)])]

∆∆SSoo = 2 mol (69.9 J/K•mol) - = 2 mol (69.9 J/K•mol) - [2 mol (130.6 J/K•mol) + [2 mol (130.6 J/K•mol) +

1 mol (205.0 J/K•mol)]1 mol (205.0 J/K•mol)]

∆So = -326.4 J/KNote that there is a Note that there is a decrease in S decrease in S because 3 mol of because 3 mol of

gas give 2 mol of liquid.gas give 2 mol of liquid.

Calculating ∆SCalculating ∆Soo for a Reaction for a Reaction

∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)∆∆SSoo = = S Soo (products) - (products) - S Soo (reactants) (reactants)

The more negative Hsyst and the lower the temperature the higher

(more positive) Ssurr 37

Temperature Dependence of Ssurroundings

system < 0 (exothermic), it adds heat to the surroundings, increasing the entropy of the surroundings (Ssurroundings > 0 )

system > 0 (endothermic), it takes heat from the surroundings, decreasing the entropy of the surroundings (Ssurroundings < 0 )

T

ΔHΔS system

gssurroundin

systemgssurroundin ΔHΔS

T

1gssurroundinΔS

K

J3K

kJsurr

syssurr

106.86 86.6S

K 298

kJ 2044

T

HS

The reaction C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) has Hrxn = -2044 kJ at 25°C.

Calculate the entropy change of the surroundings.

combustion is largely exothermic, so the entropy of the surroundings should increase (inc in # gas mol)

Hsystem = -2044 kJ, T = 298 K

Ssurroundings, J/K

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

ST, H

T

ΔHΔS sys

surr

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25 O(liq) @ 25 ooCC

∆Sosystem = -326.4 J/K

K 298

J/kJ) kJ)(1000 (-571.7 - = gssurroundin

oSK 298

J/kJ) kJ)(1000 (-571.7 - = gssurroundin

oS

Calculating ∆SCalculating ∆Soo for the surroundings for the surroundings

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

Can calculate ∆HCan calculate ∆Hoosystemsystem = ∆H = ∆Hoo

rxnrxn = -571.7 kJ = -571.7 kJ (2 mol H(2 mol H2200(l)(l)- 0)- 0)

T

ΔHΔS system

gssurroundin

2 H2 H22(g) + O(g) + O22(g) (g) 2 H 2 H22O(liq) @ 25O(liq) @ 25ooCC

∆∆SSoosystemsystem = -326.4 J/K = -326.4 J/K

∆∆SSoosurroundingssurroundings = +1917 J/K = +1917 J/K

∆∆SSoouniverse universe = +1591 J/K= +1591 J/K

The entropy of theThe entropy of the

universe is increasing,universe is increasing,

so the reaction is so the reaction is

spontaneous ( product-spontaneous ( product-

favored)favored). .

Suniverse = Ssystem + Ssurroundings

Given Given SSoosurrsurr , , SSoo

syssys and T, and T,

determine determine SSoouniv univ and predict and predict

if the reaction will be if the reaction will be spontaneous.spontaneous.

Spontaneous or Not?

originally: originally: Suniverse = Ssystem + Ssurroundings

but but

originally: originally: Suniverse = Ssystem + Ssurroundings

but but

Suniverse = Ssystem – system/T

system Ssystem Spontaneous?ExothermicHsys < 0

Less orderSsys > 0

Spontaneous under all conditions; Suniv > 0

ExothermicHsys < 0

More orderSsys < 0

Favorable at low T

EndothermicHsys > 0

Less orderSsys > 0

Favorable at high T

EndothermicHsys > 0

More orderSsys < 0

Not spontaneous under any conditions Suniv < 0

T

ΔHΔS sys

surr

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Without doing any calculations, determine the sign of Ssys and Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous.

2CO(g) + O2(g) 2CO2(g) rxn = -566.0 kJ

2NO2(g) O2(g) + 2NO(g) rxn = +113.1 kJ

Suniverse = Ssystem + (--sys/T) /T) Suniverse = Ssystem + (--sys/T) /T)

Ssystem = (-); 3 mol gas form 2 mol gas

Ssurr = (+); spontaneous @ low T

Ssystem = (+); 2 mol gas form 3 mol gas

Ssurr = (-); spontaneous @ high T

Ssurr = (--sys/T) /T)

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Without doing any calculations, determine the sign of Ssys and Ssurr for each reaction. Predict under what temperatures (all T, low T, or high T) the reaction will be spontaneous.

2H2(g) + O2(g) 2H2O(g) rxn = -483.6 kJ

CO2(g) C(s) + O2(g) rxn = +393.5 kJ

Suniverse = Ssystem + (--sys/T) /T) Suniverse = Ssystem + (--sys/T) /T)

Ssystem = (-); 3 mol gas form 2 mol gasSsurr = (+) ; spontaneous @ low T

Ssystem = (-); complicated gas forms a solid & gas

Ssurr = (-); nonspontaneous @ all T

Ssurr = (--sys/T) /T)

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At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings, if Ho

rxn = -127 kJ and Sorxn = 314 J/K.

Plan: set Sorxn = So

surr and solve for T;convert kJ to J

sysosys

o

surro ΔS

TΔH

ΔS

rxn implies system!!!

TΔSΔH

syso

syso

TJ/K 314

)J000,127(

Ans: T = +404 K

Chapter 17 Free Energy and Thermodynamics Lesson 1.

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Transcript of Chapter 17 Free Energy and Thermodynamics Lesson 1.