Lecture 8-9

Entropy and Free Energy:Predicting the direction of spontaneous

changeThe approach to Chemical equilibrium

Absolute entropy and the third law of thermodynamics

To define the entropy of a compound in absolute terms it is necessaryto define a reference value.We can define a zero of entropy at 0 K.According to the third law of thermodynamics the entropy of a perfectcrystal at T = 0 K is zero.The standard entropy S298

0 of a substance is defined as the molar entropyat T = 298 K and 1 bar pressure. Units are J mol-1 K-1. S0

298 values are termed absolute or third law entropies.

05

04

03

02

01

0298 SSSSSS

The entropy of a substance increases on heating.There are sudden increases in entropy on melting at Tm and vaporization at Tb.

When CP,m/T is plotted against T the entropy changes due to heating the solid, liquid andgas are given by the shaded areas.

KKTCSS mPT 298

/ln,0298

0

Assume that substance remains insame phase between T = 298 K andT.

Standard entropy at temp. T.

Entropy change in chemical reactions

rxSprSS jjr0298

0298

0298

Sum of standard entropiesof products

Sum of standard enrropiesof reactants

Stoichiometric coefficient(from balanced equation)

KKTCSS mPrTr 298

/ln,0298

0

rxCprCC mPjmPjmP ,,,

Reaction entropyAt temp. T

See Chemistry3 worked example 15.5, p.718. Chemistry3 section 15.4, pp.716-722.

Gibbs Energy Change

J.W. Gibbs, 1839-1903

We now discuss the way in which the spontaneityof a process may be determined.

System

Surroundings

Kotz, section 19.5, 19.6pp.871-879.Chemistry3, section 15.5,pp.722-730.

From 2nd law of thermodynamics

surrSsysStotS

Assume process occurs at const.T and P.

Tsysq

TsurrqsurrS revrev

Hence

TsysHsurrS

sysHsysq Prev

,

Now 2nd law thermodynamics implies that

0 surrSsysStotS

TsysHsysStotS

Multiplying across by -T sysGsysSTsysHtotST

where we define the Gibbs Energy asSTHG

For a spontaneous process at const T,P 0

0

sysGtotS

H

ST

ST

H

At low temp T theenthalpy change part Of G has a greatermagnitude than TS.The sign of Gdepends on the signof H.

At high temp T the enthalpychange part of G has asmaller magnitude thanTS. The sign of Gdepends on the sign of S.

G < 0 for a spontaneous process.H < 0 (exothermic change) makes favourable contribution to spontaneity.H > 0 (endothermic change) makes unfavourable contribution toSpontaneity.S < 0, entropy decreases, makes unfavourable contribution toSpontaneity.S > 0, entropy increases, makes favourable change to spontaneity.

G < 0 : reaction or process is spontaneousG > 0 : reaction or process is not spontaneousG = 0 : reaction or process is at equilibrium

Standard Gibbs energy change of formation fG0298 is defined as the change in Gibbs

energy when 1 mol of a compound is formed at P = 1 bar and at T = 298 K from itsconstituent elements in their standard states.

Standard Gibbs energy of reaction rG0298 is computed from f G0 for the

reactants and products using the following rule.

)()( 0298

0298

0298 rxGprGG fjfjr

See Chemistry3 worked example 15.8, pp.728-729.

Once the standard Gibbs energy of reaction rG0298 is known at 298 K

then it is possible to compute the correspondingGibbs energy of reaction at any other temp T using the definition ofthe Gibbs energy function.

0 0 0r T r T r TG H T S

0 0298 ,

0298 , 298

r T r P m

r P m

H H C T

H C T

KKTCSS mPrTr 298

/ln,0298

01

rxCprCC mPjmPjmP ,,,

2

See Chemistry3

Worked example 15.9,pp.729-730.

More of this type of calculation in SF Thermodynamics.

Chemical equilibrium.

• What is chemical equilibrium ?•How much product will form under a given setof starting conditions ?

• What is the composition of a reaction mixture when a chemical reaction has attained equilibrium?• What is the effect of temperature on the composition of a reaction mixture at equilibrium?

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

Chemical Equilibrium.

• Reactant concentrations decrease with time ; product concentrations increase with time .

• After a long enough time reactant and product concentrations attain steady, time invariant values.

• A state of chemical equilibrium has been attained.

• If a plot of reaction rate versus time is examined wenote that the rate of the forward reaction decreasesand the rate of the reverse reaction increaseswith increasing time, until, at long times they becomeequal. At this stage the reaction rates no longer changewith time and the reaction is said to be at equilibrium.

forward reaction

reverse reactionEquilibrium

Haber Process:Ammonia synthesis

Kinetics

Kinetic definition ofEquilibrium.

Kinetics applies to the speed of a reaction,

the concentration of product that appears

(or of reactant that disappears) per unit time.

Kinetics versus Equilibrium.

Equilibrium applies to the extent of a reaction, the concentration of product that has appeared after an unlimited time, or once no further change occurs.

A system at equilibrium is dynamic on the molecular level;

no further net change is observed

because changes in one direction are balanced by changes in the other.

At equilibrium: rate forward step = rate reverse step

N2O4 (g) 2 NO2 (g)

colourless brown

Reaching Equilibrium on the Macroscopic and Molecular Level

NO2

N2O4

Properties of an equilibrium Reaction

Equilibrium systems are dynamic (in constantchemical change) and reversible (chemical changecan be approached from either direction).

PLAY MOVIE

2622242

2242262

)(2)(2)()(

ClOHCoOHClOHCoOHClOHCoClOHCo

Pink Blue

Blue Pink

+

3 22 2 26 5

Fe H O SCN Fe SCN H O H O

PLAY MOVIE PLAY MOVIE

• After a period of time, the concentrations of reactants and products are constant.

• The forward and reverse reactions continue after equilibrium is attained.

Examples of Chemical equilibria : Phase Change

PLAY MOVIE

2 2H O s H O

Chemical Equilibrium :a kinetic definition.

• Countless experiments with chemical systems have shown that in a state of equilibrium, the concentrations of reactants and products no longer change with time.

• This apparent cessation of activity occurs because under such conditions, all reactions are microscopically reversible.

• We look at the dinitrogen tetraoxide/nitrogen oxide equilibrium whichoccurs in the gas phase.

t

t

ON

NO

42

2

Equilibriumstate

Kineticregime

NO2

N2O4

timeco

ncen

tratio

n

N2O4 (g) 2 NO2 (g)

colourless brown

eq

eq

ON

NO

42

2

Concentrations varywith time

Concentrations timeinvariant

Kinetic analysis.

22

42

NOkR

ONkR

Equilibrium:

K

kk

ONNO

NOkONk

RR

eq

eqeq

42

22

2242

Valid for any time t

t

t

Equilibriumconstant

rateforward = ratereverse

kforward[reactants]m = kreverse[products]n

= = K the equilibrium constantkforward

kreverse

[products]n

[reactants]m

The values of m and n are those of the coefficients in the balanced chemical equation.

The rates of the forward and reverse reactions are equal, NOT the concentrations of reactants and products.

This is also known as the LAW OF MASS ACTION.

forward

reverse

reactants n productsrate [reactants]

rate [products]

mforward

nreverse

mk

k

Initial and Equilibrium Concentrations for theN2O4-NO2 System at 100°C

Initial Equilibrium Ratio[N2O4] [N2O4] [N2O4][NO2] [NO2] [NO2]2

0.1000 0.0000 0.0491 0.1018 0.211

0.0000 0.1000 0.0185 0.0627 0.212

0.0500 0.0500 0.0332 0.0837 0.211

0.0750 0.0250 0.0411 0.0930 0.210Constant values

Equilibrium constant K

•The value of the ratio of initial concentrationsvaries widely but always gives the same valuefor the ratio of equilibrium concentrations.•The individual equilibrium concentrations aredifferent in each case but this ratio ofequilibrium concentrations is constant.

The size of the equilibrium constant indicates whether the reactantsor the products are favoured .

Reactants favouredwhen Kc is small

Products favouredwhen Kc is large

Reactants and productsare in almost equalabundance when Kc nearunity

The Reaction Quotient, QIn general, ALL reacting chemical

systems are characterized by their REACTION QUOTIENT, Q.

a A + b B p P + q Q

If Q = K, then system is at equilibrium.

p q

t ta b

t t

P QQ

A B

THE EQUILIBRIUM CONSTANTTHE EQUILIBRIUM CONSTANTFor any type of chemical equilibrium of the type

a A + b B p P + q Qthe following is a CONSTANT (at a given T)

If K is known, then we can predict concentrations. of products or reactants in the reaction mixture at equilibrium and hence the yield of the reaction.

p q

eq eqa b

eq eq

P QK

A B

Equilibriumconstant

Product concentrations

Reactant concentrations

Relationship between Gibbs Energy and Equilibrium Constant.

We now derive an expression which relates the change in Gibbs energy for areaction as a function of the composition of the reaction mixture at any stagein the reaction.

qQpPbBaA

Hence after some algebra and simplification the change in Gibbs energy for reaction can be computed.

BAQPr bGaGqGpGG

We can define Gibbs energy in terms ofthe activity ak of the species k. kkk aRTGG ln0

BBAA

QQPPr

aRTGbaRTGa

aRTGqaRTGpG

lnln

lnln00

00

QRTGaaaa

RTGG rbB

aA

qQ

pP

rr ln..

ln 00

Reaction quotient Q 00000BAQPr bGaGqGpGG

If the reaction is allowed to proceedTo equilibrium then we replace Q byThe equilibrium constant K and setrG = 0 by definition.

KRTG

KRTG

r

r

ln

0ln0

0

Activity = generalised concentration

0

0

ln

lnr r

r

G G RT Q

G RT K

ln ln(ln ln )

rG RT K RT QRT Q K

lnrQG RTK

If Q/K <1 then Q < K and rG is negative, thereaction tends to proceed in forward directionand the reaction is said to be product favoured.If Q/K > 1 then Q > K and rG is positive. Here thereaction does not proceed spontaneously in theforward direction and is said to be reactant favoured.If Q = K then since ln 1 = 0 rG = 0 and the reactionis at equilibrium.

This is the mostuseful form of theequation for interpretation.

a A + b B c C + d D

QRTG

BADCRTGG ba

dc

ln

ln

0

0

• Under non-equilibrium conditionsGibbs energy change is :

Q = reaction quotient ba

dc

BADCQ

• Q defines reactant and product concentration ratio (i.e. reaction composition) at any stage inchemical transformation.

• When G = 0 at constant T and P wehave equilibrium . Hence Q = Kc.

KRTG

BADC

RTG beq

aeq

deq

ceq

ln0

ln0

0

0

KRTG ln0

RTGK

0

exp

KQ KQ Gibbs energy change for reaction mixture

These are very importantrelations!

Expression showshow G varieswith compositionof reaction mixture.

Product- or Reactant Favored Processes

Product-favoredK > 1

Reactant-favoredK < 1

K comes from thermodynamics.ΔG˚ < 0: reaction is product favoredΔG˚ > 0: reaction is reactant-favored

KRTG ln0

If K > 1, then ΔG˚ is negative.If K < 1, then ΔG˚ is positive.

Gibbs energy and chemical equilibrium.

G

Qln0GG

0G

EquilibriumQ=K

RP RP

PR PR

Q small, Q<K[P]<<[R]G negative

Q large, Q>K[P]>>[R]G positive

Standard stateQ=1lnQ=0

QRTGG ln0 Reaction spontaneousIn forward direction

Reaction not spontaneousIn forward direction

Extent of reaction

0 1

Tpr

GG,

reverse reactionspontaneous

forward reactionspontaneous

Reaction Gibbs energyTp

rGG

,

extent of reaction =

0 1

rG is the slope of the G versus graph at any degree of advancement of the chemical reaction.

0 Gr0 Gr0 Gr

1K 1K1K

Key stages in the Haber-Bosch synthesis of ammonia.

Operating conditions dictated by a balance between kineticsand thermodynamics.Yield of NH3 is high when pressure P is large and temperature Tis low. Rate of formation of NH3 is low when T is low.High pressure and continuous removal of NH3 used to increase yield.Temperature is raised and catalyst employed to enhance rate of NH3 formation.

NH3 synthesis is exothermic (H0 = - 91.8 kJmol-1).Hence K decreases as T increases.

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

Fritz Haber1868-1934Nobel Prize, 1918

Carl Bosch1874-1940Nobel Prize, 1931

Effect of Temperature on Kcfor Ammonia Synthesis

T (K) Kc

200. 7.17 x 1015

300. 2.69 x 108

400. 3.94 x 104

500. 1.72 x 102

600. 4.53 x 100

700. 2.96 x 10 -1

800. 3.96 x 10 -2

TasKC

How can we explain this ?

• We need to be able topredict the way that Kvaries with temperatureT.

• This is given by the van’t Hoff equation.

Temperature dependence of equilibrium constant:van’t Hoff equation.

KRTG ln0

000 STHG KRTSTH ln00 R

SRTHK

00

ln

• We assume that H0 and S0 areindependent of temperature T overthe temperature range of interest.

• Assume that K = K1 when T = T1 andK = K2 when T = T2. Note that T2 > T1.

RS

RTHK

0

1

0

1ln

RS

RTHK

0

2

0

2ln

21

0

1

212

11lnlnlnTTR

HKKKK

21

0

1

2 11expTTR

HKK

21

0

1

2 11expTTR

HKK

This can be used to understandthe temperature dependence ofthe equilibrium constant.

21

0

1

2 11expTTR

HKK

21

0

1

2 11expTTR

HKK

Now T2>T1 so the term 1/T1-1/T2 is positivesince 1/T2 < 1/T1. Also the term ln{K2/K1} depends on the sign of H0.

• Endothermic reaction: H0 is positive, the exponential term is also positive, and so K2/K1 >1 and K2 > K1.•The equilibrium constant for an endothermic process increases with temperature. Increase in T favoursproducts.• Exothermic reaction: H0 is negative, and K2 < K1. •The equilibrium constant for an exothermic process decreaseswith an increase in temperature. Increase in T favours reactants.

Percent Yield of Ammonia vs. Temperature (°C)

0

10

20

30

40

50

60

70

350400

450500

550600

650

200250

300350

400450

500550

NH

3 yi

eld

/ mol

%

Temperature / 0 C

Pressure / atm

Best NH3 yield at low T and high P.

High P : expensive plant

Low T : slow reactionkinetics

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

340atm/P140600C/T400 0

Optimizing Ammonia Synthesis

Thermodynamic criteria ofspontaneity.

If the reaction is exothermic (rH0 < 0)and rS0 > 0 then rG0 < 0 and K > 1 atall temperatures.

If the reaction is exothermic (rH0 < 0)and rS0 < 0 then rG0 < 0 and K > 1provided that T < rH0/rS0 .

If the reaction is endothermic (rH0 > 0)and rS0 > 0 then rG0 < 0 and K > 1provided that T > rH0/rS0 .

If the reaction is endothermic (rH0 > 0)and rS0 < 0 then rG0 < 0 and K > 1 atno temperature.

KRTGr ln0

000 STHG rrr

products dominantat equilibrium

reactants dominantat equilibrium

01 0 GifK r

reaction thermodynamicallyfeasible

01 0 GifK r

reaction notthermodynamicallyfeasible

Chemical Equilibrium Problems.Suppose that 0.150 mol PCl5 is placed in a reaction vessel of volume 500 cm3 andallowed to reach equilibrium with its decomposition products phosphorous trichlorideand chlorine at 2500C. If the equilibrium constant Kc is 1.80, determine the composition of the reaction mixture at equilibrium.

PCl5(g) PCl3(g) + Cl2(g) 5

23

PClClPClKc

Initial concentration of PCl5 = 0.150 mol / 0.500 L = 0.3 M. Let an amount x of PCl5 be used up in reaction to form products.

Species PCl5 PCl3 Cl2

Initial concentration

0.3 0 0

Change in concentration

-x +x +x

Equilibrium concentration

0.3 - x x x

2.06-and262.0

2)54.0.(1.48.18.1

054.08.1

8.13.0

2

2

2

5

23

x

xx

xx

PClClPClKc

We choose the positive root of the quadraticequation and so x = 0.262.

MxCl

MxPClMxPCl

262.0262.0

038.0262.03.03.0

2

3

5

Phosphorous(V) chloride Phosphorous(III) chloride

The formation of NO from N2 and O2 contributes to air pollution whenevera fuel is burnt in air at a high temperature as in a gasoline engine.At 1500 K the equilibrium constant K = 1 x 10-5. Suppose a sample of airhas [N2]= 0.80 mol/L and [O2]=0.20 mol/L before any reaction occurs.Calculate the equilibrium concentrations of reactants and products after the mixture has been heated to 1500 K.

2 2( ) ( ) 2 ( )N g O g NO g

N2/M O2/M NO /M

Initial 0.80 0.20 0

Change -x -x +2x

Equilibrium 0.80-x 0.20-x 2x

ICE Table

2 2 25

2 2

2 6

4

2 41.0 10[ ].[ ] 0.8 0.2 0.8 0.2

4 1.6 106.3 10

NO x xKN O x x

xx

42

42

3

[ ] 0.80 0.8 6.3 10 0.8

[ ] 0.20 0.20 6.3 10 0.8

[ ] 2 1.3 10

N x M

O x M

NO x M

We assume that x << 1, i.e. lessthan 10% of initial reactant concentration [R]0 then ([R]0-x)≈[R]0Approximation valid when equilibriumConstant K is small and << 1.

x = amount reacted

Under certain conditions nitrogen and oxygen react to form dinitrogen oxide N2O.Suppose that a mixture of 0.482 mol N2 and 0.933 mol O2 is placed in a reaction vessel of volume 10 dm3 and allowed to form N2O at a temperature for which Kc= 2 x 10-13.Determine the composition of the reaction mixture at equilibrium.

2 N2(g) + O2(g) 2 N2O(g) 2

22

22

ONONKc smallvery102 2

13 ONKc

0

0933.010933.0

0482.010482.0N

ionsconcentratInitial

2

32

32

ON

Mdm

molO

Mdm

mol

Species [N2] [O2] [N2O]

Initialconcentration

0.0482 0.0933 0

Change inconcentration

-2x -x +2x

Equilibriumconcentration

0.04-2x 0.0933-x 2x

xx

xON

ONKc

0933.020482.02

2

2

22

2

22

When rearranged this expression yieldsa cubic equation in x.Since Kc is small we can assume that x isalso very small and assume :

0933.00933.00482.020482.0

xx

9

2

2

2

103.34

0933.00482.0

0933.00482.02

C

C

Kx

xK

Hence our approximationis OK

Hence at equilibrium:

MxON

MxOMxN

92

2

2

106.62

0933.00933.00482.020482.0

Le Chatelier’s PrincipleTemperature, catalysts, and changes in concentration affect equilibria.The outcome is governed by LE CHATELIER’S PRINCIPLE“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

Henri Le Chatelier1850-1936

When a reactant is added to a reaction mixture at equilibrium, the reaction tends to form products. When a reactant is removed, more reactant tends to form.When a product is added, the reaction tends to form reactants. When a product is removed, moreproduct is formed.

KQ KQ

eq

eq

RP

K

RPQ

Note : only Q responds to additionof R or P ; K remains the same.

Le Chatelier’s Principle

• Change T– change in K – therefore change in P or concentrations at

equilibrium• Use of catalyst: reaction comes more

quickly to equilibrium. K not changed.• Add or take away reactant or product:

– K does not change– Reaction adjusts to new equilibrium

“position”

Effect of Various Disturbanceson an Equilibrium System

Disturbance Net Direction of Reaction Effect on Value of KConcentration

Increase [reactant] Toward formation of product NoneDecrease [reactant] Toward formation of reactant None

Pressure (volume)Increase P Toward formation of lower

amount (mol) of gas NoneDecrease P Toward formation of higher

amount (mol) of gas NoneTemperature

Increase T Toward absorption of heat Increases if H0rxn> 0

Decreases if H0rxn< 0

Decrease T Toward release of heat Increases if H0rxn< 0

Decreases if H0rxn> 0

Catalyst added None; rates of forward and reverse reactions increaseequally . None