1

Lecture 8

Entropy and Free Energy:Predicting the direction of spontaneous 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.yThe 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.

2

05

04

03

02

01

0298 SSSSSS Δ+Δ+Δ+Δ+Δ=

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.

3

Entropy change in chemical reactions

( ) ( )∑ ∑−=Δ rxSprSS jjr0298

0298

0298 νν

Sum of standard enrropiesof reactants

Sum of standard entropiesof products

Stoichiometric coefficient(from balanced equation)

⎟⎠⎞

⎜⎝⎛Δ+Δ=Δ

KKTCSS mPrTr 298

/ln,0298

0

Reaction entropyAt temp. T

( ) ( )∑∑ −=Δ rxCprCC mPjmPjmP ,,, νν

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

Gibbs Energy Change

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

Surroundings

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

From 2nd law of thermodynamics

J.W. Gibbs, 1839-1903

System( ) ( ) ( )surrSsysStotS Δ+Δ=Δ

Assume process occurs at const.T and P.

( ) ( ) ( )Tsysq

TsurrqsurrS revrev −==Δ

Hence ( ) ( )sysHsysq Prev Δ=Multiplying across by -T( ) ( ) ( ) ( )sysGsysSTsysHtotST Δ=Δ−Δ=Δ−Hence ( ) ( )

( ) ( )TsysHsurrS

yyq Prev

Δ−=Δ

,

Now 2nd law thermodynamics implies that

( ) ( ) ( ) 0>Δ+Δ=Δ surrSsysStotS

( ) ( ) ( )TsysHsysStotS Δ

−Δ=Δ

( ) ( ) ( ) ( )sysGsysSTsysHtotST Δ=ΔΔ=Δwhere we define the Gibbs Energy as

STHG Δ−Δ=Δ

For a spontaneous process at const T,P ( )( ) 0

0<Δ>Δ

sysGtotS

4

At low temp T theenthalpy change part Of ΔG has a greatermagnitude than TΔS.

STΔ−

STΔ−

gThe sign of ΔGdepends on the signof ΔH.

HΔHΔ

At high temp T the enthalpychange part of ΔG has asmaller magnitude thanTΔS. The sign of ΔGdepends on the sign of ΔS.

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

5

ΔG < 0 : reaction or process is spontaneousΔG > 0 : reaction or process is not spontaneousΔG = 0 : reaction or process is at equilibrium

Standard Gibbs energy change of formation ΔfG0298 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 ΔrG0298 is computed from Δf G0 for the

reactants and products using the following rule.

∑∑ Δ−Δ=Δ )()( 0298

0298

0298 rxGprGG fjfjr νν

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

Once the standard Gibbs energy of reaction ΔrG0298 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 SΔ = Δ − Δ

( )

0 0298 ,

0

r T r P mH H C TΔ = Δ + Δ Δ

⎟⎠⎞

⎜⎝⎛Δ+Δ=Δ

KKTCSS mPrTr 298

/ln,0298

01

2

( )0298 , 298r P mH C T= Δ + Δ −

( ) ( )∑∑ −=Δ rxCprCC mPjmPjmP ,,, ννSee Chemistry3

Worked example 15.9,pp.729-730.

More of this type of calculation in SF Thermodynamics.

6

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

h h i l ti h tt i d ilib i ?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.forward reaction

reverse reactionEquilibrium

Kinetics

• 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.

reverse reaction

Haber Process:• If a plot of reaction rate versus time is examined we

note 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.

Haber Process:Ammonia synthesis

Kinetic definition ofEquilibrium.

7

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 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

Reaching Equilibrium on the Macroscopic and Molecular Level

N2O4 (g) 2 NO2 (g)

colourless brown

NO2

N2O4

8

Properties of an equilibrium Reaction

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

2622242

2242262

)(2)(2)()(

ClOHCoOHClOHCoOHClOHCoClOHCo

→++→

Pink Blue

Blue Pink

PLAY MOVIE

+

( ) ( )( )3 22 2 26 5

Fe H O SCN Fe SCN H O H O+ +−+ +

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

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

PLAY MOVIE PLAY MOVIE

9

Examples of Chemical equilibria : Phase Change

( ) ( )2 2H O s H O

PLAY MOVIE

Chemical Equilibrium :a kinetic definition.

• Countless experiments with chemical systems have shown that in a state of equilibrium, the concentrations of

[ ] ↑ [ ]

Concentrations varywith time

Concentrations timeinvariant

qreactants 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

entra

tion

N2O4 (g) 2 NO2 (g)

[ ][ ] eq

eq

ON

NO

42

2

∞→t

N2O4

time

conc

e2 4 (g) 2 (g)

colourless brown

Kinetic analysis.

[ ][ ]22

42

NOkR

ONkR

′=

=

Equilibrium:

[ ] [ ][ ][ ] K

kk

ONNO

NOkONk

RR

eq

eqeq

=′

=

′=

=

42

22

2242

Valid for any time t

∞→t

Equilibriumconstant

10

rateforward = ratereverse

kforward[reactants]m = kreverse[products]n

k [products]n

forward

reverse

reactants n productsrate [reactants]

rate [products]

mforward

nreverse

mk

k

=

=

= = K the equilibrium constantkforward

kreverse

[products]n

[reactants]m

This is also known as the LAW OF MASS ACTION.

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.

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

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

2 4 2 4 2 42 2 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.

11

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

Reactants favoured Products favouredReactants and productsReactants favouredwhen Kc is small

Products favouredwhen Kc is large

pare in almost equalabundance when Kc nearunity

The Reaction Quotient, QThe Reaction Quotient, QIn general, ALL reacting chemical In general, ALL reacting chemical

systems are characterized by their systems are characterized by their systems are characterized by their systems are characterized by their REACTION QUOTIENT, QREACTION QUOTIENT, Q..

a A + b B a A + b B FF p P + q Qp P + q Q

[ ] [ ]p qP Q

If Q = K, then system is at equilibrium.If Q = K, then system is at equilibrium.

[ ] [ ][ ] [ ]

t ta b

t t

P QQ

A B=

12

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

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

If K is known, then we can predict concentrations. of products or If K is known, then we can predict concentrations. of products or

[ ] [ ][ ] [ ]

p q

eq eqa b

eq eq

P QK

A B=

Equilibriumconstant

Product concentrations

Reactant concentrations

, p p, p preactants in the reaction mixture at equilibrium and hence the reactants in the reaction mixture at equilibrium and hence the yield of the reaction. yield of the reaction.

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 +→+

We can define Gibbs energy in terms of aRTGG ln0 +=

Activity = generalised concentration

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

( ) ( )BAQPr bGaGqGpGG +−+=Δ

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

( ) ( )( ) ( )

QQPPr

RTGbRTG

aRTGqaRTGpG

ll

lnln00

00 +++=Δ

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

( ) ( )BBAA aRTGbaRTGa lnln 00 +−+−

QRTGaaaa

RTGG rbB

aA

qQ

pP

rr ln..

ln 00 +Δ=⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+Δ=Δ

Reaction quotient Q( ) ( )00000BAQPr bGaGqGpGG +−+=Δ

KRTG

KRTG

r

r

ln

0ln0

0

−=Δ

=+Δ

13

a A + b B c C + d D

[ ] [ ][ ] [ ]BA

DCRTGG ba

dc

ln0

⎭⎬⎫

⎩⎨⎧

+Δ=Δ

• Under non-equilibrium conditionsGibbs energy change is :

KQ < KQ >Gibbs energy change for reaction mixture

Expression showshow ΔG varieswith composition[ ] [ ]

QRTG

BA ba

ln0 +Δ=⎭⎬

⎩⎨

Q = reaction quotient [ ] [ ][ ] [ ]ba

dc

BADCQ =

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

with compositionof reaction mixture.

chemical transformation.

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

[ ] [ ][ ] [ ]

KRTG

BADC

RTG beq

aeq

deq

ceq

ln0

ln0

0

0

+Δ=

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

+Δ= KRTG ln0 −=Δ ⎥⎦

⎤⎢⎣

⎡ Δ−=

RTGK

0

exp

These are very importantrelations!

ProductProduct-- or Reactant Favored or Reactant Favored ProcessesProcesses

K comes from thermodynamics.K comes from thermodynamics.∆G˚ < 0: reaction is ∆G˚ < 0: reaction is product favoredproduct favored∆G˚ > 0: reaction is ∆G˚ > 0: reaction is reactantreactant--favoredfavored

KRTG ln0 −=Δ

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

KRTG ln−=Δ

ProductProduct--favoredfavoredK > 1K > 1

ReactantReactant--favoredfavoredK < 1K < 1

14

Gibbs energy and chemical equilibrium.

GΔ

Equilibrium Q large Q>K

Reaction not spontaneousIn forward direction

Qln0GG Δ=Δ

0=ΔG

qQ=K

Q small, Q<K[P]<<[R]ΔG negative

Q large, Q>K[P]>>[R]ΔG positive

Standard stateStandard stateQ=1lnQ=0

QRTGG ln0 +Δ=ΔReaction spontaneousIn forward direction

Tr

GG ⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=Δξ forward reactionTp,⎠⎝ ∂ξ

reverse reactionspontaneous

forward reactionspontaneous

Extent of reaction ξ

0=ξ 1=ξ

15

Reaction Gibbs energyTp

rGG

,⎟⎟⎠

⎞⎜⎜⎝

⎛∂∂

=Δξ

extent of reaction = ξ

1<<K 1>>K1=K

0=Δ Gr0=Δ Gr0=Δ Gr

0=ξ 1=ξ

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

16

Key stages in the Haber-Bosch synthesis of ammonia.

NH3 synthesis is exothermic (ΔH0 = - 91.8 kJmol-1).Hence K decreases as T increases.

Fritz HaberFritz Haber18681868--19341934Nobel Prize, 1918Nobel Prize, 1918

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

Carl BoschCarl Bosch18741874--19401940Nobel Prize, 1931Nobel Prize, 1931

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.

17

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

↑↓ TasKC

How can we explain this ?

• We need to be able topredict the way that K

600. 4.53 x 100

700. 2.96 x 10 -1

800. 3.96 x 10 -2

varies with temperatureT.

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

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

KRTG ln0 −=Δ

000 STHG Δ−Δ=ΔKRTSTH ln00 −=Δ−Δ R

SRTHK

00

ln Δ+

Δ−=

STHG Δ−Δ=Δ

• We assume that ΔH0 and ΔS0 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 Δ+

Δ−=

RS

RTHK

0

2

0

2ln Δ+

Δ−=

⎫⎧Δ⎫⎧ 0 11HK

⎭⎬⎫

⎩⎨⎧

−Δ

=⎭⎬⎫

⎩⎨⎧

=−211

212

11lnlnlnTTR

HKKKK

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

18

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 ΔH0.

• Endothermic reaction: ΔH0 is positive, h i l i l i i d 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: ΔH0 is negative, and K2 < K1. •The equilibrium constant for an •The equilibrium constant for an exothermic process decreaseswith an increase in temperature. Increase in T favours reactants.

Percent Yield of Ammonia vs. Temperature (°C)

19

70

Best NH3 yield at low T and high P.

340atm/P140600C/T400 0

≤≤≤≤

Optimizing Ammonia Synthesis

10

20

30

40

50

60

350

NH

3 yi

eld

/ mol

%

Low T : slow reactionkinetics

0350

400450

500550

600650

200250

300350

400450

500550 Temperature /

0 C

Pressure / atmHigh P : expensive plant N2 (g) + 3 H2 (g) 2 NH3 (g)

20

Thermodynamic criteria ofspontaneity.

If the reaction is exothermic (ΔrH0 < 0)and ΔrS0 > 0 then ΔrG0 < 0 and K > 1 atall temperatures.

KRTGr ln0 −=Δ

000 STHG rrr Δ−Δ=Δ

reaction If the reaction is exothermic (ΔrH0 < 0)and ΔrS0 < 0 then ΔrG0 < 0 and K > 1provided that T < ΔrH0/ΔrS0 .

If the reaction is endothermic (ΔrH0 > 0)and ΔrS0 > 0 then ΔrG0 < 0 and K > 1provided that T > ΔrH0/ΔrS0 .

f h d h ( H0 0)

products dominantat equilibrium 01 0 <Δ>> GifK r

thermodynamicallyfeasible

0If the reaction is endothermic (ΔrH0 > 0)and ΔrS0 < 0 then ΔrG0 < 0 and K > 1 atno temperature.

reactants dominantat equilibrium

01 0 >Δ<< 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)[ ][ ][ ]

23

PClClPClKc =5(g) PCl3(g) Cl2(g) [ ]5PCl

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

-x +x +x

[ ][ ][ ]

054.08.1

8.13.0

2

2

5

23

=−+

=−

==

xx

xx

PClClPClKc

Phosphorous(V) chloride Phosphorous(III) chloride

concentration

Equilibrium concentration

0.3 - x x x

( )2.06-and262.0

2)54.0.(1.48.18.1 2

=−−±−

=x

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

====

=−=−=

21

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 25 2 4NO x x

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

N2/M O2/M NO /M

Initial 0 80 0 20 0

ICE Table

[ ] ( )( )( )

5

2 2

2 6

4

41.0 10[ ].[ ] 0.8 0.2 0.8 0.2

4 1.6 106.3 10

NO x xKN O x x

xx

−

−

−

= × = =− − ×

×

×

We assume that x << 1, i.e. lessthan 10% of initial reactant concentration [R]0 then ([R]0-x)≈[R]0Approximation valid when equilibriumInitial 0.80 0.20 0

Change -x -x +2x

Equilibrium 0.80-x 0.20-x 2x

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

−

−

−

= − − ×

− − ×

×

Approximation 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 13 ONK −×=

[ ]

9330

0482.010482.0N

ionsconcentratInitial

32 ==

l

Mdm

mol

[ ] [ ]22 ON[ ] smallvery102 2ONKc ×=[ ]

[ ] 0

0933.010933.0

2

32

=

==

ON

Mdm

molO

Species [N2] [O2] [N2O]

Initialconcentration

0.0482 0.0933 0

Change in 2x 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 :

Change inconcentration

-2x -x +2x

Equilibriumconcentration

0.04-2x 0.0933-x 2x0933.00933.0

0482.020482.0≅−≅−

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

−×≅=

≅−=≅−=

22

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

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.

KQKQ

Henri Le Chatelier1850-1936

KQ >KQ <

[ ][ ][ ][ ]eq

eq

RP

K

RPQ

=

=

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

Le Chatelier’s PrincipleLe Chatelier’s Principle

• Change T– change in K 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:y p

– K does not change– Reaction adjusts to new equilibrium

“position”

23

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 ΔH0rxn> 0

D if ΔH0 0Decreases if ΔH0rxn< 0

Decrease T Toward release of heat Increases if ΔH0rxn< 0

Decreases if ΔH0rxn> 0

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