Chapter 14 Chemical Equilibrium

This chemical engineer is testing a process for the formation of new liquid fuels from coal and petroleum. Such methods may transform the generation of energy worldwide.

Dynamic equilibrium

Physical processes: solid-liquid vapor-liquid….

Chemical reactions: reactantsproducts

Dynamic equilibria are responsiveto changes in the conditions.

Watch animation…chemical reaction reaching equilibrium equilibrium being dynamic

Assignment for Chapter 14

14.23,14.41.14.51.14.55

Figure 14.1 In the synthesis of ammonia, the molar concentrations of N2, H2, and NH3 change with time until there is no further net change and the concentrations settle into values corresponding to a mixture in which all three substances are present.

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

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

No chemical reaction is ‘complete’,No chemical reaction is ‘impossible’.

Figure 14.2 When we plot the rates of the forward and reverse reactions for the formation of ammonia on one graph, we can see that as the forward rate decreases, the reverse rate increases, until they are equal. At this point, the reaction is at equilibrium and the rates remain constant.

ReactantsProducts

Figure 14.3 In an experiment showing that equilibrium is dynamic, a reaction mixture in which N2 (pairs of blue spheres), D2 (pairs of yellow spheres), and ND3 have reached equilibrium is mixed with one with the same concentrations of N2, H2 (pairs of gray spheres), and NH3. After some time, the concentrations of nitrogen, hydrogen, and ammonia are found to be the same, but the D atoms are distributed among the hydrogen and ammonia molecules.

N2(g)+3D2(g)2ND3(g)

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

N2(g)+2HD (g)+D2(g)2NHD2(g)

N2(g)+H2(g)+2HD(g)2NH2Dg)

Chemical reaction is dynamic, forward and reverse reactions taking place all the time.

Try Yourself to Define the Equilibrium Constant?

aA+bB+cC+…pP+qQ+rR+…

Kc=[P]+[Q]+[R]+…-[A]-[B]-[C]-…?

Kc=p[P]+q[Q]+r[R]+…-a[A]-b[B]-c[C]-…?

Kc=[P][Q][R]…/[A][B][C]…?

Kc=pqr…[P][Q][R]…/abc…[A][B][C]…?

Equilibrium ConstantStoichiometric coefficients

Stoichiometric coefficients[Molar concentrations of products] [Molar concentrations of reactants] cK

aA+bB+cC+…pP+qQ+rR+…

...][][][...][][][

cba

rqp

CBARQP

cK

Cato Guldberg & Peter Waage (1864)

‘Justification’ of the Definition of Equilibrium Constant

Stoichiometric coefficients

Stoichiometric coefficients[Molar concentrations of products] [Molar concentrations of reactants] cK

aA+bB+cC+…pP+qQ+rR+…

[ ][ ]...[ ][ ]...[ ][ ]... [ ] [ ] [ ] ...[ ][ ]...[ ][ ]...[ ]... [ ] [ ] [ ] ...

p q r

a b cP P Q Q R R P Q R

c A A B B C A B CK

A+A+…B+B+…+C+C+…P+P+..+Q+Q+…+R+R+…

][][][

322

22

2

23

)(2)()(2

OSOSO

cK

gSOgOgSO

Is the same for all experiment no matter what the initial compositions are.

Example

O(g)6H4NO(g)(g)5O(g)4NH 223

52

43

62

4

][O][NHO][H[NO]cK

Classroom Exercise

- - -2 3

3 2

(a) 3ClO (aq) 2ClO (aq) Cl (aq)(b) 2O (g) 3O (g)

- 2 -3

- 32

32

23

[ClO ] [Cl ][ClO ]

[O ][O ]

( )

( )

c

c

a K

b K

Write the equilibrium constant for the following reactions:

• Multiply a reaction:

• Reverse a reaction:

2541

[HI]]][I[H

3

22

32][I][H

[HI]2

22

]][I[H[HI]

1

22

108.1

(g)I(g)H2HI

109.254

4HI(g)(g)I2(g)2H

54

2HI(g)(g)I(g)H

222

22

22

4

22

2

c

c

c

K

K

K

Using smallest possible stoichiometric coefficients to write equilibrium constants

Classroom Exercise• Write down the equilibrium constant of the

following reaction:

H2+D22HD

Suppose the equilibrium constant of above reaction at 500 K is 3.6, what is the equilibrium constant of its reverse reaction?

Kc=[HD]2/[H2][D2]

Reverse reaction: Kc-1=0.28

Multi-Step Reactions

""'

]][Cl[PCl][PCl

]][Cl[PCl][PCl

][Cl[P]][PCl

][Cl[P]][PCl

52

523

523

32

][Cl[P]][PCl

52

]][Cl[PCl][PCl"

523

][Cl[P]][PCl'

32

23

5

23

53

22

23

52

2

25

52

2

25

23

5

32

2

23

(g)2PCl(g)Cl52P(g) ________________________

(g)PCl(g)Cl(g)PCl (g)PCl(g)Cl(g)PCl (g)2PCl(g)Cl32P(g)

:first two theof sum theisreaction thirdThe

(g)2PCl(g)Cl52P(g)

(g)PCl(g)Cl(g)PCl

(g)2PCl(g)Cl32P(g)

ccc

c

c

c

c

KKK

K

K

K

K

Equilibrium constant depends on reaction conditions

Reaction Rate and Equilibrium

2

'

[ ][ ][ ]

2

2

2A C D

Suppose elementary reactions:

A A C D Rate [A]

C D A A Rate [C][D]

At equilibrium: [A] [C][D] (equilibrium constant equals the

ratio of the forward to th

C Dc A

'

'

kc k

K

k

k

k kK

e reverse reaction rates)

Figure 14.4 The equilibrium constant for a reaction is equal to the ratio of the rate constants for the forward and reverse elementary reactions that continue in a state of dynamic equilibrium. (a) A relatively large forward rate constant means that the forward rate can match the reverse rate even though only a small amount of reactants is present. (b) Conversely, if the reverse rate constant is relatively large, then the forward and reverse rates are equal when only small amounts of products are present.

Multi-Step Reactions

1'1

2'2

1 2' '

1 2

k1,a 1,b 1,a 1,bk

k2,a 2,b 2,a 2,bk

a b a b

R R ... P P ...

R R ... P P ...

...______________________________________ R R ... P P ...

...k kc k k

K

Homogeneous Equilibria: All products and reactants are of the same phase.

Heterogeneous Equilibria:Reacting systems with more than one phase.

Molar concentration of pure solid and liquid is a constant, independent of the amount present . It is ignored in the calculation of equilibrium constant.

22

2

]][OH[Ca

(aq)2OH(aq)Ca

cK

(s)Ca(OH)2Ignored!

Another way of understanding: the concentration of pure solid/liquidIs always 100%1. Since 1a =1, it does not affect Kc.

+ 2-2Ag S(s) 2Ag (aq)+S (aq)

?cK

Ignored!

2 2[Ag ] [S ]cK

Classroom Exercise

Gaseous Equilibria

X.roduct reactant/p gaseousfor introducemay weThus

][CO

(g)COCaO(s)(s)CaCO22CO

Vn

2

23

Xp

RTP

c

PK

K CO

. andbetween ipRelationsh pc K K

[ ] [ ][ ] [ ]

[ ] [ ][[ ] ] [[ ] ][[ ] ] [[ ] ] [ ] [ ]

aA(g) bB(g) cC(g) dD(g)

aA(g) bB(g) cC(g) dD(g)

( )

( )change in the numbe

c dC Da bA B

c d

a b

c d c dc dC Da b a b a bA B

P Pp P P

C Dc A B

P P C D c d a bC RT D RTp P P A RT B RT A B

nc

K

K

K RT

K RTn

r of the gas phase molcules.

N2(g)+3H2(g)2NH3(g)T=500 K, equilibrium concentrations:[NH3]=0.796 mol/L, [N2]=0.305 mol/L, [H2]=0.324 mol/L.

2 233 3

2 2

[ ] 0.796[ ][ ] 0.305 0.324

2 1 3

5

61.1.

( ) 61.1 (0.08206 500)

1.027 10

NHc N H

np c

K

K K RT

Classroom Exercise

2 233 3

2

[ ] 0.3[ ] 0.25

2 3

1

5.76.

( ) 5.76 (0.08206 300)

1.42 10

SOc O

np c

K

K K RT

2S(s)+3O2(g)2SO3(g)T=300K, equilibrium concentrations: [O2]=0.25 mol/L, [SO3]=0.3 mol/L. calculate the equilibrium constant Kc and Kp.

Gaseous Equilibria

• Equilibrium constants for gaseous reactions can be written by using either molar concentrations or partial pressures.

The equilibrium constant is the ratio of the concentrations or partial pressures of the products to those of the reactants, each concentration raised to a power equal to its stoichiometric coefficient in the balanced equation.

Summary of Equilibrium Constants

( ) ,

change in the number of the gas phase molcules.

np cK RT K

n

1

2

3

1 2 3

Reaction 1 + Reaction 2 + Reaction 3 +_______________________ Overall reaction = Reaction 1+Reaciton 2+Reaction 3+...

overall

KKK

K K K K

1'1

2'2

1 2' '

1 2

k1,a 1,b 1,a 1,bk

k2,a 2,b 2,a 2,bk

a b a b

R R ... P P ...

R R ... P P ...

...______________________________________ R R ... P P ...

...k kc k k

K

Using Equilibrium Constants

A. The Extent of ReactionB. The Direction of ReactionC. Equilibrium Tables

The Extent of Reaction

• K>1000, product dominant;• 0.001<K<1000, neither reactants nor

products dominate equilibrium;• K<0.001,reactants dominant.

Figure 14.6 The size of the equilibrium constant indicates whether the reactants (blue squares) or the products (yellow squares) are favored. Note that reactants are favored when Kc is small (left), products are favored when Kc is large (right), and reactants and products are in almost equal abundance when Kc is close to 1 (middle). Here, for simplicity, we compare reactions with Kc 102 and Kc 102.

Calculating equilibrium concentration

• H2(g)+Cl2(g)2HCl(g)

2

2 2

31c

[

17

HCl]2 2[H

2

][Cl ]

162

172

2

mol/L; mol/L;

K 4.0 10 at300K. [HCl] ?Solution:

[H ][Cl ]

10 times higher than [Cl

[H ] [C

[H

l ]

] and [H ]!

1.0 10

Cl] 0.2

1 0 0

8

. 1

c cK K

Classroom Exercise• Suppose that the equilibrium molar concentrations of H2 and Cl2 at

300 K are both 1.0×10-16 mol/L. What is the equilibrium molar concentration of HCl, given Kc= 4.0×1031?

• H2(g)+Cl2(g)2HCl(g)

2

2 2

31c

31 32[HCl]2 2[H ][Cl ]

1

1

2

62

62

2

[H

mol/L; K 4.0 10 at 300K. [HCl] ?Solution:

[H ][Cl ] 4Cl] 0.63,.0 10 1.0 10

almo

[

st 10 times higher than [C

H ] [Cl ] 1.0 1

l ] and

0

[H ]!c cK K

Using K to determine a partial pressure

5 2 3

5 3 2PCl (g) PCl (g) Cl (g) 25 at 298K.

0.0021 atm, 0.48 atm, ?p

PCl Cl PCl

K

P P P

3 2 5

35 2

:

0.11PCl Cl PCl

PCl Cl

P P Pp PCl pP P

Solution

K P K

Classroom Exercise:Using K to determine a partial pressure

2

22NOCl(g) 2NO(g) Cl (g) 0.018 at 500 K.

0.11 atm, 0.84 atm, ?p

NO Cl NOCl

K

P P P

2 2 2 22 22

0.11 atm 0.84 atm0.018 atm

:

0.75 atmCl ClNO NO

pNOCl

P P P Pp NOCl KP

Solution

K P

The Direction of Reaction

Reaction quotient:

(equilibrium)

molar concentrations of products at any stagemolar concentrations of reactants at any stagepartial pressures of products at any stagepartial pressures of reactants at any stage

c

p

c c

Q

Q

Q K

(equilibrium)p pQ K

2 2

2

2 2

2 2H (g) I (g) 2HI(g)At certain stage: 0.1 atm, 0.2 atm, 0.4 atm,

then 0.8

A reaction has a tendency to form products if and to form reactants if .

HI

H I

H I HI

Pp P P

P P P

Q

Q K Q K

Example

Figure 14.7 The relative sizes of the reaction quotient Q and the equilibrium constant K indicate the direction in which a reaction mixture tends to change. The arrows point from reactants to products when Q, K (left) or from products to reactants when Q K (right). There is no tendency to change once the reaction quotient has become equal to the equilibrium constant.

Quiz

5 3 2PCl (g) PCl (g) Cl (g) 25 at 298K.p K

3 2

5

0.08 0.480.0021

:

18.3PCl Cl

PCl

P Pp pP

Solution

Q K

5 2 30.0021 atm, 0.48 atm, 0.08 atmPCl Cl PClP P P

At certain stage, it is measured that

Will the reaction be moving to formation of more product or not?

The reaction will be moving to formation of more product.

A table that shows the initial concentrations, the changes needed to reach equilibrium, and the final equilibrium compositions.

Equilibrium Tables

Species

Draw an equilirium table:

PCl PCl Cl5 3 2

Step 1: Initial concentration 0.30 0 0

Step 2: Change in the molar concentration

Step 3: equilibrium concentration 0 30

- x x x

. - x x

21.8 0.54 0 0.262(positive root).0.30

x

x xK x x xc x

5

oc

Problem: 0.15 mol PCl is placed in reaction vessel of volume 500 mL and allowed to reach equilibrium with its

decomposition products at 250 C and K 1.8. What is the composition of the equilibrium

3 2

5

[PCl ][Cl ]5 3 2 c [PCl ]

mixture?

PCl (g) PCl (g) Cl (g) K

5The initial concentration [PCl ] 0.150 mol/0.5L 0.30 mol/L

o

2 2

A mixture of 0.002 mol/L hydrogen gas and 0.002 mol/L iodine

is allolwed to form hydroegn iodide at 773 C by the reactionH (g)+I (g) 2HI(g). At equilibrium, the concentration of HI is0.002 mol/L. Cal

culate the equilibrium constant at this temperature.

Classroom Exercise

Species

Draw an equilirium table:

H I HI

Step 1: Initial concentration 0.002 0.002 0

2 2

Step 2: Change in the molar concentration 2

Step 3: equilibrium concentration 0 002 0.002

- x x x

. - x

0.002

(2 )4.0(0.002 )(0.002 )

2 20.00220.001

0.001

x

xKc x x

x

2 2The initial concentration [H ] 0.002 mol/L, [I ] 0.002 mol/L

Le Chatelier’s Principle

When a stress is applied to a system in dynamic equilibrium, the equilibrium tends to adjust to minimize the effect of the stress.

Henry Louis LE CHATELIER (1850-1936)

Use of Le Chatelier Principle (I)

• Adding a reactant or removing a product

reaction tends to form products.• Adding a product or removing a reactant

more reactant tends to form.

Use of Le Chatelier Principle (II)

• Compression of a reaction mixture the

reaction that reduces the number of gas-phase molecules.

• Increasing by introducing an inert gas no effect on the equilibrium!

Commercial ammonia synthesis vessel

Introducing inert gas (yellow) has no effect on the equilibrium composition.

More generally, introducing anything that does not react with any productand reactant of a reaction would not change the equilibrium composition of the reaction. (A catalyst can increase reaction rate but not equilibrium constant.)

Cheating equilibrium (1)

Cheating equilibrium (2)

Use of Le Chatelier Principle (III)

• Raising the temperature of an exothermic

reaction reaction tends to form more reactants.

• Raising the temperature of an endothermic reaction reaction tends to form more products.

(a) Endothermic reaction (b) Exothermic reaction

Radar image of Venus: high partial pressure of carbon dioxide

Catalysts

• A catalyst can increase the rate of a chemical reaction.

• A catalyst has no effect on the equilibrium composition of a reaction mixture.

Haber’s Achievements

Fritz Haber (1868-1934，Nobel Prize for Chemistry, 1916)

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

Assignment for Chapter 14

14.23,14.41.14.51.14.55