Bài soạn Chapter 14 Chemical Equilibrium

Reaction Dynamics • when a reaction starts, the reactants are consumed and products are made  forward reaction = reactants  products  therefore the reactant concentrations decrease a

Hemoglobin

• protein (Hb) found in red blood

cells that reacts with O2

enhances the amount of O2 that

can be carried through the blood

stream

Hb + O2  HbO2

the Hb represents the entire protein

– it is not a chemical formula

the  represents that the reaction

is in dynamic equilibrium

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

• the relative amounts of Hb, O2, and HbO2 at

equilibrium are related to a constant called the

equilibrium constant, K

the larger the value of K, the more product is found at

equilibrium

• changing the concentration of any one of these

necessitates changing the other concentrations to

reestablish equilibrium

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

combine the Hb and O2

together to make more

HbO2

in the cells, with low

concentration of O2, the

equilibrium shifts to

break down the HbO2

and increase the amount

of free O2

HbO2

O2

in lungs

O2

in cells

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

HbF

Hb

Fetal Hemoglobin, HbF HbF + O2  HbFO2

transferred to the fetal

hemoglobin from the

Oxygen Exchange between

Mother and Fetus

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

Reaction Dynamics

• when a reaction starts, the reactants are consumed and

products are made

 forward reaction = reactants  products

 therefore the reactant concentrations decrease and the product

 reverse reaction = products  reactants

 assuming the products are not allowed to escape

 as product concentration increases, the reverse reaction rate increases

• processes that proceed in both the forward and reverse

direction are said to be reversible

 reactants  products

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

Notice that equilibrium does not mean

that the concentrations are equal!

Once equilibrium is established, the rate

of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red

Initially, only the forward reaction takes place.

As the forward reaction proceeds

it makes products and uses reactants.

Because the reactant concentration decreases, the forward reaction slows.

As the products accumulate, the reverse reaction speeds up.

Eventually, the reaction proceeds

in the reverse direction as fast as

it proceeds in the forward direction.

At this time equilibrium is established.

Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

Chemistry, Julia Burdge, 2 nd e., McGraw Hill.

Tro, Chemistry: A Molecular Approach 11

Dynamic Equilibrium

reaction accelerates, eventually they reach the same rate

• dynamic equilibrium is the condition where the rates of the forward and reverse reactions are equal

concentrations of all the chemicals remain constant

because the chemicals are being consumed and

made at the same rate

Tro, Chemistry: A Molecular Approach 12

H2(g) + I2(g)  2 HI(g)

at time 0, there are only reactants

in the mixture, so only the forward reaction can take place

[H2] = 8, [I2] = 8, [HI] = 0

at time 16, there are both reactants and products in the mixture, so both the forward reaction and reverse reaction can take place

[H2] = 6, [I2] = 6, [HI] = 4

Tro, Chemistry: A Molecular Approach 13

H2(g) + I2(g)  2 HI(g)

at time 32, there are now more products than reactants in the mixture − the forward reaction has slowed down as the reactants run out, and the reverse reaction accelerated

[H2] = 4, [I2] = 4, [HI] = 8

at time 48, the amounts of products and reactants in the mixture haven’t changed – the forward and reverse reactions are proceeding at the same rate – it has reached equilibrium

Tro, Chemistry: A Molecular Approach 14

Since the [HI] at equilibrium

is larger than the [H2] or [I2],

we say the position of equilibrium favors products

As the reaction proceeds,

the [H2] and [I2] decrease

and the [HI] increases

Since the reactant

concentrations are

decreasing, the

forward reaction

rate slows down

And since the

no longer change

At equilibrium, the forward reaction rate is the same as the reverse reaction rate

Tro, Chemistry: A Molecular Approach 15

position of equilibrium favors the products

• other reactions reach equilibrium when only a small

percentage of the reactant molecules are consumed –

we say the position of equilibrium favors the reactants

• What occurs in a reaction at

equilibrium.

moving up is the same as the

number of people moving

down, the number of people

on each floor remains

constant, and the two

populations are in

equilibrium.

forward and reverse reactions

are the same.

Chapter Seven 16

Tro, Chemistry: A Molecular Approach 18

Equilibrium Constant

• even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them

• the relationship between the chemical equation and the

concentrations of reactants and products is called the Law

of Mass Action

• for the general equation aA + bB  cC + dD, the Law of

Mass Action gives the relationship below

 the lowercase letters represent the coefficients of the balanced

chemical equation

 always products over reactants

• K is called the equilibrium constant

Tro, Chemistry: A Molecular Approach 19

Writing Equilibrium Constant

Expressions

• for the reaction aA (aq)

+ bB (aq)  cC (aq) + dD (aq) the

equilibrium constant expression

O



K

Tro, Chemistry: A Molecular Approach 20

• when the value of K eq >> 1, we know that when the reaction reaches equilibrium there will be many more product molecules present than reactant molecules

 the position of equilibrium favors products

• when the value of K eq << 1, we know that when the reaction reaches equilibrium there will be many more reactant molecules present than product molecules

 the position of equilibrium favors reactants

Tro, Chemistry: A Molecular Approach 21

A Large Equilibrium Constant

Tro, Chemistry: A Molecular Approach 22

A Small Equilibrium Constant

Tro, Chemistry: A Molecular Approach 23

Relationships between K

and Chemical Equations

• when the reaction is written backwards, the

equilibrium constant is inverted

for the reaction aA + bB  cC + dD

the equilibrium constant expression

is:

for the reaction cC + dD  aA + bB

the equilibrium constant expression is:

1

K

Tro, Chemistry: A Molecular Approach 24

Relationships between K

and Chemical Equations

• when the coefficients of an equation are multiplied by a factor, the equilibrium constant is raised to that factor

C original





for the reaction aA + bB  cC

the equilibrium constant expression

is:

for the reaction 2aA + 2bB  2cC

the equilibrium constant expression is:

2 new

B A

C

B A

c

b a

c

K

Tro, Chemistry: A Molecular Approach 25

Relationships between K

and Chemical Equations

• when you add equations to get a new equation, the

equilibrium constant of the new equation is the product

of the equilibrium constants of the old equations

for the reactions (1) aA  bB and

(2) bB  cC the equilibrium constant

expressions are:

for the reaction aA  cC

the equilibrium constant expression is:

2 1

Writing Equilibrium Expressions ani mation

Tro, Chemistry: A Molecular Approach 26

Kbackward = 1/Kforward, Knew = Koldn

Ex 14.2 – Compute the equilibrium constant at 25°C for

3

1 1







K K

NH3(g)  0.5 N2(g) + 1.5 H2(g)  

5

2 1 8 2

1 2

10 2

5 '

10 7

3

1 '

Equilibrium Constants for Reactions

Involving Gases

• the concentration of a gas in a mixture is

proportional to its partial pressure

• therefore, the equilibrium constant can be

expressed as the ratio of the partial pressures of the gases

• for aA(g) + bB(g)  cC(g) + dD(g) the

equilibrium constant expressions are

P P

P

P K

B A

D

C p







Tro, Chemistry: A Molecular Approach 29

n is the difference between the number of

moles of reactants and moles of products

Tro, Chemistry: A Molecular Approach 30

Deriving the Relationship

Law Gas

Ideal the

RT

RT V

n P

A

A A

A

[A]

ng substituti









B A

a

d c d

c

b a

d c

RT

K RT

P P

RT

P P

RT

P RT

P

RT

P RT

P K

A

D C

B A

D C

c

b a

d c

P P

P

P K

B A

D

C p

since there are more moles of reactant than product, Kc should be

larger than Kp, and it is

K mol

L atm

12

p c

10 4

.

5 K

298 08206

0

10 2

2 NO(g) + O2(g)  2 NO2(g)

n = 2  3 = -1

Tro, Chemistry: A Molecular Approach 33

Heterogeneous Equilibria

• pure solids and pure liquids are materials whose

concentration doesn’t change during the course of a

reaction

 its amount can change, but the amount of it in solution

doesn’t

 because it isn’t in solution

• because their concentration doesn’t change, solids and liquids are not included in the equilibrium constant

expression

• for the reaction aA(s) + bB(aq)  cC(l) + dD(aq) the

equilibrium constant expression is:  

d

K

B D

c 

Tro, Chemistry: A Molecular Approach 34

Heterogeneous Equilibria

 

2 CO

CO p

2

2 c

2

COCO

P

P K

K





The amount of C is different, but the amounts of CO and CO2remains the same

Therefore the amount of

C has no effect on the position of equilibrium.

Tro, Chemistry: A Molecular Approach 35

Calculating Equilibrium Constants from Measured Equilibrium Concentrations

• the most direct way of finding the equilibrium constant

is to measure the amounts of reactants and products in

a mixture at equilibrium

 actually, you only need to measure one amount – then use

stoichiometry to calculate the other amounts

• the equilibrium mixture may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same

 as long as the temperature is kept constant

 the value of the equilibrium constant is independent of the initial amounts of reactants and products

2 2

2

eq 

K

50 1]

[0.11][0.1

[0.78]2



50 055]

[0.055][0.

[0.39]2



50 165]

[0.165][0.

[1.17]2



50 33]

[0.53][0.0 [0.934]2



Calculating Equilibrium Concentrations

• Stoichiometry can be used to determine the equilibrium concentrations of all reactants and products if you know initial concentrations and one equilibrium concentration

• suppose you have a reaction 2 A(aq) + B(aq)  4 C(aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0 You then measure the equilibrium

-½(0.50)

Tro, Chemistry: A Molecular Approach 38

2 CH4(g)  C2H2(g) + 3 H2(g) at 1700°C if the initial

Construct an ICE

table for the reaction

for the substance

equilibrium 0.035

2 CH4(g)  C2H2(g) + 3 H2(g) at 1700°C if the initial

use the known

0035

0

CH

HH

C

2

3

2 4

3 2 2

2 c







K

Tro, Chemistry: A Molecular Approach 40

The following data were collected for the reaction

2 NO2(g)  N2O4(g) at 100°C Complete the table and determine values of Kp and Kc for each experiment.

Tro, Chemistry: A Molecular Approach 41

The following data were collected for the reaction

2 NO2(g)  N2O4(g) at 100°C Complete the table and determine values of Kp and Kc for each experiment.

Expt 1 Expt 2 initial [N2O4] 0 0.0200

4 0310

0

00452

0

15 0 0327

0 7 4

7

4 0172

0

0014

0

) 73 3 08206

0 (

]

[NO

] O [N

2 2

Expt C,

1,2 Expt P,

2 1

Expt C,

2 -

1 C

P

2 2

4

2 C

K K

K

Tro, Chemistry: A Molecular Approach 42

The Reaction Quotient

• if a reaction mixture, containing both

reactants and products, is not at

equilibrium; how can we determine

which direction it will proceed?

• the answer is to compare the current

concentration ratios to the equilibrium

constant

• the concentration ratio of the products

(raised to the power of their

coefficients) to the reactants (raised to

the power of their coefficients) is

called the reaction quotient, Q

 a  b

d c

Q

B A

d c

P P

P

P Q

B A

D

C p







Tro, Chemistry: A Molecular Approach 43

The Reaction Quotient:

Predicting the Direction of Change

• if Q > K, the reaction will proceed fastest in the reverse

direction

 the [products] will decrease and [reactants] will increase

• if Q < K, the reaction will proceed fastest in the

forward direction

 the [products] will increase and [reactants] will decrease

• if Q = K, the reaction is at equilibrium

 the [products] and [reactants] will not change

• if a reaction mixture contains just reactants, Q = 0, and

the reaction will proceed in the forward direction

• if a reaction mixture contains just products, Q = ∞, and

the reaction will proceed in the reverse direction

Q, K, and the Direction of Reaction

If Q = K, equilibrium; If Q < K, forward; If Q > K, reverse

Ex 14.7 – For the reaction below at equilbrium, which direction

will it proceed if PI2 = 0.114 atm, PCl2 = 0.102 atm & PICl =

0.355 atm

for I2(g) + Cl2(g)  2 ICl(g), Kp = 81.9 direction reaction will proceed

102

0 114

0

355

0

p

2

Cl I

2 ICl p

2 2

P Q

since Q (10.8) < K (81.9), the reaction will proceed to the right

Tro, Chemistry: A Molecular Approach 49

Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial

Concentrations or Pressures

• first decide which direction the reaction will proceed

 compare Q to K

• define the changes of all materials in terms of x

 use the coefficient from the chemical equation for the coefficient of x

 the x change is + for materials on the side the reaction is proceeding

Tro, Chemistry: A Molecular Approach 50

[I2] [Cl2] [ICl]initial 0.100 0.100 0.100change

equilibrium

25°C, Kp = 81.9 If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

0 100

0

100

0

p

2

Cl I

2 ICl p

2 2

P Q

since Qp(1) < Kp(81.9), the reaction is proceeding forward

Tro, Chemistry: A Molecular Approach 51

[I2] [Cl2] [ICl]initial 0.100 0.100 0.100change

equilibrium

25°C, Kp = 81.9 If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

represent the change

in the partial

pressures in terms of

x

sum the columns to

find the equilibrium

Cl I

2 ICl p

100

0

2 100

.

0 100

0 100

0

2 100

.

0 9

81

2 2

x

x x

x

x

P P

P K

Tro, Chemistry: A Molecular Approach 52

[I2] [Cl2] [ICl]initial 0.100 0.100 0.100change

equilibrium

25°C, Kp = 81.9 If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

substitute into the

x x

x

x x

x

9 81 2

100

0 100

0

9

.

81

2 100

0 9

81 100

0

9

.

81

2 100

0 100

0

9

.

81

100

0

2 100

0 100

0

2 100

0 9

x x

0

05 11 805

0

9 81 2

100 0 100

0 9 81

Tro, Chemistry: A Molecular Approach 53

[I2] [Cl2] [ICl]initial 0.100 0.100 0.100change

equilibrium

25°C, Kp = 81.9 If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

substitute x into the

0

atm 027

0 0729

0 100

0 100

0 0729

0 100

0 100

100

0 2

100

Tro, Chemistry: A Molecular Approach 54

[I2] [Cl2] [ICl]initial 0.100 0.100 0.100change

equilibrium

-0.0729

25°C, Kp = 81.9 If the initial partial pressures are all

0.100 atm, find the equilibrium concentrations

p

Cl I

2 ICl p

2 2

P K

Kp(calculated) = Kp(given) within significant figures

Tro, Chemistry: A Molecular Approach 55

placed into a 2.00 L flask and heated, what will be

Tro, Chemistry: A Molecular Approach 56

placed into a 2.00 L flask and heated, what will be

[I2] [I]

initial 0.500 0

change -x +2x

equilibrium 0.500- x 2x

since [I]initial = 0, Q = 0

and the reaction must proceed forward

2

2 c

4 500

0 10

76 3

500

0

2 10

76 3

I I

x x

x x