Chemical Equilibrium ________

(1)

________

Chemical Equilibrium

Focusing on Acid-Base Systems

Unit 4

(2)

Chapter 15:

(3)

Chapter 15.1: Reaction Graphs & Equilibrium

is a homogeneous system ( equilibrium) system has same phase throughout

Rxn on graph:

NOTE: #s of molecules don't matter for equilibrium! Equilibrium is reached when:

Rates of rxn are equal Conc. Remains constant

There are no visible changes (eg. colour)

(4)

Equilibrium: the extent of a reaction

In stoichiometry we talked about theoretical yields, and the many reasons actual yields may be lower.

Another critical reason actual yields may be lower is the reversibility of chemical

reactions: some reactions may produce only 70% of the product you may calculate they ought to produce.

Equilibrium looks at the extent of a chemical

reaction.

(5)

A chemical system that is separate from its surroundings where no

matter can enter or leave is called a closed system.

When a bottle of carbonated

beverage is opened, the pressure on the system changes and dissolved gas is allowed to leave the system.

The equilibrium has been disturbed.

(6)

Closed Systems at Equilibrium

Evidence from many chemical reactions occurring in a

closed system has shown us that after some reactions appear to have stopped, there is a mixture of reactants and

products present.

Na₂SO₄(aq) + CaCl₂(aq) CaSOforward ₄(s) + 2 NaCl(aq) reverse

We assume that any closed system with no observable changes occurring is in a state of dynamic equilibrium.

The forward reaction (collisions between reactants to form products) and the reverse reaction (collisions between

products to form reactants) are occurring simultaneously and at the same rate.

(7)

Equilibrium is a state in which there are no observable changes as time goes by.

Chemical equilibrium is achieved when:

• the rates of the forward and reverse reactions are equal and

• the concentrations of the reactants and products remain constant

Physical equilibrium H₂O ^(l)

Chemical equilibrium N₂O₄ ^(g)

H₂O ^(g)

2NO₂ ^(g)

(8)

phase equilibrium

solubility equilibrium

2 2

H O(l)  H O(g)

2+ 2

4 4

CuSO (s)  Cu (aq) + SO (aq)

^

(9)

Rate of sale of cookies

=

Rate of replacing

cookies

(10)

Chemical Reaction Equilibrium

2 2

H (g) + I (g)  2 HI(g),

t

 448 C 

(11)

1 – start with reactants only

2 – start with a mixture of reactants and product 3 – start with products only

No matter the starting conditions, the system reaches a state of dynamic equilibrium each time.

2 2

H (g) + I (g)  2 HI(g),

t

 448 C 

(12)

The rate of the forward reaction decreases as the

number of reactant molecules decreases (fewer collisions).

The rate of the reverse reaction increases as the

number of product molecules increases (more collisions).

Dynamic equilibrium is reached when the rate of the forward reaction is equal to the rate of the reverse reaction.

2 2

H (g) + I (g)  2 HI(g),

t

 448 C 

(13)

Percent Yield

Percent yield provides a way to refer to the chemicals present in equilibrium systems.

78%

The maximum possible yield is calculated using

stoichiometry, assuming all of the reactant molecules are used up to form products.

2 2

H (g) + I (g)  2 HI(g),

t

 448 C 

(14)

Based on percent yield, equilibrium systems fall under one of the above classifications.

(15)

The Concept of Equilibrium

• Consider colorless frozen N₂O₄. At room temperature, it decomposes to brown NO₂:

N₂O₄(g)  2NO₂(g).

• At some time, the color stops changing and we have a mixture of N₂O₄ and NO₂.

• Chemical equilibrium is the point at which the rate of the

forward reaction is equal to the rate of the reverse reaction.

At that point, the concentrations of all species are constant.

• Using the collision model:

– as the amount of NO₂ builds up, there is a chance that two NO₂ molecules will collide to form N₂O₄.

– At the beginning of the reaction, there is no NO₂ so the reverse reaction (2NO₂(g)  N₂O₄(g)) does not occur.

(16)

The Concept of Equilibrium

• As the substance warms it begins to decompose:

N₂O₄(g)  2NO₂(g)

• When enough NO₂ is formed, it can react to form N₂O₄:

2NO₂(g)  N₂O₄(g).

• At equilibrium, as much N₂O₄ reacts to form NO₂ as NO₂ reacts to re-form N₂O₄

• The double arrow implies the process is dynamic.N₂O₄(g) 2NO₂(g)

(17)

N₂O₄ ^(g) 2NO₂ ^(g)

Start with NO₂ Start with N₂O₄ Start with NO₂ & N₂O₄

equilibrium

(18)

constant

(19)

ICE Tables

I – initial C – change

E – equilibrium

ICE tables are used for quantitative

calculations involving chemical equilibrium systems that are not quantitative (i.e. < 99.9%

yield).

(20)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L) Initial 1.00

Change

Equilibrium

H₂ (g) + I₂ (g) 2HI (g)

(21)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00

Change

Equilibrium

H₂ (g) + I₂ (g) 2HI (g)

(22)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change Equilibrium

H₂ (g) + I₂ (g) 2HI (g)

(23)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change

Equilibrium 0.22

H₂ (g) + I₂ (g) 2HI (g)

(24)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

(25)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

(26)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

(27)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78 - 0.78

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

(28)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78 - 0.78

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

0.78 x 2 = 0.78 1

(29)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78 - 0.78 + 1.6

Equilibrium 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

0.78 x 2 = 1.6 1

(30)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78 - 0.78 + 1.6

Equilibrium 0.22 0.22

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

0.78 x 2 = 1.6 1

(31)

Set up ICE Table

Conc. [H₂ (g)]

(mmol/L)

[I₂ (g)]

(mmol/L)

[HI (g)]

(mmol/L)

Initial 1.00 1.00 0.00

Change - 0.78 - 0.78 + 1.6

Equilibrium 0.22 0.22 1.6

1.00 – x = 0.22 0.78

H₂ (g) + I₂ (g) 2HI (g)

0.78 x 1 = 0.78 1

0.78 x 2 = 1.6 1

(32)

 Read pgs. 676 – 687

 pg. 682 Practice #’s 3 – 7

Homework:

(33)

(34)

The Equilibrium Constant, K

_c

Consider the following generic reaction equation for a system at equilibrium:

a, b, c, d – coefficients A, B, C, D – chemical formulas

The equilibrium law expression allows us to calculate the value of the equilibrium constant, K_c.

If K_c > 1, then the products are favoured at equilibrium.

If K_c < 1, then the reactants are favoured at equilibrium.

products reactants

The greater the value of the equilibrium constant, the more the products are favoured at equilibrium.

A + B C + D

a b



c d

   

C D

A B

c d

c a b

K



(35)

First, write the balanced equation with whole-number coefficients.

 

   

2 2

2

NO (g)

NO(g) O (g)

Kc



2 2

2 NO(g) + O (g)  2 NO (g)

(36)

Using Equilibrium Constant (Law) more examples 2 important rules:

1. [H₂O_(l)] is constant!

Does not change EVER!

DO NOT INCLUDE!

2. also solids (s) are considered constant ALSO DO NOT INCLUDE!

Eg) 6HCl _(aq) + 1Fe₂O ₃ _(s) ↔^2FeCl₃ _(aq)^{+ 3H}₂^O_(l)

K = [C]^c [D]^d = [FeCl ₃] ² [H₂O] ³ [A]^a [B]^b [HCl ] ⁶ [Fe₂O₃] ³

Actually is = [FeCl ₃] ² [HCl ] ⁶

(37)

Calculating "K"

E.g.2) 1CO_(g) + 2H₂ _(g) ↔^1CH₃^OH_(aq)

At 773°C the value for the equilibrium constant is 0.398 mol²/L². The equilibrium concentration for CO_(g) & H₂ _(g) are 0.105 mol/L &

0.250 mol/L respectively. What is the equilibrium conc. of methanol?

K = [C]^c [D]^d = [CH₃OH] = [CH₃OH] = 0.398 mol² [A]^a [B]^b [H₂ ]² [CO] [0.250] ² [0.105] L²

[CH₃OH] = 2.61 x 10 ^-3 mol L

(38)

formation decomposition

reciprocal

K_c = 40 (given)

K_c = 0.025

reciprocal

2 2

H (g) + I (g)  2 HI(g) 2 HI(g)  H (g) + I (g)2 2

 

  

2

2 2

HI(g)

H (g) I (g)

Kc 

  

 

2 2

2

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

Kc 

1

c 40 K 

(39)

Equilibrium law expressions do NOT include solids or liquids because their concentrations are fixed – the

chemical amount (number of moles) per unit volume is a constant value.

 NH (g) HCl(g)

3

  

K

c



4 3

NH Cl(s)  NH (g) + HCl(g)

(40)

Ions in solution must be represented as single entities.

Equilibrium constant expressions are always written from the net ionic form of reaction equations. Spectator ions are not included.

2+

Zn (aq) Cu (aq)

Kc

    

 

 

2+ 2+

Zn(s) + Cu (aq)  Cu(s) + Zn (aq)

(41)

Predicting Final Equilibrium Concentrations

RICE Tables

Are used to help determine changes in equilibrium after a reaction.

R = Ratio I = Initial C = Change

E = Equilibrium

Mmmm rice!

& butter chicken!

(42)

Example 1:

In a 1.00 L reaction vessel, hydrogen and iodine react to form hydrogen iodide. If

0.100 mol of each gas are present initially

and at equilibrium is 0.020 mol of iodine

are present calculate the K for the reaction?

(43)

H

_{2 (g)}

+ I

_{2 (g)}

 2HI

_(g)

R I C E

H

₂

I

₂

HI

1 1 2

0.100 0.100 0

-0.08 -0.08 +0.16

0.02 0.020 0.16

Ratio, Initial, Change, Equilibrium [H

₂

] [I

₂

]

K = [HI]

²

= [0.020] [0.020]

[0.160]

²

= 64

(44)

Example 2:

Carbon monoxide and water react in an equilibrium to form carbon dioxide and hydrogen gas. Each of the reactants is present initially with a concentration of 0.10 mol/L and the value for K for the reaction is 4.06. What are the final

concentrations of reactants and products?

(45)

CO

_(g)

+ H

₂

O

_(g)

 CO

_{2 (g)}

+ H

_{2 (g)}

CO H

₂

O CO

₂

1 1 1

0.10 0.10 0

-x -x +x

0.10 - x 0.10 - x x R

I C E

H

₂

1 0 +x

x

= 4.06

= [0.10 - x]

²

[x]

²

K = [CO

₂

][H

₂

] [CO][H

₂

O]

K value tells us that reaction favors the products.

(46)

_x_ = 2.01 [0.10 - x]

x = 0.201 - 2.01x 3.01x = 0.201

x = 0.0668

CO & H

₂

O: 0.10 - 0.0668 = 0.033 M

CO

₂

& H

₂

: x = 0.0668 = 0.067 M

(47)

Predicting Final Equilibrium Concentrations

Go over the example above on p. 686-7.

(48)

 Read pgs. 676 – 689

 pg. 688 Practice #’s 1,3 – 11

Homework:

(49)