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(1)

CHAPTER 13:

CHEMICAL

EQUILIBRIUM

A

Reversible

Reaction

Can Occur In Either Direction

CO(

g

) + H

2

O(

g

)

CO

2

(

g

) + H

2

(

g

)

The forward reaction

proceeds to the right.

The reverse reaction

proceeds to the left.

A System is At

Equilibrium

When

the Rate of the Forward Reaction

Equals the Rate of the Reverse Reaction

H

2

+ I

2

2HI

H

2

+ I

2

2HI

H

2

+ I

2

2HI

The Net Concentrations of Reactants and Products

Do NOT Change At Equilibrium

Equilibrium

Is Established When

the

Rate of the Forward Reaction

Equals

the Rate of the Reverse Reaction

H

2

+ I

2

2HI

What can be said about concentrations at

equilibrium?

When you open the valve on the flasks above, the hydrogen and iodine gases will react according to the equilibrium equation below. By visually observing the flasks, how do

(2)

What can be said about concentrations at

equilibrium?

Forward reaction:

Reverse reaction:

What can be said about concentrations at

equilibrium?

Forward & reverse

reactions at the same time:

A system is at

equilibrium

when the rates of the

forward and reverse reactions are the same.

Both the forward and reverse reactions are

elementary

reactions, so we can write their rate

laws from the balanced equation.

rate forward =

k

f

[N

2

O

4

] and rate reverse =

k

r

[NO

2

]

2

N

2

O

4

(

g

)

2NO

2

(

g

)

N

2

O

4

(

g

)

2NO

2

(

g

)

Starting with N2O4

N

2

O

4

(

g

)

2NO

2

(

g

)

Starting with NO2

Important Things to Remember

About Equilibrium

1.

Equilibrium is a dynamic state – both the

forward and reverse reactions continue to

occur, although there is no net change in

reactant and product concentrations over time.

2.

At equilibrium, the rates of the forward and

reverse reactions are equal.

(3)

The Equilibrium Constant,

K

rate forward = rate reverse kf[N2O4] eq = kr[NO2]2eq

The subscript “eq” denotes a concentration at equilibrium.

Rearranging

The ratio of two constants (kf/kr) is also a constant:

equilibrium expression

N2O4(

g

)

2NO2(

g

)

2

k k

2 eq f r 2 4 eq

NO N O

2

Kc 2 eq 2 4 eq NO N O

equilibrium constant

The

Equilibrium Constant

,

K

,

Relates the Equilibrium Concentrations

of Reactants and Products.

The value of Keq varies with temperature and units are usually omitted.

a

A +

b

B

c

C +

d

D

equilibrium

constant

=

K

=

[C]

c

[D]

d

[A]

a

[B]

b

=

[products]

[reactants]

Brackets, [ ], are used to symbolize concentration in

moles

per

liter

(mol/L).

When Writing an Equilibrium Constant,

K

, the

Coefficient

Becomes the

Exponent

N

2

(

g

) + O

2

(

g

)

2 NO(

g

)

equilibrium

constant

=

K

=

[N

2

] [O

2

]

[NO]

2

4NH

3

+ 3O

2

2N

2

+ 6H

2

O

When Writing an Equilibrium Constant,

K

, the

Coefficient

Becomes the

Exponent

K

eq

=

N

2

[ ]

2

H

2

O

[

]

6

NH

3

[

]

4

O

2

[ ]

3

The Size of

K

Indicates Whether Reactants or

Products are Favored at Equilibrium

When

K

is around 1

(0.01 <

K

< 100),

[products]

[reactants]

Both are similar

in magnitude.

both reactants

and

products

are present in

similar amounts

.

The Size of

K

Indicates Whether Reactants or

Products are Favored at Equilibrium

When

K

is much greater than 1,

When

K

is much

less than 1

,

[products]

[reactants]

The numerator is larger.

[products]

[reactants]

The denominator is larger.

equilibrium lies to the

right

and favors the

products

.

(4)

The Size of

K

Indicates Whether Reactants or

Products are Favored at Equilibrium

Very small

Very large

Reaction proceeds hardly at all

Reaction proceeds nearly to completion Appreciable concentrations

of both reactants and products are present at equilibrium

K

eq

10–3 1 103

2 H

2

(

g

) + O

2

(

g

)

2 H

2

O(

g

)

K

= 2.9 x 10

82

The product is favored because

K

> 1

.

The equilibrium lies to the

right

.

How To Calculate the Equilibrium Constant for

a Reaction

Given the following information, calculate K for the

reaction between the reactants A

2

and B

2

.

[A

2

] = 0.25 M

[B

2

] = 0.25 M

[AB] = 0.50 M

A

2

+ B

2

2 AB

How To Calculate the Equilibrium Constant for

a Reaction

STEP 1:

Write the expression for the equilibrium constant from

the balanced equation.

A

2

+ B

2

2 AB

[AB]

2

[A

2

][B

2

]

K

=

How To Calculate the Equilibrium Constant for

a Reaction

STEP 2:

Substitute the given concentrations in the equilibrium

expression and calculate K.

[AB]

2

[A

2

][B

2

]

K

=

=

[0.50]

2

[0.25][0.25]

=

0.25

0.0625

=

4.0

The units of

K

are usually omitted.

Use Equilibrium Concentrations and the Equilibrium Constant

Expression to Calculate the Value of the Equilibrium Constant

Calculate the value of the equilibrium constant for

the reaction:

If the concentrations at equilibrium are

[A] = 2.0 M

[B] = 1.5 M

[C] = 0.010 M

A + 2B

C

K

eq

=

[ ]

C

A

[ ]

[ ]

B

2

=

0.010

(

)

2.0

( )

( )

1.5

2

= 0.0022

Use Equilibrium Concentrations and the Equilibrium Constant

Expression to Calculate the Value of the Equilibrium Constant

Carbonyl chloride (COCl

2

), also called phosgene, is a highly

poisonous gas that was used on the battlefield in World War I. It is

produced by the reaction of carbon monoxide with chlorine gas:

CO(

g

) + Cl

2

(

g

)



COCl

2

(

g

)

The equilibrium concentrations are:

[CO] = 0.012

M

[Cl

2

] = 0.054

M

[COCl

2

] = 0.14

M

(5)

When the species in a reversible chemical reaction are not all

in the same phase, the equilibrium is

heterogeneous

.

CaCO

3(s)

CaO

(s)

+ CO

2(g)

¢

K

=

[

CO

2

]

[

CaO

]

CaCO

3

[

]

K

eq

=

[

CO

2

]

Pure Solids and Pure Liquids do not appear in

Equilibrium Expressions

(

s

)

not included

in the equilibrium expression

(

l

)

not included

in the equilibrium expression

(

g

) included in the equilibrium expression

(

aq

) included in the equilibrium expression

CO

2(g)

+ C

(s)

2CO

(g)

K

eq

=

[ ]

CO

2

CO

2

[

]

K

eq

=

1

Hg

22+

[

]

[ ]

Cl

- 2

Hg

22+(aq)

+ 2Cl

(aq)

Hg

2

Cl

2(s)

Worked Example 15.3

Strategy Use the law of mass action to write the equilibrium expressions for each reaction. Only gases and aqueous species appear in the expression. Write equilibrium expressions for each of the following reactions:

(a) CaCO3(s) ⇌ CaO(s) + CO2(g)

(b) Hg(l) + Hg2+(aq) Hg 22+(aq)

(c) 2Fe(s) + 3H2O(l) ⇌ Fe2O3(s) + 2H2(g)

(d) O2(g) + 2H2(g) ⇌ 2H2O(l)

Solution (a) Kc = [CO2] (b) Kc =

[Hg22+]

[Hg2+]

(d) Kc =

1 [O2][H2]2

(c) Kc = [H2]2

Think About It Like writing equilibrium expressions for homogeneous equilibria, writing equilibrium expressions for heterogeneous equilibria becomes second nature if you practice. The importance of developing this skill now cannot be overstated. Your ability to understand the principles and to solve many of the problems in and Chapters 16 to 19 depends on your ability to write equilibrium expressions correctly and easily.

When a chemical equation is reversed,

the new equilibrium expression is

the reciprocal of the original equilibrium expression.

C + D A + B

K

eq

=

C

[ ]

[ ]

D

A

[ ]

[ ]

B

A + B C + D

¢

K

eq

=

A

[ ]

[ ]

B

C

[ ]

[ ]

D

=

1

K

eq

When a chemical equation is

multiplied

by a constant,

the concentrations of the new equilibrium expression

are raised to a power equal to the constant.

2A + 2B 2C + 2D

A + B C + D

¢

K

eq

=

C

[ ]

2

D

[ ]

2

A

[ ]

2

B

[ ]

2

=

C

[ ]

[ ]

D

A

[ ]

[ ]

B

æ

è

çç

ö

ø

÷÷

2

=

K

2eq

K

eq

=

C

[ ]

[ ]

D

A

[ ]

[ ]

B

(6)

Worked Example 15.4

Strategy Begin by writing the equilibrium expressions for the reactions that are given. Then, determine the relationship of each equation’s equilibrium expression to the equilibrium expression of the original equations, and make the corresponding change to the equilibrium constant for each.

Kc = and Kc =

The following reactions have the indicated equilibrium constants at 100°C:

(1) 2NOBr(g) ⇌ 2NO(g) + Br2(g) Kc = 0.014

(2) Br2(g) + Cl2(g) ⇌ 2BrCl(g) Kc = 7.2

Determine the value of Kc for the following reactions at 100°C:

(a) 2NO(g) + Br2(g) ⇌ 2NOBr(g) (d) 2NOBr(g) + Cl2(g) ⇌ 2NO(g) + 2BrCl(g)

(b) 4NOBr(g) ⇌ 4NO(g) + 2Br2(g) (e) NO(g) + BrCl(g) ⇌ NOBr(g) + Cl2(g)

(c) NOBr(g) ⇌ NO(g) + Br2(g)

1 2 1

2

[NO]2[Br 2]

[NOBr]2

[BrCl]2

[Br2][Cl2]

Worked Example 15.4 (cont.)

Solution (a) This equation is the reverse of original equation 1. Its equilibrium expression is the reciprocal of that for the original equation:

Kc = = 1/0.014 = 71

(b) This is original equation 1 multiplied by a factor of 2. Its equilibrium expression is the original expression squared:

Kc = = (0.014)2 = 2.0×10-4

(c) This is the original equation 1 multiplied by . Its equilibrium expression is the square root of the original

Kc = or Kc = = (0.014)1/2 = 0.12

[NOBr]2

[NO]2[Br 2]

[NO]2[Br 2]

[NOBr]2

2

1 2

[NO]2[Br 2]

[NOBr]2

1/2

2 2 2

] [

] [ ] [

NOBr Br NO

Worked Example 15.4 (cont.)

Solution (d) This is the sum of the original equations 1 and 2. Its equilibrium expression is the product of the two individual expressions:

Kc = = (0.014)(7.2) = 0.10

(e) Probably the simplest way to analyze this reaction is to recognize that it is the reverse of the reaction in part (d), multiplied by . Its equilibrium expression is the square root of the reciprocal of the expression in part (d):

Kc = = (1/0.10)1/2 = 3.2

[NO]2[BrCl]2

[NOBr]2[Cl 2]

[NO]2[BrCl]2

[NOBr]2[Cl 2]

1/2

1 2

Worked Example 15.4 (cont.)

Think About It The magnitude of an equilibrium constant reveals whether products or reactants are favored, so the reciprocal relationship between Kc values

of forward and reverse reactions should make sense. A very large Kc value means

that products are favored. In the reaction of hydrogen ion and hydroxide ion to form water, the value of Kc is very large, indicating that the product, water, is

favored.

H+(aq) + OH-(aq) H

2O(l) Kc = 1.0×1014 (at 25°C)

Simply writing the equation backward doesn’t change the fact that water is the predominate species. In the reverse reaction, therefore, the favored species is on the reactant side:

H2O(l) ⇌ H+(aq) + OH-(aq) Kc = 1.0×10-14 (at 25°C)

As a result, the magnitude of Kc should correspond to reactants being favored;

that is, it should be very small.

When an equilibrium expression contains only gases,

we can write an alternative form of the expression in which

the concentrations of gases are expressed

in partial pressures (atm).

we can write as

K

c

=

or

K

P

=

The relationship between

K

c and

K

P

can be expressed as

where Δ

n

= moles of gaseous products – moles of gaseous reactants.

N

2

O

4

(

g

)

2NO

2

(

g

)

[NO

2

]

2

[N

2

O

4

]

(

P

NO2

)

2

P

N2O4

K

P

=

K

c

[(0.08206 L∙atm/K∙mol)×

T

]

Δn

Equilibrium Expressions Involving Pressures

PV

=

nRT

P

=

n

V

æ

è

ç

ö

ø

÷

RT

=

CRT

C = moles of gas per volume

(7)

Equilibrium Expressions Involving Pressures

K

eq

=

Z

[ ]

4

A

[ ]

[ ]

B

=

C

Z4

C

A

C

B

=

P

Z

RT

æ

è

ç

ö

ø

÷

4

P

A

RT

æ

è

ç

ö

ø

÷

P

B

RT

æ

è

ç

ö

ø

÷

A + B

4Z

C

=

P

RT

Equilibrium Expressions Involving Pressures

=

P

Z

RT

æ

è

ç

ö

ø

÷

4

P

A

RT

æ

è

ç

ö

ø

÷

P

B

RT

æ

è

ç

ö

ø

÷

=

P

Z4

P

A

( )

( )

P

B

´

1

RT

æ

è

ç

ö

ø

÷

4

1

RT

æ

è

ç

ö

ø

÷

1

RT

æ

è

ç

ö

ø

÷

Equilibrium Expressions Involving Pressures

K

eq

=

P

Z4

P

A

( )

( )

P

B

´

1

RT

æ

è

ç

ö

ø

÷

2

K

eq

=

K

p

´

( )

RT

-2

K

p

Equilibrium Expressions Involving Pressures

K

p

=

K

eq

´

( )

RT

2

K

eq

=

K

p

´

( )

RT

-2

Where Dn = sum of product coefficients – sum of reactant coefficients

Worked Example 15.5

Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures.

Write KP expressions for (a) PCl3(g) + Cl2(g) ⇌ PCl5(g),

(b) O2(g) + 2H2(g) ⇌ 2H2O(l), and (c) F2(g) + H2(g) ⇌ 2HF(g).

Solution(a) All the species in this equation are gases, so they will all appear in the KP expression.

KP =

(b) Only the reactants are gases. KP =

(c) All species are gases. KP =

(PPCl5)

(PPCl3)(PCl2)

1 (PO2)(PH2)2

(PHF)2

(PF2)(PH2)

Think About It It isn’t necessary for every species in the reaction to be a gas– only those species that appear in the equilibrium expression.

Worked Example 15.6

Strategy Use KP = Kc[(0.08206 L∙atm/K∙mol)×Tn. Be sure to convert

temperature in degrees Celsius to kelvins. Using Δn = moles of gaseous products – moles of gaseous reactants, Δn = 2(NO2) – 1(N2O4) = 1. T = 298K.

The equilibrium constant, Kc, for the reaction

N2O4(g) ⇌ 2NO2(g)

is 4.63×10-3 at 25°C. What is the value of K

P at this temperature.

Solution

KP = Kc ×T

= (4.63×10-3)(0.08206 × 298)

= 0.113 0.08206 L∙atm

K∙mol

Think About It Note that we have essentially disregarded the units of R and T so that the resulting equilibrium constant, KP, is unitless.

(8)

The

Reaction Quotient

,

Q

, has the same form as the

Equilibrium Constant, K, except that the concentrations are

not the equilibrium concentrations

(usually

initial concentrations

are used to calculate Q).

   

   

c d

a b

Q

c

C D

A B

a

A +

b

B

c

C +

d

D

The Equilibrium Constant

The value of the reaction quotient,

Q,

changes as the reaction

progresses

N2O4(

g

)

2NO2(

g

)

To predict the direction in which a reaction will proceed

when we start with a mixture of reactants and products,

calculate the value of the reaction quotient Q

and compare it to the value of the equilibrium constant K

There are three possibilities:

1.

Q

<

K

The ratio of initial concentrations of products to

reactants is too small. To reach equilibrium, reactants must be

converted to products. The

system proceeds in the forward

direction

.

2.

Q

=

K

the initial concentrations are equilibrium concentrations.

The

system is at equilibrium

.

3.

Q > K

the ratio of initial concentrations of products to reactants

is too large. To reach equilibrium products must be converted

to reactants. The

system proceeds in the reverse direction

.

Worked Example 15.7

Strategy Use the initial concentrations to calculate Qc, and then compare Qc

with Kc.

Qc = = = 0.61

At 375°C, the equilibrium constant for the reaction

N2(g) + 3H2(g) ⇌ 2NH3(g)

is 1.2. At the state of a reaction, the concentrations of N2, H2, and NH3 are

0.071 M, 9.2×10-3M, and 1.83×10-4M, respectively. Determine whether the

system is at equilibrium, and if not, determine in which direction it must proceed to establish equilibrium.

[NH3]i2

[N2]i[H2]i3

(1.83×10-4)2

(0.071)(9.2×10-3)3

Strategy The calculated value of Qc is less than Kc. Therefore, the reaction is

not at equilibrium and must proceed to the right to establish equilibrium. Think About It In proceeding to the right, a reaction consumes reactants and produces more products. This increases the numerator in the reaction quotient and decreases the denominator. The result is an increase in Qc until it is equal to Kc, at which point equilibrium

will be established.

Equilibrium concentrations can be calculated from initial

concentrations if the equilibrium constant is known.

Kc = 24.0 (200°C)

Initial concentration (M) 0.850 0 Change in concentration (M)

Equilibrium concentration (M)

–x

0.850 – x +x

x

cis-Stilbene trans-Stilbene

cis-Stilbene trans-Stilbene

Equilibrium concentrations can be calculated from initial

concentrations if the equilibrium constant is known.

Initial concentration (M) 0.850 0 Change in concentration (M)

Equilibrium concentration (M)

–x

0.850 – x

+x

x

cis-Stilbene trans-Stilbene

Use the equilibrium concentrations, defined in terms of

x

, in the

equilibrium expression:

trans

K cis c

-stilbene -stilbene

 

24.0 0.850

 

x x

(9)

Equilibrium concentrations can be calculated from initial

concentrations if the equilibrium constant is known.

Initial concentration (M) 0.850 0 Change in concentration (M)

Equilibrium concentration (M)

–0.816

0.8500.816

+0.816

0.816 cis-Stilbene trans-Stilbene

Calculate the equilibrium concentrations of

cis-

and

trans-

stilbene:

[

cis

-stilbene] = (0.850 –

x

)

M

= (0.850 – 0.816)

M

= 0.034

M

[

trans

-stilbene] =

x

M

= 0.816

M

Calculate the value of the equilibrium constant for the reaction

if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.

A + 2B

C + 2D

Example Problem 1

A + 2B

C

+ 2D

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

1.20 1.70 0.000 0.000

0.100

Calculate the value of the equilibrium constant for the reaction

if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.

A + 2B

C + 2D

Example Problem 1

A + 2B

C

+ 2D

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

1.20 1.70 0.000 0.000

0.100

0.100 0.200 –0.200

–0.100

Calculate the value of the equilibrium constant for the reaction

if 1.20 mol of A and 1.70 mol of B are dissolved in 1.00 L of solution, whereupon 0.100 mol of C is produced at equilibrium.

A + 2B

C + 2D

Example Problem 1

A + 2B

C

+ 2D

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

1.20 1.70 0.000 0.000

0.100

0.100 0.200 –0.200

–0.100

1.10 1.50 0.200

Example Problem 1

A + 2B

C

+ 2D

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

1.20 1.70 0.000 0.000

0.100

0.100 0.200 –0.200

–0.100

1.10 1.50 0.200

K

eq

=

[ ]

C

[ ]

D

2

A

[ ]

[ ]

B

2

=

0.100

(

)

(

0.200

)

2

1.10

(

)

(

1.50

)

2

= 0.00162

A + B

2Z

For the reaction shown below, Keq = 4.0 x 10–4.

Calculate the equilibrium concentration of Z if 0.500 mol of A and 0.500 mol of B are dissolved in 1.00 L of solution and allowed to come to equilibrium.

A + B

2Z

Example Problem 2

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.500 0.500 0.000

(10)

A + B

2Z

For the reaction shown below, Keq = 4.0 x 10–4.

Calculate the equilibrium concentration of Z if 0.500 mol of A and 0.500 mol of B are dissolved in 1.00 L of solution and allowed to come to equilibrium.

A + B

2Z

Example Problem 2

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.500 0.500 0.000

–x –x 2x

0.500–x 0.500–x 2x

A + B

2Z

Example Problem 2

Initial concentration (M)

Change due to reaction (M)

Equilibrium concentration (M)

0.500 0.500 0.000

–x –x 2x

0.500–x 0.500–x 2x

K

eq

=

Z

[ ]

2

A

[ ]

[ ]

B

=

2

x

( )

2

0.500

-

x

(

)

(

0.500

-

x

)

= 4.0

´

10

-4

Example Problem 2

K

eq

=

[ ]

Z

2

A

[ ]

[ ]

B

=

2

x

( )

2

0.500

-

x

(

)

(

0.500

-

x

)

= 4.0

´

10

-4

2

x

( )

2

0.500

-

x

(

)

2

= 4.0

´

10

-4

2

x

0.500

-

x

æ

è

ç

ö

ø

÷

2

= 4.0

´

10

-4

Example Problem 2

2

x

0.500

-

x

æ

è

ç

ö

ø

÷

2

= 4.0

´

10

-4

2

x

0.500

-

x

= 4.0

´

10

-4

= 0.020

2

x

= 0.500

(

-

x

)

(

0.020

)

= 0.0100

-

0.020

x

2

x

= Z

[ ]

= 0.0100 M

Worked Example 15.8

Strategy Insert the starting concentrations that we know into the equilibrium table:

Kc for the reaction of hydrogen and iodine to produce hydrogen iodide, H2(g) + I2(g) ⇌ 2HI(g)

is 54.3 at 430°C. What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and I2?

Initial concentration (M) 0.240 0.240 0

Change in concentration (M)

Equilibrium concentration (M)

H2 + I2 ⇌ 2HI

Worked Example 15.8 (cont.)

Solution We define the change in concentration of one of the reactants as x. Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium. According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x). Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the equilibrium concentrations.

Initial concentration (M) 0.240 0.240 0

Change in concentration (M) –xx +2x

Equilibrium concentration (M) 0.240 – x 0.240 – x 2x

(11)

Worked Example 15.8 (cont.)

Solution Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression and solve for x.

Kc =

54.3 = =

=

x = 0.189

Use the calculated value of x, we can determine the equilibrium concentration of each species as follows:

[H2] = (0.240 – x) M = 0.051 M

[I2] = (0.240 – x) M = 0.051 M

[HI] = 2x = 0.378 M [HI]2

[H2][I2]

(2x)2

(0.240 – x)(0.240 – x) (2x)2

(0.240 – x)2

3 .

54 2x

0.240 – x

Think About It Always check your answer by inserting the calculated concentrations into the equilibrium expression:

The small difference between the calculated Kc and the one given in

the problem statement is due to rounding. = 54.9 ≈ Kc

[HI]2

[H2][I2]

(0.378)2

(0.051)2

=

Worked Example 15.9

Strategy Using the initial concentrations, calculate the reaction quotient, Qc,

and compare it to the value of Kc (given in the problem statement of Worked

Example 15.8) to determine which direction the reaction will proceed to establish equilibrium. Then, construct an equilibrium table to determine the equilibrium concentrations.

For the same reaction and temperature as in Worked Example 15.8, calculate the equilibrium concentrations of all three species if the starting concentrations are as follows: [H2] = 0.00623 M, [I2] = 0.00414 M, [HI] = 0.0424 M.

[HI]2

[H2][I2]

(0.0424)2

(0.00623)(0.00414)

= = 69.7

Worked Example 15.9 (cont.)

Strategy Therefore, Qc > Kc, so the system will proceed to the left (reverse) to

reach equilibrium. The equilibrium table is

Initial concentration (M) 0.00623 0.00414 0.0424

Change in concentration (M)

Equilibrium concentration (M)

H2 + I2 ⇌ 2HI

Worked Example 15.9 (cont.)

Solution Because we know the reaction must proceed from right to left, we know that the concentration of HI will decrease and the concentrations of H2 and

I2 will increase. Therefore, the table should be filled in as follows:

Next, we insert these expressions for the equilibrium concentration into the equilibrium expression and solve for x.

Initial concentration (M) 0.00623 0.00414 0.0424

Change in concentration (M) +x +x –2x

Equilibrium concentration (M) 0.00623 + x 0.00414 + x 0.0424 – 2x

H2 + I2 ⇌ 2HI

[HI]2

[H2][I2]

Kc =

(0.0424 – 2x)2

(0.00623 + x)(0.00414 + x) 54.3 =

Worked Example 15.9 (cont.)

Solution It isn’t possible to solve this equation the way we did in Worked Example 15.8 (by taking the square root of both sides) because the concentrations of H2 and I2 are unequal. Instead, we have to carry out the multiplications.

54.3(2.58×10-5 + 1.04×10-2x + x2) = 1.80×10-3 – 1.70×10-1x + 4x2

Collecting terms we get

50.3x2 + 0.735x – 4.00×10-4 = 0

This is a quadratic of the form ax2 + bx + c = 0. The solution for the quadratic

equation [Appendix 1] is

x =

Here we have a = 50.3, b = 0.735, and c = -4.00×10-4, so

x =

a ac b b 2 4 2   ) 3 . 50 ( 2 ) 10 00 . 4 )( 3 . 50 ( 4 ) 735 . 0 ( 735 .

0 2 4

Worked Example 15.9 (cont.)

Solution x = 5.25×10-4 or x = –0.0151

Only the first of these values, 5.25×10-4, makes sense because concentration

cannot be a negative number. Using the calculated value of x, we can determine the equilibrium concentration of each species as follows:

[H2] = (0.00623 + x) M = 0.00676 M

[I2] = (0.00414 + x) M = 0.00467 M

[HI] = (0.0424 – 2x) M = 0.0414 M

Think About It Checking this result gives

[HI]2

[H2][I2]

(0.0414)2

(0.00676)(0.00467)

= = 54.3

(12)

Worked Example 15.10

Strategy Construct an equilibrium table to determine the equilibrium partial pressures.

A mixture of 5.75 atm of H2 and 5.75 atm of I2 is contained in a 1.0-L vessel at 430°C. The equilibrium constant (KP) for the reaction

H2(g) + I2(g) ⇌ 2HI(g)

at this temperature is 54.3. Determine the equilibrium partial pressures of H2, I2, and HI.

Initial partial pressure (atm) 5.75 5.75 0

Change in partial pressure (atm) –xx +2x

Equilibrium partial pressure (atm) 5.75 – x 5.75 – x 2x

H2 + I2 ⇌ 2HI

Worked Example 15.10 (cont.)

Solution Setting the equilibrium expression equal to KP,

54.3 =

Taking the square root of both sides of the equation gives

=

The equilibrium partial pressures are PH2 = PI2 = 5.75 – 4.52 = 1.23 atm, and

PHI = 9.04 atm.

(2x)2

(5.75 – x)2

3 .

54 2x

5.75 – x 2x 5.75 – x 7.369(5.75 – x) = 2x

42.37 – 7.369x = 2x 42.37 = 9.369x

x = 4.52 7.369 =

Think About It Plugging the calculated partial pressures into the equilibrium expression gives

The small difference between this result and the equilibrium constant given in the problem is due to rounding.

= 54.0 (PHI)2

(PH2)(PI2)

(9.04)2

(1.23)2

=

Le Châtelier

s Principle

is a General Rule

Used to Explain the Effect of a

Change in Reaction Conditions on Equilibrium

If a chemical system at equilibrium is disturbed or

stressed, the system will react in a direction that

counteracts the disturbance or relieves the stress.

The direction of shift is usually stated as

to the left

or

to the right.

Le Chatelier’s Principle

states that when a stress is

applied to a system at equilibrium, the system will

respond by shifting in the direction that minimizes the

effect of the stress.

Stress refers to any of the following:

The addition of a reactant or product.

The removal of a reactant or product.

A change in volume of the system, resulting in a change

in concentration or partial pressure of the reactants and

products.

A change in temperature.

What Happens When the

Concentration

of a

Reactant or Product is

Changed

?

2 CO(

g

) + O

2

(

g

)

2 CO

2

(

g

)

What happens if [CO(

g

)] is

increased

?

The concentration of O

2

(

g

) will

decrease

.

The concentration of CO

2

(

g

) will

increase

.

What Happens When the

Concentration

of a

Reactant or Product is

Changed

?

2 CO(

g

) + O

2

(

g

)

2 CO

2

(

g

)

What happens if [CO

2

(

g

)] is

increased

?

The concentration of CO(

g

) will

increase

.

(13)

What Happens When the

Concentration

of a

Reactant or Product is

Changed

?

What happens if a product is

removed

?

The concentration of ethanol will

decrease

.

•The concentration of the other product (C

2

H

4

)

will

increase

.

Application of Le Châtelier

s Principle

Hydrogen sulfide (H

2

S) is a contaminant commonly found in

natural gas. It is removed by reaction with oxygen to produce

elemental sulfur.

2 H

2

S(

g

) + O

2

(

g

)



2 S(

s

) + 2 H

2

O(

g

)

For each of the following, determine whether the equilibrium

will shift to the right, shift to the left, or neither.

a)

Addition of O

2

(

g

)

b)

Removal of H

2

S(

g

)

c)

Removal of H

2

O(

g

)

d)

Addition of S(

s

)

What Happens When the

Temperature

of a

Reaction is

Changed

?

When the temperature is

increased

, the reaction

that

absorbs heat is favored

.

An

endothermic

reaction absorbs heat, so increasing

the temperature favors the forward reaction.

What Happens When the

Temperature

of a

Reaction is

Changed

?

Conversely, when the

temperature is decreased

,

the

reaction that adds heat is favored

.

An

exothermic

reaction releases heat, so

increasing

the temperature

favors the

reverse

reaction.

Factors That Affect Chemical Equilibrium

Changes in volume and concentration do not change the value of the

equilibrium constant.

A change in temperature can alter the value of the equilibrium

constant.

Heat + N2O4(

g

)

2NO2(

g

)

Δ

H

°

= 58.0

kJ/mol

Because the processes is endothermic, adding heat shifts the equilibrium

toward products

Factors That Affect Chemical Equilibrium

For any endothermic reaction, heat is a reactant:

Adding heat shifts the reaction towards products,

K

c

increases

Removing heat shifts the reaction towards reactants,

K

c

decreases.

For any exothermic reaction, heat is a product:

Adding heat shifts the reaction towards reactants,

K

c

decreases

Removing heat shifts the reaction towards products,

K

c

increases

heat + reactants

products

Δ

H

°

> 0

kJ/mol

(14)

What Happens When the

Pressure

of a

Reaction is

Changed

?

When

pressure increases

, equilibrium shifts in

the

direction that decreases the number of moles

in order to decrease pressure.

What Happens When the

Pressure

of a

Reaction is

Changed

?

When

pressure decreases

, equilibrium shifts in the

direction that increases the number of moles

in

order to increase pressure.

Worked Example 15.11

Strategy Use Le Châtelier’s principle to predict the direction of shift in each case. Remember that the position of the equilibrium is only changed by the addition or removal of a species that appears in the reaction quotient expression.

Because sulfur is a solid, it does not appear in the expression.

Hydrogen sulfide (H2S) is a contaminant commonly found in natural gas. It is removed by reaction with oxygen to produce elemental sulfur.

2H2S(g) + O2(g) ⇌ 2S(s) + 2H2O(g)

For each of the following scenarios, determine whether the equilibrium will shift to the right, shift to the left, or neither: (a) addition of O2(g), (b) removal of H2S(g), (c) removal of H2O(g), and (d) addition of S(s).

[H2O]2

[H2S][O2]

Qc =

Worked Example 15.11 (cont.)

Solution

Changes in concentration of any of the other species will cause a change in the equilibrium position. Addition of a reactant or removal of a product that appears in the expression Qc will shift the equilibrium to the right:

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

Removal of a reactant or addition of a product that appears in the expression Qc will shift the equilibrium to the left:

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

(a) Shift to the right (d) No change

[H2O]2

[H2S][O2]

Qc =

addition addition

removal

removal removal

addition

(b) Shift to the left (c) Shift to the right Think About It In each case, analyze the effect the change will have on the value of Qc. In part (a), for example, O2 is added, so its

concentration increases. Looking at the reaction quotient expression, we can see that a larger concentration of oxygen corresponds to a larger overall denominator – giving the overall fraction a smaller value. Thus, Q will temporarily be smaller that K and the reaction will have to shift to the right, consuming some of the added O2

(along with some of the H2S in the mixture) to reestablish

equilibrium.

When

volume decreases

, equilibrium shifts in the

direction that decreases the number of moles of gas

in order to decrease pressure.

N

2

O

4

(

g

)

2NO

2

(

g

)

Equilibrium mixture:

[N2O4] = 0.643 M [NO2] = 0.0547 M

 

 2   c

2 4

NO

N O 

  ..

. K

2 2

3

0 0547 4 65 10 0 643

 

 2  

c c

2 4

NO

N O 

  ..  

.

Q K

2 2

3

0 1094 9 31 10 1 286

Volume decreases by half, concentrations are initially

doubled: [N2O4] = 1.286 M [NO2] = 0.1094 M

The reaction shifts to the left. N2O4(g) ⇌ 2NO2(g)

Worked Example 15.12

Strategy Determine which direction minimized the number of moles of gas in the reaction. Count only moles of gas.

For each reaction, predict in what direction the equilibrium will shift when the volume of the reaction vessel is decreased.

(a) PCl5(g) ⇌ PCl3(g) + Cl2(g)

(b) 2PbS(s) + 3O2(g) ⇌ 2PbO(s) + 2SO2(g)

(c) H2(g) + I2(g) ⇌ 2HI(g)

Solution (a) We have 1 mole of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the left.

(b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, so it will shift to the right.

(c) 2 moles of gas on each side, so no shift.

(15)

Summary of Le Chatelier’s Principle

Summary of Le Chatelier’s Principle

Component added

Shift to

opposite

side

Component removed

Shift to

same

side

A + B C

+ D

Application of Le Châtelier

s Principle

For an exothermic reaction, indicate which direction the

equilibrium will shift for each of the indicated changes:

N

2(g)

+ O

2(g)

2NO

(g)

First where do we write

heat

?

Application of Le Châtelier

s Principle

i.

Some N

2

is removed.

ii. The temperature is decreased.

iii. Some NO is added.

iv. Some O

2

is removed.

v. A catalyst is added.

vi. The temperature is increased and some O

2

is removed.

N

2(g)

+ O

2(g)

2NO

(g)

+

heat

Application of Le Châtelier

s Principle

For an exothermic reaction, indicate which direction the

equilibrium will shift for each of the indicated changes:

i.

Decreasing the concentration of H

2

.

ii. Increasing the concentration of C

6

H

6

.

iii. Decreasing the temperature.

iv. Increasing the pressure by decreasing the volume of the

container.

C

6

H

6(g)

+ 3H

2(g)

C

6

H

12(g)

+

heat

Le Châtelier

s Principle and Respiration

Hb + O

2

HbO

2

Lung [O

2

] high

(16)

Le Châtelier

s Principle and Respiration

Hb + O

2

HbO

2

Muscle [O

2

] low

Reaction shifts left

More Applications of Le Châtelier

s Principle

Co(H

2

O)

62+(aq)

+ 4Cl

–(aq)

+ heat 4CoCl

42–(aq)

+ 6H

2

O

(l)

Room temperature

More Applications of Le Châtelier

s Principle

Co(H

2

O)

62+(aq)

+ 4Cl

–(aq)

+ heat 4CoCl

42–(aq)

+ 6H

2

O

(l)

Heating the solution

Reaction shifts right

More Applications of Le Châtelier

s Principle

Co(H

2

O)

62+(aq)

+ 4Cl

–(aq)

+ heat 4CoCl

42–(aq)

+ 6H

2

O

(l)

Cooling the solution

Reaction shifts left

CoCl

42-

+ 6H

2

O

Co(H

2

O)

62+

+ 4Cl

-

+ heat

blue

pink

How does a catalyst effect equilibrium?

Catalysts act by lowering the activation energy of a reaction,

which occurs to the same extent for both the forward and

reverse reactions.

References

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