Chemical Equilibriumk1

k-1

a b1

c d-1

a1

a A + b B c C + d D

rate of the forward reaction = k [A] [B]

rate of the reverse reaction = k [C] [D]

At equilibrium:

rate of the forward reaction = rate of the reverse reaction

k [A] [B b c d-1] = k [C] [D]

Chemical Equilibrium

a b c d1 -1

c d1

a b-1

c d

eq a b

k [A] [B] k [C] [D]

k [C] [D]

k [A] [B]

[C] [D]K

[A] [B]

Chemical Equilibrium

• When equilibrium, a dynamic state, is reached the concentrations of A, B, C, and D are invariant (i.e., no longer change)

• The reaction continues in a dynamic state, but the concentrations remain the same.

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

Chemical EquilibriumConcentration versus time

[C] and [D]

[A] and [B]

at equilibrium

0 2 4 6 8 10 12 14 16 18 200

0.2

0.4

0.6

0.8

1

1.2

time (minutes)

conc

entr

ation

, M

Chemical EquilibriumRate versus time

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4

time (minutes)

Rate

c d-1k [C] [D]

at equilibrium

a b c d1 -1k [A] [B] = k [C] [D]

a b1k [A] [B]

Consider the Reaction between Hydrogen and Iodine

If the initial concentrations of hydrogen and iodine gases are 0.0175 M, calculate the equilibrium concentrations of the reactants and products if the equilibrium constant is 55.64 at 425oC.

2

(g)

2(g) 2(g)

HI = 55.64

H I

0.0175 0.0175 2

2(g) 2(g) (g)H + I 2HIx x x

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

2

(g)

2(g) 2(g)

HI = 55.64

H I

2

2

-4 2

2 2

2

2x = 55.64

0.0175-x 0.0175-x

4x = 55.64

3.06 x 10 - 0.035 x + x

4x = 0.0170 - 1.95 x + 55.64 x

0 = 0.0170 - 1.95 x + 51.64 x

2-b ± b -4acx =

2a

a = 51.64; b = -1.95; and c = 0.0170

2-b ± b -4acx =

2a

If x = 1.37 x 10-2 M

x = 2.41 x 10-2 M is not possible, because it would lead to negative matter!

-2 -2

1.95 ± 3.80-4 (51.64)(0.0170)x =

2(51.64)

1.95 ± 0.54x =

103.28

x = 1.37 x 10 M or 2.41 x 10 M

-2(g)HI = 2x = 2.73 x 10 M

-32(g)H = (0.0175 - x) M = (0.0175 - 0.0137) M = 3.8 x 10 M

-32(g)I = (0.0175 - x) M = (0.0175 - 0.0137) M = 3.8 x 10 M

The equilibrium expression does not include solids

For example, the equilibrium expression for the following reaction:

2(g) 25 oeq

2(g)

SO = K = 4.2 x 10 at 25 C

O

8 2 (g) 2 (g)

1 S + O SO

8

The equilibrium expression when water is involved

For example,

- +(aq) 2 (l) (aq) 3 (aq)HCOOH + H O HCOO + H O

- +(aq) 3 (l)

2 (l) a

(aq)

HCOO H O = K H O = K

HCOOH

Kc versus Kp

2

3 (g) 8 oc

2 (g) 2 (g)

NHK = = 3.5 x 10 at 25 C

N H

3 (g)

2 (g) 2 (g)

2NH o

p 2 2N H

PK = = ? at 25 C

P P

2(g) 2 (g) 3 (g)N + 3 H 2 NH

From PV = nRT

n pwhere = concentration =

V RT

2c p

cp 2

K = K x (RT)

KK =

(RT)

3(g)

3(g)

2(g) 2(g) 2(g) 2(g)

3(g)

2(g) 2(g)

2NH

2 22 2NH3 (g)

c 2 2 2 2N H N H2 (g) 2 (g)

42 2

2NH 2

c 2 2N H

2c P

P 1PNH (RT) (RT)

K = = = x 1P P P x PN H

x (RT)(RT) (RT)

PK = x (RT)

P x P

K = K x (RT)

2c P

cP P2

8

P 2

5

P 2

5P

K = K x (RT)

KK = K

(RT)

3.5 x 10K =

(0.08205 x 298)

3.5 x 10K =

(0.08205 x 298)

K = 5.9 x 10

Consider the Following Reaction:

c 2 (g)K = CO 2

2

COc 2 (g)

CO pc

c p

PK = CO =

RTP K

K = RT RT

(RT) K = K

2

2

COc 2 (g)

CO pc

c p

PK = CO =

RTP K

K = RT RT

(RT) K = K

3 (s) (s) 2 (g)CaCO CaO +CO

Consider the Following Reaction:

4 2+3 (aq) aq

c 23 4 (aq)

NH CuK =

Cu(NH )

2 2+3 4 (aq) (aq) 3 (aq)Cu(NH ) Cu + 4 NH

Manipulating the Equilibrium Expression:

(s) 2 (g) (g)

1C + O CO

2

(g) 23 o1 1

22 (g)

COK = = 4.6 x 10 at 25 C

O

Manipulating the Equilibrium Expression:

2

(g) 2 47 o2 1

2 (g)

COK = = K = 2.1 x 10 at 25 C

O

(s) 2 (g) (g)2 C + O 2CO

Manipulating the Equilibrium Expression:

If Keq equals 3.5 x 108 for

Calculate Keq for

2 (g) 2 (g) 3 (g)N + 3 H 2NH

3 (g) 2 (g) 2 (g)

1 3NH N + H

2 2

Manipulating the Equilibrium Expression:

3 (g) 2 (g) 2 (g)

1 3NH N + H

2 2

-58

1 = 5.3 x 10

3.5 x 10

Finding K for a Reaction from Other K values

+ -3 2 (aq)

eq 2

3 (aq)

Ag(NH ) ClK =

NH

+ -(s) 3 (aq) 3 2 (aq) (aq)AgCl + 2 NH Ag(NH ) + Cl

Finding K for a Reaction from Other K values

+3 2 7

f 2+ (aq) 3 (aq)

Ag(NH )K = = 1.6 x 10

Ag NH

+ - -10sp (aq) (aq)K = Ag Cl = 1.8 x 10

+ - (s) (aq) (aq)AgCl Ag + Cl

+ +(aq) 3 (aq) 3 2 (aq)Ag + 2 NH Ag(NH )

Finding K for a Reaction from Other K values

+ + -3 2 3 2 (aq)+ -

(aq) (aq)2 2+(aq) 3 (aq) 3 (aq)

+ -3 2 (aq)

f sp 2

3 (aq)

+ -3 2 (aq)7 -10

Ag(NH ) Ag(NH ) Cl x Ag Cl =

Ag NH NH

Ag(NH ) ClK x K =

NH

Ag(NH ) Cl1.6 x 10 x 1.8 x 10 =

N

2

3 (aq)

+ -3 2 (aq)-3

eq2

3 (aq)

H

Ag(NH ) Cl2.9 x 10 = = K

NH

Calculate Kp at 226.8oC for Reaction (4) given Reactions (1), (2), and (3):

112 (g) 2 (g) (g) P

-412 (g) (g) P

-152 (g) (g) P

(g) (g) (g) P

(1) H + Br 2 HBr K = 7.9 x 10

(2) H 2 H K = 4.8 x 10

(3) Br 2 Br K = 2.2 x 10

(4) H + Br HBr K

= ?

Kp at 226.8oC for

(g) (g) (g)H + Br HBr

1

2

3

112 (g) 2 (g) (g) P

-412 (g) (g) P

-152 (g) (g) P

(g) (g) (g)

(1) H + Br 2 HBr K = 7.9 x 10

(2) H 2 H K = 4.8 x 10

(3) Br 2 Br K = 2.2 x 10

(4) H + Br HBr

4PK = ?

Kp at 226.8oC for

(g) (g) (g)H + Br HBr

1 4

2 3

2 2

(g) 2 (g) 2 (g) (g)

2 2 2 22 (g) 2 (g) (g) (g) (g) (g)

2P P

P P

HBr H Br HBr x x =

H Br H Br H Br

1 1K x x = K

K K

4 4

2

(g) (g) 2P P2 2

(g) (g) (g) (g)

HBr HBr = = K = K

H BrH Br

Kp at 226.8oC for

4

1133

P-41 -15

7.9 x 10 = K = 2.7 x 10

4.8 x 10 x 2.2 x 10

(g) (g) (g)H + Br HBr

The Reaction Quotient, Q

• Q is obtained by using a similar formula as K, but Q may not be at equilibrium. The reaction is at dynamic equilibrium when Q = Keq

• If Q is greater than Keq , then the reaction is not at equilibrium and must adjust so that product(s) will go to reactant(s) until a state of dynamic equilibrium is reached

• If Q is less than Keq , then the reaction is not at equilibrium and must adjust so that more reactant(s) will go to product(s) until a state of dynamic equilibrium is reached

Application of Q

For the following equilibrium:

Keq = 2.5 at 25oC is the reaction at equilibrium when[CH3CH2CH2CH3] = 0.97 M and [(CH3)2CHCH3] = 2.18M?If not, calculate the equilibrium concentrations.

3 2 2 3 3 2 3CH CH CH CH (CH ) CHCH

Application of Q

3 2 3

3 2 2 3

eq

(CH ) CHCH = Q

CH CH CH CH

2.18 M = Q

0.97 M2.25 = Q

Therefore, Q is less than K

Application of QSince Q (2.25) is less than Keq (2.5)

2.18 + x = 2.5

0.97 - xx = 0.07

Therefore, the equilibrium concentration of CH3CH2CH2CH3 is 0.90 M (0.97 – 0.07) and the equilibrium concentration of (CH3)2CHCH3 is 2.25M (2.18 + 0.07)

CH3CH2CH2CH3 (CH3)2CHCH3

0.97 M - x 2.18 + x

Another Application of QFor the following equilibrium:

Keq = 1.7 x 10-3 is the reaction at equilibrium when[N2] = 0.50 M , [O2] = 0.25 M, and [NO] = 0.0042 M?If not, calculate the equilibrium concentrations.

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

Another Application of QFor the following equilibrium:

2

2 2

2

-4

NOQ =

N O

0.0042Q = = 1.41 x 10

0.50 0.25

Q (0.000141) is less than Keq (0.0017)

Another Application of Q

2-3

eq

(0.0042+2x)K = 1.7 x 10 =

(0.50 -x)(0.25 - x)

x = 5.1 x 10-3

N2 (g) + O2 (g) NO (g)2

0.50 M - x 0.25 - x 0.0042+2x

Another Application of Q

x = 5.1 x 10-3

-3 2

2-3

(0.0042+2 x 5.1 x 10 ) =

(0.50 -0.0051)(0.25 - 0.0051)

(0.0144) = 1.7x10

(0.495)(0.245)

Let’s calculate the equilibrium constant for the following reaction at 1000 K:

1.00 mol of SO2 and 1.00 mol of O2 are placed in a 1.00 L flask. When equilibrium is achieved 0.925 mol of SO2 is formed.

2 (g) 2 (g) 3 (g)2 SO + O 2 SO

Let’s calculate the KP for the following reaction at 800 K:

A tank contains hydrogen sulfide gas at 10.00 atm at 800 K. When equilibrium is achieved, the partial pressure of S2

is 2.0 x 10-2 atm.

What is Kc for this reaction?

2 (g) 2 (g) 2 (g)2 H S 2 H + S

Let’s calculate the Kc for the following reaction:

If 0.050 mol of trans-1,2-diodocyclohexane is dissolved in carbon tetrachloride to make a 1.00 L solution. At equilibrium the concentration of I2 is 0.035 M.

What is Kp for this reaction?

H

H

II

+ I2

Kc for the following reaction is 1.0 x 10-5 at 1227oC:

Calculate the equilibrium concentrations if 0.80 mole of N2 and 0.20 mole of O2 are placed initially in a 2.00 L flask at 1227oC.

Calculate Kp at 1227oC, and calculate the equilibrium pressures at N2, O2, and NO at 1227oC.

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

Le Chatelier’s Principle

When a system in dynamic equilibrium encounters stress, the system will adjust itself in order to relieve the stress.

The Stresses

• Change the temperature• Change the concentration of reactants or

products• Change the volume• Change the pressure (only when there is a

difference between the number of moles of gaseous reactants and products)

Changing the Temperature

• Adjustment in the equilibrium is determined by whether the reaction is endothermic or exothermic.

• If the reaction is exothermic, an increase in temperature shifts the equilibrium toward the reactants. A decrease in temperature shifts the reaction toward the products.

Changing the Temperature

• If the reaction is endothermic, an increase in temperature shifts the equilibrium toward the products. A decrease in temperature shifts the reaction toward the reactants.

• Changes in temperature result in changes in the value of the equilibrium constant.

oreactionΔH

-RT

eq

oreaction

eq

K = A e

ΔHln K = - + ln A

RT

ToC

Keq ln Keq 104(1/T)

25 4.5 x 10-31 -69.9 33.6

500 2.43 x 10-21 -47.5 12.9

1000 6.07 x 10-19 -41.9 7.891500 6.74 x 10-18 -39.5 5.64

2000 2.60 x 10-17 -38.2 4.40

2500 6.15 x 10-17 -37.3 3.61

For example, the following data were obtained for changes in the equilibriumConstant with changes in temperature for the reaction

2 (g) 2 (g) (g)

1 1 N + O NO

2 2

0 5 10 15 20 25 30 35 40

-80

-70

-60

-50

-40

-30

-20

-10

0

f(x) = − 1.08320586451514 x − 33.5018092298579

104(1/T)

Ln K

reaction

4 reaction

4reaction

4reaction

ΔHslope = -

RΔH

-1.0832 x 10 K = -R

J-1.0832 x 10 K x -8.314 = ΔH

K-molJ

9.00 x 10 = ΔHmol

Changes in Concentration of Reactants and Products

Increasing the concentration of the reactants creates a situation where Q is less than K; therefore, the system will adjust itself to produce more product so that Q will equal K, i.e., the equilibrium will shift toward the product.

Changes in Concentration of Reactants and Products

Decreasing the concentration of the reactants creates a situation where Q is greater than K; therefore, the system will adjust itself to produce more reactants so that Q will equal K, i.e., the equilibrium will shift toward the reactants.

An Example of the Effect of Concentration Change on an Equilibrium Reaction

3 2 2 3 3 2 3CH CH CH CH (CH ) CHCH

If at 25oC, the equilibrium concentrations of n-butane and Isobutane are respectively 0.500 M and 1.25 M, what would be the equilibrium concentrations if 1.50 mol of n-butane is addedto the equilibrium mixture?

3 2 3eq

3 2 2 3

eq

(CH ) CHCHK =

CH CH CH CH

1.25K = = 2.5

0.50

CH3CH2CH2CH3 (CH3)2CHCH3

1.75 M - x 1.25 M + x

eq

1.25 mol + x molVK = = 2.5

1.75 mol - xmolV

1.25 + x = 2.5

1.75 - x1.25 + x = 2.5 1.75 - x

1.25 + x = 4.375 - 2.5 x

3.125 = 3.5 x

0.89 = x

3 2 2 3

3 2 3

CH CH CH CH 1.75 M -0.89 M = 0.86 M

(CH ) CHCH 1.25 M + 0.89 M = 2.14 M

Changing the Volume

• If the number of moles of products equals the number of moles of reactants, then a change in volume of the reaction vessel will have no affect on the equilibrium.

• If the number of moles of product is greater than the number of moles of reactants, then an increase in volume of the reaction vessel will shift the equilibrium toward the formation of product. A decrease in the volume will shift the equilibrium toward the reactants.

Changing the Volume

• If the number of moles of reactant is greater than the number of moles of products, then an increase in the volume of the reaction vessel will shift the equilibrium toward the reactants. A decrease in volume will shift the equilibrium toward the products.

Effect of Pressure Change on a Gaseous Reaction

• If the number of moles of gaseous products equals the number of moles of gaseous reactants, then pressure changes have no effect on the equilibrium reaction.

• If the number of moles of gaseous products is greater than the number of moles of gaseous reactants, then an increase in pressure shifts the equilibrium toward the reactants. A decrease in pressure shifts the equilibrium toward the products.

Effect of Pressure Change on a Gaseous Reaction

If the number of moles of gaseous reactants is greater than the number of moles of gaseous products, then an increase in pressure shifts the equilibrium toward the products. A decrease in pressure shifts the equilibrium toward the reactants.