1

Chemical EquilibriumChemical Equilibrium

Chapter 14Chapter 14

Henri L. le Chatlier1850-1936.Adapted thermodynamics toequilibria; formulated theprinciple known by his name.

2

The Concept of EquilibriumThe Concept of Equilibrium• Consider colorless N2O4. At room temperature, it decomposes to brown NO2:

N2O4(g) 2NO2(g).

(colorless) (brown)

N2O4(g) 2NO2(g)

The double arrow implies a dynamic equilibrium

At some point, the forward and reverse rates become the same. This is equilibrium.

Furthermore, it is a dynamic equilibrium.

As NO2 is produced, it reacts back to form N2O4:

N2O4(g) 2NO2(g)

3N2O4(g) 2NO2(g)

4

The Concept of EquilibriumThe Concept of Equilibrium

• Consider

Forward reaction: A B Rate = kf[A]

Reverse reaction: B A Rate = kr[B]

• At equilibrium kf[A] = kr[B].

• For an equilibrium we write A B•As the reaction progresses

–[A] decreases to a constant,

–[B] increases from zero to a constant.

–When [A] and [B] are constant, equilibrium is achieved.

5

The Concept of EquilibriumThe Concept of EquilibriumConsider reaction: A(g) B(g)

At beginning: PA = PAo

PB = 0

As reaction proceeds, PA decreases and PB increases

Equilibrium occurs when there is no further change in concentration as time progresses.

PA

PB

oAP

6

The Concept of EquilibriumThe Concept of EquilibriumAlternatively:

As reaction proceeds,

until the two rates become the same. This is equilibrium.Or, when kf[A] = kr[B], equilibrium is achieved.

forward rate, kf[A] decreases

reverse rate, kr[B] increases

7

The Equilibrium ConstantThe Equilibrium Constant

Consider

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

•If we start with a mixture of N2 and H2 (in any proportions), the reaction will reach equilibrium with a constant set of pressures of N2, H2 and NH3.

•If we start with just NH3 and no N2 or H2 , the reaction will proceed in reverse and N2 and H2 will be produced until equilibrium is achieved, with a constant set of pressures of N2, H2 and NH3.

When dealing with gases, partial pressures can conveniently be used.

8

The Equilibrium ConstantThe Equilibrium Constant• No matter the starting composition of reactants and products, the same ratio of concentrations (for solutions) or pressures (for gases) is

achieved at equilibrium.• For a general reaction

ba

qp

eqKBA

QP

where Keq is the equilibrium constant (in terms of

concentration or pressure)

the equilibrium constant expression is

aA + bB pP + qQ

9

The Magnitude of Equilibrium ConstantsThe Magnitude of Equilibrium Constants• The equilibrium constant, Keq, is the ratio of products to

reactants.

• Therefore, the larger the Keq the more products are present at equilibrium.

• Conversely, the smaller the Keq the more reactants are present at equilibrium.

• If Keq >> 1, then products dominate at equilibrium and equilibrium lies to the right.

• If Keq << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.

• Keq = kf/kb

For A B, Keq = [B][A]

10

N2O4(g) 2NO2(g)Calculating the equilibrium constant for:

[NO2]2

Kc = [N2O4]

11

An equilibrium is established by placing 2.00 moles of frozen An equilibrium is established by placing 2.00 moles of frozen NN22OO44(g) in a 5.00 L and heating the flask to 407 K. It was (g) in a 5.00 L and heating the flask to 407 K. It was

determined that at equilibrium the amount of the NOdetermined that at equilibrium the amount of the NO22(g) is (g) is

1.31 mole. What is the value of the equilibrium constant?1.31 mole. What is the value of the equilibrium constant? [NO2]2

Kc = [N2O4]

N2O4(g) 2 NO2(g)

[Initial] (mol/L) 0.400 M 0 M[change][Equilibrium] 0.262

+ x = 0.262-½ x = -0.1310.269

K = (0.262)2/0.269 = 0.255

Set up a table.Let x = [NO2 ] produced.

given 2.00/5.00 = 0.400 M

Given 1.31/5.00 = 0.262 M

given

Usemolarities

in table

12

N2O4(g) 2NO2(g)We just calculated that at 407 Kelvin , Kc = 0.255 for

For the reverse reaction, calculate the new K by taking the reciprocal: Kc = 1/0.255 = 3.92

Proof:

Kc = [N2O4]/[NO2]2 (0.269)/(0.262)2 =3.92

(g( )NO2 g N2O4 )2

For the “half” reaction, calculate the square root:Kc = (0.255)1/2 = 0.505

Proof: Kc = [NO2]/[N2O4]1/2 =(0.262)/(0.269)1/2 = = 0.505

½ N 2O4(g)

NO2(g)

Be careful! Kc values must beevaluated for the reaction as written!

13

Heterogeneous EquilibriaHeterogeneous Equilibria• When all reactants and products are in one phase, the

equilibrium is homogeneous.• If one or more reactants or products are in a different

phase, the equilibrium is heterogeneous.• Consider:

CaCO3(s) CaO(s) + CO2(g)

–experimentally, the amount of CO2 does not depend on the amounts of CaO and CaCO3. Solids and pure liquids are not considered in the equilibrium constant expression. Thus,

solids

Kc = [CO2]

14

Calculating Equilibrium ConstantsCalculating Equilibrium Constants

For 2 NO(g) + Cl2(g) 2NOCl(g), calculate Keq given that:PNO=0.095 atm; PCl2=0.171 atm and PNOCl= 0.28 atmat equilibrium.

2Cl2NO

2NOCl

eq PP

PK

= (0.28)2/(0.095)2(0.171)= (0.0784)/(0.009025)(0.171)= (0.0784)/(0.00154)= 50.8

15

Calculating Equilibrium ConstantsCalculating Equilibrium Constants1.374 g H2 and 70.31 g Br2 heated in 2-L vessel at 700 K are allowed to react. At equilibrium, the vessel contains 0.566 g H2.

Calculate Kc .

The reaction is: H2 (g) + Br2 (g) 2 HBr (g)

What is known?start: 1.374 g 70.31 g 0at equil 0.566 g ? ?

Start by changing everything from grams to moles:H2: 1.374 g = 0.687 mol; at equil H2: 0.566 g = 0.283 molBr2: 70.31 g = 0.440 mol

Now divide by 2 to give molarities (the vessel is 2-L)Initial [H2]= 0.344 mol; Final [H2] = 0.142Initial [Br2] = 0.220

16

Set up table (values are molarities): H2 (g) + Br2 (g) 2 HBr (g)Initial 0.344 0.220 0ChangeFinal 0.142

Now fill in changes: H2 (g) + Br2 (g) 2 HBr (g)Initial 0.344 0.220 0Change -0.202 -0.202 +0.404Final 0.142

Now calculate final values: H2 (g) + Br2 (g) 2 HBr (g)Initial 0.344 0.220 0Change -0.202 -0.202 +0.404Final 0.142 0.018 0.404

Kc = [HBr]2/[H2][Br2]= (0.404)2/(0.142)(0.018)= 63.8

4th step

1st step

2nd step

3rd step

17

Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction• We define Q, the reaction quotient, for a general

reactionaA + bB(g) pP + qQ

ba

qp

QBA

QP

where [A], [B], [P], and [Q] are molarities at any time.

•Q = Keq only at equilibrium.

as

18

Applications of Equilibrium ConstantsApplications of Equilibrium ConstantsPredicting the Direction of ReactionPredicting the Direction of Reaction• If Q > Keq then the reverse reaction must occur to

reach equilibrium (i.e., products are consumed, reactants are formed, the numerator in the equilibrium constant expression decreases and Q decreases until it equals Keq).

• If Q < Keq then the forward reaction must occur to reach equilibrium.

19

Le Châtelier’s PrincipleLe Châtelier’s PrincipleLe Châtelier’s Principle: if a system at equilibrium is disturbed, the

system will shift to remove the disturbance.

Three types of disturbances:

(1) Amount of material. If you add a reactant or remove a product, the equilibrium shifts to the right. If you remove a reactant or add a product, the equilibrium shifts to the left.

(2) Pressures. By increasing the pressure you shift the equilibrium in the direction of the lesser number of gas moles; by decreasing the pressure you shift the equilibrium in the direction of the greater number of gas moles.

(3) Temperature. By increasing the temperature, you shift an exothermic reaction to the left, and an endothermic reaction to the right. By decreasing the temperature, you shift an exothermic reaction to the right, and an endothermic reaction to the left.

20

Le Châtelier’s PrincipleLe Châtelier’s Principle• Consider the production of ammonia (the Haber process)

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

•As the temperature decreases, the amount of ammonia at equilibrium increases (reaction goes to right).

•As the pressure increases, the amount of ammonia present at equilibrium increases (reaction goes to right).

The following have been observed for this exothermic reaction:

+ heat

•If ammonia is removed, the equilibrium shifts to the right.

•If hydrogen is added, the equilibrium shifts to the right.

•If hydrogen is removed, the equilibrium shifts to the left.

21

Le Châtelier’s PrincipleLe Châtelier’s PrincipleThe Effect of CatalystsThe Effect of Catalysts• A catalyst lowers the activation energy barrier for the

reaction.• Therefore, a catalyst will decrease the time taken to

reach equilibrium.• A catalyst does not effect the composition of the

equilibrium mixture.• Catalysis is a kinetics matter, not one of equilibrium.• Catalysis speeds up a reaction, but the reaction

arrives at the same equilibrium point (it just gets there faster).

22

• A catalyst changes the rate of a chemical reaction, by lowering the activation energy Ea., but the ΔH is not changed.

ΔH