Chemical Equilibrium

•  Concept of Equilibrium •  Equilibrium Constant •  Equilibrium expressions •  Applications of equilibrium constants •  Le Chatelier’s Principle

The decomposition of N2O4 is a reversible process, meaning the products of the reaction can react to form reactants. The system is in equilibrium when the rates of the forward reaction and the reverse reaction are the same.

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

N2O4(g) ⇌ 2NO2(g)

The Concept of Equilibrium

The Concept of Equilibrium

Starting with N2O4

Starting with NO2

Reaction Rates

Some important things to remember about equilibrium are: Equilibrium is a dynamic state—both the forward and reverse reactions continue to occur, although there is no net change in reactant and product concentration over time. At equilibrium, the rates of the forward and reverse reactions are equal. Equilibrium can be established starting with only reactants, with only products, or with any mixture of reactants and products.

The Equilibrium Constant

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

=kk

2 eqf

r 2 4 eq

NON O

[ ][ ]

2

=K 2 eqc

2 4 eq

NON O

equilibrium constant

The Equilibrium Constant Note the relationship between the equilibrium constant and the balanced chemical equation:

[ ][ ]

2

=K 2 eqc

2 4 eq

NON O N2O4(g) ⇌ 2NO2(g)

Equilibrium Expressions When a reversible chemical equation is manipulated, it is also necessary to make appropriate changes in the equilibrium expression and the equilibrium constant.

Practice-Eq. Expression

Write the equilibrium expression for Kc, for the following reactions: (a) 2O3 (g) ⇌ 3O2 (g)

(b) 2NO (g) + Cl2 (g) ⇌ 2NOCl (g)

(c) Ag+ (aq) + 2NH3 (aq) ⇌ Ag(NH3)2+ (aq)

Determine the vale of the equilibrium constant for the following reaction:

N2(g) + O2(g) ⇌ 2NO2(g) Given: N2(g) + O2(g) ⇌ 2NO(g) Kc1 = 4.3 x 10–25

2NO(g) + O2(g) ⇌ 2NO2(g) Kc2 = 6.4 x 109

Practice Problems

In an analysis of the following reaction at 100°C,

Br2(g) + Cl2(g) ⇌ 2BrCl(g) The equilibrium concentrations were found to be [Br2] = 2.3 x 10–3 M, [Cl2] = 1.2 x 10–2 M, [BrCl] = 1.4 x 10–2 M. Write the equilibrium expression and calculate the equilibrium constant for this reaction at 100°C.

Practice-Eq. Expression

In an analysis of the following reaction at 100°C,

Br2(g) + Cl2(g) ⇌ 2BrCl(g) The equilibrium concentrations were found to be [Br2] = 2.3 x 10–3 M, [Cl2] = 1.2 x 10–2 M, [BrCl] = 1.4 x 10–2 M. Write the equilibrium expression and calculate the equilibrium constant for this reaction at 100°C.

(a) calculate K for the reverse reaction.

(b) calculate the equilibrium constant for the reaction:

½ Br2(g) + ½ Cl2(g) ⇌ BrCl(g)

Practice-Eq. Expression

Equilibrium Expressions When the species in a reversible chemical reaction are not all in the same phase, the equilibrium is heterogeneous. Only gaseous species and aqueous species appear in equilibrium expressions, pure solids and pure liquids do not.

CO2(g) + C(s) ⇌ 2CO(g)

2Fe(s) + 3H2O(l) ⇌ Fe2O3(s) + 2H2(g)

[ ][ ]

2

=Kc2

COCO

[ ]2=Kc 2H

The Equilibrium Constant The equilibrium constant gives the extent a reaction will proceed at a particular temperature. Three outcomes are possible: 1) The reaction will go essentially to completion and the equilibrium

mixture will consist predominately of products.

Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(g) Kc = 1.5 x 107 (at 25°C)

Large Kc, product favored

The Equilibrium Constant The equilibrium constant gives the extent a reaction will proceed at a particular temperature. Three outcomes are possible: 2) The reaction will not occur to any significant degree, and the

equilibrium mixture will consist predominantly of reactant.

N2(g) + O2(g) ⇌ 2NO(g) Kc = 4.3 x 10–25 (at 25°C)

3) The reaction will proceed a significant degree but will not go to

completion, and the equilibrium mixture will contain comparable amounts of both reactants and products.

Small Kc, reactant favored

The Equilibrium Constant The reaction quotient (Qc) is a fraction with product concentrations in the numerator and reactant concentrations in the denominator. Each concentration is raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation. The law of mass action:

aA + bB ⇌ cC + dD applies to not only elementary reactions, but also to more complex reactions.

[ ] [ ][ ] [ ]

= =c d

a bQ Kc cC DA B

(at equilibrium)

The Equilibrium Constant The value of the reaction quotient, Q, changes as the reaction progresses

N2O4(g) ⇌ 2NO2(g)

The Equilibrium Constant At any point during the progress of a reaction:

[ ] [ ][ ] [ ]

=c d

a bQcC DA B

aA + bB ⇌ cC + dD

Write the reaction quotient for the following reaction:

CH4(g) + 2H2S(g) ⇌ CS2(g) + 4H2(g)

Practice-Reaction Quotient

Relationship between Q and K The equilibrium expression may be used to predict the direction of a reaction and to calculate equilibrium concentrations. Predictions are made based on comparisons between Qc and Kc. 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.

Practice- using K to solve problems The equilibrium constant, Kc, for the formation of nitrosyl chloride from nitric oxide and chlorine,

2NO(g) + Cl2(g) ⇌ 2NOCl(g) Is 6.5 x 104 at 35°C. In which direction will the reaction proceed to reach equilibrium if the starting concentrations of NO, Cl2, and NOCl are 1.1 x 10–3 M, 3.5 x 10–4 M, and 1.9 M respectively?

Practice- Using K to solve problems 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

cis-Stilbene trans-Stilbene

cis-Stilbene trans-Stilbene ⇌

Practice- Using K to solve problems Consider the following reaction:

H2(g) + F2(g) ⇌ 2HF(g) 3.000 mol H2 and 6.000 mol F2 are mixed in a 3.000 L flask. If the equilibrium constant is 115 at room temperature, calculate the equilibrium concentrations of all three species.

Practice- Using K to solve problems 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 the initial concentrations are: [H2] = 0.00623 M, [I2] = 0.00414 M, [HI] = 0.0424 M

Practice- Using K to solve problems Consider the following reaction:

N2O4 (g) ⇌ 2NO2 (g) Kp = 0.25 A flask containing only N2O4(g) at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures.

Factors That Affect Chemical Equilibrium

Le Châtelier’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

Changes in Concentration Consider the Haber process at 700 K:

N2(g) + 3H2(g) ⇌ 2NH3(g)

At equilibrium:

[N2] = 2.05 M [H2] = 1.56 M [NH3] = 1.52 M

Applying stress by the addition of N2 to give the following concentrations:

[N2] = 3.51 M [H2] = 1.56 M [NH3] = 1.52 M

[ ][ ][ ]

( )( )( )

3c

2

NHN H

= = =.

.. .

K2 2

3 32

1 520 297

2 05 1 56

[ ][ ][ ]

( )( )( )

3c c

2

NHN H

= = = ≠.

.. .

Q K2 2

3 32

1 520 173

3 51 1 56

The reaction shifts to the right. N2(g) + 3H2(g) ⇌ 2NH3(g)

Changes in Concentration N2(g) + 3H2(g) ⇌ 2NH3(g)

Changes in Concentration Addition of a reactant or removal of a product will cause an equilibrium to shift to the right.

Addition of a product or removal of a reactant will cause an equilibrium to shift to the left.

Changes in Volume and Pressure When volume is decreased, the equilibrium is driven toward the side with the smallest number of moles of gas.

N2O4(g) ⇌ 2NO2(g) Equilibrium mixture:

[N2O4] = 0.643 M

[NO2] = 0.0547 M

[ ][ ]

( )( )

2c

2 4

NON O

−= = = ×.

..

K2 2

30 05474 65 10

0 643[ ][ ]

( )( )

2c c

2 4

NON O

−= = = × ≠.

..

Q K2 2

30 10949 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)

Changes in Temperature 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) ΔH° = 58.0 kJ/mol

Because the processes is endothermic, adding heat

shifts the equilibrium toward products

Changes in Temperature For any endothermic reaction, heat is a reactant: Adding heat shifts the reaction towards products, Kc increases Removing heat shifts the reaction towards reactants, Kc decreases. For any exothermic reaction, heat is a product: Adding heat shifts the reaction towards reactants, Kc decreases Removing heat shifts the reaction towards products, Kc increases

heat + reactants ⇌ products ΔH° > 0 kJ/mol

reactants ⇌ products + heat ΔH° < 0 kJ/mol

Key Points The Concept of Equilibrium The Equilibrium Constant

Calculating Equilibrium Constants Magnitude of the Equilibrium Constant

Equilibrium Expressions Heterogeneous Equilibria Manipulating Equilibrium Expressions

Using Equilibrium Expressions to Solve Problems Predicting the Direction of a Reaction Calculating Equilibrium Concentrations

Factors that Affect Chemical Equilibrium Addition or Removal of a Substance Changes in Volume and Pressure Changes in Temperature