Chemical Equilibrium Chapter 13

+ Sections 16.1 & 16.2 + A brief revisit of Chp. 17

Henry Louis Le Châtelier (1850−1936) devised what came to be known as “Le Châtelier’s principle,” which is used by chemists to predict the effect a changing condition has on a system in chemical equilibrium.

Chemical Equilibrium: An Introduction!Many chemical reactions stop far short of completion and some only occur to a slight extent.!

Chemical Equilibrium:!•  Concentrations of all reactants & products remain constant with time.!

•  Not static, but highly dynamic.!

•  Rate of forward reaction = Rate of reverse reaction!

Chemical Equilibrium: An Introduction!a A + b B c C + d D

Equilibrium expression:!

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K =[C]c[D]d

[A]a[B]b

K or Keq ≡ equilibrium constant![ ] ≡ molar concentration!*Note: K is customarily unitless.!

Chemical Equilibrium: An Introduction!a A + b B c C + d D

Equilibrium expression:!

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K =[C]c[D]d

[A]a[B]b

Ø  Solids and pure liquids are NOT included in the equilibrium expression – only aqueous and gaseous reactants and products are included.!

Ø  K > 1 ⇒ products are, forward reaction is, favored.!Ø  K < 1 ⇒ reactants are, reverse reaction is, favored.!Ø  The magnitude (size) of K and the time required to reach

equilibrium are NOT directly related.!

Chemical Equilibrium: An Introduction!CaCO3(s) CaO(s) + CO2(g)!

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K = CO2[ ]

The Equilibrium Condition!Consider the following reaction!

H2O(g) + CO(g) H2(g) + CO2(g)!

•  closed vessel!•  high T!•  rapid reaction!

The Equilibrium Condition!

Consider the following reaction:!

H2O(g) + CO(g) H2(g) + CO2(g)!

The Equilibrium Condition!H2O(g) + CO(g) H2(g) + CO2(g)!

If we inject some H2O(g), what will happen to the forward reaction?!

Example 13.2, page 611!

!The production of ammonia, known as the Haber process, is shown above. The following equilibrium concentrations were observed for the Haber process at 127°C.!

[NH3] = 3.1 × 10−2 mol L−1!

[N2] = 8.5 × 10−1 mol L−1!

[H2] = 3.1 × 10−3 mol L−1!

(a)  Calculate the value of the equilibrium constant, K, at 127°C for this reaction.!

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

Example 13.2, page 611!!

!

The production of ammonia, known as the Haber process, is shown above. The following equilibrium concentrations were observed for the Haber process at 127°C.!

[NH3] = 3.1 × 10−2 mol L−1!

[N2] = 8.5 × 10−1 mol L−1!

[H2] = 3.1 × 10−3 mol L−1!

(b)  Calculate the value of the equilibrium constant, K′, at 127°C for the reaction!

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

Example 13.2, page 611!!

!

The production of ammonia, known as the Haber process, is shown above. The following equilibrium concentrations were observed for the Haber process at 127°C.!

[NH3] = 3.1 × 10−2 mol L−1!

[N2] = 8.5 × 10−1 mol L−1!

[H2] = 3.1 × 10−3 mol L−1!

(c) Calculate the value of the equilibrium constant, K″, at 127°C for the reaction given by the equation!

12 N2(g) + 3

2 H2(g) NH3(g)

The Aliases of K!

I.  2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g)!

All reactants and products are in the gas phase, so K can be expressed in terms of concentration (subscript c).!

The Aliases of K!

II.  2 H2S(g) + 3 O2(g) 2 H2O(g) + 2 SO2(g)!

All reactants and products are in the gas phase, so K can be expressed in terms of partial pressures (subscript p).!

The Aliases of K!III.  PbCl2(s) Pb2+(aq) + 2 Cl−(aq)!

Since solids are left out of the equilibrium expression, K = Ksp, known as the “solubility product constant.” (Chapter 16)!

Example 16.2, page 762!Bismuth(III) sulfide, Bi2S3, has a solubility of 1.0 × 10−15 M at 25°C.!

(a)  Write the solubility-product constant expression for Bi2S3.!

(b)  Calculate the value of the solubility-product constant, Ksp, for Bi2S3 at 25°C.!

The Aliases of K!IV.  HC2H3O2(aq) H+(aq) + C2H3O2

−(aq)!

Ka ≡ acid dissociation constant#

The Aliases of K!V.  NH3(aq) + H2O(ℓ) NH4

+(aq) + OH−(aq)!

Kb ≡ base dissociation constant#

The Aliases of K!VI.  H2O(ℓ) H+(aq) + OH−(aq)!

Kw describes the ionization constant of water.#

The Aliases of K!

K has many aliases, but they all take the same form and tell you the same thing: the relative amounts of products and reactants at equilibrium.#

The Reaction Quotient ( Q )

•  Determined in the same way as K, but initial conditions are used instead of equilibrium conditions.

•  Used to predict the direction in which a reaction will proceed from a given set of initial conditions.

Q < K

Q > K

Q = K

reaction proceeds forward, products generated

reaction proceeds backward, reactants generated

reaction is at equilibrium

The Reaction Quotient ( Q )

Adding Chemical Equations

If a reaction can be expressed as the sum of two or more reactions, K for the overall reaction is the product of the equilibrium constants of the individual reactions.

If Reaction 3 = Reaction 1 + Reaction 2, then

K3 = K1 · K2

Page 645, Problem 47

At a particular temperature, 12.0 mol SO3 is placed into a 3.0 L rigid container, and the SO3 decomposes by the reaction

2 SO3(g) 2 SO2(g) + O2(g)

At equilibrium, 3.0 mol of SO2 is present. Calculate the value of the equilibrium constant, K, for this reaction.

Page 645, Problem 49 An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction

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

At equilibrium, the concentrations are [H2] = 5.0 M, [N2] = 8.0 M, and [NH3] = 4.0 M. What were the concentrations, in mol L−1, of nitrogen gas and hydrogen gas that were reacted initially?

Connection to Thermochemistry (Chapter 17 revisited)

The equilibrium position represents the lowest free energy value available to a particular reaction system.

ΔG = ΔG° + RT ln Q

At equilibrium, ΔG = 0 and Q = K.

ΔG° = − RT ln K

Connection to Thermochemistry

Conventional Definitions of Standard States!(Page 265)!•  The standard state of a gaseous substance is a

pressure of exactly 1 atm.!

•  For a substance present in a solution, the standard state is a concentration of exactly 1 M.!

•  Amount of NH3 present at equilibrium is favored by conditions of low T and high P .!

•  At low T, reaction is too slow to be feasible.!

N2(g) + 3 H2(g) 2 NH3(g) ΔH° = − 92.6 kJ

We must study both the thermodynamics and the kinetics of a reaction before we really understand the factors that control it.!

N2(g) + 3 H2(g) 2 NH3(g) ΔH° = − 92.6 kJ

Le Châtelier’s Principle!

If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.!

Le Châtelier’s Principle!

I.  ΔConcentration: Adding or Removing a Component

If a component (reactant or product) is added to a reaction system at equilibrium (at constant T and P or constant T and V), the equilibrium position will shift in the direction that lowers [that component].

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

Example 13.13, page 635 Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As4O6, which is then reduced using carbon:

As4O6(s) + 6 C(s) As4(g) + 6 CO(g)

Predict the direction of the shift in the equilibrium position in response to each of the following changes in conditions.

(a) Addition of carbon monoxide.

Example 13.13, page 635 Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As4O6, which is then reduced using carbon:

As4O6(s) + 6 C(s) As4(g) + 6 CO(g)

Predict the direction of the shift in the equilibrium position in response to each of the following changes in conditions.

(b) Removal of carbon

Example 13.13, page 635 Arsenic can be extracted from its ores by first reacting the ore with oxygen (called roasting) to form solid As4O6, which is then reduced using carbon:

As4O6(s) + 6 C(s) As4(g) + 6 CO(g)

Predict the direction of the shift in the equilibrium position in response to each of the following changes in conditions.

(c) Removal of As4(g)

Le Châtelier’s Principle!

II.  Changes in Pressure and Volume

•  The addition of an inert gas increases the total pressure but has no effect on the concentrations or partial pressures of the reactants or products.

•  When the volume of the container holding a gaseous system is reduced, the system responds by reducing its own volume. This is done by decreasing the total number of gaseous molecules in the system.

2 NO2(g) N2O4(g) brown colorless!

(a) Brown NO2(g) and colorless N2O4(g) in equilibrium in a syringe. (b) The volume is suddenly decreased, giving the greater concentration of both N2O4 and NO2 (indicated by the darker brown color). (c) A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts the brown NO2(g) to colorless N2O4(g) as predicted by Le Châtelier’s principle, since in the equilibrium shown above the product side has the smaller number of molecules.

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

Example 13.14, page 637 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.

The preparation of liquid phosphorus trichloride by the reaction

P4(s) + 6 Cl2(g) 4 PCl3(ℓ)

Example 13.14, page 637 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.

The preparation of gaseous phosphorus pentachloride according to the equation

PCl3(g) + Cl2(g) PCl5(g)

Example 13.14, page 637 Predict the shift in equilibrium position that will occur for each of the following processes when the volume is reduced.

The reaction of phosphorus trichloride with ammonia:

PCl3(g) + 3 NH3(g) P(NH2)3(g) + 3 HCl(g)

Le Châtelier’s Principle!

III.  Change in Temperature

Opening remarks:

•  Previously discussed changes may alter the equilibrium position, but they DO NOT alter the equilibrium constant.

•  However, the value of K changes with temperature.

•  Le Châtelier’s Principle does not predict the size of ΔK, but it does correctly predict the direction of the change.

Le Châtelier’s Principle!

III.  Change in Temperature

•  Increase in T favors the endothermic direction.

•  Decrease in T favors the exothermic direction.

Alternate method:

①  Treat energy as a reactant (endothermic) or product (exothermic).

②  Predict shift in the same way as when an actual reactant or product is added or removed.

100°C 0°C!

N2O4(g) 2 NO2(g) ΔH° = 58 kJ colorless brown!

Example 13.15, page 639 For each of the following reactions, predict how the value of K changes as the temperature is increased.

(a)  N2(g) + O2(g) 2 NO(g) ΔH° = 181 kJ

(b)  2 SO2(g) + O2(g) 2 SO3(g) ΔH° = − 198 kJ

The End