Chapter 18 Chemical Equilibrium

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Mark S. CracoliceEdward I. Peters

Mark S. Cracolice • The University of Montana

Chapter 18Chemical Equilibrium

Collision Theory of ReactionsCollision Theory of Gas-phase Reactions

A chemical reaction can occur only when two molecules collide with a kinetic energy at least equal to certain energy Ea ,

called activation energy of the reaction.

The success of a collision also depends the relative orientation of molecules. This direction-dependence is called the steric

requirement of the reaction.

Collision Theory of Reactions

a) Sufficient energy proper orientation

b) Proper orientationnot sufficient energy

c) Sufficient energy poor orientation

Chemical reaction is the overall effect of collisionsbetween reacting molecules

Collision Theory of Reactions

A conversion of kinetic energy to potential energyoccurs during formation of an intermediate complex that

can either go on to form products or fall apart into the unchanged reactants.

This can be shown by a graph that traces the energy of the system before, during, and after the collision.

Energy Changes During a Reaction

Energy Changes During a Reaction

Transition State ComplexIn the transition state complex, the original bonds have

weakened, whereas the new bonds are only partially formed .

Activation Energy Ea:The difference between the energy of the transition statecomplex and the reactant energy.

Rate of a Chemical ReactionThree important factors influencethe speed of chemical reactions:

TemperatureThe higher the temperature, the faster the rate of reaction.

CatalysisA catalyst increases the rate of reaction.

Concentration of ReactantsThe greater the concentration, the greater the rate of reaction.

Effect of Temperature on the Distribution of Energy

Effect of Temperature on Reaction Rates

Kinetic energy distribution curves at two temperatures explainthe effect of temperature on reaction rates.

Ea, the activation energy, is the same at both temperatures.Only the fraction of the particles in the sample represented by the area beneath the curve to the right of Ea is able to react.

The fraction of molecules that is able to react increases rapidly as the temperature is raised.

Effect of Catalyst on the Rate of a Chemical Reaction: Change in Activation Energy

A catalyst speeds up a reaction by providing a new pathway that has a lower activation energy.

Effect of Change in Activation Energy on the Rate of a Chemical Reaction

Effect of Change in Activation Energy on the Rate of a Chemical Reaction

The fraction of molecules that collide with kinetic energy that is at least equal to the activation energy,

Ea , is bigger in a catalyzed reaction because the activation energy barrier is lowered.

Since Ea‘ < Ea

The catalyzed reaction rate is faster.

Effect of Concentration on Reaction Rate

Reaction rate depends on the frequency of effective collisions:

The more particles there are in a given volume,the more frequently collisions will occur andthe more rapidly the reaction will take place.

Development of Equilibrium

For a reversible reaction in a closed system, the equilibrium is established when

the forward reaction rate is equal to the reverse reaction rate.

Development of Equilibrium

If the system is not in equilibrium, the concentration of the species in the faster reaction will decrease, and thus the reaction will become slower; the concentration of the species in the slower reaction will increase, and thus the reaction will become faster. Opposite rates will eventually become equal,and an equilibriumwill be established.

The Equilibrium Constant

Consider the reaction H2(g) + I2(g) 2 HI(g)

At equilibrium the following ratio is a constant

K is called equilibrium constant

K][I ][H

[HI]

22

2

The Equilibrium ConstantFor the general equilibriuma A + b B c C + d D

When writing an equilibrium constant expression,use only the concentrations of gases, (g),

or dissolved substances, (aq).Do not include solids, (s), or liquids, (l).

The Equilibrium ConstantEquilibrium Constant, K

For any equilibrium at a given temperature, the ratio of the product of the concentrations of the species on

the right side of the equilibrium equation, each raised to a power equal to its coefficient in the equation, to the corresponding product of the

concentrations on the left side of the equation, each raised to a power equal to its coefficient in the

equation, is a constant.

The equilibrium constant is bothequation-dependent and temperature-dependent.

Significance of the Value of KExample:Is the forward reaction favored, the reverse reaction favored, or

are appreciable quantities of all species present at equilibrium in the following reaction?

HC2H3O2(aq) H+(aq) + C2H3O2–(aq)

K = 1.8 × 10–5.

Solution:Since K is very small, the reverse reaction is favored.

Significance of the Value of KConsider the general reaction: Reactants Products

If the equilibrium constant is very large (K > 100),[Products] > [Reactants], so the forward reaction is favored.

If the equilibrium constant is very small (K < 0.01),[Products] < [Reactants], so the reverse reaction is favored.

If the equilibrium constant is neither larger nor small,[Products] ≈ [Reactants], so appreciable quantities of all species

are present at equilibrium.

][Reactants[Products] =K

Le Chatelier’s PrincipleLe Chatelier’s Principle

If a system is in equilibrium, any change imposed on the system tends to shift the equilibrium in a direction

that tends to counteract the initial change.

Le Chatelier’s principle only suggests an outcome; it does not provide an explanation.

Le Chatelier’s Principle The Pressure (Volume) Effect

A gas-phase equilibrium responds to compression-a reduction in volume of the reaction vessel.

If a gaseous equilibrium is compressed, the equilibrium will be shifted in the direction of formation of fewer molecules, thus minimizes the increase in pressure.

If the system is expanded, the shift will be in the direction of formation of more molecules.

3 H2 (g) + N2 (g) ↔ 2 NH3 (g) More molecules less molecules

Le Chatelier’s Principle The Pressure (Volume) Effect

3 H2 (g) + N2 (g) ↔ 2 NH3 (g)

If the pressure increases, the equilibrium will be shifted to the right (less molecules).

If the pressure decreases, the shift will be to the left.

To increase the yield of ammonia, industrial process uses pressures of 250 atm or higher.

Le Chatelier’s PrincipleThe Temperature Effect

If a reaction is exothermic, the reverse reaction is endothermic.

If the temperature increases, the equilibrium will be shifted to the direction of consuming heat

(endothermic, to the left for NH3 reaction lelow). If the temperature decreases, the shift will be in

direction of producing heat (exothermic, to the right).

3 H2 (g) + N2 (g) ↔ 2 NH3 (g) + 92kJ

Le Chatelier’s PrincipleThe Temperature Effect

The left tube at 25 0C contains very little brown gas compared to the tube on the right at 80 0C

N2O4 (g) + heat ↔ 2 NO2 (g) colorless brown

Le Chatelier’s Principle The Concentration Effect

Let us consider the reaction equilibrium:

3H2 (g) + N2 (g) ↔ 2 NH3 (g)

If H2 is added to the reaction chamber, the shift will be in the forward direction to counteract the increase in the number of hydrogen molecules thus producing more NH3. If H2 is removed, the equilibrium will shift to the reverse direction to increase the H2 concentration.

Le Chatelier’s Principle The Concentration Effect

Let us consider the reaction equilibrium:

3H2 (g) + N2 (g) ↔ 2 NH3 (g)

If H2 is added (increase of [H2]), the shift will be in the forward direction to increase in the concentration [NH3] and decrease the concentrations [H2] and [N2].Adding an inert gas has no effect on the equilibrium, although the total pressure increases.

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23

][H ][N][NH K

Solubility Equilibria

The equation for dissolving AgCl , a low-solubility compound, is

 AgCl (s) ↔ Ag+ ( aq) + Cl- (aq)

 This equilibrium is characterized by the solubility product constant Ksp

 Ksp = [Ag+ ] [Cl-]

Solubility Equilibria

Calculation of solubility product from the solubility: The chloride ion concentration of a saturated solution of silver chloride is 1.3 x 10-5 M . Calculate the solubility product for silver chloride .

  Ksp = [Ag+ ] [Cl-] In saturated solution of pure silver chloride the

concentration of [Ag+ ] and [Cl-] are equal. Therefore  

Ksp = [Ag+ ] [Cl-] =(1.3 x 10-5) x(1.3 x 10-5) = 1.7 x 10-10

Solubility Equilibria

Solubility and Solubility Product.For compounds of similar structure, the smaller the solubility product, the smaller the solubility. For example the solubility of silver bromide ( Ksp = 5.2 x 10-13 ) is lower than the solubility of silver chloride.Common Ion Effect

  Suppose that a soluble chloride, such as NaCl were to be added to the saturated solution of silver chloride. According to the Le Chatelier’s principle the equilibrium would shift the equilibrium in the reverse direction, reducing the solubility of silver chloride.

.

Ionization Equilibria: Weak acidWeak acids ionize only slightly when dissolved in water.

For a general weak acid HA,

HA(aq) H+(aq) + A–(aq)

Major species: HA(aq)

Minor Species: H+(aq) + A–(aq)

Ionization Equilibria: Weak acidThe ionization of a weak acid is usually so small that it is

negligible compared with the initial concentration of the acid.

We assume that all ionization concentrations are negligiblewhen subtracted from the initial concentration.

In other words, the initial concentration of the weak acid is also the final concentration after the acid ionizes.

Ionization Equilibria: Weak acidExample:Find the pH of 0.1 M nitrous acid. Ka = 4.5 × 10–4.

Solution:HNO2(aq) H+(aq) + NO2

–(aq)

Let x = [H+] = [NO2–]; [HNO2] = 0.1 M

0.1(x) (x) 10 4.5 =

][HNO][NO ][H = K 4-

2

-2

+

a

Ionization Equilibria: Weak acidFind the pH of 0.1 M nitrous acid. Ka = 4.5 × 10–4.

Solution:

x2 = (0.1) (4.5 × 10–4)

x = [H+] = 7 × 10–3

pH = – log [H+] = – log (7 × 10–3) = 2.2

Ionization Equilibria: Buffer SolutionBuffer Solution

A solution that resists changes in pH becauseit contains relatively high concentrations of both

a weak acid and a weak base.

The acid reacts with any added OH–;The base reacts with any added H+.

Ionization Equilibria: Buffer SolutionDetermine the pH of a solution that is 0.25 M in HAc and 0.35 M

in NaAc. Ka = 1.8 × 10–5.NaAc(aq) Na+(aq) + Ac–(aq)

HAc(aq) H+(aq) + Ac–(aq)

pH = – log [H+] = – log (1.3 × 10–5) = 4.89

[HAc]][Ac ][H = K

—+

a

5—5—-a

+ 10 1.3 = M 0.35M 0.25 10 1.8 =

][Ac[HAc] K = ][H

Ionization Equilibria: Buffer SolutionExample:

Determine the acid-to-base concentration ratio that will yield a buffer solution with a pH of 4.50 if the acid has Ka = 1.0 × 10–5.

Solution:HA(aq) H+(aq) + A–(aq)[H+] = antilog (–pH) = antilog (–4.50) = 3.2 × 10–5 M

[HA]][A ][H = K

—+

a [HA]][A =

][HK –

+a

3.2 = 10 1.010 3.2 =

K][H =

][A[HA]

5—

—5

a

+

—

Homework• Homework: 25, 27, 29, 31, 33, 35, 37, 41, 43, 47, 61, 63, 71,

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