chapter 14 chemical equilibrium chemistry a molecular approach 3rd edition

Chapter 14Chemical Equilibrium

CHEMISTRY A Molecular Approach

3rd Edition

Definition of EQUILIBRIUM

• When the rate of the forward reaction is equal to the rate of the backward reaction.

• Rate of production of product is the same as the rate of the reformation of the reactants.

• The double arrow implies the process is dynamic.

• Consider: A B

Forward reaction: A B Rate = kf[A]

Reverse reaction: B A Rate = kr[B]

• At equilibrium kf[A] = kr[B].

The Concept of EquilibriumThe Concept of Equilibrium

• For an equilibrium we write • 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.

• Alternatively:– kf[A] decreases to a constant,

– kr[B] increases from zero to a constant.

– When kf[A] = kr[B] equilibrium is achieved.


A B

The Equilibrium ConstantThe Equilibrium Constant

• For a general reaction

the equilibrium constant expression for everything in solution is

where Keq is the equilibrium constant.

aA + bB cC + dD

ba

dc

cKBA

DC


• For a general reaction in the gas phase

the equilibrium constant expression is

where Keq is the equilibrium constant.

aA + bB cC + dD

ba

dc

p PP

PPK

BA

DC

Kp and Kc

• Kp = Kc(RT)n or Kc = Kp

RTn

• Where n = # moles products minus the # mole of reactants

2 TYPES OF EQUILIBRIUM

• 1. HOMOGENEOUS EQUILIBRIUM – all components in the same state

• 2. HETEROGENEOUS EQUILIBRIUM – components in different states

• DIFFERENCES IN K EXPRESSION

• In heterogeneous equilibria, pure solids and pure liquids are not included in the equilibrium expression. Since the concentration of pure liquids and solids is constant, they will be given a value of 1.

• For example:

Heterogeneous EquilibriaHeterogeneous Equilibria

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

HETEROGENEOUS EQUILIBRIA

• We ignore the concentrations of pure liquids and pure solids in equilibrium constant expressions.

• CaCO3 (s) CaO (s) + CO2 (g)

• Keq = 1/[PCO2]


• Kc is based on the molarities (Kp on partial pressure) of reactants and products at equilibrium.

• We generally omit the units of the equilibrium constant.

• Note that the equilibrium constant expression has products over reactants.

• The same equilibrium is established no matter how the reaction is begun.

The Magnitude of Equilibrium Constants

• K = [products] ; therefore

• [reactants]

• If K >> 1; MORE products present at equil. than reactants; Equilibrium lies to the right

• If K << 1; LESS products present at equil. than reactants; Equilibrium lies to the left

The Equilibrium ConstantThe Equilibrium ConstantThe Magnitude of Equilibrium

Constants• If K << 1, then reactants dominate at equilibrium and the

equilibrium lies to the left.

The Equilibrium ConstantThe Equilibrium ConstantThe Direction of the Chemical

Equation and Keq

• An equilibrium can be approached from any direction.• Example:

• has

N2O4(g) 2NO2(g)

46.642

2

ON

2NO

P

PK p

The Direction of the Chemical Equation and Keq

• In the reverse direction:


46.61

155.02NO

ON

2

42 P

PKeq

2NO2(g) N2O4(g)

Other Ways to Manipulate Chemical Equations and Keq Values

• The reaction

has

which is the square of the equilibrium constant for


2N2O4(g) 4NO2(g)

2ON

4NO

42

2

P

PKc

N2O4(g) 2NO2(g)

Other Ways to Manipulate Chemical Equations and Keq Values

• Equilibrium constant for the reverse direction is the inverse of that for the forward direction.

• When a reaction is multiplied by a number, the equilibrium constant is raised to that power.

• The equilibrium constant for a reaction which is the sum of other reactions is the product of the equilibrium constants for the individual reactions.


Sample Problems

• The equil. constant for the ff. rxn. is 0.013.

• 2NO (g) + Br2 (g) 2NOBr(g)

• 1. Does the rxn favor the reactants or products?

• 2. Calculate Keq for: a. 2NOBr(g) 2NO (g) + Br2 (g)

• b. 4NO(g) + 2Br2 4NOBr(g)

Sample Problem

• Write the equilibrium expression for:

• NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

Predicting the Direction of Reaction• We define Q, the reaction quotient, for a general reaction

as

• Q = K only at equilibrium.

Applications of Applications of Equilibrium ConstantsEquilibrium Constants

aA + bB cC + dD

ba

dc

PP

PPQ

BA

DC

Predicting the Direction of Reaction• If Q > K 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 K).

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

• Please note that Q can only be calculated when initial concentrations are given.

Applications of Applications of Equilibrium ConstantsEquilibrium Constants

Problem

• N2(g) + 3H2(g) 2NH3(g), Keq = 4.51 x 10-5

• Indicate whether the mixture is at equilibrium at 450 oC and in which direction the mixture must shift to achieve equilibrium if:

• A.) 105 atm NH3, 35 atm N2, 495 atm H2

• B.) 26 atm NH3, 42 atm H2, 202 atm N2

• C.) 105 atm NH3, 55 atm H2, 5.0 atm N2

Calculating K

• 1. Balance the reaction.

• 2. Determine whether initial or equilibrium concentrations are given.

• 3. If equil. concentrations are given, K can be calculated directly using the equation K= [product]n/[reactants]n.

• 4. If initial concentrations are given, use the “ICE method”.

• Proceed as follows:– Tabulate initial and equilibrium concentrations (or partial pressures)

given.– If an initial and equilibrium concentration is given for a species, calculate

the change in concentration.– Use stoichiometry on the change in concentration line only to calculate the

changes in concentration of all species.– Deduce the equilibrium concentrations of all species.

• Usually, the initial concentration of products is zero. (This is not always the case.)

Calculating Equilibrium Calculating Equilibrium ConstantsConstants

Calculating K

• H2(g) + F2(g) 2HF(g)

• The equilibrium constant for the reaction above is 115 at a certain temperature. In a particular experiment, 3.00 mol of each component was added to a 1.5 L flask. Calculate the equilibrium concentration for all species.

Problem

• At 327 oC, the equilibrium concentrations are [CH3OH] = 0.15 M, [CO] = 0.24 M and [H2] = 1.1 M for the reaction:

•

• CH3OH (g) CO (g) + 2H2 (g)

Calculate Kc at this temperature.

Problem

• Consider the following rxn. at a certain temperature:

• Fe (s) + 3 O2 (g) 2 Fe2O3 (s)

• An equil. mixture contains 1.0 mole Fe, 0.0010 mole O2 and 2.0 mole Fe2O3 all in a 2.0 L container. Calculate Kc for this reaction.

Problem

• A 1.00 L flask was filled with 2.00 mole gaseous SO2 and 2.00 mole of NO2 and heated After equil. was reached, it was found that 1.30 mole gaseous NO was present. Calculate the value of Kc for:

• SO2 (g) + NO2 (g) SO3 (g) + NO (g)

Problem

• At a particular temperature, Kc = 4.0 x 10-7 for the reaction:

N2O4 (g) 2NO2 (g)

In an expt. 1.0 mole of N2O4 is placed in a 1.00 -L vessel. Calculate the concentrations of N2O4 and NO2 when this reaction reaches equilibrium.

Use of Approximation

• Approximations can be used if the value of K is so small that the terms (-x) or (+x) will barely have an effect.

• Validity of approximation: < 5%

• ______[x]______ = < 5%[initial concentration]

• Le Châtelier’s Principle:

• If a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.

Le ChLe Châtelier’s Principleâtelier’s Principle

Effects of Volume and Pressure Changes

Effect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.


Effect of Temperature Changes• Adding heat (i.e. heating the vessel) favors away from

the increase:– if H > 0, adding heat favors the forward reaction,

– if H < 0, adding heat favors the reverse reaction.

• Removing heat (i.e. cooling the vessel), favors towards the decrease:– if H > 0, cooling favors the reverse reaction,

– if H < 0, cooling favors the forward reaction.


The Effect of Catalysis• 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.


• CH4 (g) + 2H2S (g) CS2 (g) + 4H2 (g) H = -47 kJ

• What will be the effect on the pressure of H2 if:

• A. Volume of the container is increased?

• B. CS2 gas is decreased?

• C. A catalyst was added?

• D. Heat was added?

• E. Some CH4 were added?

Problem

• For the reaction at a particular temperature:

• 3 H2 (g) + N2 (g) 2 NH3 (g)

• The equil. concentrations are [H2] = 5.0 M, [N2] = 8.0 M and [NH3] = 4.0 M. What are the initial concentrations of H2 and N2?

Sample Problem

• NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)

• Enough ammonia is dissolved in 5.00 L of H2O at 25 oC to produce a 0.0124 M NH3 solution. Analysis of the equilibrium mixture shows the OH- concentration to be 0.000464 M. Calculate Keq at 25 oC for the reaction.

Sample Problem

• SO2Cl2(g) SO2(g) + Cl2(g)

• At equilibrium, the partial pressures of SO2Cl2 = 3.31 atm, SO2 = 1.59 atm and Cl2 = 0.98 atm.

• Calculate the value of the equilibrium constant Kp.

Sample Exercise

• N2(g) + 3H2(g) 2NH3(g), Kp = 1.45 x 10-5

• In an equilibrium mixture of the three gases at 500 oC, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in the equilibrium mixture?

Quiz

• SO2Cl2(g) SO2(g) + Cl2(g) Kc = 0.078

• In an equilibrium mixture of the 3 gases, the concentration of SO2Cl2 = 0.108 M and SO2 = 0.052 M.

• A. Write the K expression.

• B. What is the partial pressure of Cl2 in the equilibrium mixture?

Problem

• A mixture of 0.10 mole NO, 0.050 mole H2 and 0.10 mole of H2O is placed in a 1.0 L vessel at 300 K. The following equilibrium is established.

• 2NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g)

• At equilibrium [NO] = 0.062. Calculate the equilibrium concentrations of H2, N2 and H2O.

• Calculate Kc.

Quiz

• 2SO2(g) + O2(g) SO3(g) Keq = 0.345

• A. Write the K expression.

• B. In an equilibrium mixture , the partial pressures of SO2 and O2 are 0.165 atm and 0.755 atm, respectively. What is the equilibrium partial pressure of SO3 in the mixture?

Problem • At 800 K, Kc = 3.1 x 10-5 for: I2 (g) 2I (g)

• If an equil. mixture in a 10.0 L vessel contains 2.67 x 10-2 grams of I (g), how many grams of I2 are in the mixture?

• ------------------------------------------------------------------• 2SO2 (g) + O2 (g) 2SO3 (g) Kp = 3.0 x 104

• In a 2.00-L vessel, the equil. mixture contains 1.57 g SO3 and 0.125 g of O2. How many grams of SO2 are in the vessel.

Problem

Br2(g) +Cl2(g) 2BrCl(g) Kc = 7.0

At 400 K, if 0.30 mol of Br2 and 0.30 mol of Cl2 are introduced into a 1.0-L container, what will be the equilibrium concentrations of Br2, Cl2 and BrCl?

Problem

• At 218 oC, Kc = 1.2 x 10-4 for the equilibrium

• NH4HS (s) NH3 (g) + H2S (g)

• Calculate the equilibrium concentrations of NH3 and H2S if a sample of solid NH4HS is placed in a closed vessel and decomposes until equilibrium is reached.

Problem

• At 25 oC, the reaction:

• CaCrO4 (s) Ca 2+ (aq) + CrO4 2- (aq)

• has an equilirbium constant Kc = 7.1 x 10-4. What are the equilibrium concentrations of Ca2+ and CrO4 2- in a saturated solution of CaCrO4?

Sample problem

• H2(g) + I2(g) 2HI(g) Keq = 51

• Predict in which direction the reaction will proceed to reach equilibrium at 448 oC if we start with 0.02 mol HI, 0.01 mol H2 and 0.03 mol I2 in a 2.00 L container.

Sample problem

• H2(g) + I2(g) 2HI(g) Keq = 50.5

• A 1.00 L flask is filled with 1.00 mol H2 and 2.00 mol I2 at 448 oC. What are the partial pressures of H2, I2 and HI at equilibrium?

Sample Problem

• 2NOCl(g) 2NO(g) + Cl2(g) K = 1.6 x 10-5

• 1.0 mole NOCl is placed in a 2.0-L flask. What are the equilibrium concentrations?

Sample Problem

• N2(g) + 3H2(g) 2NH3(g), Keq = 4.51 x 10-5

• Indicate whether the mixture is at equilibrium at 450 oC and in which direction the mixture must shift to achieve equilibrium if:

• A.) 105 atm NH3, 35 atm N2, 495 atm H2

• B.) 26 atm NH3, 42 atm H2, 202 atm N2

• C.) 105 atm NH3, 55 atm H2, 5.0 atm N2

• Le Châtelier’s Principle:

• If a system at equilibrium is disturbed, the system will move in such a way as to counteract the disturbance.


Effects of Volume and Pressure Changes• Le Châtelier’s Principle: if pressure is increased the

system will shift to counteract the increase.• That is, the system shifts to remove gases and decrease

pressure.• An increase in pressure favors the direction that has

fewer moles of gas.• In a reaction with the same number of product and

reactant moles of gas, pressure has no effect.


Effects of Volume and Pressure Changes

Effect of Temperature Changes• The equilibrium constant is temperature dependent.• For an endothermic reaction, H > 0 and heat can be

considered as a reactant.• For an exothermic reaction, H < 0 and heat can be

considered as a product.


Effect of Temperature Changes• Adding heat (i.e. heating the vessel) favors away from

the increase:– if H > 0, adding heat favors the forward reaction,

– if H < 0, adding heat favors the reverse reaction.

• Removing heat (i.e. cooling the vessel), favors towards the decrease:– if H > 0, cooling favors the reverse reaction,

– if H < 0, cooling favors the forward reaction.


The Effect of Catalysis• 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.


• CH4 (g) + 2H2S (g) CS2 (g) + 4H2 (g) H = -47 kJ

• What will be the effect on the pressure of H2 if:

• A. Volume of the container is increased?

• B. CS2 gas is decreased?

• C. A catalyst was added?

• D. Heat was added?

• E. Some CH4 were added?

chapter 14 chemical equilibrium chemistry a molecular approach 3rd edition

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