Chapter 9. Chemical Equilibrium
9.1 The Nature of Chemical Equilibrium
-Approach to Equilibrium
[Co(H2O)6]2+ + 4 Cl-
[CoCl4]2- + 6 H2O
Characteristics of the Equilibrium State
example) H2O(l)
1.
2.
3.
4.
H2O(g)
They display no macroscopic evidence of change.
They are reached through spontaneous processes.
They show a dynamic balance of forward and reverse
processes.
They are the same regardless of direction of approach.
Steady states: macroscopic concentrations of species are not
changing with time, even though the system in not at
equilibrium.
9.2 The Empirical Law of Mass Action
aA + bB
cC + dD
KC and KP: empirical equilibrium constant
Law of mass action:
1) The numerical value of KC or KP is an inherent property of the
chemical reaction itself and does not depend on the specific
initial concentrations of reactants and products.
2) The magnitude gives direct information about the nature of the
equilibrium state or position of the reaction.
Law of Mass Action for Gas-Phase Reactions
• A deeper study shows that instead of inserting just the
partial pressure for each reactant or product, we must insert
the value of the partial pressure relative to a specified
reference pressure Pref.
Example 9.1
•
Write equilibrium expressions for the following gasphase chemical equilibria.
(a) 2 NOCl(g)
2 NO(g) + Cl2(g)
(b) CO(g) + 0.5 O2(g)
CO2(g)
Law of Mass Action for Reactions in Solution
Example 9.2
•
Household laundry bleach is a solution of sodium hypochlorite
(NaOCl) prepared by adding gaseous Cl2 to a solution of sodium
hydroxide:
Cl2(aq) + 2 OH-(aq)
ClO-(aq) + Cl-(aq) + H2O(l)
The active bleaching agent is the hypochlorite ion, which can
decompose to chloride and chlorate ions in a side reaction that
competes with bleaching:
3 ClO-(aq)
2 Cl-(aq) + ClO3-(aq)
Write the equilibrium expression for the decomposition reaction.
Law of Mass Action for Reactions Involving Pure
Substances and Multiple Phases
1.
H2O(l)
H2O(g)
PH2O = K
Experiments show that as long as some liquid qater is in the
container, the pressure of water vapor at 25oC is 0.03126 atm.
2.
I2(s)
I2(aq)
[I2] = K
Experiments show that the position of the equilibrium (given by the
concentration of I2 dissolved at a given temperature) is independent of
the amount of solid present, as long as there is some.
3.
CaCO3(s)
CaO(s) + CaCO3(aq)
PCO2 = K
Experiments show that the pressure of CO2 is constant, independent
of the amount of solid present, as long as there is some.
General Procedure for writing the mass action law
for these more complex reactions
1.
2.
3.
4.
Gases enter the equilibrium expression as partial pressures,
measured in atmospheres.
Dissolved species enter as concentrations, in moles per liter.
Pure solids and pure liquids do not appear in equilibrium
expressions; neither does a solvent taking part in a chemical
reaction, provided the solution is dilute.
Partial pressures and concentrations of products appear in the
numerator, and those of reactants in the denominator; each is raised
to a power equal to its coefficient in the balanced chemical equation
for the reaction.
This procedure gives the dimensionless thermodynamic equilibrium
constant K because each species has entered the equilibrium
expression relative to its standard reference state.
Example 9.3
•
Hypochlorous acid (HOCl) is produced by bubbling chlorine through
an agitated suspension of mercury(II) oxide in water. The chemical
equation for this process is
2 Cl2(g) + 2 HgO(s) + H2O(l)
HgO×HgCl2(s) + 2 HOCl(aq)
Write the equilibrium expression for this reaction.
The preceding discussion has shown the procedures for setting
up the mass action law for broad classes of chemical reactions.
However, numerous fundamental questions about chemical
equilibrium are not answered by these procedures.
Why should the law of mass action exist in the first place, and
why should it take the particular mathematical form shown here?
Why should the equilibrium constant take a unique value for
each individual chemical reaction?
What factors determine that value?
Why does the value of the equilibrium constant change slightly
when studied over broad ranges of concentration?
Why should the equilibrium constant depend on temperature,
and can a quantitative explanation be provided for this
dependence?
9.3 Thermodynamic Description of the Equilibrium State
• In this section we use thermodynamics to demonstrate why the
mass action law takes its special mathematical form and why
the thermodynamic equilibrium constant K is a dimensionless
quantity.
• Thermodynamics views a chemical reaction as a process in
which atoms “flow” from reactants to products.
• If the reaction is spontaneous and is carried out at constant T
and P, thermodynamics requires that ΔG < 0 for the process.
• When a chemical reaction has reached equilibrium, ΔG = 0.
Reactions among Ideal Gases
• Dependence of Gibbs Free Energy of a Gas on Pressure
• The Equilibrium Expression for Reactions in the Gas
Phase
Prove that for the general reaction
aA + bB
cC + dD
Example 9.4
• The ΔGo of the chemical reaction
3 NO(g) Æ N2O(g) + NO2(g)
was calculated in Example 8.10. Now calculate the
equilibrium constant of this reaction at 25oC.
Reactions in Ideal Solutions
Example 9.5
•
Calculate ΔGo and equilibrium constant at 25oC for the chemical
reaction
3 ClO-(aq)
2 Cl-(aq) + ClO3-(aq)
whose equilibrium expression we developed in Example 9.2.
Reactions Involving Pure Solids and Liquids
and Multiple Phases: The Concept of Activity
•
•
•
The mass action law for homogeneous reactions in ideal gases and
ideal solutions was written in Section 9.2 by straightforward inspection
of the balanced equation for the reaction under study.
If one or more of the reactants was a solid or liquid in its pure state,
the procedure was less obvious, because “concentration” has no
meaning for a pure species.
This apparent difficulty is resolved by the concept of activity, which is
a convenient means for comparing the properties of a substance in a
general thermodynamic state with its properties in a specially selected
reference state.
Example 9.6
•
The compound urea, important in biochemistry, can be prepared in
aqueous solution by the following reaction:
CO2(g) + 2 NH3(g)
CO(NH2)2(aq) + H2O(l)
(a) Write the mass action law for this reaction. (b) Calculate ΔGo for this
reaction at 25oC. (c) Calculate K for this reaction at 25oC.
9.4 The Law of Mass Action for Related and
Simultaneous Equilibria
• Relationships among Equilibrium Expressions
• Relationships among Equilibrium Expressions (continued)
• Relationships among Equilibrium Expressions (continued)
Example 9.7
•
The concentration of the oxides of nitrogen are monitored in airpollution reports. At 25 C, the equilibrium constant for the reaction
NO(g) + ½ O2(g)
NO2(g)
is
K1 = 1.3 x 106
and that for
is
½ N2(g) + ½ O2(g)
K2 = 6.5 x 10-16
NO(g)
Find the equilibrium constant K3 for the reaction
N2(g) + 2 O2(g)
NO2(g)
Consecutive Equilibria;
Hemoglobin and Oxygen Transport
9.5 Equilibrium Calculations for Gas-Phase
and Heterogeneous Reactions
• Problem Solving Technique for two classes of problems.
- Evaluating Equilibrium Constants from Reaction Data
- Calculating Equilibrium Compositions When K is Known
Evaluating Equilibrium Constants from Reaction Data
Example 9.8: Phosgene, COCl2, forms from CO and Cl2 according to the
equilibrium
CO(g) + Cl2(g)
COCl2(g)
At 600 C, a gas mixture of CO and Cl2 is prepared that has initial partial
pressures (before reaction) of 0.60 atm for CO and 1.10 atm for Cl2. After
the reaction mixture has reached equilibrium, the partial pressure of
COCl2(g) at this temperature is measured to be 0.10 atm. Calculate the
equilibrium constant for this reaction. The reaction is carried out ina vessel
of fixed volume.
For a more complex reaction,
2 C2H6(g) + 7 O2(g)
4 CO2(g) + 6 H2O(g)
Example 9.9
Graphite (a form of solid carbon) is added to a vessel that contains CO2(g)
at a pressure of 0.824 atm at a certain high temperature. The pressure
rises due to a reaction that produces CO(g). The total pressure reaches
an equilibrium value of 1.366 atm. (a) Write a balanced equation for
the process. (b) Calculate the equilibrium constant.
Calculating Equilibrium Composition When K is
Known
Example 9.10
Suppose H2(g) and I2(g) are sealed in a flask at T = 400 K with partial
pressures PH2 = 1.320 atm and PI2 = 1.140. At this temperature H2
and I2 do not react rapidly to form HI(g), although after a long
enough time they would produce HI(g) at its equilibrium partial
pressure. Suppose, instead, that the gases are heated in the sealed
flask to 600 K, a temperature at which they quickly reach
equilibrium:
H2(g) + I2(g)
2 HI(g)
The equilibrium constant for the reaction is 92.6 at 600 K:
(a) What are the equilibrium values of PH2, PI2, and PHI at 600 K?
(b) What percentage of the I2 originally present has reacted when
equilibrium is reached?
• Example 9.11
Hydrogen is made from natural gas (methane) for immediate consumption
in industrial processes, such as ammonia production. The first step is
called the “steam reforming of methane”
CH4(g) + H2O(g)
CO(g) + 3 H2(g)
The equilibrium constant for this reaction is 1.8 x 10-7 at 600 K. Gaseous
CH4, H2O, and CO are introduced into an evacuated container at 600
K, and their initial partial pressures (before reaction) are 1.40 atm, 2.30
atm, and 1.60 atm, respectively. Determine the partial pressure of
H2(g) that will result at equilibrium.
•
Suppose the data for a gas-phase equilibrium are given in terms of
concentrations rather than partial pressures. In such cases all
concentrations can be converted to partial pressures before the
calculations are carried out, or the equilibrium expression can be
rewritten in terms of concentration variables.
• Example 9.12
At elevated temperatures, PCl5 dissociates extensively according to
PCl5(g)
PCl3(g) + Cl2(g)
At 300 C, the equilibrium constant for this reaction is K = 11.5. The
concentrations of PCl3 and Cl2 at equilibrium in a container at 300 C
are both 0.01100 mol L-1. Calculate [PCl5].
9.6 The Direction of Change in Chemical
Reactions: Empirical Description
• The Reaction Quotient
• Example 9.13
The reaction between nitrogen and hydrogen to produce ammonia
N2(g) + 3 H2(g)
2 NH3(g)
is essential in making nitrogen-containing fertilizers. This reaction has an
equilibrium constant equal to 1.9 x 10-4 at 400 C. Suppose that 0.10
mol of N2, 0.040 mol of H2, and 0.020 mol of NH3 are sealed in a 1.00
L vessel at 400 C. In which direction will the reaction process?
• Example 9.14
Solid ammonium chloride is in equilibrium with ammonia and HCl gases:
NH4Cl(s)
NH3(g) + HCl(g)
The equilibrium constant at 275 C is 1.04 x 10-2. We replace 0.980 g of
solid NH4Cl into a closed vessel with volume 1.00 L and heat to 275
C. (a) In what direction does the reaction proceed? (b) What is the
partial pressure of each gas at equilibrium? (c) What is the mass of
solid NH4Cl at equilibrium?
External Effects on K: Principle of Le Chatelier
• A principle stated by Henri Le Chatelier in 1884
A system in equilibrium that is subjected to a stress will react
in a way that tends to counteract the stress.
Le Chatelier’s principle provides a way to predict
qualitatively the direction of change of a system under an
external perturbation. It relies heavily on Q as a predictive
tool.
Effect of Changing the Concentration of a Reactant or Product
• Example 9.15
An equilibrium gas mixture of H2(g), I2(g), and HI(g) at 600 K has partial
pressures of 0.4756 atm, 0.2056 atm, and 3.009 atm, respectively.
This is essentially the final equilibrium state of Example 9.10. Enough H2 is
added to increase its partial pressure to 2.000 atm at 600 K before any
reaction take place. This mixture then once again reaches equilibrium at
600 K. What are the final partial pressures of the three gases?
Effect of Changing the Volume
Effect of Changing the Temperature
Maximizing the yield of a reaction
9.7 The Direction of Change in Chemical
Reactions: Thermodynamic Explanation
• Here, we identify thermodynamic factors that determine
the magnitude of K. We also provide a thermodynamic
criterion for predicting the direction in which a reaction
proceeds from a given initial condition.
The Magnitude of the Equilibrium Constant
Free Energy Changes and the Reaction Quotient
Temperature Dependence of Equilibrium Constants
• Example 9.16
Calculate K for the equilibrium of Example 9.4 at T = 400 K,
assuming ΔHo to be approximately independent of
temperature over the range from 298 to 400 K.
Exmple 9.4
The ΔGo of the chemical reaction
3 NO(g) Æ N2O(g) + NO2(g)
was calculated in Example 8.10. Now calculate the equilibrium constant
of this reaction at 25oC.
Temperature Dependence of Vapor Pressure
• Example 9.17
The ΔHvap for water is 40.66 kJ mol-1 at the normal boiling point, Tb =
373 K. Assuming ΔHvap and ΔSvap are approximately independent of
temperature from 50 C to 100 C, estimate the vapor pressure of water
at 50 C (323 K).
9.8 Distribution of a Single Species between Immiscible
Phases: Extraction and Separation Processes
• Extraction Processes
Example 9.18
An aqueous solution has an iodine concentration of 2.00 x 10-3
M. Calculate the percentage of iodine remaining in the
aqueous phase after extraction of 0.100 L of this aqueous
solution with 0.050 L of CCl4 at 25 C.
Chromatographic Separations
•
Chromatography
Column Chromatography
Gas-Liquid Chromatography