AP Chemistry Chapter 15 Equilibrium
Chapter 15. Chemical Equilibrium
15.1 The Concept of Equilibrium
• Chemical equilibrium is the point at which the concentrations of all species are constant.
• A dynamic equilibrium exists when the rates of the forward and reverse reactions are equal.
• The double arrow D implies that the process is dynamic.
15.2 The Equilibrium Constant
Equilibrium-Constant Expression
• law of mass action.
• For a general reaction aA + bB D dD + eE
• The equilibrium-constant expression is given by:
• Where Kc is the equilibrium constant.
• The subscript “c” indicates that molar concentrations were used to evaluate the constant.
• Note that the equilibrium constant expression has products in the numerator and reactants in the denominator.
Equilibrium Constants in Terms of Pressure, Kp
• When the reactants and products are gases the equilibrium constant is Kp where “p” stands for pressure.
• For the reaction:
aA + bB D dD + eE
• They can be interconverted using the ideal gas equation and our definition of molarity:
PV = nRT thus P = (n/V)RT
• If we express volume in liters the quantity (n/V) is equivalent to molarity.
• Thus the partial pressure of a substance, A, is given as:
PA =(nA/V)RT = [A]RT
• We can use this to obtain a general expression relating Kc and Kp:
Kp = Kc(RT)∆n
• Where ∆n = (moles of gaseous products) – (moles of gaseous reactants).
• The numerical values of Kc and Kp will not differ if ∆n = 0.
(Note: See p. 635 in textbook for info about why we do not use units for K)
Sample Exercise 15.1 (p. 632)
Write the equilibrium expression for Keq for these three reactions:
a) 2 O3(g) D 3 O2(g)
b) 2 NO(g) + Cl2(g) D 2 NOCl(g)
c) Ag+(aq) + 2 NH3(g) D Ag(NH3)2+(aq)
Practice Exercise 15.1
Write the equilibrium expression for Keq for these three reactions:
a) H2(g) + I2(g) D 2 HI(g)
b) Cd2+(aq) + 4 Br-(aq) D CdBr42-(aq
Evaluating Kc
• The value of Keq does not depend on initial concentrations of products or reactants.
• The equilibrium expression depends on stoichiometry.
• It does not depend on the reaction mechanism.
• The value of Keq varies with temperature.
• We generally omit the units of the equilibrium constant.
Sample Exercise 15.2 (p. 634)
In the synthesis of ammonia from nitrogen and hydrogen,
N2(g) + 3 H2(g) D 2 NH3(g)
Kc = 9.60 at 300oC. Calculate Kp for this reaction at this temperature.
(4.34 x 10-3)
Practice Exercise 15.2
For the equilibrium 2 SO3(g) D 2 SO2(g) + O2(g), Kc is 4.08 x 10-3 at 1000 K.
Calculate the value for Kp.
(0.335)
15.3 Interpreting and Working with Equilibrium Constants
The Magnitude of Equilibrium Constants
• The larger Keq the more products are present at equilibrium.
• The smaller Keq the more reactants are present at equilibrium.
• If Keq > 1, then products dominate at equilibrium and equilibrium lies to the right.
• If Keq < 1, then reactants dominate at equilibrium and the equilibrium lies to the left.
Sample Exercise 15.3
The following diagrams represent three different systems at equilibrium, all in the same size containers.
a) Without doing any calculations, rank the three systems in order of increasing equilibrium constant, Kc.
b) If the volume of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kc for each system.
Practice Exercise 15.3
The equilibrium constant for the reaction H2(g) + I2(g) D 2 HI(g) varies with temperature as follows:
Kp = 792 at 298 K; Kp = 54 at 700 K.
Is the formation of HI favored more at the higher or lower temperature?
Sample Exercise 15.4 (p. 637)
The equilibrium constant for the reaction of N2 with O2 to form NO equals Kc = 1 x 10-30 at 25oC.
N2(g) + O2(g) D 2 NO(g)
Using this information, write the equilibrium constant expression and calculate the equilibrium constant for the following reaction:
2 NO(g) D N2(g) + O2(g)
Practice Exercise 15.4
For the formation of NH3 from N2 and H2, N2(g) + H2(g) D 2 NH3(g), Kp = 4.34 x 10-3 at 300oC.
What is the value of Kp for the reverse reaction?
(2.30 x 102)
Relating Chemical Equations and Equilibrium Constants
• The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction.
• The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number.
• The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.
Sample Exercise 15.5 (p. 638)
Given the following information,
HF(aq) D H+(aq) + F-(aq) Kc = 6.8 x 10-4
H2C2O4(aq) D 2 H+(aq) + C2O42-(aq) Kc = 3.8 x 10-6
determine the value of Kc for the following reaction:
2 HF(aq) + C2O42-(aq) D 2 F-(aq) + H2C2O4(aq)
(0.12)
Practice Exercise 15.5
Given the following information at 700 K,
H2(g) + I2(g) D 2HI(g) Kp = 54.0
N2(g) + 3 H2(g) D 2 NH3(g) Kp = 1.04 x 10-4
determine the value of Kp (at 700 K)
2 NH3(g) + 3 I2(g) D 6 HI(g) + N2(g)
(1.51 x 109)
15.4 Heterogeneous Equilibria
• Equilibria in which one or more reactants or products are present in a different phase are called heterogeneous equilibria.
• If a pure solid or pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium constant expression.
• Note: Although the concentrations of these species are not included in the equilibrium expression, they do participate in the reaction and must be present for an equilibrium to be established!
• systems where the solvent is involved as a reactant or product and the solutes are present at low concentrations.
• Example: H2O(l) + CO32–(aq) D OH–(aq) + HCO3–(aq)
Kc = [OH–][HCO3–] / [CO32–]
• Here the concentration of water is essentially constant and we can think of it as a pure liquid.
Sample Exercise 15.6 (p. 640)
Write the equilibrium-constant Kc for each of the following reactions:
a) CO2(g) + H2(g) D CO(g) + H2O(l) Kc =
b) SnO2(s) + 2 CO(g) D Sn(s) + 2 CO2(g) Kc =
Practice Exercise 15.6
Write the equilibrium-constant expressions for each of the following reactions:
a) Cr(s) + 3 Ag+(aq) D Cr3+(aq) + 3 Ag(s) Kc =
b) 3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g) Kp =
Sample Exercise 15.7 (p. 641)
Each of the following mixtures was placed in a closed container and allowed to stand. Which of these mixtures is capable of attaining the equilibrium
CaCO3(s) D CaO(s) + CO2(g)
a) pure CaCO3
b) CaO and a pressure of CO2 greater than the value of Kp
c) Some CaCO3 and a pressure of CO2 greater than the value of Kp
d) CaCO3 and CaO
Practice Exercise 15.7
When added to Fe3O4(s) in a closed container, which one of the following substances – H2(g), H2O(g), O2(g) - will allow equilibrium to be established in the reaction
3 Fe(s) + 4 H2O(g) D Fe3O4(s) + 4 H2(g)
15.5 Calculating Equilibrium Constants
Sample Exercise 15.8 (p. 642)
A mixture of hydrogen and nitrogen in a reaction vessel is allowed to attain equilibrium at 472oC. The equilibrium mixture of gases was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data calculate the equilibrium constant, Kp, for
N2(g) + 3 H2(g) D 2 NH3(g)
(2.79 x 10-5)
Practice Exercise 15.8
An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25oC: [HC2H3O2] = 1.65 x 10-2 M; [H+] = 5.44 x 10-4 M; and [C2H3O2-] = 5.44 x 10-4 M. Calculate the equilibrium constant, Kc, for the ionization of acetic acid at 25oC. The reaction is
HC2H3O2(aq) D H+(aq) + C2H3O2-(aq)
(1.79 x 10-5)
Proceed as follows:
1. Tabulate all the known initial and equilibrium concentrations of the species that appear in the equilibrium-constant expression.
2. For those species for which both the initial and equilibrium concentrations are known, calculate the change in concentration that occurs as the system reaches equilibrium.
3. Use the stoichiometry of the reaction (that is, use the coefficients in the balanced chemical equation) to calculate the changes in concentration of all the other species in the equilibrium.
4. From the initial concentrations and the changes in concentration, calculate the equilibrium
concentrations. These are then used to evaluate the equilibrium constant.
Sample Exercise 15.9 (p. 643)
A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 at 448oC is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10-3 M. Calculate Kc at 448oC for the reaction taking place, which is
H2(aq) + I2(g) D 2 HI(g)
(1.81 x 10-5)
Practice Exercise 15.9
Sulfur trioxide decomposes at high temperature in a sealed container:
2 SO3(g) D 2 SO2(g) + O2(g)
Initially the vessel is charged at 1000 K with SO3(g) at a partial pressure of 0.500 atm. At equilibrium the SO3 partial pressure is 0.200 atm. Calculate the value of Kp at 1000 K.
(0.338)
15.6 Applications of Equilibrium Constants
Predicting the Direction of Reaction
• For a general reaction: aA + bB D cC + dD
• We define Q, the reaction quotient, as:
• Where [A], [B], [C], and [D] are molarities (for substances in solution) or partial pressures (for gases) at any given time.
• Note: Q = Keq only at equilibrium.
• If Q < Keq then the forward reaction must occur to reach equilibrium.
• If Q > Keq then the reverse reaction must occur to reach equilibrium.
• Products are consumed, reactants are formed.
• Q decreases until it equals Keq.
Calculating Equilibrium Concentrations
• The same steps used to calculate equilibrium constants are used to calculate equilibrium concentrations.
• Generally, we do not have a number for the change in concentration.
• Therefore, we need to assume that x mol/L of a species is produced (or used).
• The equilibrium concentrations are given as algebraic expressions.
Sample Exercise 15.10 (p. 645)
At 448oC the equilibrium constant, Kc, for the reaction H2(g) + I2(g) à 2 HI(g) is 51.
Predict how the reaction will proceed to reach equilibrium at 448oC if we start with 2.0 x 10-2 mol of HI, 1.0 x 10-2 mole of H2, and 3.0 x 10-2 mol of I2 in a 2.00-L container.
(Q = 1.3, so reaction must proceed from left to right)
Practice Exercise 15.10
At 1000 K the value of Kp for the reaction 2 SO3(g) D 2 SO2(g) + O2(g) is 0.338. Calculate the value for Qp, and predict the direction in which the reaction will proceed toward equilibrium if the initial partial pressures of reactants are PSO3 = 0.16 atm; PSO2 = 0.41 atm; PO2 = 2.5 atm.
(Qp = 16; Qp > Kp, so reaction will proceed from right to left)
Sample Exercise 15.11 (p. 646)
For the Haber process, N2(g) + 3 H2(g) D 2 NH3(g), Kp = 1.45 x 10-5 at 500oC. In an equilibrium mixture of the three gases at 500oC, 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 this equilibrium mixture?
(2.24 x 10-3 atm)
Practice Exercise 15.11
At 500 K the reaction PCl5(g) D PCl3(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?
(1.22 atm)
Sample Exercise 15.12 (p. 646)
A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448oC. The value of the equilibrium constant, Kc, for the reaction
H2(g) + I2(g) D 2 HI(g)
at 448oC is 50.5. What are the partial pressures of H2, I2, and HI in the flask at equilibrium?
([H2] = 0.065 M, [I2] = 1.065 M, [HI] = 1.87 M)
Practice Exercise 15.12
For the equilibrium, PCl5(g) D PCl3(g) + Cl2(g), the equilibrium constant, Kp, has the value 0.497 at 500 K. A gas cylinder at 500 K is charged with PCl5(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature?
(PPCl5 = 0.967 atm, PPCl3 = PCl2 = 0.693 atm)
15.7 Le Châtelier’s Principle
• If a system at equilibrium is disturbed by a change in temperature, a change in pressure, or a change in the concentration of one or more components, the system will shift its equilibrium position in such a way as to counteract the effects of the disturbance.