Chapter 15 - Chemical Equilibrium
15.1
At room temperature colorless N2O4 decomposes to brown NO2.
N2O4(g)  2NO2(g)
(colorless) (brown)
• We expect the forward reaction rate to slow over time and the rate of the reverse reaction rate to increase
over time.
• At some time, the color stops changing which indicates and we have an equilibrium mixture of N2O4 and
NO2.
• Chemical equilibrium is the point at which the concentrations (partial pressures of gases) of all species are
constant.
•
A dynamic equilibrium exists when the rates of the forward and reverse reactions are equal.
• The reaction does NOT stop, but there is no NET change in the reactant or product concentrations.
• Use a double arrow () to imply that the process is an equilibrium.
Haber process: industrial preparation of ammonia from nitrogen and hydrogen:
N2(g) + 3H2(g)  2NH3(g)
• Process is carried out at high temperature (500oC) and pressure (200 atm).
• Much of the NH3 produced industrially is used as a fertilizer.
• If we start with a mixture of nitrogen and hydrogen (in any proportions), the reaction will reach
equilibrium with constant concentrations of nitrogen, hydrogen and ammonia.
• However, if we start with just ammonia and no nitrogen or hydrogen, the reaction will proceed and N2
and H2 will be produced until equilibrium is achieved.
• No matter what the starting composition of reactants and products is, the equilibrium mixture contains
the same relative concentrations of reactants and products.
• Equilibrium can be reached from either direction.
Consider a simple reaction.
A(g)  B(g)
• Assume that both the forward and reverse reactions are elementary processes.
We can write rate expressions for each reaction.
Forward reaction:
A  B
Rate =
Reverse reaction:
Rate =
kf = rate constant (forward reaction)
B  A
kr = rate constant (reverse reaction)
In equilibrium the forward and reverse rates are equal:
For our original example:
N2O4(g)  2NO2(g)
Forward rate = kf[N2O4]
Reverse rate = kr[NO2]2
• At equilibrium:
•
Rearranging:
This is called the Equilibrium constant!
“c” because it’s based on concentrations
For gaseous substances we can use the ideal gas equation to convert between concentration and pressure:
PV = nRT so M = (n/V) = (P/RT)
For substances A and B:
[A] = (PA/RT) and [B] = (PB/RT)
Ratefwd = kf PA/RT and Raterev = krPB/RT
At equilibrium: kf PA/RT = krPB/RT
Rearranging, we get:
( PB / RT )

( PA / RT )
•
This is called the Equilibrium constant! “p” because it’s based on pressure
15.2 The Equilibrium Constant
We can write an expression for the relationship between the concentration of the reactants and products at equilibrium.
•
•
•
For a general reaction: aA + bB  cC + dD
The equilibrium expression (law of mass action)
is given by:
:
Kc=
Kp=
*equilibrium concentration/pressure of (g) and (aq) only! (more on this later!)
• Kc and Kp are equilibrium constants (Keq).
Converting between Kp and Kc
Δn = (sum of coefficients of gaseous product) – (sum of coefficients of gaseous reactant)
R is the gas constant (0.0821)
T must be Kelvin
When Δn is zero 
•
K depends on ___________________________
eq
•
K varies with ___________________________
eq
•
K does NOT depend on __________________________________________________
eq
•
K does NOT depend on __________________________________________________
eq
•
We generally ________________________ of K
eq
The Magnitude of Equilibrium Constants
• The equilibrium constant, Keq, is the ratio of products to reactants.
– the larger Keq the more products are present at equilibrium.
– the smaller Keq the more reactants are present at equilibrium.
•
If Keq >> 1, then ___________________________ at equilibrium and “equilibrium lies to the right”
•
If Keq << 1, then ___________________________ at equilibrium and “equilibrium lies to the left”
Manipulating Chemical Equations and Keq
An equilibrium can be approached from any direction.
Consider the reaction:
•
N2O4(g)  2NO2(g)
The equilibrium constant (at 100oC) is:
K eq
P 2 NO2

 6.49
PN 2O4
However, when we write the equilibrium expression for the reverse reaction: 2NO2(g)  N2O4(g)
•
K eq 
PN2O4
P 2 NO2
 0.154
•
The equilibrium constant for a reaction in one direction is the reciprocal of the equilibrium constant of
the reaction in the opposite direction.
A + B  2C K
2C  A + B
•
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.
A + B  2C K
2A + 2B  4C
•
+
The equilibrium constant (at 100oC) is:
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.
A + B  2C
K1
C + A  D
K2
2A + B  C+D
Homogeneous equilibria - Equilibria in which all reactants and products are present __________________
Heterogeneous equilibria - Equilibria in which one or more reactants or products are present __________
______________________________________________
Consider the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
Experimentally, the amount of CO2 does not depend on the amounts of CaO and CaCO3. Why?
• The concentration of a pure solid or pure liquid equals its density divided by its molar mass.
• Neither density nor molar mass is a variable.
Since the concentrations of solids and pure liquids ___________________________________________
___________________________________________________________________________________
Another way to see it:
For the decomposition of CaCO3:
• CaO and CaCO3 are pure solids and have constant concentrations.
Canceling
•
•
Since the concentration of a pure solid or pure liquid is not included in the equilibrium constant expression. We
anticipate that the amount of CO2 formed will not depend on the amounts of CaO and CaCO3 present.
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!
#1 Suppose that the gas phase reactions A → B and B → A are both elementary processes with rate constants of
3.8  10-2 s-1 and 3.1  10-1 s-1 respectively.
(a) What is the value of the equilibrium constant for the equilibrium A  B?
(b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?
#2 Write the expression for Kc for the following reactions. In each case indicate whether the reaction is
homogeneous or heterogeneous.
(a) 3 NO(g)  N2O(g) + NO2(g)
(b) CH4(g) + 2 H2S(g)  CS2(g) + 4 H2(g)
(c) Ni(CO)4(g)  Ni(s) + 4 CO(g)
(d) HF(aq)  H+1(aq) + F-1(aq)
(e) 2 Ag(s) + Zn+2(aq)  2 Ag+1(aq) + Zn(s)
#3 When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or
mostly products?
(a) N2 + O2  2 NO
Kc = 1.5  10-10
(b) 2 SO2 + O2  2 SO3
Kp = 2.5  109
#4 If Kc = 0.042 for PCl3(g) + Cl2(g)  PCl5(g) at 500 K, what is the value of Kp for this reaction at this
temperature?
#5 At 1000 K, Kp = 1.85 for the reaction: SO2(g) + ½ O2(g)  SO3(g)
(a) What is the value of Kp for the reaction SO3(g)  SO2(g) + ½ O2(g)?
(b) What is the value of Kp for the reaction 2 SO2(g) + O2(g)  2 SO3(g)?
(c) What is the value of Kc for the reaction in part b?
(d) Does this reaction favor SO2 and O2 or does it favor SO3 (at this temperature)?
#6 Mercury (I) oxide decomposes into mercury and oxygen: 2 Hg2O(s)  4 Hg(l) + O2(g)
(a) Write the equilibrium constant expression for this reaction in terms of partial pressures
(b) Explain why we normally exclude pure solids and liquids from equilibrium constant expressions.
#7 Gaseous hydrogen iodide is placed in a closed container at 425oC, where it partially decomposes to
hydrogen and iodine:
2 HI(g)  H2(g) + I2(g)
At equilibrium, it is found that
[HI] is 3.53  10-3 M, [H2] is 4.79  10-4 M, and [I2] is 4.79  10-4 M.
What is the value of Kc at this temperature?
#8 The equilibrium: 2 NO(g) + Cl2(g)  2 NOCl(g) is established at 500 K.
An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for
NO, Cl2, and NOCl respectively.
Calculate Kp for this reaction at 500 K.
Calculating Equilibrium Constants from initial concentrations:
• Tabulate initial and equilibrium concentrations (or partial pressures) for all species in the equilibrium.
• If an initial and an equilibrium concentration is given for a species, calculate the change in concentration.
• Use the coefficients in the balanced chemical equation to calculate the changes in concentration of all
species.
• Deduce the equilibrium concentrations of all species.
• Use these equilibrium concentrations to calculate the value of the equilibrium constant.
#9 A mixture of 0.10 mol of NO and 0.050 mol of H2, and 0.10 mol of H2O is placed in a 1.0 L vessel at 300 K.
The following equilibrium is established:
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
At equilibrium, [NO] = 0.062 M.
(a) Calculate the equilibrium concentrations of H2, N2, and H2O
(b) Calculate Kc
#10
A mixture of 0.2000 mol CO2, 0.1000 mol H2, and 0.1600 mol H2O is placed in a 2.000 L vessel. The
following equilibrium is established at 500 K:
CO2(g) + H2(g)  CO(g) + H2O(g)
(a) Calculate the initial partial pressures of CO2, H2, and H2O
(b) At equilibrium, PH2O = 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO
(c) Calculate Kp for the reaction
HW #1 15(1, 2, 16, 18, 22, 26, 28, 32, 34, 38, 40)