Chapter Fifteen
Chemical Equilibrium
1
The Concept of Equilibrium
2
Dynamic Equilibrium
Opposing processes occur at equal rates
Forward and reverses reaction proceed at equal rates
No outward change is observed
Ratio of reactants to products is constant.
A double sided arrow ( ) is used
Often temperature dependent
Physical equilibrium
Equilibrium between phases
H2O(l)  H2O(g)
Chemical equilibrium
Equilibrium between reactants and products
N2O4(g)  2NO2(g)
colorless brown
3
Equilibrium Constant Expression, Kc
At equilibrium, [N2O4] and [NO2] are constant
N2O4 2NO2
Rate (forward) = Rate (reverse)
k1 [N2O4] = k-1 [NO2]2
k1 [ NO2 ]2

 Kc
k 1 [ N 2O4 ]
4
Initial vs. Equilibrium Concentrations
Proof of Kc Calculation
5
The Equilibrium Constant Expression, Kc
For a reaction of the type: aA  bB  ....  cC  dD  ...
c
d
[C ] [ D]
The equilibrium expression is: K c 
a
b
[ A] [ B ]
Equilibrium Constant Expression:
Concentrations of the products appear in the numerator
Concentrations of the reactants appear in the denominator.
Exponents match coefficients in chemical equation.
Units are not included!
Also known as a Mass Action Expression
Chemical Equation must be balanced to calculate Kc
K is based on HOW the equation is balanced
6
Chemical Reactions and Kc
a. SO2(g) + ½O2(g)  SO3(g)
[ SO 3 ]
Kc 
1/ 2
[ SO 2 ][ O 2 ]
b. 2O2(g) + O2(g)  2SO3(g)
[ SO3 ]2
Kc 
2
[ SO2 ] [O2 ]
Equilibrium constants change if reaction balanced differently.
Numerical values for Kc are different
Kc (reaction b) = [Kc (reaction a)] 2
You must know how the reaction was balanced
7
Rules for Modifying A Chemical Equation
aA + bB  cC + dD
Reversing a chemical equation: Invert Kc
[ A]a [ B ]b
1
cC  dD  aA  bB
Kc 

c
d
[C ] [ D]
Kc
Multiplying coefficients by n: Raise Kc to n power
na
nb
[ A] [ B]
n
n (cC  dD  aA  bB ) K c 
 Kc
nc
nd
[C ] [ D]
Adding equations: Multiply Kcs
eE  fF
aA  bB  cC  dD
eE  aA  bB  fF  cC  dD
[C ]c [ D ] d [ F ] f
 K1  K 2
Kc 
a
b
e
[ A] [ B ] [ E ]
8
Meaning of Equilibrium Constants
aA  bB  cC  dD
At Equilibrium
[C ]c [ D]d
Kc 
[ A]a [ B ]b
More products than reactants
More reactants than products
As K goes to infinity reaction goes to completion
As K goes to zero, no reaction occurs
9
At a given temperature, 0.100 moles of NO were added to a
2.00 L vessel. At equilibrium, 0.044 moles of NO were
remaining. What is the value of Kc?
Balanced Equation and Equilibrium Constant Expression:
2NO(g)  N2(g) + O2(g)
Kc = [N2] [O2]
[NO]2
Find equilibrium molarities of reactants and products
molesN2 = molesO2 = ½ moles NO used up in reaction
molesN2 = molesO2 = ½ (0.100-0.044)= 0.028moles
[N2] = [O2] = 0.028moles/2.00L = 0.014M
[NO]= 0.044moles/2.00L = 0.022M
Calculate Kc
Kc = [N2] [O2] = [0.014][0.014] = 0.40
[NO]2
[0.022]2
10
Ways of Expressing
Equilibrium Constants
11
Homogeneous Equilibria, Kc and Kp
All products and reactants are in an aqueous or gas
phase for ANY equilibrium constant
Kc
Kp
All chemicals in units of molarity, moles/L
Can be either gas or aqueous solutions
2CO(g) + O2(g)2CO2(g)
All chemicals in gas phase
Partial pressure in atmospheres
Relationship between Kc and Kp
Kp = Kc(RT) n
Derived from ideal gas law
R is the gas constant, T is temperature in kelvin
n = moles gaseous products - moles gaseous reactants
Kp=Kc if moles gas of product = moles gas of reactant
Otherwise, a decrease in moles of gas decreases pressure
12
Calculate Kp for the following reactions
2CO(g) + O2(g)2CO2(g)
Kc= 12800
ngas = 
2NO(g)  N2(g) + O2(g)
Kc= 12800
ngas = 0
Kp = Kc(RT) ngas
Kc
R
T
= 12800
= 0.0821L atm/mol K
= 1500 oC= 1773K
ngas = Moles productsgas phase - Moles reactantsgas phase
0 . 0821 Latm
K p  12800 x (
x1773 K ) 1  12800 x (. 00687 )  87 . 9
molK
If ngas= 0 : Kc = Kp
13
Heterogeneous Equilibrium and Keq
Are the equilibrium constants for the 2 reactions the same?
SO3(g) + H2O(g)  H2SO4(g) SO3(g) + H2O(l)  H2SO4(aq)
K1 =
[H2SO4(g)]
[SO3(g)][H2O(g)]
K2 =
[H2SO4(aq)]
[SO3(g)][H2O(l)]
K1 equals K2 if [H2SO4(g)]=[H2SO4(aq)] & [H2O(g)]=[H2O(l)]
[H2SO4(g)]
[H2SO4(aq)]
[H2O(g)]
[H2O(l)]
variable
variable
variable
56M
dependent on partial pressure
# moles dissolved in water
dependent on partial pressure
Density =1 g/mL = 1000 g/L
[H2O(l)] =1000g/L x mol/18g= 56mol/L
The two equilibrium constants are not the same
The phase of the material matters!
14
Equilibria Involving Mixed Phases
Use the subscript “eq” to designate general K
Keq = Kc= K=, etc. Value depends on equation
Do not include solids and liquids in Keq expression
Keq = [H2SO4(aq)]
SO3(g) + H2O(l)  H2SO4(aq)
[SO3(g)]
Kp for liquid-vapor equilibrium is vapor pressure of liquid
H2O(l)  H2O(g)
Kp = P(H2O)
Liquid will not be included
Reactions must show phase for each reactant and product.
15
Equilibria Involving Pure Solids And Liquids
CaCO3(s)CaO(s) + CO2(g)
Kp= PCO2
PCO2 does not depend on the amount of CaCO3 or CaO
16
Exam 1 Feb. 23rd 7:30AM-9:15AM
Covers Chapters 14 and 15
8 questions, 3 parts per question (see practice exam)
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Request preferences immediately
Check seat before exam so you can get started
Bring your ID to the exam
TAs will check IDs once exam is started
Exam will be on your seat with your name on it
Sit in assigned seat and start exam
17
Alternate Exam: Feb. 24th, 6:30AM-7:50AM
If you do not take the exam on Tuesday…….
Ideally email me by 5pm on the 22nd so I know you are coming
If you just miss the exam, email me as soon as possible
No guarantee I’ll have an extra exam if I don’t know you are coming!!!
If you do not take exam 1 on either day
It is your dropped exam, no further makeups given
DSS Exams
I need DSS paperwork by next Tuesday (1 week before exam)
I will try to get a time and date between Tues-Fri that works for all
Once I know your schedules, I’ll get back to you with options
18
What Does the Equilibrium Constant
Tell Us?
19
Relative Meaning Of Equilibrium Constants
aA  bB  ....  cC  dD  ...
Large value of Kc or Kp
[C ]c [ D]d
Kc 
[ A]a [ B ]b
[C] & [D] very large or [A] & [B] very small.
Reaction goes to completion
Equilibrium favors products.
Small value of Kc or Kp
[C] & [D] very small or [A] & [B] very large
Reaction proceeds slowly or not at all (NR: no reaction)
Equilibrium favors reactants
Does not guarantee reaction will occur
Thermodynamically favored, but may be kinetically
controlled.
20
The Reaction Quotient, Qc or Qp
Represents Kc or Kp at non-equilibrium conditions
[C ]c [ D]d
aA  bB  ....  cC  dD  ...
Qc 
a
b
Predicts direction of reaction to get to equilibrium. [ A] [ B ]
Qc < K
Concentration of products is less than at equilibrium
Reaction occurs in the forward direction
Qc > Kc
Concentration of products is greater than at equilibrium
Reaction proceeds in the reverse direction
Qc = Kc
Concentration of products equals that at equilibrium
No macroscopic changes are observed
21
For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium
concentrations were found to be 1.0×10-3 M in both hydrogen and nitrogen
and 0.020 M in ammonia. Now add 0.010 M nitrogen. Which direction does
the reaction shift?
N2(g) + 3H2(g)  2NH3(g)
22
For the reaction of hydrogen with nitrogen to form ammonia, the equilibrium
constant Kc = 4.0×108 at 25 oC. The equilibrium concentrations were found
to be 1.0×10-3 M in both hydrogen and nitrogen and 0.020 M in ammonia.
Now add 0.010 M nitrogen. Which direction does the reaction shift?
N2(g) + 3H2(g)  2NH3(g)
[ NH 3 ]2
Kc 
[ N 2 ][ H 2 ]3
Fill in Concentrations:
[N2] = 1.0×10-3 M + 0.010M= 0.011 M
[ 0 . 020 ] 2
7
[NH3] = 0.020 M
Qc 

3
.
6
x
10
3 3
-3
[
0
.
011
][
1
.
0
x
10
]
[H2] = 1.0×10 M
Compare Qc to Kc:
3.6×107< 4.1×108
The reaction goes toward products
Adding Nitrogen forces reaction to the right
Qc<Kc
23
Solving Equilibrium Problems
24
Use of Tabulated Equilibrium Constants
1. Write the properly balanced chemical reaction.
This gives the stoichiometry and the species involved in the reaction
2. Set up a table of concentrations for all components
Use reaction stoichiometry to express all unknown concentrations in
terms of a single reactant or product, x.
3. Write out the equilibrium constant expression for the
reaction
This is the equation for K
4. Substitute these concentrations into the equation for K
5. Solve the equation for the unknown concentration, x.
6. Substitute the value you calculated for x into the
expressions for the other equilibrium concentrations.
25
The decomposition of BrCl to bromine and chlorine has a Kc, of 0.14 at 350 K.
If the initial concentration of BrCl is 0.062 M, what are the equilibrium
concentrations of all components?
1.
Equation
2BrCl(g)
2.
Table
BrCl
.062
– 2x
0.062-2x
Initial
Change
Equilibrium
 Br2(g) +
Br2
0
+x
+x
3. Equilibrium Constant Expression
K
c

[ Br 2 ][ Cl 2 ]
[ BrCl ] 2
4. Substitute equilibrium values
[ x ][ x ]
0 . 14 
[ 0 . 062  2 x ] 2
Cl2(g)
Cl2
0
+x
+x
5. Solve for x


[ x ]2
1/ 2
0.14  
2
[0.062  2 x] 
[ x]
0.374 
[0.062  2 x]
x = 0.0232-0.748x
6. Substitute back into table
[BrCl] = 0.062–2x = 0.062–2(0.013) = 0.036 M
[Br2] = [Cl2] = x = 0.013M
1/ 2
x = 0.013
26
Carbon monoxide reacts with water at 1000 oC to give carbon dioxide and hydrogen with
Kc = 0.58. A reaction was started with the following composition: CO2, 0.2M M; H2 1.20 M;
H2O, 0.50 M, and CO, 1.0 M.
What are the equilibrium concentrations of all components?
1.
Equation
CO(g) + H2O(g)  CO2(g) + H2(g)
2.
Table
Initial
Change
Equilibrium
CO
1.0
–x
1.0-x
H2 0
0.50
-x
0.50-x
3. Equilibrium Constant Expression
CO2
0.2
+x
0.2+x
H2
1.2
+x
1..2+x
Kc 
4. Substitute equilibrium values
5. Solve for x:
Need quadratic equation
6. Substitute back into table
[CO] = 1.00 – 0.022 = 0.98M
[CO2] = 0.20+ 0.022 = 0.22M
0.58 
[ CO 2 ][ H 2 ]
[ CO ][ H 2 O ]
[0.20  x][1.2  x]
[1.00  x][0.50  x]
x = 0.022
[H2O] = 0.50 – 0.022 = 0.48M
[H2] = 1.20 + 0.022 = 1.22M
27
Solving Quadratic Equations
For equations: aX2 + bX + c = 0
From the previous problem:
Solving for X
x = 0.022 or –5.4
Only x = 0.022 makes sense
x = – 5.4 gives negative
concentrations
28
Factors that Affect
Chemical Equilibrium
29
Le Châtelier’s Principle
When stress is applied to a system at equilibrium, the system will
shift to reduce the applied stress and reestablish equilibrium
2NO2(g) 1 N2O4(g)
Addition of pressure causes reaction to shift towards products
Number of moles decreases (decreases the stress).
30
General Rules for Le Châtelier’s Principle
5 types of stresses will affect the equilibrium
Change in Components or Concentrations
Change in Partial Pressure
Change in External Pressure
Change in Total Volume
Change in Temperature
31
Change in Components or Partial Pressures
If the concentration of a reactant increases:
Fe3+(aq) + SCN-(aq)FeSCN2+(aq)
yellow
clear
red
[ FeSCN 2 ]
Kc 
[ Fe3 ][ SCN  ]
Denominator of the Qc increases
Qc < Kc
To equalize Qc and Kc,
Concentration of other reactants decrease
Concentration of the products increases
Equilibrium is pushed right.
32
Change in Temperature
Exothermic Reaction
Endothermic Reaction
aA  bB  cC  dD  Heat
Heat  aA  bB  cC  dD
Increase T: K decreases
Decrease T: K increases
Increase T: K increases
Decrease T: K Decreases
2NO2(g)  N2O4(g)
Exothermic Reaction
Heat is released as a product
reactant
red
clear
Heat is needed as a
ΔH=-58kJ/mol
Cold: more N2O4
K increases
Hot: More NO2
K decreases
33
Change in External Pressure or Volume
3 A  2 B  1C  1D
When pressure is applied (increased)
Equilibrium shifts to produce smaller # of
moles of gas
When pressure is decreased
Equilibrium shifts to produce larger # of
moles of gas
3 A  2 B  3C  2 D
Reaction with same # of moles of gas on
both sides
Changes in external pressure do not affect
equilibrium
34
Adding A Catalyst or An Inhibitor
Catalysts
The role of a catalyst is to lower activation energy.
Catalysts and inhibitors do not affect the equilibrium
Catalysts cause a reaction to reach equilibrium faster
Inhibitors
Inhibitors slow down the reaction to prevent equilibrium
from being reached as quickly
Inhibitors may slow the reaction rate to the point that
there appears to be no reaction, but it is really just
infinitely slow.
35
Which way does the reaction shift if…
N2O3(g)  NO(g) + NO2(g)
Ho = +39.7 kJ/mol
NO is added?
Shifts left to use up additional chemical
Volume of reaction vessel reduced?
Increased pressure: shifts left to reduce # moles
The total internal pressure is increased by adding He
gas?
No shift: Total pressure on system increases, but no change in
partial pressures of reactants or products.
The temperature is increased?
This an endothermic reaction, heat is absorbed.
Increasing the temperature adds heat to the system.
Reaction shifts right (endothermic) to remove heat.
A catalyst is added?
No shift: A catalyst does not affect the equilibrium composition
36
Summary
Equilibrium is a balance between products and reactants
Use stoichiometry to determine reactant or product ratios, but NOT
reactant to product ratios.
Capital K is used to represent the equilibrium constant
Products over reactants raised to their stoichiometric coefficients
Calculated from balanced equation, subscript designates units used, Kc,
Kp
No units used in final written K
Equlibrium Calculations and Reaction Quotients
ICE tables used to manipulate initial and equilibrium concentrations.
Factors influencing K
Concentration of chemicals and temperature affect all equilibria
T, P, V changes affect Kp in gases
Catalysis
Effect of catalysts and inhibitors, reasons they are used.
37