Download Equilibrium Constant

Equilibrium
Unit 4
Chapters 17, 18, 19, 20
Chapter 17


Equilibrium – when two opposite reactions
occur simultaneously and at the same rate
Dynamic equilibrium – equilibrium can be
shifted if one or more of the reactants is
changed in any way. The equilibrium will
“react”
Chemistry Connections

Connection to Kinetics
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Equilibrium affects reaction rates, especially for
multi-step reactions!
Factors that affect equilibrium affect reaction rates
Connection to Thermo

If the reaction rate slows, the energy to be released
or absorbed may be released more slowly or
quickly….can be dangerous or take
FOREVER!...when setting up a reaction MUST
keep this in mind!
Equilibrium Basics – Kinetics!

For a reversible reaction with a one step
mechanism:
aA + bB  cC + dD
Rate = kf[A]a[B]b = kR[C]c[D]d
Definition of equilibrium: Rf = Rr
At Equilibrium
kf[A]a[B]b = kR[C]c[D]d
Kf = [C]c[D]d
kR [A]a[B]b
kf / kR = k c

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kc is the equilibrium constant (mass action
expression)
The constant can be determined by the ratio of the
constants or the concentrations!
Example 17-1, 17-2, 17-3
Equilibrium Constant

kc
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Just as in kinetics, intermediates are not allowed,
so substitution may be necessary!
Must be found experimentally or by means of
equilibrium concentrations from thermodynamic
data.
Also varies with temperature, and constant at a
given temperature (just like kinetic rate constants)
Independent of initial concentration
Equilibrium Constant

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Equilibrium constants are expressed as a ratio
of “activities” rather than concentrations in
order to be dimensionless.
Activity:
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Pure liquid or solid = 1 (if it is pure, amt. does not
affect)
Solution = the molar concentration
Ideal gas = partial pressure in atm
Reaction Quotient

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Reaction quotient – calculated the same way
as kc, but the reaction is NOT YET at
equilibrium. (mass action expression)
If reaction quotient is calculated, comparing it
to kc will allow prediction of the direction of
the reaction.
Comparing Quotient to K
aA + bB  cC + dD
[C]c[D]d = Q
[A]a[B]b
When kc > Q, reaction moves forward
When kc < Q, reaction moves reverse
Reaction Quotient


Forward reaction, [C]c[D]d = kc
[A]a[B]b
Reverse reaction, [A]a[B]b = 1/kc = kc’
[C]c[D]d
 Reversing the reaction, kc must be raised to -1
power.
Manipulation of kc
 Multiply the equation by ½
a
b
c
d
A B  C  D
2
2
2
2
c
2
d
2
a
2
b
2
[C ] [ D ]
 Kc* =
[ A] [ B ]
 [C ]c [ D] d
= 
 [ A] a [ B]b





1
2
= kc1/2
 Whatever factor (n) is applied to the
balanced equation, the new kc* is kc raised
to the power of n
 Example 17-4, 17-5
What is kc used for?
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Generally used to calculate concentrations at
equilibrium of various reactants and products
Example 17-6
Example 17-7
Le Chatelier’s Principle
O la la
Le Chatlier’s Principle (1884)

If a change in conditions (stress) is applied to a
system at equilibrium, the system responds to
reduce the stress and reach a new state of
equilibrium.
Le Chatelier’s Principle
Major stresses include:
1.
Change in Concentration (or pressure for
gasses)
PV = nRT
P = (n/V) RT
n/V = molar concentration if V is in L
 At constant T, P  concentration so it can be
used in its place in calculations.
Example

H2 (g) + I2 (g)  2HI (g)
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What happens when you add H2?
What happens when you remove H2?
What happens when you compress the container?
Example 17-10
Le Chatelier’s Principle
2.
Change in Volume – same moles, but smaller V,
affects gas concentrations, but not liquid, solid, or
solution.
2NO2 (g)  N2O4 (g)
What will happen if we decrease the volume of the
container?
You can increase the pressure by pumping in an inert
gas, which does not increase the concentration.
Le Chatelier’s Principle
Effect of Temperature Change – depends
on “thermicity” of reaction
Thermicitiy is considering heat as a reactant or a
product and writing it into the equation.
3.
C (s) + O2 (g)  CO2 (g) + 393.5kJ
If I add heat, which way will equilibrium move?
Le Chatelier’s Principle
4.
Catalysts –
Le Chatelier’s Principle
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Catalysts decrease the activation energy of
both the forward and reverse reaction,
therefore, no net effect on equilibrium, only on
the rate of reaction.
Example Problem 17-8
Example Problem 17-9
Example Problem 17-10
The Haber Process
A Classic Example of Equilibrium
Haber Process
N2 (g) + 3H2 (g)  2NH3 (g) + 92.22kJ
Is the enthalpy (H) Favorable?
Is the entropy (S) Favorable?
Will this reaction occur spontaneously?
Example Problem 17-8
Mole % NH3 in Equilibrium Mixture
oC
Kc
25
3.6e8
209
650
51
82
98
467
0.50
4
25
80
758
0.014
0.5
5
13
10atm 100atm 1000at
m
near 0 near 0 near 0
The Haber Process
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Problems:
N2 (g) + 3H2 (g)  2NH3 (g) + 92.22kJ
1.
2.
Reaction is very slow at low
temperatures, where Kc is highest
Increasing the temperature increases
the rate, but decreases the yield
because heat is produced.
The Haber Process
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Solutions:
1.
2.
3.
4.
5.
Increase the Temperature (generally 450oC)
Increase the Pressure (generally 200-1000atm)
Use excess Reactants
Remove the Ammonia as it is produced
Use a catalyst (solid Iron Oxide, Potassium
Oxide, and Aluminum Oxide)
Partial Pressures and the
Equilibrium Constant
kp
Partial Pressures

For gas phase equilibria, the equilibrium
constant (keq or kc) may be expressed in terms
of the partial pressure (kp)
Partial Pressure
2Cl2 (g) + 2H2O (g)  4HCl (g) + O2 (g)
4
4
[ HCl ] [O2 ]
kc 
2
2
[Cl2 ] [ N 2O]
kp 
( PHCl ) ( PO )
2
2
( PCl 2 ) ( PN 2 O )
2
Partial Pressure

If Kc or Kp are given and the other is wanted,
you can use the relationship between them to
calculate (PV = nRT)
Scary Math Proof
4
 PHCl   PO2 

 

4
 RT   RT 
PHCl  PO2




kc 

2
2
2
PCl2 PH 2O
 PCl2   PH 2O 

 

 RT   RT 
 
  
2
 1

RT

x
 1

 RT
5



4



Scary Math Proof Explained
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Kc = Kp (1/RT) for this reaction
Kc R T = Kp
For every reaction:
Kc = Kp(RT)-n
Kp = Kc(RT)n
Example Problem 17-11, 17-12, 17-13, 17-14
Heterogeneous Equilibria
Heterogeneous Equilibria
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Remember:
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The activities of pure solids and liquids are 1
Solvents of very dilute solutions are treated as pure
liquids with activity of 1
Heterogeneous Equilibria
(Examples)

Write Kc and Kp expressions:
1.
2.
CaCO3 (s)  CaO (s) + CO2 (g)
kc = [CO2] kp = (PCO2)
H2O (l) + CaF2 (s)  Ca2+ (aq) + 2F1- (aq)
kc = [Ca2+][F-]2
3.
kp = undefined (no gas)
3Fe(s) + 4H2O (g)  Fe3O4 (s) + 4H2 (g)
kc = [H2]4/[H2O]4 kp = (PH2)4/(PH2O)4
Relating Equilibrium to
Thermodynamics
Relation to Thermo
Grxn = Gorxn + RTlnQ
R = 8.314 J / k mol
Can use (+ 2.303RT logQ)
At equilibrium, Grxn = 0 and Q = k
k can be related directly to Grxn
Relation to Thermo
0 = Gorxn + RTlnk
Gorxn = -RTlnk
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K is the thermodynamic equilibrium constant
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kc where solutions are involved
kp where gases are involved
Finding K at different
Temperatures
Van’t Hoff Equation

Knowing Ho and kc at T1 allows the
ESTIMATION of the kc at T2
ln(kT1/kT2) = Ho (T2-T1)
R (T2T1)
ln(k2/k1) = (H/R) ((1/T1) – (1/T2))