17 Equilibrium

1
Equilibrium The Extent of Chemical Reactions
The Dynamic Nature of the Equilibrium State
The Reaction Quotient and the Equilibrium
Constant Expressing Equilibria with Pressure
Units Relation Between Kc and Kp
Reaction Direction Comparing Q and K How
to Solve Equilibrium Problems Reaction
Conditions and the Equilibrium State
Le Châteliers Principle
2
Use of the double arrow!

• For reactions that do not go to completion

• Size of arrow can represent extent of reaction

3
Implication of the double arrow

• Two reactions are occurring One forward and one
  reverse
• The two reactions establish a dynamic equilibrium
• rate of forward rxn. is equal to reverse rxn.
• concentration of reactants and products is
  constant
• Not equal consider shopping mall
• Reaction is considered to be reversible
• Are all reactions reversible?

4
The Range of Equilibrium Constants
A) Small K
N2 (g) O2 (g) 2 NO(g)
K 1 x 10 -30
B) Large K 2 CO(g) O2 (g)
2 CO2 (g) K 2.2 x 1022
C) Intermediate K 2 BrCl(g)
Br2 (g) Cl2 (g) K 5
5
Basic Concepts

• Graphical representation of the rates for the
  forward and reverse reactions for this general
  reaction
• a A b B ? c C d D

6
Reaction Quotient

• Products over reactants
• Exponent is same as coefficient
• Omit pure liquids and solids
• Omit the solvent in dilute soln.s

7
Equilibrium
WHEN FORWARD AND REVERSE RATE ARE EQUAL!
8
Example 16 - 2

• What is the equilibrium expression for the Haber
  synthesis of ammonia?
• N2 3 H2 ? 2 NH3

9
The State of Equilibrium
For the Nitrogen dioxide - dinitrogen tetroxide
equilibrium
2 NO2 (g, brown) N2O4 (g,
colorless)
At equilibrium ratefwd raterev
ratefwd kfwdNO22 raterev krev N2O4
kfwdNO22 krev N2O4
K
See pg 677 for a 2-step mechanism
10
Initial and Equilibrium Concentrations at 100C
2 NO2 (g, brown) N2O4 (g,
colorless)
Initial Equilibrium
Ratio
N2O4
N2O4
N2O4
NO2
NO2
NO22
0.1000 0.0000 0.0491 0.1018
0.211 0.0000 0.1000
0.0185 0.0627 0.212
0.0500 0.0500 0.0332
0.0837 0.211 0.0750 0.0250
0.0411 0.0930
0.210
11
2 NO2 (g, brown) N2O4 (g,
colorless)
12
Pure Solids and Liquids

• 56M vs 1x10-7

s
13
Removal of CO2
CaCO3(s) ? CaO (s)CO2 (g)
14
Example 16 - 4

• Write the equilibrium constant expression for the
  reaction of iron metal with strong aqueous acid,
  and indicate the concentration units for each
  reagent.
• 2 Fe (s) 6 H3O (aq) ? 2 Fe3 (aq) 6 H2O (l)
  3 H2 (g)

15
Writing the Reaction Quotient from the Balanced
Equation
Problem Write the reaction quotient for each of
the following reactions (a) Write the reaction
quotient for decomposition of potassium
chlorate KClO3 (s)
KCl(s) O2 (g) (b) The equilibrium
constant for the combustion of butane in oxygen
C4H10 (g) O2 (g) CO2
(g) H2O(g) Plan We first balance the
equations, then construct the reaction
quotient Or equilibrium constant as described by
equation. Solution
16
Writing the Equilibrium Constant for
an Overall ReactionI
Problem Oxygen gas combines with nitrogen gas
in the internal combustion engine to produce
nitric oxide, which when out in the atmosphere
combines with additional oxygen to form nitrogen
dioxide.
(1) N2 (g) O2 (g) 2 NO(g)
Kc1 4.3 x 10-25 (2) 2 NO(g) O2 (g)
2 NO2 (g) Kc2 6.4 x 109
(a) Show that the overall Kc for this reaction
sequence is the same as the product of the
Kcs for the individual reactions. (b) Calculate
Kc for the overall reaction. Plan We first
write the overall reaction by adding the two
reactions together and write the Kc. We then
multiply the individual Kcs for the total K.
overall N2 (g) 2 O2 (g) 2 NO2
(g)
17
Writing the Reaction Quotient for an Overall
ReactionII
(a) cont.
NO2
Kc (overall)
N2O22
(b) K Kc1 x Kc2 (4.3 x 10-25)(6.4 x 109)
2.8 x 10-15
18
Ways of Expressing the K (and Q)
Form of Chemical Equation Form of Q
B
Reference reaction A B
K(ref) Reverse reaction
B A K
Reaction as sum of two steps
A
1 A
K(ref) B
(1) A C
Koverall K1 x K2 K(ref) Koverall K1 x K2
x
(2) C B
A B
C B B
K(ref)
A C A
19
Lab Examples
20

• Solution with Fe3, SCN-, and FeSCN
• Add NaF turned Clear
• Removal of FeSCN
• How? What did we observe?
• Add Fe3 - turned red
• So SCN Still around!
• Add SCN - no change
• So no Fe3 around to react

21
Equilibrium - Le Chateliers

• How will the equilibrium shift?

• Look at the Reaction Quotient

If Q lt K the rxn shifts right If Q K the rxn is
at equil. If Q gt K the rxn shifts left
22
Reaction Direction and the Relative Sizes of Q
and K
Fig. 17.5
23
Equilibrium - Le Chateliers

• When a system that is at equilibrium is subject
  to a stress that disturbs the system, the
  system will proceed in a direction that will
  bring the system back to equilibrium by
  relieving that stress.

24
Equilibrium - Le Chateliers
Shifts left!

• If we add product?

Shifts right!

• If we add reactant?

Shifts right!

• If we add reagent that reacts with B?

• If we increase the temperature?

Shifts left!

• If exothermic HEAT is a product
• If endothermic HEAT is a reactant

Shifts right!
25
From Lab Cu2 NH3

• Add HCl -
• Dark Blue turned Cloudy then Clear Light Blue

26
The Effect of a Change in ConcentrationI
Given an equilibrium equation such as
If one adds ammonia to the reaction mixture at
equilibrium, it will force the reaction to go to
the right producing more product. Likewise, if
one takes ammonia from the equilibrium mixture,
it will force the reaction back to produce more
reactants by recombining H2 and HCN to give more
of the initial reactants, CH4 and NH3.
Forces equilibrium to produce more product.
Remove NH3
Forces the reaction equilibrium to go back to
the left and produce more of the reactants.
27
The Effect of a Change in ConcentrationII
If to this same equilibrium mixture one decides
to add one of the products to the equilibrium
mixture, it will force the equilibrium
back toward the reactant side and increase the
concentrations of reactants. Likewise, if one
takes away some of the hydrogen or hydrogen
cyanide from the product side, it will force the
equilibrium to replace it.
Forces equilibrium to go toward the reactant
direction.
Add H2
Remove HCN
Forces equilibrium to make more produce and
replace the lost HCN.
28
Temperature

• Cold water
• got darker
• Hot water
• got lighter

29
The Effect of a Change in Temperature
Only temperature changes will alter the
equilibrium constant, and that is why we always
specify the temperature when giving the value of
Kc. The best way to look at temperature effects
is to realize that temperature is a component of
the equation, the same as a reactant, or product.
For example, if you have an exothermic reaction,
heat (energy) is on the product side of the
equation, but if it is an endothermic reaction,
it will be on the reactant side of the equation.
O2 (g) 2 H2 (g) 2 H2O(g)
Energy Exothermic
Electrical energy 2 H2O(g) 2 H2
(g) O2 (g) Endothermic
A temperature increase favors the endothermic
direction and a temperature decrease favors the
exothermic direction.
A temperature rise will increase Kc for a system
with a positive H0rxn A temperature rise will
decrease Kc for a system with a negative H0rxn
30
Effect of Various Disturbances on an Equilibrium
System
Disturbance Net Direction of Reaction
Effect on Value of K
Concentration Increase reactant Toward
formation of product None Decrease
reactant Toward formation of reactant
None Pressure (volume) Increase P
Toward formation of lower
amount (mol) of gas
None Decrease P Toward
formation of higher
amount (mol) of gas
None Temperature Increase T Toward
absorption of heat Increases if H0rxngt 0

Decreases if
H0rxnlt 0 Decrease T Toward release
of heat Increases if H0rxnlt 0

Decreases if H0rxngt
0 Catalyst added None rates of forward
and reverse
reactions increase equally None
31
The Effect of a Change in Pressure (Volume)
Pressure changes are mainly involving gases as
liquids and solids are nearly incompressible. For
gases, pressure changes can occur in three
ways Changing the concentration of a
gaseous component Adding an inert gas
(one that does not take part in the reaction)
Changing the volume of the reaction
vessel When a system at equilibrium that
contains a gas undergoes a change in pressure as
a result of a change in volume, the equilibrium
position shifts to reduce the effect of the
change. If the volume is lower (pressure is
higher), the total number of gas
molecules decrease. If the volume is higher
(pressure is lower), the total number of gas
molecules increases.
32
Predicting the Effect of Temperature and Pressure
Problem How would you change the volume
(pressure) or temperature in the following
reactions to increase the chemical yield of the
products? (a) 2 SO2 (g) O2 (g) 2
SO3 (g) H0 197 kJ (b) CO(g) 2 H2 (g)
CH3OH(g) H0 -90.7 kJ (c) C(s)
CO2 (g) 2 CO(g) H0 172.5
kJ Plan For the impact of volume (pressure), we
examine the reaction for the side with the most
gaseous molecules formed. For temperature, we
see if the reaction is exothermic, or
endothermic. An increase in pressure will force a
reaction toward fewer gas molecules. Solution
To get a higher yield of the products you should
(a) Increase the pressure, and increase the
temperature. (b) Increase the pressure, and
decrease the temperature. (c) Decrease the
pressure, an increase in the temperature
will increase the product yield.
33
Thermodynamics and Keq

• Keq is related to reaction favorability.
• If DGorxn lt 0, reaction is product-favored.
• DGorxn is the change in free energy as reactants
  convert completely to products.
• But systems often reach a state of equilibrium in
  which reactants have not converted completely to
  products.
• How to describe thermodynamically ?

34
?Grxn versus ?Gorxn
Under any condition of a reacting system, we can
define ?Grxn in terms of the reaction quotient, Q
?Grxn ?Gorxn RT ln Q
If ?Grxn lt 0 then reaction proceeds to right If
?Grxn gt 0 then reaction proceeds to left
35
DG and the Equilibrium Constant

• At equilibrium, DG0 and QK. Then

36
Thermodynamics and Keq
Keq is related to reaction favorability and so to
DGorxn. The larger the negative value of DGorxn
the larger the value of K.
DGorxn - RT lnK
where R 8.31 J/Kmol
37
Thermodynamics and Keq
DGorxn - RT lnK

• Calculate K for the reaction
• N2O4 ? 2 NO2 DGorxn 4.8 kJ
• DGorxn 4800 J - (8.31 J/K)(298 K) ln K

K 0.14
When ?Gorxn gt 0, then K lt 1 - reactant
favored When ?Gorxn lt 0, then K gt1 - product
favored
38
The Equilibrium Constant

• At a given temperature 0.80 mole of N2 and 0.90
  mole of H2 were placed in an evacuated 1.00-liter
  container. At equilibrium 0.20 mole of NH3 was
  present. Calculate Kc for the reaction.

39
The Equilibrium Constant
40
The Equilibrium Constant
41
The Equilibrium Constant
42
The Equilibrium Constant
43
Predicting Reaction Direction and Calculating
Equilibrium Concentrations I
Problem Two components of natural gas can react
according to the following chemical equation
In an experiment, 1.00 mol CH4, 1.00 mol CS2,
2.00 mol H2S, and 2.00 mol H2 are mixed in a 250
mL vessel at 960C. At this temperature, Kc
0.036. (a) In which direction will the reaction
go? (b) If CH4 5.56 M at equilibrium, what
are the concentrations of the other
substances? Plan To find the direction, we
calculate Qc using the calculated concentrations
from the data given, and compare it with Kc. (b)
Based upon (a), we determine the sign of each
component for the reaction table and then use the
given CH4 at equilibrium to determine the
others. Solution
44
Predicting Reaction Direction and Calculating
Equilibrium Concentrations II
CS2 H24
4.00 x (8.00)4
Qc
64.0
CH4 H2S2
4.00 x (8.00)2
Comparing Qc and Kc Qc gt Kc (64.0 gt 0.036, so
the reaction goes to the left. Therefore,
reactants increase and products decrease their
concentrations. (b) Setting up the reaction
table, with x CS2 that reacts, which
equals the CH4 that forms.
Concentration (M) CH4 (g) 2 H2S(g)
CS2(g) 4 H2(g)
Initial 4.00
8.00 4.00 8.00 Change
x 2x
-x -4x Equilibrium
4.00 x 8.00 2x 4.00
- x 8.00 - 4x
Solving for x at equilibrium CH4 5.56 M
4.00 M x
x 1.56 M
45
Predicting Reaction Direction and Calculating
Equilibrium Concentrations (p. 735)III
x 1.56 M CH4
Therefore
H2S 8.00 M 2x 8.00 M 2(1.56 M) 11.12 M
CS2 4.00 M - x 4.00 M - 1.56 M 2.44 M
H2 8.00 M - 4x 8.00 M - 4(1.56 M) 1.76 M
CH4 1.56 M
46
Using the Quadratic Formula to Solve for the
Unknown
Given the Reaction between CO and H2O
Concentration (M) CO(g) H2O(g)
CO2(g) H2(g)
Initial 2.00
1.00 0
0 Change -x
-x x
x Equilibrium 2.00-x
1.00-x x x
We rearrange the equation 0.56x2 - 4.68x
3.12 0
ax2 bx c 0
CO 1.27 M H2O 0.27 M CO2 0.73 M
H2 0.73 M