CHAPTER 16 Chemical Equilibria

1
CHAPTER 16 Chemical Equilibria
Chemistry Chemical Reactivity Kotz Treichel
5th edition
Review of concepts Connection between reaction
rates and equilibria Changes in concentration as
a reaction proceeds and the final concentrations
achieved upon reaching equilibria Reaction
Quotient (Q) and Equilibrium Constant
(K) Equilibria involving pure solids, liquids and
solvents Expressions for gases and the
relationship between Kp and Kc Meaning of K
(product-favored/reactant-favored)
2
Concepts (2)
Meaning of Q and Example 16.2 Determining
equilibrium constants (two approaches) Using
equilibrium constants in calculations Example
16.5 Use of the quadratic equation Approximations
involving a small value of K and a large initial
concentration of reactant Value of K when the
reaction is reversed Value of K when the equation
is balanced using different coefficients Value
of K for an overall reaction
3
Concepts (3)
le Chatelier's principle concentration/pressure
effects le Chatelier's principle temperature
effects Catalysts and equilibria Review of the
Haber Process
4
Connection between reaction rates and equilibria
Forward Rxn A ? B rate kfA
Reverse Rxn B ? A rate krB
Where kf is the rate constant for the forward
reaction and kr is the rate constant for the
reverse reaction.
5
Connection between reaction rates and equilibria
(2)
At equilibrium, the two rates are equal and so
the following relationship can be derived.
If kf gt kr, B gt A at equilibrium If kr gt kf,
A gt B at equilibrium
6
Connection between reaction rates and equilibria
(3)
Handout on dynamic equilibrium and water dipping
analogy. Graphical representations for changes in
concentration as a function of time and changes
in reaction rates as a function of time.
7
Equilibrium Constant/Reaction Quotient
H2(g) I2(g) ? 2HI(g)
Figure 16.2 Changes in concentration as a
function of time. Note the same result can be
obtained if we start with an appropriate amount
of HI (it will decompose to form H2 and I2).
8
Equilibrium Constant/Reaction Quotient (2)
At equilibrium, the concentrations do not change
and the following expression yields a constant.
The constant is called the equilibrium constant
and the expression is referred to as the
equilibrium constant expression. The equilibrium
constant depends upon the temperature and is
unitless.
9
Equilibrium Constant/Reaction Quotient (3)
The equilibrium constant can be expressed using
molarities (Kc) or, for gases, pressures (Kp).
10
Equilibrium Constant/Reaction Quotient (4)
A more general approach involves the same type of
expression but refers to the resulting ratio as
the reaction quotient (Q). For Q, the
concentrations may or may not be equilibrium
values.
aA bB ? cC dD
reaction quotient expressions
When Q K, the system is at equilibrium.
11
The Effect of Initial Concentrations on the
Magnitude of K
N2O4(g) ? 2NO2(g)
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
12
Experiment 25
Fe3(aq) SCN-(aq) ? FeNCS2(aq)
Source
13
Reactions Involving Solids, Liquids and Solvents
If pure solids and pure liquids appear in the
reaction, those chemicals do not appear in the
equilibrium constant expression. The
concentration of such chemicals is constant and
is determined by the molar mass and density.
S(s) O2(g) ? SO2(g)
Similarly, if a chemical is the solvent,
typically its concentration does not change much
during the reaction and it is considered to be
constant.
NH3aq) H2O(l) ? NH4(aq) OH-(aq)
14
Relationship Between Kc and Kp (Closer Look pg
661)
aA bB ? cC dD
(PC)c (PD)d
Cc Dd
Kp --------------
Kc --------------
(PA)a (PB)b
Aa Bb
Kp Kc (RT)?n
?n change in moles (product side - reactant
side)
15
Meaning of Equilibrium Constants
CO(g) Cl2(g) ? COCl2(g)
Reaction is product favored. Equilibrium lies
far to the Right.
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
16
Meaning of Equilibrium Constants
See figure on page 663.
17
Meaning of Reaction Quotients
Butane ? Isobutane
Isobutane KButane Figure 16.4 slope of
line equals K can see if system is at
equilibrium based on the ratios of the
concentrations.
18
Meaning of Reaction Quotients (2)
General conditions Q lt K the system is not at
equilibrium and some of the reactants will be
converted to products. Q gt K the system is not
at equilibrium and some of the procudts will be
converted to reactants. Q K the system is at
equilibrium
Example 16.2 NO2 to N2O4 K 171 at 298 K. If
NO2 is 0.015 M and N2O4 is 0.025 M, is system
at equilibrium and, if not, in what direction
does it proceed to achieve equilibrium?
19
16.3 Determining Equilibrium Constants
Two approaches the first is the easiest and
requires that you have all of the equilibrium
concentrations or pressures.
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
20
Determining Equilibrium Constants
2SO2(g) O2(g) ? 2SO3(g)
-0.925 M
0.925 M
-0.463 M
0.075 M
0. 537 M
21
Example 16.5
1.00 mol of H2 and I2 are placed in a 0.500-L
container and allowed to react. Given that the
value of Kc is 55.64, calculate the equilibrium
concentrations of H2, I2 and HI.
-X M
-X M
2X M
22
Example 16.5 (2)
Since there is one equation and one unknown, you
can solve for X. Multiplying the terms and
rearranging would yield a form that looks like
this ax2 bx c 0 (solve using
quadratic equation)
Since the starting quantities of H2 and I2 are
identical, a simplified expression that does not
required the quadratic equation can be used.
23
Example 16.5 (3)
H2 I2 (2.00 - X) M 0.42 M HI (2X) M
3.16 M
24
Using the Quadratic Equation
X 2.73 or 1.58 ( H2 2.00 - X gt 0 )
25
16.5 More About Balanced Equations and K
N2O4(g) ? 2NO2(g) Kp 6.46
x 2 2N2O4(g) ? 4NO2(g) Kp ?
This equals Kp for the original equation squared.
( 6.462 41.7 at 100?C )
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
26
16.5 More About Balanced Equations and K (2)
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
27
16.5 More About Balanced Equations and K (3)
2NOBr(g) Cl2(g) ? 2NO(g) 2BrCl(g)
Source Brown, Lemay, Bursten, Chemistry The
Central Science 8th ed.
28
16.5 More About Balanced Equations and K (4)
1) AgCl(s) ? Ag(aq)
Cl-(aq)
2) Ag(aq) 2NH3(aq) ? Ag(NH3)2(aq)
AgCl(s) 2NH3(aq) ? Ag(NH3)2(aq)
Cl-(aq)
29
16.5 More About Balanced Equations and K (5)
Knet K1K2
30
16.5 More About Balanced Equations and K
(summary)
1. The equilibrium constant of a reaction in the
reverse direction is the inverse of the
equilibrium constant of the reaction in the
forward direction. (discussed before)
2. 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.
3. The equilibrium constant of a net reaction
made up of two or more steps is the product of
the equilibrium constants for the individual
steps.
31
Le Chatelier's Principle
If a system at equilibrium is disturbed by a
change in temperature, pressure, or the
concentration of one of the components, the
system will shift its equilibrium position so as
to counteract the effect of the disturbance.
The expressions "Shift Right" or "Shift Left" are
sometimes used to indicate how the reaction must
proceed in order for equilibrium to be restored.
This relates to the earlier exercises we did
relating Q and K.
32
Le Chatelier's Principle Concentration Changes
N2(g) 3H2(g) ? 2NH3(g)
If reactant (N2 or H2) is added or product (NH3)
is removed, the value of Q is altered and Q lt
Keq If product (NH3) is added or reactant (N2 or
H2) is removed, Q is altered and Q gt Keq
See Figures
33
The meaning of the terms shift left and shift
right
A Typical Chemical Reaction
A(aq) ? B(aq) reactants on the left
? products on the right
When equilibrium is established, the rate of the
forward reaction equals the rate of the reverse
reaction kfA
krB
The value of Keq and the final equilibrium
concentrations of A and B can vary depending upon
the reaction.
34
shift left and shift right (2)
Regardless of the values, the rate of conversion
of A to B is the same as the rate of conversion
of B to A. To the outside observer, there is no
net change in the concentration of either
chemical
You can think of it like a balanced see-saw.
B right side chemical
A left side chemical
35
shift left and shift right (2)
When A is added to the equilibrium mixture, the
balance is thrown off. An adjustment is made to
restore equilibrium.
36
shift left and shift right (3)
In summary When the equilibrium shifts to the
right, it means that the rate of conversion of
chemicals on the left- hand side of the equation
(reactants) to chemicals on the right-hand side
of the equation (products) is higher than the
reverse process.
When there is a shift to the right, chemicals on
the left-hand side of the equation decrease and
chemicals on the right-hand side of the equation
increase.
Shifting the equilibrium to the left implies the
opposite response.
37
Le Chatelier's Principle Volume/Pressure Changes
At constant temperature, reducing the volume of a
gaseous equilibrium mixture causes system to
shift in the direction that reduces the number of
moles of gas.
N2O4(g) ? 2NO2(g)
For gas PVnRT PMRT
?V, ?P (e.g. going from 2 L to 1 L reduces V by
half and doubles P) ?V, ?P (opposite effect)
38
Le Chatelier's Principle Volume Changes
The partial pressures given above represent
equilibrium values at 472?C. If volume is cut in
half, the pressures/concentrations of all
chemicals double.
39
Another Way to Look at Volume Changes
Consider the generic reaction A(g) ?
2B(g)
You can examine the effect of a concentration
change by using a factoring method to see how the
concentration change alters the value of Q
relative to K. If X is the concentration of
chemical A at equilibrium and Y is the
concentration of chemical B at equilibrium, then
the following expression can be derived
40
Another Way to Look at Volume Changes (2)
If the volume is decreased by a factor of 2 and
both concentrations are simultaneously increased
by a factor of 2, the following scenario results
Since Q gt K, there is too much product and not
enough reactant, so the equilibrium will shift to
the left thus decreasing the total number of
moles of chemical in the mixture.
41
Another Way to Look at Volume Changes (3)
Conversely, if the mixture is expanded so that
the volume is increased by a factor of 2 and both
concentrations are simultaneously decreased by a
factor of 2, the following scenario results
Since Q lt K, there is too much reactant and not
enough product, so the equilibrium will shift to
the right thus increasing the total number of
moles of chemical in the mixture.
42
Another Way to Look at Volume Changes (4)
Another way to look at this phenomenon is to
realize that when you alter the pressures or
concentrations of chemicals involved in the
equilibrium, the higher exponents amplify the
effect. If you raise the pressures, the side of
the fraction associated with the higher exponents
gets disproportionately larger. If you lower the
pressures, the side of the fraction associated
with the higher exponents gets disproportionately
smaller. If the exponents are the same, the two
sides of the fraction are affected the same and
no shift in the equilibrium is observed.
43
Another Way to Look at Volume Changes (5)
(original concentrations)
increase concentrations
decrease concentrations
larger exponent
Q
44
Experiment 25 Volume Changes
Fe3(aq) SCN-(aq) ? FeNCS2(aq)
How would the equilibrium shift if water was
added to a solution containing all of those
chemicals (e.g. increased volume from 1 L to 2
L)?
Expect a shift to the Left
45
Le Chatelier's Principle Temperature Changes
Unlike other changes, changing temperature
changes the value of the equilibrium constant (Kc)
Co(H2O)62(aq) 4Cl-(aq) ? CoCl42-(aq) 6H2O(l)
pink
blue
?H gt 0
Endothermic Reactants heat ?
Products Exothermic Reactants ? Products
heat
46
Example See Study Questions 32 and 33 for similar
Consider following equilibrium
N2O4(g) ? 2NO2(g) ?H 58.0 kJ
How will the following changes affect the
equilibrium? (shift right, shift left or no
change)
a) add N2O4
Right
Right
b) remove NO2
c) increase total pressure by adding N2
No Change
d) increase the volume
Right
e) decrease temperature
Left
47
Catalysts
Catalysts speed up the rate of chemical reactions.
A catalyst increases the rate at which
equilibrium is achieved but does not change the
composition of the equilibrium mixture (i.e.
value of K is not changed)
A catalyst decreases the Ea for both the forward
and reverse reactions. The result is that the
equilibrium constant is not changed
Summary of the Haber Reaction N2(g) 3H2(g)
? 2NH3(g) ?H -92 kJ Figure 16.8