Download Chapter 17 lecture notes on Chemical Equilibria

17 Chemical Equilibria
Consider the following reaction:
aA + bB ---> cC + dD
As written is suggests that reactants A + B will be used up in
forming products C + D. However, what we learned in the section
on thermodynamics is that a reaction might not occur to completion
to leave only C + D. Instead it can form any proportion of A and B
to C and D, including remaining primarily as the products.
17-1
Consider a system in which every compound is present:
A better way of considering the reaction is to recognize that from a
starting point with A and B placed in a closed thermodynamic
system, a reaction proceeds to the point in time at which four
compounds, A, B, C and D come to exist simultaneously. This is
shown in the next graph.
17-2
17-3
Extracting kinetics from the plot:
This plot provides us with enormous information about both the
kinetics and the thermodynamics of the reaction. From a kinetic
perspectives we know that the reaction proceeds at a rate determined
by
- 1 d[A]
rate =   =
a dt
1
d[C]
 
c
dt
It changes rapidly at first and then levels off later in time.
17-4
Kinetics when nothing seems to be happening:
What is of greater interest in this chapter is the rate of reaction at the
point where there appears to be no change in concentration of
reactants and products. In fact, at that point, we can consider two
reactions to be occurring, a forward reaction
aA + bB ----> c C + dD
and a reverse reaction
cC + dD ----> aA + bB
17-5
Deriving the Equilibrium Constant, Keq, from Rate Constant, k.
To simplify things, consider reactions following simple one step
bimolecular mechanisms. Then we have rate laws of the following
form:
Rate = kf[A][B]
forward reaction
Rate = kr[C][D]
reverse reaction
and
17-6
A little bit of math and presto: Keq
Obviously from the plot, at the point where all of the curves flatten
out, the rates at which A, B, C and D are formed are the same. Thus
we can equate the two rate laws
kf[A][B] = kr [C][D]
and rearranging
Kc =
kf
__
kr
=
17-7
[C][D]
______
[A][B]
The constant we will use the rest of the semester.
The ratio, Kc, is a very important value know as the
constant.
equilibrium
How important is it? Coming to understand and apply
this constant will be the primary focus of THE REST OF THE
COURSE. We will come to learn that though Kc is derived from the
kinetics of the reaction, it is an important thermodynamic constant
which is directly related to the most important of thermodynamic
state functions, the Gibbs free energy, ∆Go.
17-8
Generalizing the equilibrium constant, Kc.
It can be shown that no matter what the mechanistic pathway of a
reaction, the equilibrium constant, Kc, is defined as
aA + bB ⇔ cC + dD
Kc
[C]c[D]d
= 
[A]a[B]b
for any reactants and products AT EQUILIBRIUM.
17-9
Example.
reaction:
What is the equilibrium expression for the following
H2(g) + I2 (g) ⇔ 2HI (g)
[HI]2
Kc = 
[H2][I2]
17-10
A new way to describe all chemical reactions,
the equilibrium expression:
Note that every reaction ultimately arrives at an equilibrium; i.e., the
point at which the forward and reverse reactions have equal rates so
that there is no change in the overall concentration of species present
in the system. Thus every reaction has an equilibrium constant, Kc.
A few hundred equilibrium constants for the work we will do in
Chapters 18-20 are found in Appendixes F, H and I of Davis.
17-11
The magnitude of Keq values
Equilibrium constants can vary enormously in magnitude. Very large
and very small numbers are possible:
Cu++ + S=
HPO4-2
⇔
CuS
⇔ H+ + PO4-3
Kc = 1.1 x 1035
Kc = 3.5 x 10-13
Note that we can rewrite an equilibrium expression as the reverse
reaction and the new Kc’ is calculated simply as the reciprocal of Kc.
17-12
Example. The equilibrium expression for the formation of CuS is
written below:
[CuS]
++
=
35
Cu + S ⇔
CuS(s)
Kf = 1.1 x 10 = 
[Cu++] [S=]
From this information, determine the equilibrium constant for the
dissociation of CuS. The dissociation of CuS is written as
[Cu++] [S=]
CuS(s) ⇔ Cu++ + S= Kd =  = 1/Kf = 9 x 10-36
[CuS]
17-13
A few points to make about Kc:
1. Temperature dependence. The value is obtained for a specific
temperature. This is not surprising since K is a thermodynamic
constant and all thermodynamic constants are strongly dependent
upon T. Once again, a standard temperature has to be selected for
recording values in the Appendix. The typical temperature used is
25oC. Later we will learn to use the van’t Hoff equation to convert
between Kc values at different temperatures.
17-14
2. In performing equilibrium calculations:
i. Pure solids and pure liquids are assigned an activity of 1
ii. For ideal solutions, the activity of samples dissolved in
solution is given in molarity
iii. For ideal gases, the activity of samples is given as partial
pressures in atm.
17-15
3. Equilibrium constants are dimensionless
17-16
.
4. Looking at K to determine reaction spontaneity.
If the value for K>1, the formation of products is favored over
reactants. If the value for K<1, the formation of reactants over
products is favored.
17-17
Example. Calculation of equilibrium constant.
One liter of an equilibrium mixture is found to contain 0.172 moles
of PCl 3, 0.086 moles of chlorine and 0.028 moles of PCl5. Calculate
Kc for the reaction.
PCl5 ⇔ PCl3 + Cl2
[PCl3][Cl2]
(0.172mole/1l)(0.086mole/1l)
Kc =  =  = 0.53
[PCl5]
(0.028mole/1l)
17-18
Example. Calculation of equilibrium constant.
The decomposition of PCl5 at a different temperature involves
introduction of one mole of PCl 5 to an evacuated 1 liter container. At
equilibrium, 0.6 moles of PCl3 is present in the container. Calculate
Kc .
PCl5
⇔
PCl3
+
Cl2
initial
1M
0M
0M
change
-0.6M
+.6M
+.6M
equilibrium
0.4M
0.6M
0.6M
(0.6M)(0.6M)
Kc =  = 0.9
(0.4M)
17-19
Example. Calculation of equilibrium constant.
In an evacuated 1 liter container, 0.8 moles of N2 and 0.9 moles of H2
are placed. At equilibrium, 0.2 mole of NH3 is present. Calculate Kc.
N2
+
3H2
⇔
2NH3
initial
0.8M
0.9M
0
change
-0.1M
-0.3M
+0.2M
equilibrium
0.7M
0.6M
0.2M
(0.2M)2
Kc =  = 0.26
(0.7M)(0.6M)3
17-20
Relating the Reaction Quotient, Q, to Kc.
The calculation of Kc is accomplished by ratioing the equilibrium
concentration of products to reactant. However, nothing stops us
from making a calculation of the ratio of products to reactants at any
non-equilibrium position along the reaction path. The ratio of
products to reactants is generally called the mass action equation, Q,
or in Davis, the reaction quotient.
Q
[C]c[D]d
= 
[A]a[B]b
Recall that at equilibrium, Q ------> K
17-21
Some rules for relating Q to K
We can calculate the value of Q anywhere along the course of the
reaction. Note that in the reaction of A and B forming C and D, the
value of Q starts off very small (reactants in denominator) compared
to the final K value. Q increases until the point that the forward and
reverse reaction rates are equal, and the system is at equilibrium; then
Kc can be calculated.
We can establish the following rules for describing where the
reaction is relative to equilibrium by comparing the value of Q to K.
Q<K
Q = K
Q >K
forward reaction dominates (example above)
system at equilibrium
reverse reaction dominates
17-22
17-23
Example. Calculating Q.
The equilibrium constant for the following reaction, Kc, is 49 at
450oC. If 0.22 mole of H2, 0.22 mole of I2 and 0.666 mole of HI
were put into an evacuated container, would the system be at
equilibrium? If not, in what direction must the reaction proceed to
reach equilibrium?
H2 + I2 ⇔ 2HI (assume one liter)
[HI]2
(0.666M)2
Q =  =  = 9.2
[H2][I2]
(0.22M)(0.22M)
Q<K ∴ rxn proceeds to right to form more [HI]
17-24
Example. The equilibrium constant, Kc, is 71.00 for the reaction
below. If 0.6 mole of SO2 and 0.2 mole NO2 are placed into an
evacuated 2.0 liter container and allowed to reach equilibrium, what
will be the concentration of each compound at equilibrium?
⇔ SO3 + NO
SO2 + NO2
Initial
0.3M
0.1M
0M
0M
Change
-xM
-xM
+xM
+xM
xM
xM
Equilibrium
(0.3-x)M
(0.1-x)M
x2
x2
Kc = 71 =  = 
(0.3-x)(0.1-x)
0.03-.4x+x2
0 = 70x2-28.4x+2.13
solve quadratic to get
x = .0993
[NO2]=0.0007M
[SO2] = 0.2M
[NO] = [SO3] = 0.1M
17-25
Example. The equilibrium constant is 49 for the reaction below. If
1.00 mole of HI is placed into a 1.0 liter container and allowed to
reach equilibrium, what will be the equilibrium concentration of each
compound?
H2 +
I2
⇔ 2HI
Initial
0M
0M
1M
Change
+xM
+xM
-2xM
Equilibrium
xM
xM
(1-2x)M
(1-2x)2
Kc =  = 49
x2
use quadratic equation [H2] = 0.11M
x = 0.11
[I2] = 0.11M
[HI] = 0.78M
17-26
Stress and LeChatlier’s Principle:
Stress is a big deal, in real life and in chemical equilibria. What do
we do in life when things change? We do what we can to relieve the
stress by moving away from it. We apply the same notion to
chemical equilibria. In a system at equilibrium, when a change
occurs a stress is imposed on the system. The system then responds
by adopting new equilibrium conditions that relieve the stress.
This is the definition of LeChatlier’s Principle: If a change in
conditions occurs to a system at equilibrium, the system responds to
relieve the stress and reach a new state of equilibrium.
What are some of the ways stress is introduced to a system?
1. Changes in concentration of system components.
2. Change in system temperature.
3. Change in volume or pressure (in gaseous systems.)
17-27
Case 1. LeChatlier’s Principle: Changes in Concentration
Consider the general reaction:
A + B ⇔ C + D
Suppose we add additional reactant A or B to the system. This means
we have more reactant than is necessary to establish equilibrium and
the system shifts to reduce the amount of A and B. This results in an
increase in the amount of C and D. The counter to this is also true. If
the system changes by adding C and D to the system, the equilibrium
shifts to the left forming more A and B.
Looking at the equilibrium constant equation gives us a mathematical
justification for the direction of the change.
[C] [D]
___________
Kc =
[A] [B]
Note for example that if we increase the amount of A, then to
maintain the equality, the equilibrium must shift so that the amount of
C and D go up as well. What happens to the amount of B? It must
go down.
17-28
Some General Rules for LeChatelier and Concentration Changes.
From a chemical perspective, we have an equilibrium system with
relative concentrations of compounds defined by Kc. We impose a
stress on the system by changing a concentration and create a nonequilibrium Q value for the system. However, because of the nonzero free energy now in the system, Q will in time return to K as ∆G
---> 0.
We can make a little table telling us how equilibrium
adjust to compensate for changes in concentration.
A
+
B
⇔
C
case 1:
Add A
B decreases C increases
case 2: A decreases
Add B
C increases
case 3:
A increases B increases
Add C
case 4:
A increases B increases C decreases
17-29
concentrations
+
D
D increases
D increases
D decreases
Add D
Example:
LeChatlier’s Principle applied to a change in
concentration.
A 2 liter vessel contains an equilibrium mixture of 1.2 mole of
COCl2, 0.6 mole of CO and 0.2 mole of Cl2. First calculate the
equilibrium constant for the reaction:
CO
initial
equilibrium:
0.3M
+
Cl2
⇔
0.1M
[COCl2]
Kc =  = 20
[CO] [Cl2]
17-30
COCl2
0.6M
An additional 0.8 mole of Cl2 is added to the vessel. Calculate the
molar concentration of CO, Cl2 and COCl2 when the new equilibrium
is established.
CO +
Cl2
⇔ COCl2
Original equilibrium
0.3M
0.1M
0.6M
Stress added
0M
+0.4M
0M
New initial conc.
0.3M
0.5M
0.6M
Change
(0.3-x)M
(0.5-x)M
(0.6+x)M
New equilibrium
0.121M
0.321M
(0.6)
1) After stress Q =  = 4
(0.3)(0.5)
Q<K so rxn proceeds forward
(0.6+x)(0.6+x)
2) K = 20 =  = 
(0.3-x)(0.5-x) 0.15-0.8x+x2
solve quadratic, 0 = 20x2 - 17x + 2.4, to get x = 0.179
17-31
0.779M
Case 2. LeChatelier’s Principle: Changes in Temperature.
Consider the following exothermic reaction.
A + B
⇔
C + D + heat
(∆H is -)
This means that heat is being generated by the reaction. If we add
additional heat to the system, the stress of the increased temperature
will make the system respond to try and to cool things down. This
means the system reaction will shift to the left to reduce the heat
generated by the exothermic process.
17-32
We can use similar sorts of reasoning to establish the following rules:
exothermic rxn
exothermic rxn
endothermic rxn
endothermic rxn
temperature
increases
temperature
decreases
temperature
increases
temperature
decreases
17-33
reaction shifts left
reaction shifts right
reaction shifts right
reaction shifts left
Examples: How will an increase in temperature affect each of the
following reactions?
2NO2(g) ⇔ N2O4
+ heat
Exothermic reaction so equilibrium shifts to the left to reduce the
temp. of the system.
H2 (g) + Cl2 (g) ⇔ 2HCl (g)
+92 kJ
Exothermic reaction so equilibrium shifts to the left to reduce the
temp. of the system.
H2 (g) + I2 (g) ⇔ 2HI
∆H = +25kJ
Endothermic reaction so equilibrium shifts to the right to reduce the
temp. of the system.
17-34
Case 3. LeChatlier’s Principle: Changes in Volume and Pressure
(application to reactions involving gases)
First note from the ideal gas law that P and V are inversely related. If
we reduce the volume of a system, this has the affect of increasing
the concentration of a gas in the system. The system will thus
respond by shifting the reaction in the direction to reduce the
concentration of gases in the system. This means the reaction shifts
toward the side that has the fewest moles of gas at equilibrium
Similarly reasoning yields the following:
V decreases (P, [ ] increase)
V increases (P, [ ] decrease)
shift reaction toward side with smaller
number of moles of gas
shift reaction toward side with larger
number of moles of gas
Note that if we assume an ideal gas, then if there are an equal number
of moles of gas on either side of the reaction, then pressure or volume
changes have no affect on the equilibrium.
17-35
Examples.
How will an increase in pressure (concentration) resulting from a
decrease in volume affect the following equilibria?
H2 (g) + I2 (g) ⇔ 2HI (g)
∆n = 2-2 = 0
therefore, no change in equilibrium
4NH3 (g) + 5O2 (g) ⇔ 4NO (g) + 6H2O (g)
∆n = 10-9 = 1,
therefore, shifts to the left to reduce moles of gas.
PCl3 (g) + Cl2 (g) ⇔ PCl5 (g)
∆n = 1-2 = -1,
therefore, shifts to the right to reduce moles of gas
2H2 (g) + O2 (g) ⇔ 2H2O (g)
∆n = -1
therefore, shifts to the right to reduce moles of gas
17-36
Equilibrium Constant, Kp, Expressed for Gases:
Taking a look at the ideal gas law:
PV = nRT
The equation suggests a way to relate partial pressure, P, of a gas in
the system to concentration, [ ] by
[ ] = n/V = P/RT
Substituting this relationship into the general equilibrium equation
yields:
aA (g) + bB(g) ⇔ cC (g) + dD (g)
(PC)c (PD)d
Kp = ____________________
(PA)a (PB)b
The relationship between Kc and Kp follows from the ideal gas law:
Kp = Kc(RT)∆n
∆n =(ngas prod - ngas react)
where
17-37
Example:
For the reaction below with a Kc = 20 at 25oC, what is Kp?
CO
+
Cl2
⇔
COCl2
∆n = 1-2 = -1
Kp = Kc(RT)∆n = 20[(0.082atml/molK)(298K)]-1
= 0.82
17-38
Example: A calculation interchanging Kc and Kp
Kc is 49 for the following reaction. If 1.0 mole H2 and 1 mole I2 are
allowed to reach equilibrium in a 3 liter vessel, how many moles of I2
are unreacted at equilibrium?
H2 (g) + I2 (g) ⇔ 2HI (g)
initial
0.33M
0.33M
change
0.33-xM
0.33-xM
+2xM
equilibrium
0.073M
0.073M
0.514M
[HI]2
4x2
Kc = 49 =  = 
[H2][I2]
0.01089-0.66x+x2
0M
45x2-32.34x+5.34 = 0
solve quadratic
x = 0.257
3(0.257) = 0.219 moles of reacted I2
17-39
Now what are the equilibrium pressures of H2, I2 and HI?
∆n = 0, therefore, Kc = Kp
[HI]2
______
[I2][H2]
PHI2
= ______
PI2 PH2
0.264(PI22) = PHI2(0.0053)
So
moles I2 = moles of H2, therefore,
(2.2(PI2) )2
49 = 
PI22
PI2 = 0.044 = PH2
And
2.2(PI2) = PHI
so
PHI = 0.0968
17-40
Relating Kc to Thermodynamic Quantities through Go:
Recall that in the last chapter we introduced the idea of the free
energy of the system, ∆Go, under standard conditions. ∆Go describes
the free energy change in the system that accompanies the conversion
of all reactants in their standard states (1M, 1 atm) to products in their
standard states (1M, 1 atm).
We now define the free energy, ∆G, which is the free energy of a
reaction at concentrations other than 1M or 1 atm. It is found from
∆Go by correcting for non-standard conditions using the reaction
quotient, Q:
∆G = ∆Go + RT ln Q
17-41
The special case of thermodynamics at equilibrium.
Remember that at equilibrium, ∆G = 0 and Q = K. This means the
above equation reduces to:
∆Go = -RT ln K
(at equilibrium)
Among other things this relationship tells us that we can calculate the
equilibrium constant, Kc, from the thermodynamic constants in
Appendix K.
17-42
Examples: Converting between Kc and Go
Case 1.
∆G of =
2C graphite + 3H2 (g) + O2 (g) D C2H5OH
1
2
-168.6
Kc = exp (
Kc = exp
∆G
kKJ
o
RT
mol
)
−168.6x103
(
)
(−8.3)(298K )
Kc = 3.5 x 1029
So a spontaneous reaction has a K > 1
17-43
Case 2.
∆G of
N2(g) D 2N(g)
= 456
kKJ
mol
Kc = exp ( −456x103 )
(8.3)(298)
Kc = 8.6 x 10 –81
So a non-spontaneous reaction has K < 1. Reaction is spontaneous in
opposite direction.
17-44
What generalization can be made about ∆Go and Kc?
If ∆Go < 0 then K>1
If ∆Go > 0 then K< 1
products favored over reactant
reactants favored over products
Isn’t it exciting to see that chemical equilibrium is another model for
describing thermodynamics? In this case we don’t measure the
change in energy, instead we measure the equilibrium concentrations.
17-45
End-of chapter equations and the Famouns Scientists who
invented them:
It is the end of the chapter and it is time to throw a hard equation at
you. Just to let you know that this is a pretty common occurrence,
consider:
end of chapter
equation name
ClausiusClapeyron
Arrhenius
van’t Hoff
if we
then for the
we get a new
know this: following change: physical constant
∆Hvap
∆T
P(vapor pressure)
∆T
∆T
Ea
∆H
17-46
rate constant
equilibrium
constant
The van’t Hoff Equation: Finding a new Kc at a new T:
Of interest in Chapter 17 is the bottom equation: the van’t Hoff
equation that allows us to determine Kc for a chemical system at any
temperature we desire.
KT2
∆Ho
1
1
ln  =  .  - 
KT1
R
T1
T2
Notice that this is the quantitative way to look at LeChatelier’s
principle applied to changes in temperature.
17-47
Example: Using the van’t Hoff equation to find a new K.
For the reaction below,
2NO2 ⇔ 2NO + O2
∆Ho = 114 kJ/mole and Kp = 4.3 x 10-13 at 25oC.
So what is Kp at 250oC?
Insert T2 = 523K, R = 8.312 J/mol K. ∆Ho = 114 kJ/mole and Kp =
4.3 x 10-13 at 25oC.
KT2 = KT1 [exp ( ∆H ( 1 − 1 ))]
R
T1
T2
exp (1n ( KT )) = exp [
∆H
R
2
KT1
-13
KT2 = (4.3 x 10 ) exp
So at 250oC,
(1 T1
114x103
(
8.314
(
1
)
T2
]
1
1
−
))
298 523
KT2 = 1.7 x 10-4
Is this consistent with what you know from LeChatelier?
17-48