A more appropriate definition of K (for the same
chemical reaction discussed previously) is
K
c

























[C]/M [D]/M




a





[A]/M [B]/M




d
b
With this definition, each concentration term is
divided by the units of molarity, M, so K is then
dimensionless.
481
A more appropriate definition of K (for the same
chemical reaction discussed previously) is
K
c

























[C]/M [D]/M




a





[A]/M [B]/M




d
b
With this definition, each concentration term is
divided by the units of molarity, M, so K is then
dimensionless. Because it is tedious to write the
above equation, most people simply write
c
d
K  [C]a[D]b
[A] [B]
to save writing time.
482
Relationship between Q and K
N2O4(g)
2 NO2(g)
483
484
Different ways of expressing the
equilibrium constant
485
Different ways of expressing the
equilibrium constant
Homogeneous equilibrium: A state of equilibrium
between reactants and products in the same phase.
486
Different ways of expressing the
equilibrium constant
Homogeneous equilibrium: A state of equilibrium
between reactants and products in the same phase.
+
Example: CH3CO2H(aq) 
CH
CO
+
H
3
2 (aq)
(aq)

487
Different ways of expressing the
equilibrium constant
Homogeneous equilibrium: A state of equilibrium
between reactants and products in the same phase.
+
Example: CH3CO2H(aq) 
CH
CO
+
H
3
2 (aq)
(aq)


[CH3CO2][H ]
Kc 
[CH3CO2H]
488
Example: For the gas phase reaction
N2O4(g) 
 2 NO2(g)
[NO2]2
Kc 
[N2O4 ]
489
Example: For the gas phase reaction
N2O4(g) 
 2 NO2(g)
[NO2]2
Kc 
[N2O4 ]
When working with gaseous reactions, it is
frequently more convenient to use different units.
490
For the preceding reaction
Kp 
2
pNO
2
pN O
2 4
491
For the preceding reaction
Kp 
2
pNO
2
pN O
2 4
The subscript p on K indicates partial pressures are
used in the mass action expression.
492
For the preceding reaction
Kp 
2
pNO
2
pN O
2 4
The subscript p on K indicates partial pressures are
used in the mass action expression.
pN O is the equilibrium partial pressure of N O .
2 4
2 4
pNO is the equilibrium partial pressure of NO .
2
2
493
For the preceding reaction
Kp 
2
pNO
2
pN O
2 4
The subscript p on K indicates partial pressures are
used in the mass action expression.
pN O is the equilibrium partial pressure of N O .
2 4
2 4
pNO is the equilibrium partial pressure of NO .
2
2
In general, Kp and Kc (for the same reaction) are not
equal.
494
Exercise: For the reaction a A(g) 
 b B(g) find a
connection between Kp and Kc. Assume the gases
can be treated as ideal gases.
495
Exercise: For the reaction a A(g) 
 b B(g) find a
connection between Kp and Kc. Assume the gases
can be treated as ideal gases.
pbB
Kp  a
pA
496
Exercise: For the reaction a A(g) 
 b B(g) find a
connection between Kp and Kc. Assume the gases
can be treated as ideal gases.
pbB
Kp  a
pA
Now from the ideal gas equation PV = n RT,
497
Exercise: For the reaction a A(g) 
 b B(g) find a
connection between Kp and Kc. Assume the gases
can be treated as ideal gases.
pbB
Kp  a
pA
Now from the ideal gas equation PV = n RT,
pA 
nART
 [A]RT
VA
pB 
nBRT
 [B]RT
VB
498
Exercise: For the reaction a A(g) 
 b B(g) find a
connection between Kp and Kc. Assume the gases
can be treated as ideal gases.
pbB
Kp  a
pA
Now from the ideal gas equation PV = n RT,
pA 
nART
 [A]RT
VA
pB 
nBRT
 [B]RT
VB
Plug these two results into the expression for Kp.
499
Hence:
Kp 
b
pB
paA
500
Hence:
Kp 
Kp 
b
pB
paA
b




[B]RT












[A]RT
a
501
Hence:
Kp 
Kp 
b
pB
paA
b




[B]RT












[A]RT
a
b RT b
 [B]a
a
[A] RT












502
Hence:
Kp 
Kp 
b
pB
paA
b




[B]RT












[A]RT
a
b RT b
 [B]a
a
[A] RT












b
b-a
[B]
 a RT
[A]








503
Hence:
Kp 
Kp 
b
pB
paA
b




[B]RT












[A]RT
a
b RT b
 [B]a
a
[A] RT












b
b-a
[B]
 a RT
[A]








b
n
[B]
where
 a RT
[A]








Δn  b  a
504
Hence:
Kp 
Kp 
b
pB
paA
b




[B]RT












[A]RT
a
b RT b
 [B]a
a
[A] RT












b
b-a
[B]
 a RT
[A]








b
n
[B]
where
 a RT
[A]




and therefore
Kp  Kc RT












Δn  b  a
n
505
Sample Problems
506
Sample Problems
Example: 2 NO(g) + O2(g) 
 2 NO2(g)
At a temperature of 230 oC the concentrations of
the various species are [NO] = 0.0542 M, [O2] =
0.127 M, and [NO2] = 15.5 M at equilibrium.
Calculate the equilibrium constant Kc at 230 oC.
507
Sample Problems
Example: 2 NO(g) + O2(g) 
 2 NO2(g)
At a temperature of 230 oC the concentrations of
the various species are [NO] = 0.0542 M, [O2] =
0.127 M, and [NO2] = 15.5 M at equilibrium.
Calculate the equilibrium constant Kc at 230 oC.
[NO2]2
Kc 
[NO]2[O2]
508
Sample Problems
Example: 2 NO(g) + O2(g) 
 2 NO2(g)
At a temperature of 230 oC the concentrations of
the various species are [NO] = 0.0542 M, [O2] =
0.127 M, and [NO2] = 15.5 M at equilibrium.
Calculate the equilibrium constant Kc at 230 oC.
[NO2]2
Kc 
[NO]2[O2]
2
(15.5)

(0.0542)2(0.127)
= 6.44 x 105
509
Example: The equilibrium constant Kp for the
oC.
reaction PCl5 (g) 
PCl
+
Cl
is
1.05
at
250
3(g)
2(g)

If the equilibrium partial pressures of PCl5 and PCl3
are 0.875 atm and 0.463 atm, respectively, what is
the equilibrium partial pressure of Cl2?
510
Example: The equilibrium constant Kp for the
oC.
reaction PCl5 (g) 
PCl
+
Cl
is
1.05
at
250
3(g)
2(g)

If the equilibrium partial pressures of PCl5 and PCl3
are 0.875 atm and 0.463 atm, respectively, what is
the equilibrium partial pressure of Cl2?
Kp 
pPCl pCl
3
pPCl
2
5
511
Example: The equilibrium constant Kp for the
oC.
reaction PCl5 (g) 
PCl
+
Cl
is
1.05
at
250
3(g)
2(g)

If the equilibrium partial pressures of PCl5 and PCl3
are 0.875 atm and 0.463 atm, respectively, what is
the equilibrium partial pressure of Cl2?
Kp 
pPCl pCl
3
pPCl
This is shorthand for
2
5
Kp 















pPCl / atm pCl / atm
3





2





pPCl / atm
5
512
Example: The equilibrium constant Kp for the
oC.
reaction PCl5 (g) 
PCl
+
Cl
is
1.05
at
250
3(g)
2(g)

If the equilibrium partial pressures of PCl5 and PCl3
are 0.875 atm and 0.463 atm, respectively, what is
the equilibrium partial pressure of Cl2?
Kp 
pPCl pCl
3
pPCl
2
5
Kp 
This is shorthand for
so that
pCl / atm 





2

























pPCl / atm pCl / atm
3





2





pPCl / atm
5





pPCl / atm Kp
5










pPCl / atm
3
513
pCl / atm 





and so
2













0.875 atm/ atm (1.05)








0.463 atm/ atm
pCl  1.98 atm
2
514
Heterogeneous Equilibria
515
Heterogeneous Equilibria
A reaction often involves reactants that are not
present in the same phase – this leads to a
heterogeneous equilibrium.
516
Heterogeneous Equilibria
A reaction often involves reactants that are not
present in the same phase – this leads to a
heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
517
Heterogeneous Equilibria
A reaction often involves reactants that are not
present in the same phase – this leads to a
heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
CaCO3(s) 
 CaO(s) + CO2(g)
518
Heterogeneous Equilibria
A reaction often involves reactants that are not
present in the same phase – this leads to a
heterogeneous equilibrium.
Example: Heating CaCO3 in a closed vessel.
CaCO3(s) 
 CaO(s) + CO2(g)
Kc 
[CaO][CO2]
[CaCO 3]
519
The “concentration” of any pure solid is the ratio of
the total number of moles present in the solid,
divided by the volume of the solid.
520
The “concentration” of any pure solid is the ratio of
the total number of moles present in the solid,
divided by the volume of the solid. If part of the
solid is removed, the number of moles of solid will
decrease – but so will its volume.
521
The “concentration” of any pure solid is the ratio of
the total number of moles present in the solid,
divided by the volume of the solid. If part of the
solid is removed, the number of moles of solid will
decrease – but so will its volume. The ratio of moles
to volume remains unchanged.
522
Consider the ratio:
grams
density  liters
molar mass grams
moles
523
Consider the ratio:
grams
density  liters
molar mass grams
moles
moles

liters
524
Consider the ratio:
grams
density  liters
molar mass grams
moles
moles

liters
density CaCO 3
[CaCO3] 
molar mass CaCO 3
525
Consider the ratio:
grams
density  liters
molar mass grams
moles
moles

liters
density CaCO 3
[CaCO3] 
molar mass CaCO 3
If the temperature is held fixed, then [CaCO3] is a
constant. Similarly for [CaO], which is also a
constant.
526
So we can rewrite
[CaO][CO2]
Kc 
[CaCO 3]
527
So we can rewrite
[CaO][CO2]
Kc 
[CaCO 3]
in the form
Kc
[CaCO3]
 [CO2]
[CaO]
528
So we can rewrite
[CaO][CO2]
Kc 
[CaCO 3]
in the form
Kc
[CaCO3]
 [CO2]
[CaO]
Now since [CaCO3] and [CaO] are both constant, the
left-hand side of the preceding equation is
constant.
529
So we can rewrite
[CaO][CO2]
Kc 
[CaCO 3]
in the form
Kc
[CaCO3]
 [CO2]
[CaO]
Now since [CaCO3] and [CaO] are both constant, the
left-hand side of the preceding equation is
constant. Now set
[CaCO3]
Kc  Kc
[CaO]
530
Hence
Kc = [CO2]
531
Hence
Kc = [CO2]
Notice that terms involving pure solids do not
appear in the final equilibrium constant
expression.
532
Hence
Kc = [CO2]
Notice that terms involving pure solids do not
appear in the final equilibrium constant
expression. This result generalizes to all chemical
reactions.
533
Hence
Kc = [CO2]
Notice that terms involving pure solids do not
appear in the final equilibrium constant
expression. This result generalizes to all chemical
reactions.
The corresponding expression for Kp for the
decomposition of CaCO3 is:
Kp = pCO2
534
The same concentration of CO2 exists in both containers
(provided the temperature is the same), even though the
amounts of CaO and CaCO3 are different.
535
Summary Comment
Concentration factors for pure solids and pure
liquids are ignored in the expression for the
equilibrium constant for a chemical reaction.
536
Multiple Equilibria
537
Multiple Equilibria
Consider the two reactions:
538
Multiple Equilibria
Consider the two reactions:

A + B C + D
C + D
 E + F
539
Multiple Equilibria
Consider the two reactions:

A + B C + D
C + D
 E + F
For the first reaction Kc  [C][D]
[A][B]
540