Multiple Equilibria
Consider the two reactions:

A + B C + D
C + D
 E + F
For the first reaction Kc  [C][D]
[A][B]
For the second reaction Kc  [E][F]
[C][D]
541
The overall reaction is given by the sum of the two
reactions:

A + B  E + F
542
The overall reaction is given by the sum of the two
reactions:

A + B  E + F
The corresponding equilibrium constant is
Kc  [E][F]  Kc Kc
[A][B]
543
The overall reaction is given by the sum of the two
reactions:

A + B  E + F
The corresponding equilibrium constant is
Kc  [E][F]  Kc Kc
[A][B]
In general: If a reaction can be expressed as the
sum of two or more reactions, the equilibrium
constant for the overall reaction is given by the
product of the equilibrium constants of the
individual reactions.
544
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
545
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
[H ][HCO 3- ]
-7
Kc 
=
4.2
x
10
[H2CO3]
546
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
[H ][HCO 3- ]
-7
Kc 
=
4.2
x
10
[H2CO3]
-  H+
HCO3(aq)
 (aq) + CO23(aq)
547
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
[H ][HCO 3- ]
-7
Kc 
=
4.2
x
10
[H2CO3]
-  H+
HCO3(aq)
 (aq) + CO23(aq)
[H ][CO23- ]
-11
Kc 
=
4.8
x
10
[HCO3- ]
548
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
[H ][HCO 3- ]
-7
Kc 
=
4.2
x
10
[H2CO3]
-  H+
HCO3(aq)
 (aq) + CO23(aq)
[H ][CO23- ]
-11
Kc 
=
4.8
x
10
[HCO3- ]
For the overall reaction
+
2H2CO3(aq) 
2
H

(aq) + CO3(aq)
549
Example:
+
HCO
H2CO3(aq) 
H
+
(aq)

3(aq)
[H ][HCO 3- ]
-7
Kc 
=
4.2
x
10
[H2CO3]
-  H+
HCO3(aq)
 (aq) + CO23(aq)
[H ][CO23- ]
-11
Kc 
=
4.8
x
10
[HCO3- ]
For the overall reaction
+
2H2CO3(aq) 
2
H

(aq) + CO3(aq)
[H ]2[CO23 - ]
-7)(4.8 x 10-11) = 2.0 x 10-17
Kc 
=
(4.2
x
10
[H2CO3]
550
What does the equilibrium constant
tell us?
551
What does the equilibrium constant
tell us?
Knowing the equilibrium constant enables us to
predict the direction a reaction will take to reach
equilibrium – if we are given the initial
concentrations of the reacting species. We can also
determine the extent of the reaction.
552
What does the equilibrium constant
tell us?
Knowing the equilibrium constant enables us to
predict the direction a reaction will take to reach
equilibrium – if we are given the initial
concentrations of the reacting species. We can also
determine the extent of the reaction.
The larger the value of K, the further the reaction
will proceed towards completion when equilibrium
is reached.
553
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
554
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
Kc = 9 x 1080 at 25 oC
555
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
Kc = 9 x 1080 at 25 oC
The very large value of Kc indicates that the
reaction goes essentially to completion in the
forward direction.
556
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
Kc = 9 x 1080 at 25 oC
The very large value of Kc indicates that the
reaction goes essentially to completion in the
forward direction.
N2(g) + O2(g) 
 2 NO(g)
557
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
Kc = 9 x 1080 at 25 oC
The very large value of Kc indicates that the
reaction goes essentially to completion in the
forward direction.
N2(g) + O2(g) 
 2 NO(g)
Kc = 4.8 x 10-31 at 25 oC
558
Examples: 2 H2(g) + O2(g) 
 2 H2O(g)
Kc = 9 x 1080 at 25 oC
The very large value of Kc indicates that the
reaction goes essentially to completion in the
forward direction.
N2(g) + O2(g) 
 2 NO(g)
Kc = 4.8 x 10-31 at 25 oC
The very small value of Kc indicates that very little
NO is formed. The reaction proceeds hardly at all
towards completion before equilibrium is reached.
559
Examples: N2(g) + 3 H2(g) 
 2 NH3(g)
560
Examples: N2(g) + 3 H2(g) 
 2 NH3(g)
Kc ~ 1 at 380 oC
561
Examples: N2(g) + 3 H2(g) 
 2 NH3(g)
Kc ~ 1 at 380 oC
When Kc is around 1, concentrations of reactants
and products are approximately the same at
equilibrium.
562
Summary
563
Summary
1. Large K
reaction proceeds
564
Summary
1. Large K
2. Small K
reaction proceeds
reaction proceeds
565
Summary
1. Large K reaction proceeds
2. Small K reaction proceeds
3. K ~ 1
conc. of reactants and products
about the same at equilibrium
566
567
K small
K large
K around 1
568
Sample Calculations
569
Sample Calculations
The following sample calculations are much more
representative of the types of problems that show
up on exams.
570
Sample Calculations
The following sample calculations are much more
representative of the types of problems that show
up on exams.
Example 1: A mixture of 0.500 moles of H2 and
0.500 moles of I2 was placed in a 1.00 liter flask at
430 oC. At this temperature the I2 is in the gas
phase. Calculate the concentrations of H2, I2, and
HI(g) at equilibrium. The equilibrium constant Kc for
oC.
the reaction H2(g) + I2(g) 
2
HI
is
54.3
at
430

(g)
571
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
572
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
H2
I2
HI
573
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
H2
I2
0.500 mol  0.500 M 0.500 mol  0.500 M
1.00 l
1.00 l
HI
0M
574
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
H2
I2
0.500 mol  0.500 M 0.500 mol  0.500 M
1.00 l
1.00 l
-x
-x
HI
0M
2x
575
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
H2
I2
0.500 mol  0.500 M 0.500 mol  0.500 M
1.00 l
1.00 l
-x
-x
HI
0M
2x
(where x is the number of mol/l of H2 or I2 that react)
576
Correct setup is important !!!!!!!!
The following is referred to as an ICE table (I = initial
conditions, C = change in conditions, E = equilibrium
conditions.)
H2
I2
0.500 mol  0.500 M 0.500 mol  0.500 M
1.00 l
1.00 l
-x
-x
HI
0M
2x
(where x is the number of mol/l of H2 or I2 that react)
0.500 – x
0.500 – x
2x
577
Now
2
[HI]
Kc 
[H2][I2]
578
Now
2
[HI]
Kc 
[H2][I2]
2
(2
x)

(0.500 - x)(0.500- x)
579
Now
2
[HI]
Kc 
[H2][I2]
2
(2
x)

(0.500 - x)(0.500- x)
At this point, always stop and look for any possible
simplifications. In this case there is a mathematical
simplification on the right-hand side of the equation
– what is it?
580
Now
2
[HI]
Kc 
[H2][I2]
2
(2
x)

(0.500 - x)(0.500- x)
At this point, always stop and look for any possible
simplifications. In this case there is a mathematical
simplification on the right-hand side of the equation
– what is it?







2
x
Kc 
0.500 - x
2







581
Now take the square root of both sides of the
equation
582
Now take the square root of both sides of the
equation







2x
Kc 
0.500 - x







583
Now take the square root of both sides of the
equation




























2x
Kc 
0.500 - x
so that
2x
7.37 
0.500 - x
584
Now take the square root of both sides of the
equation




























2x
Kc 
0.500 - x
so that
2x
7.37 
0.500 - x
3.685 - 7.37 x  2x
585
Now take the square root of both sides of the
equation




























2x
Kc 
0.500 - x
so that
2x
7.37 
0.500 - x
3.685 - 7.37 x  2x
3.685  9.37x
586
Now take the square root of both sides of the
equation




























2x
Kc 
0.500 - x
so that
2x
7.37 
0.500 - x
3.685 - 7.37 x  2x
3.685  9.37x
x  3.685  0.393
9.37
587
At equilibrium:
588
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
589
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
[I2] = 0.500 – 0.393 = 0.107 M
590
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
[I2] = 0.500 – 0.393 = 0.107 M
[HI] = 2 x = 0.786 M
591
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
[I2] = 0.500 – 0.393 = 0.107 M
[HI] = 2 x = 0.786 M
Check:
[HI]2
Kc 
[H2][I2]
592
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
[I2] = 0.500 – 0.393 = 0.107 M
[HI] = 2 x = 0.786 M
Check:
[HI]2
Kc 
[H2][I2]
2
(0.786)
Kc 
 54.0
(0.107)(0.107)
593
At equilibrium:
[H2] = 0.500 – 0.393 = 0.107 M
[I2] = 0.500 – 0.393 = 0.107 M
[HI] = 2 x = 0.786 M
Check:
[HI]2
Kc 
[H2][I2]
2
(0.786)
Kc 
 54.0
(0.107)(0.107)
Note that this value is very close, but not exactly
the same as the given Kc, due to round-off errors in
the mathematical solution.
594
If the reaction in the problem went 100 % to
completion, the amount of HI formed would be
1.00 moles. Because of the equilibrium process,
only 0.786 moles of HI are formed.
595
If the reaction in the problem went 100 % to
completion, the amount of HI formed would be
1.00 moles. Because of the equilibrium process,
only 0.786 moles of HI are formed.
Thus, the maximum yield for this reaction is
0.786 x 100  78.6 %
1.00
at the temperature of the experiment.
596
Example 2: At the start of a reaction of H2(g) and I2(g)
there are 0.0218 mol of H2, 0.0145 mol of I2, and
0.0783 mol of HI(g) present in a 3.50 l reaction vessel
at 430 oC. At this temperature Kc = 54.3.
Calculate the concentration of the three gases at
equilibrium.
597
Example 2: At the start of a reaction of H2(g) and I2(g)
there are 0.0218 mol of H2, 0.0145 mol of I2, and
0.0783 mol of HI(g) present in a 3.50 l reaction vessel
at 430 oC. At this temperature Kc = 54.3.
Calculate the concentration of the three gases at
equilibrium. The reaction is:
H2(g) + I2(g) 
 2 HI(g)
598
Example 2: At the start of a reaction of H2(g) and I2(g)
there are 0.0218 mol of H2, 0.0145 mol of I2, and
0.0783 mol of HI(g) present in a 3.50 l reaction vessel
at 430 oC. At this temperature Kc = 54.3.
Calculate the concentration of the three gases at
equilibrium. The reaction is:
H2(g) + I2(g) 
 2 HI(g)
Note: Initially we do not know if more HI will form
or not. Let y denote the amount of H2 that reacts. If
y turns out to be negative, it will mean that the
reaction goes from right to left, i.e. more H2 is
formed.
599
The ICE table is as follows:
H2
I2
HI
600