Chem 112 Class Guide: Calculating K (Chapter 15, Sections 5-6)
Chapter 15, Sections 5 - 6 Learning Goals:
Upon completion of Chapter 15, Sections 5-6, you should be able to determine the following:
Calculate any variable in an equilibrium expression given equilibrium values.
Calculate any variable in an equilibrium expression given one equilibrium value and one or more
initial values through an ICE chart.
Chapter Reading Guide: Chapter 15, Sections 5-6
Section 5: CALCULATING EQUILIBRIUM CONSTANTS
Read Chapter 15.5
You may be given the initial concentrations of one or two species in a reaction. Any initial
concentration not given may be assumed to be zero. You may then be given one
equilibrium condition. Based upon the difference between the given initial condition and
the given equilibrium condition, you can figure out how much that species changes in
concentration. Then by looking at stoichiometric coefficients, you can calculate the change
for each species involved in the equilibrium. Once the changes of all species have been
solved for, the equilibrium concentrations of all species can be calculated. K can then be
calculated based on these equilibrium conditions.
Example:
Consider the reaction: 2 CO2 (g) ↔ 2 CO (g) + O2 (g). The initial concentration of CO2 is 0.5
M and the equilibrium concentration of O2 is 0.1 M. Calculate the equilibrium concentration
of all three species and then calculate Kc.
The first step is to construct an ICE chart, where I = initial concentrations, C = change in
concentrations, and E = equilibrium concentration. We use the stoichiometric coefficients
to help us do this:
I
C
E
2 CO2 (g) ↔
0.5
-2X
0.5 – 2X
2 CO (g) +
0
+2X
2X
O2 (g)
0
+X
X
The 2 in front of the X’s for CO2 and CO come from the stoichiometric coefficients. We use
X because we don’t yet know how much CO2 is decreasing as the reaction proceeds, or how
much CO and O2 we are obtaining. However we can figure it out from the equilibrium value
of O2.
We know that at equilibrium, the concentration of O2 = 0.1 M. Therefore, since the
equilibrium concentration of O2 is also X, X must equal 0.1 M. We can then fill out the ICE
chart with this value in place:
I
C
E
2 CO2 (g) ↔
0.5
-2 (0.1)
0.5 – 0.2 = 0.3
2 CO (g) +
0
+2 (0.1)
0.2
O2 (g)
0
+0.1
0.1
Now that we know our equilibrium values, we can calculate K:
CO O2   0.2 0.1  0.04
Kc 
2
2
CO2 
0.3
2
2
Try Practice Exercise 15.9
Section 6: APPLICATIONS OF EQUILIBRIUM CONSTANTS
Read Chapter 15.6, but you can skip page 627 and 628 (through Sample Exercise 15.10). We
will cover these two pages next class.
It is possible that you will be given initial conditions but no equilibrium conditions (in this
case, you would also be given K). In this case, you know that the initial concentrations will
change (reactants decrease and products increase), but you do not know how much. When
filling out the change row in your ICE chart, you will use “x” to define an unknown change.
Remember to take into consideration stoichiometric coefficients. Your equilibrium
concentrations will then have “x” in them. Given K, solve for x. It may be necessary to use
the quadratic equation.
b  b2  4ac
x
2a
Once x has been determined, calculate the appropriate equilibrium concentrations.
Example:
Consider the reaction: PCl5 (g) ↔ PCl3 (g) + Cl2 (g). The initial pressure of PCl5 is 1.0 atm
and the equilibrium constant Kp = 0.497. Calculate the equilibrium pressure values of all
three species.
As before, we begin by constructing an ICE chart:
I
C
E
PCl5 (g) ↔
1.0
-X
1.0 – X
PCl3 (g) +
0
+X
X
Cl2 (g)
0
+X
X
Unlike before, however, we do not have an equilibrium value of anything so we cannot
directly solve for X. We must use our KP value and solve for X using the quadratic equation:
KP 
 PCl3 Cl2    X  X   0.497
 PCl5 
1  X 
Multiplying out this equation will give us:
X 2  0.497* 1  X  ; Further simplifying will give us:
X 2  0.497  0.497 X , and now get everything to the same side of the equation:
X 2  0.497 X  0.497  0 ; From here we use the quadratic equation
x
b  b2  4ac
2a
And plug in our values for a, b and c:
X
0.497  0.497 2   4*1* 0.497 
2*1

0.497  1.5
 0.50
2
We can only ever have a positive value for X so we are able to ignore -1.5 because 0.497 – 1.5 = -2.
Now that we know X = 0.50, we can calculate the equilibrium values of the species (given
the E line from our ICE chart above):
PCl5 = 1.0 – X = 1.0 – 0.5 = 0.5 atm;
PCl3 = X = 0.5 atm;
Cl2 = X = 0.5 atm
Try Practice exercise 15.11
Try Practice exercise 15.12
Learning Resources
Chapter Learning Goals
Chapter 15, Sections 5-6 Learning Goals
Pre Class Assignment: This assignment must be completed prior to the next class. Check your syllabus
for the exact due date and time.
Complete the pre class assignment
(http://berks.psu.edu/clt/chem112/CalculatingK_HW.docx)
Submit a copy to the dropbox located in ANGEL called “Pre Class Assignment Submission:
Calculating K”
End of Chapter Problems:
Practice with these problems if you are having difficulty with any of the concepts covered in this
class guide AFTER we have met in class. If you cannot easily complete these problems, seek
help from your instructor, your mentor, or the learning center.
Chapter 15: 37, 39, 47, 49, 51