Chem 2 - Chemical Equilibrium V: ICE Tables and Equilibrium Calculations

1. Chemical Equilibrium (Pt. 5)
ICE Tables and Equilibrium
Calculations
By Shawn P. Shields, Ph.D.
This work is licensed by Dr. Shawn P. Shields-Maxwell under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0
International License.

2. Recall: The Law of Mass Action
For the reaction
𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃
The relationship between the value of
the equilibrium constant K and the
concentrations of reactants and
products is
𝐊 =
𝐂 𝐜
𝐃 𝐝
𝐀 𝐚 𝐁 𝐛
𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬
𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

3. Calculating Equilibrium Concentrations and
Partial Pressures
It is possible to calculate the
concentrations (or partial pressures) of
reactants and products at equilibrium.
If we have the value for the equilibrium
constant K, then we can use an “ICE”
table to figure it out!

4. ICE Tables provide a way to organize given
information and variables for equilibrium
calculations.
ICE Tables and Equilibrium Concentrations
2 NO2 N2O4
I (Initial)
C (Change)
E (Equilibrium)

5. Fill in the ICE Table with what is known
(or not known) about the initial conditions,
the direction the reaction will proceed
(change) and the conditions at equilibrium.
2 NO2 N2O4
I
C
E

6. Use the Equilibrium Constant Expression
and K to Solve Equilibrium Problems
We can use the relationship between the
value of the equilibrium constant K and
the initial concentrations (partial
pressures) of reactants and products
to solve for the concentrations (or
partial pressures) of reactants and
products at equilibrium. HOW?

7. Use the Equilibrium Constant Expression
and K to Solve Equilibrium Problems
For the reaction
the equilibrium constant
expression is
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐
2 NO2 N2O4

8. Using the Equilibrium Constant Expression
and K to Solve Equilibrium Problems
Use an ICE table to fill in the
equilibrium constant expression.
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐
2 NO2 N2O4
Let’s do an
example…

9. A flask contains 1.66 atm NO2(g) initially. At
some temperature, the equilibrium constant K
is 0.125. Calculate the equilibrium partial
pressures of the two gases.
2 NO2 N2O4
I
C
E
Example Problem: Filling in the ICE Table

10. A flask contains 1.66 atm NO2(g) initially. At
some temperature, the equilibrium constant K
is 0.125. Calculate the equilibrium partial
pressures of the two gases.
2 NO2 N2O4
I
C
E
Example Problem: Filling in the ICE Table
1.66 0

11. Central Concept: There has to be
at least “some” of everything at
equilibrium!
If one of the reactants or products in
the reversible reaction initially has
“zero”, then the reaction shifts in that
direction!
Which Way Will the Reaction Shift to
Reach Equilibrium?

12. NOTE:
The change (x) must be multiplied by the
coefficient for the reactant or product.
2 NO2 N2O4
I
C
E
The Change Line in the ICE Table
1.66 0
 2x + 1x

13. “Add up” the Initial and Change lines and
enter the values (equations) into the
Equilibrium line.
The Equilibrium Line in the ICE Table
2 NO2 N2O4
I
C
E
1.66 0
 2x + 1x
1.66  2x 0 + 1x
usually just
entered as
“x”

14. Substitute the values in the equilibrium line
on the ICE table into the equilibrium
constant expression.
Using the Equilibrium Constant Expression
2 NO2 N2O4
I
C
E
1.66 0
 2x + 1x
1.66  2x x
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐
𝐊 =
𝐱
𝟏. 𝟔𝟔 − 𝟐𝐱 𝟐
Solve for
“x”

15. Recall, K has a value of 0.125
Plug this in for K in the expression, then
solve for x
Solving for Equilibrium Partial Pressures
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐
𝟎. 𝟏𝟐𝟓 =
𝐱
𝟏. 𝟔𝟔 − 𝟐𝐱 𝟐
Solve for
“x”

16. Solving for x (algebra review)
𝟎. 𝟏𝟐𝟓 =
𝐱
𝟏. 𝟔𝟔 − 𝟐𝐱 𝟐
Expand
𝟏. 𝟔𝟔 − 𝟐𝐱 𝟏. 𝟔𝟔 − 𝟐𝐱
𝟐. 𝟕𝟓𝟓𝟔 − 𝟑. 𝟑𝟐𝒙 − 𝟑. 𝟑𝟐𝒙 + 𝟒𝒙 𝟐
𝟏. 𝟔𝟔 − 𝟐𝐱 𝟐
𝟎. 𝟏𝟐𝟓 = 𝒙
Multiply both
sides by the
denominator
Plug this
back into
the equation
𝟐. 𝟕𝟓𝟓𝟔 − 𝟔. 𝟔𝟒𝒙 + 𝟒𝒙 𝟐
simplify

17. Solving for x
Plug the values (with sign)
into the quadratic equation
𝟐. 𝟕𝟓𝟓𝟔 − 𝟔. 𝟔𝟒𝒙 + 𝟒𝒙 𝟐
(𝟎. 𝟏𝟐𝟓) = 𝒙
Multiply each
term by
Rearrange to
quadratic form
(subtract x from
both sides and
collect terms)
𝟎. 𝟑𝟒𝟒 − 𝟎. 𝟖𝟑𝒙 + 𝟎. 𝟓𝒙 𝟐
= 𝒙
𝟎. 𝟑𝟒𝟒 − 𝟏. 𝟖𝟑𝒙 + 𝟎. 𝟓𝒙 𝟐
= 𝟎
abc

18. The Quadratic Equation
Plug the values
(with sign) into the
quadratic equation
(not shown)
𝟎. 𝟑𝟒𝟒 − 𝟏. 𝟖𝟑𝒙 + 𝟎. 𝟓𝒙 𝟐
= 𝟎
abc
𝐱 =
−𝐛 ± 𝐛 𝟐 − 𝟒𝐚𝐜
𝟐𝐚
x = 0.199 and x = 3.46

19. Choosing a Reasonable Value for x
𝟎. 𝟑𝟒𝟒 − 𝟏. 𝟖𝟑𝒙 + 𝟎. 𝟓𝒙 𝟐
= 𝟎
abc
𝐱 =
−𝐛 ± 𝐛 𝟐 − 𝟒𝐚𝐜
𝟐𝐚
x = 0.199 and x = 3.46
Not physically
reasonable!

20. Substitute the value for x in the equilibrium
line on the ICE table.
The Final Equilibrium Partial Pressures
2 NO2 N2O4
I
C
E
1.66 0
 2x + 1x
1.66  2(0.199) 0.199
The equilibrium
partial pressure for
NO2 is 1.26 atm
The equilibrium
partial pressure for
N2O4 is 0.199 atm

21. Substitute the equilibrium partial pressures
into the equilibrium constant expression to
calculate K.
Check Your Answer
2 NO2 N2O4
I
C
E
1.66 0
 2x + 1x
1.26 0.199
𝐊 =
𝑷 𝑵 𝟐 𝑶 𝟒
𝑷 𝑵𝑶 𝟐
𝟐
𝐊 =
𝟎. 𝟏𝟗𝟗
(𝟏. 𝟐𝟔) 𝟐
𝐊 = 𝟎. 𝟏𝟐𝟓

22. Next up,
Heterogeneous Equilibria
(Pt 6)