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.

Recall: The Law of Mass Action

For the reaction

The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is

𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬

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!

ICE Tables provide a way to organize given information and variables for equilibrium calculations.

ICE Tables and Equilibrium Concentrations

2 NO2 N2O4I (Initial)C (Change)E (Equilibrium)

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 N2O4ICE

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?

Use the Equilibrium Constant Expression and K to Solve Equilibrium Problems

For the reaction

the equilibrium constant expression is

𝐊=𝑷𝑵𝟐𝑶𝟒

❑

𝑷𝑵𝑶𝟐

𝟐

2 NO2 N2O4

Using the Equilibrium Constant Expression and K to Solve Equilibrium Problems

Use an ICE table to fill in the equilibrium constant expression.𝐊=

𝑷𝑵𝟐𝑶𝟒

❑

𝑷𝑵𝑶𝟐

𝟐

2 NO2 N2O4

Let’s do an example…

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 N2O4ICE

Example Problem: Filling in the ICE Table

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 N2O4ICE

Example Problem: Filling in the ICE Table

1.66 0

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?

NOTE: The change (x) must be multiplied by

the coefficient for the reactant or product.

2 NO2 N2O4ICE

The Change Line in the ICE Table

1.66 0 2x + 1x

“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 N2O4ICE

1.66 0 2x + 1x

1.66 2x 0 + 1xusually just entered as

“x”

Substitute the values in the equilibrium line on the ICE table into the equilibrium constant expression.

Using the Equilibrium Constant Expression

2 NO2 N2O4ICE

1.66 0 2x + 1x

1.66 2x x

𝐊=𝑷𝑵𝟐𝑶𝟒

❑

𝑷𝑵𝑶𝟐

𝟐

𝐊=𝐱❑

(𝟏.𝟔𝟔−𝟐𝐱 )𝟐

Solve for “x”

Recall, K has a value of 0.125Plug this in for K in the expression, then solve for x

Solving for Equilibrium Partial Pressures

𝐊=𝑷𝑵𝟐𝑶𝟒

❑

𝑷𝑵𝑶𝟐

𝟐

𝟎 .𝟏𝟐𝟓=𝐱❑

(𝟏 .𝟔𝟔−𝟐𝐱 )𝟐

Solve for “x”

Solving for x (algebra review)

𝟎 .𝟏𝟐𝟓=𝐱❑

(𝟏 .𝟔𝟔−𝟐𝐱 )𝟐

Expand

(𝟏 .𝟔𝟔−𝟐𝐱 ) (𝟏 .𝟔𝟔−𝟐𝐱 )

𝟐 .𝟕𝟓𝟓𝟔−𝟑 .𝟑𝟐 𝒙−𝟑 .𝟑𝟐𝒙+𝟒 𝒙𝟐

(𝟏 .𝟔𝟔−𝟐𝐱 )𝟐𝟎.𝟏𝟐𝟓=𝒙

Multiply both sides by the denominator

Plug this back into

the equation𝟐 .𝟕𝟓𝟓𝟔−𝟔 .𝟔𝟒𝒙+𝟒 𝒙𝟐

simplify

Solving for x

Plug the values (with sign) into the quadratic equation

𝟐 .𝟕𝟓𝟓𝟔−𝟔 .𝟔𝟒 𝒙+𝟒 𝒙𝟐(𝟎 .𝟏𝟐𝟓)=𝒙Multiply each

term by

Rearrange to quadratic form

(subtract x from both sides and collect terms)𝟎 .𝟑𝟒𝟒−𝟎 .𝟖𝟑 𝒙+𝟎 .𝟓𝒙𝟐=𝒙

𝟎 .𝟑𝟒𝟒−𝟏 .𝟖𝟑𝒙+𝟎 .𝟓 𝒙𝟐=𝟎abc

The Quadratic Equation

Plug the values (with sign) into the quadratic equation(not shown)

𝟎 .𝟑𝟒𝟒−𝟏 .𝟖𝟑𝒙+𝟎 .𝟓 𝒙𝟐=𝟎abc

𝐱=−𝐛±√𝐛𝟐−𝟒𝐚𝐜𝟐𝐚

x = 0.199 and x = 3.46

Choosing a Reasonable Value for x

𝟎 .𝟑𝟒𝟒−𝟏 .𝟖𝟑𝒙+𝟎 .𝟓 𝒙𝟐=𝟎abc

𝐱=−𝐛±√𝐛𝟐−𝟒𝐚𝐜𝟐𝐚

x = 0.199 and x = 3.46 Not physically reasonable!

Substitute the value for x in the equilibrium line on the ICE table.

The Final Equilibrium Partial Pressures

2 NO2 N2O4ICE

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

Substitute the equilibrium partial pressures into the equilibrium constant expression to calculate K.

Check Your Answer

2 NO2 N2O4ICE

1.66 0 2x + 1x

1.26

0.199

𝐊=𝑷𝑵𝟐𝑶𝟒

❑

𝑷𝑵𝑶𝟐

𝟐

𝐊=𝟎 .𝟏𝟗𝟗❑

(𝟏 .𝟐𝟔)𝟐

𝐊=𝟎 .𝟏𝟐𝟓

Next up, Heterogeneous Equilibria

(Pt 6)