Top Banner
By: Oscar Garcia
8

Chemistry ICE Box Equilibrium Problems

Jun 23, 2015

Download

Education

Oscar Garcia

This .pptx explains in a very simple way how to solve Chemical Equlibrium problems using the ICE box method.
-Really Useful
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Chemistry ICE Box Equilibrium Problems

By: Oscar Garcia

Page 2: Chemistry ICE Box Equilibrium Problems

An ice box is a table that is used for equilibrium calculations when we are given starting concentrations for chemicals.

The starting concentrations have to be changed to equilibrium concentrations before they can be used in an equilibrium expression.

The I stands for initial, the C for change and the E for equilibrium.

An ice box looks like this: Molarity H+ Cl- HCl

Initial 1.0M 0.5M 0M

Change -0.3M -0.3M +0.3M

Equilibrium

0.7M 0.2M 0.3M

Page 3: Chemistry ICE Box Equilibrium Problems

A 10.0L bulb is filled with 4.0 mol of NO(g), 2.0 mol of O2(g). The gases then reached equilibrium according to the equation 2NO(g) + O2(g) ↔ 2NO2(g). At Equilibrium, the bulb only contained 2.8 mol of NO(g). Calculate the Keq for this reaction.

Since we aren’t given all of the equilibrium concentrations, we need to use an ice box. First we calculate the initial concentrations and fill them into the ice box. Since we weren’t given a concentration for the product, we assume it starts at 0M. Then with the one given equilibrium concentration, we find the change in concentrations. The concentrations change according to the mol ratio in the balanced equation, so O2 will change half as much as NO. Generally, reactants are used up and products are produced so the reactants will have negative changes and the products will have positive ones.

Then we calculate the Keq. Keq = (0.12)2/(0.28)2(0.14) = 1.3 Since the Keq is greater than one, the products are favoured. If it was less

than one the reactants would be favoured.

NO O2 NO2

0.4M 0.2M 0M

-0.12 -0.06 +0.12

0.28M 0.14M 0.12M

Page 4: Chemistry ICE Box Equilibrium Problems

You can also use variables to represent the change in concentrations in an ice box.

You know you need to use variables if the starting concentrations are given but no equilibrium concentrations are given.

For the following reaction, find all equilibrium concentrations if you started off with 0.5M of H2 and 0.5M of I2 0.5M and Keq = 32. H2(g) + I2(g) ↔ 2HI(g)

You start by setting up your equilibrium expression. 32 = (2x)2/(0.5-x)2

Solve for x and you get x = 0.37. Then you go back to the equilibrium concentrations and plug in x to get 0.13M for H2 and I2 and 0.74M for HI.

H2 I2 HI

0.5M 0.5M 0M

-x -x +2x

0.5-x 0.5-x 2x

Page 5: Chemistry ICE Box Equilibrium Problems

• Sometimes when using variables, it doesn’t work out so that you can take the square root of every term in the equilibrium expression. For these questions you would need to use the quadratic formula: x = -b ± √(b2-4ac)

2a • Find all equilibrium concentrations for the following reaction if you

begin with 0.2M of PCl5 and the equilibrium constant is 1.3. PCl5(g) ↔ PCl3(g) + Cl2(g)

• Again you set up you equilibrium expression. 1.3 = (x)2/(0.2-x) • This equilibrium expression is a quadratic so you need the quadratic

formula to solve for x. Once you solve, you get x = -1.48 and x = 0.176. x = -1.48 gives you negative concentrations so you need to discard it which leaves x = 0.176 as your answer. Plug it back into the equilibrium concentrations and you get 0.024M for PCl5 and 0.176M for PCl3 and Cl2.

PCl5 PCl3 Cl2

0.2M 0M 0M

-x +x +x

0.2-x x x

Page 6: Chemistry ICE Box Equilibrium Problems

Ice boxes are often needed for finding the ionization constants for acids or bases.

You can use ice boxes to calculate the pH or ionization percentage of an acid.

Example: An unknown acid HX has a concentration of 10M and its Ka = 0.25. Find its pH.

For this example, you would need the quadratic formula to solve for x but in some cases you can disregard the –x in the ice box.

The rule is: if the molarity of the acid divided by the Ka is greater than 1000, you can drop the –x because it does not make a significant difference in the calculation. If the number is less than 1000, you need to keep the –x and use the quadratic equation to solve for x.

HX H+ X-

10M 0M 0M

-x +x +x

10-x x x

Page 7: Chemistry ICE Box Equilibrium Problems

Ice boxes are used often for equilibrium calculations and ionization constants for acids and bases.

You know you need an ice box when: 1) You are given starting concentrations instead of

equilibrium calculations. 2) You are dealing with ionization constants for weak acids

or bases (Ka or Kb)

The ICE in ice box stands for Initial, Change, Equilibrium. Only equilibrium concentrations are used in an equilibrium

expression. Always make sure the chemical equation is balanced before

you find the changes in concentrations because they are related to the mol ratio.

Page 8: Chemistry ICE Box Equilibrium Problems

Information provided by Mr. Therrien’s Chemistry 12 booklet.

+ = H

Cl H+

Cl-

I 1M

0M

0M

C -.5 +.5

+.5

E 0.5

0.5

0.5