ice box calculations n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Ice Box Calculations PowerPoint Presentation
Download Presentation
Ice Box Calculations

Loading in 2 Seconds...

play fullscreen
1 / 8

Ice Box Calculations - PowerPoint PPT Presentation


  • 504 Views
  • Uploaded on

Ice Box Calculations. By: Clifford Beaton. Introduction. An ice box is a table that is used for equilibrium calculations when we are given starting concentrations for chemicals.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Ice Box Calculations' - kenisha


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
ice box calculations

Ice Box Calculations

By: Clifford Beaton

introduction
Introduction
  • 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:
example 1
Example 1
  • 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.
example 2
Example 2
  • 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.
example 3
Example 3
  • 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.
ice boxes for acids and bases
Ice Boxes for Acids and Bases
  • 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.
remember
Remember
  • 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.
the end
THE END
  • Information provided by Mr. Therrien’s Chemistry 12 booklet.

+ =