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Chem 2 - Chemical Equilibrium V: ICE Tables and Equilibrium Calculations
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Transcript of Chem 2 - Chemical Equilibrium V: ICE Tables and Equilibrium Calculations
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
๐ฉ๐ซ๐จ๐๐ฎ๐๐ญ๐ฌ๐ซ๐๐๐๐ญ๐๐ง๐ญ๐ฌ
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 โICEโ 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
๐=๐ท๐ต๐๐ถ๐
โ
๐ท๐ต๐ถ๐
๐
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.๐=
๐ท๐ต๐๐ถ๐
โ
๐ท๐ต๐ถ๐
๐
2 NO2 N2O4
Letโs do an exampleโฆ
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 โsomeโ of everything at equilibrium!If one of the reactants or products in the reversible reaction initially has โzeroโ, 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
โAdd upโ 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
โxโ
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
๐=๐ท๐ต๐๐ถ๐
โ
๐ท๐ต๐ถ๐
๐
๐=๐ฑโ
(๐.๐๐โ๐๐ฑ )๐
Solve for โxโ
Recall, K has a value of 0.125Plug this in for K in the expression, then solve for x
Solving for Equilibrium Partial Pressures
๐=๐ท๐ต๐๐ถ๐
โ
๐ท๐ต๐ถ๐
๐
๐ .๐๐๐=๐ฑโ
(๐ .๐๐โ๐๐ฑ )๐
Solve for โxโ
Solving for x (algebra review)
๐ .๐๐๐=๐ฑโ
(๐ .๐๐โ๐๐ฑ )๐
Expand
(๐ .๐๐โ๐๐ฑ ) (๐ .๐๐โ๐๐ฑ )
๐ .๐๐๐๐โ๐ .๐๐ ๐โ๐ .๐๐๐+๐ ๐๐
(๐ .๐๐โ๐๐ฑ )๐๐.๐๐๐=๐
Multiply both sides by the denominator
Plug this back into
the equation๐ .๐๐๐๐โ๐ .๐๐๐+๐ ๐๐
simplify
Solving for x
Plug the values (with sign) into the quadratic equation
๐ .๐๐๐๐โ๐ .๐๐ ๐+๐ ๐๐(๐ .๐๐๐)=๐Multiply each
term by
Rearrange to quadratic form
(subtract x from both sides and collect terms)๐ .๐๐๐โ๐ .๐๐ ๐+๐ .๐๐๐=๐
๐ .๐๐๐โ๐ .๐๐๐+๐ .๐ ๐๐=๐abc
The Quadratic Equation
Plug the values (with sign) into the quadratic equation(not shown)
๐ .๐๐๐โ๐ .๐๐๐+๐ .๐ ๐๐=๐abc
๐ฑ=โ๐ยฑโ๐๐โ๐๐๐๐๐
x = 0.199 and x = 3.46
Choosing a Reasonable Value for x
๐ .๐๐๐โ๐ .๐๐๐+๐ .๐ ๐๐=๐abc
๐ฑ=โ๐ยฑโ๐๐โ๐๐๐๐๐
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
๐=๐ท๐ต๐๐ถ๐
โ
๐ท๐ต๐ถ๐
๐
๐=๐ .๐๐๐โ
(๐ .๐๐)๐
๐=๐ .๐๐๐
Next up, Heterogeneous Equilibria
(Pt 6)