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Chem 2 - Chemical Equilibrium III: The Equilibrium Constant Expression and the Law of Mass Action (LOMA)

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Chem 2 - Chemical Equilibrium III: The Equilibrium Constant Expression and the Law of Mass Action (LOMA)

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Chemical Equilibrium (Pt. 3)
The Equilibrium Constant
Expression and the Law of
Mass Action (LOMA)
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.
At equilibriumโ€ฆ
Macroscopic observables have
stopped changing.
The forward and reverse reaction
rates are equal.
Recall: Equilibrium Constant K
and Rate Constants
2 NO2 N2O4
k1[NO2]2 = k๏€ญ1[N2O4]
๐Š =
๐ค ๐Ÿ
๐คโˆ’๐Ÿ
=
๐ ๐Ÿ ๐Ž ๐Ÿ’
๐๐Ž ๐Ÿ
๐Ÿ
The equilibrium constant K is the ratio of
the forward and reverse rate constants.
k1
k๏€ญ1
The Equilibrium Constant Expression
Experiments done by Guldburg and
Waage (1864-1879) demonstrated
the ratio of products to reactants is
always constant
under a certain set of experimental
conditions.
(This is a simplified statement.)
The Law of Mass Action
Consider the generalized reaction
aA + bB โ‡Œ cC + dD
Reactants
A and B
Products
C and D
coefficients
The Law of Mass Action
For the reaction
๐š๐€ + ๐›๐ โ‡Œ ๐œ๐‚ + ๐๐ƒ
The relationship between the value of
the equilibrium constant K and the
concentrations of reactants and
products is
๐Š =
๐‚ ๐œ
๐ƒ ๐
๐€ ๐š ๐ ๐›
๐ฉ๐ซ๐จ๐๐ฎ๐œ๐ญ๐ฌ
๐ซ๐ž๐š๐œ๐ญ๐š๐ง๐ญ๐ฌ

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Chem 2 - Chemical Equilibrium III: The Equilibrium Constant Expression and the Law of Mass Action (LOMA)

  • 1. Chemical Equilibrium (Pt. 3) The Equilibrium Constant Expression and the Law of Mass Action (LOMA) 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. At equilibriumโ€ฆ Macroscopic observables have stopped changing. The forward and reverse reaction rates are equal.
  • 3. Recall: Equilibrium Constant K and Rate Constants 2 NO2 N2O4 k1[NO2]2 = k๏€ญ1[N2O4] ๐Š = ๐ค ๐Ÿ ๐คโˆ’๐Ÿ = ๐ ๐Ÿ ๐Ž ๐Ÿ’ ๐๐Ž ๐Ÿ ๐Ÿ The equilibrium constant K is the ratio of the forward and reverse rate constants. k1 k๏€ญ1
  • 4. The Equilibrium Constant Expression Experiments done by Guldburg and Waage (1864-1879) demonstrated the ratio of products to reactants is always constant under a certain set of experimental conditions. (This is a simplified statement.)
  • 5. The Law of Mass Action Consider the generalized reaction aA + bB โ‡Œ cC + dD Reactants A and B Products C and D coefficients
  • 6. The Law of Mass Action For the reaction ๐š๐€ + ๐›๐ โ‡Œ ๐œ๐‚ + ๐๐ƒ The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is ๐Š = ๐‚ ๐œ ๐ƒ ๐ ๐€ ๐š ๐ ๐› ๐ฉ๐ซ๐จ๐๐ฎ๐œ๐ญ๐ฌ ๐ซ๐ž๐š๐œ๐ญ๐š๐ง๐ญ๐ฌ
  • 7. The Law of Mass Action What happens if we reverse the reaction? ๐œ๐‚ + ๐๐ƒ โ‡Œ ๐š๐€ + ๐›๐ The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is still products over reactants! ๐Š = ๐€ ๐š ๐ ๐› ๐‚ ๐‚ ๐ƒ ๐ ๐ฉ๐ซ๐จ๐๐ฎ๐œ๐ญ๐ฌ ๐ซ๐ž๐š๐œ๐ญ๐š๐ง๐ญ๐ฌ
  • 8. The Law of Mass Action ๐š๐€ + ๐›๐ โ‡Œ ๐œ๐‚ + ๐๐ƒ This relationship is true no matter the initial distribution (relative amounts) of reactants and products! ๐Š = ๐‚ ๐œ ๐ƒ ๐ ๐€ ๐š ๐ ๐› ๐ฉ๐ซ๐จ๐๐ฎ๐œ๐ญ๐ฌ ๐ซ๐ž๐š๐œ๐ญ๐š๐ง๐ญ๐ฌ
  • 9. ๐ ๐๐Ž ๐Ÿ ๐ ๐ ๐Ÿ ๐Ž ๐Ÿ’ โ€œkinetics โ€ Rememberโ€ฆ Once we reached the โ€œequilibriumโ€ part of the experiment, we ended up with 0.50 atm NO2 and 0.25 atm N2O4 โ€ฆ And it didnโ€™t matter if we started with all products or all reactants! Time (s) P (atm) 0.75 0.50 0.25 1.0 ๐ ๐๐Ž ๐Ÿ ๐ ๐ ๐Ÿ ๐Ž ๐Ÿ’
  • 10. Equilibrium Constant K is Unitless Why is the value for K unitless? ๐š๐€ + ๐›๐ โ‡Œ ๐œ๐‚ + ๐๐ƒ Each concentration in the equilibrium constant expression is divided by a โ€œstandard concentrationโ€ of 1.0 M. ๐Š = ๐‚ ๐œ ๐Ÿ. ๐ŸŽ ๐ƒ ๐ ๐Ÿ. ๐ŸŽ ๐€ ๐š ๐Ÿ. ๐ŸŽ ๐ ๐› ๐Ÿ. ๐ŸŽ All molarity units cancel out!
  • 11. Equilibrium Constant K is Unitless ๐š๐€ + ๐›๐ โ‡Œ ๐œ๐‚ + ๐๐ƒ When we divide by a standard concentration, we are left with โ€œeffective concentrationsโ€ or โ€œactivitiesโ€ (a). ๐Š = ๐’‚ ๐‘ช ๐œ ๐’‚ ๐‘ซ ๐ ๐’‚ ๐‘จ ๐š ๐’‚ ๐‘ฉ ๐› โ€œactivityโ€ can be though of as the โ€œpresenceโ€ of the reactant or product
  • 12. Gases and the Equilibrium Constant Expression ๐š๐€ ๐’ˆ + ๐›๐ ๐’ˆ โ‡Œ ๐œ๐‚ ๐’ˆ + ๐๐ƒ(๐’ˆ) The Law of Mass Action holds for gases in equilibrium, as well. ๐‘ท ๐‘ช ๐œ represents the partial pressure of gas C, etc. ๐Š = ๐‘ท ๐‘ช ๐œ ๐‘ท ๐‘ซ ๐ ๐‘ท ๐‘จ ๐š ๐‘ท ๐‘ฉ ๐›
  • 13. Gases and the Equilibrium Constant Expression ๐š๐€ ๐’ˆ + ๐›๐ ๐’ˆ โ‡Œ ๐œ๐‚ ๐’ˆ + ๐๐ƒ(๐’ˆ) Each partial pressure is divided by a standard pressure (1 atm), so the K for gaseous systems is also unitless! ๐Š = ๐‘ท ๐‘ช ๐œ ๐‘ท ๐‘ซ ๐ ๐‘ท ๐‘จ ๐š ๐‘ท ๐‘ฉ ๐› Partial pressures for gases are given in โ€œactivitiesโ€, just like solutions.
  • 14. Example 1: The Equilibrium Constant Expression 2 NO2 N2O4 Write the equilibrium constant expression for the following reaction:
  • 15. Example 1 Solution: The Equilibrium Constant Expression 2 NO2 N2O4 Write the equilibrium constant expression for the following reaction: ๐Š = ๐ ๐Ÿ ๐Ž ๐Ÿ’ ๐๐Ž ๐Ÿ ๐Ÿ The coefficient for N2O4 is 1
  • 16. Calculating the Value of the Equilibrium Constant 2 NO2 N2O4 The value for the equilibrium constant K can be calculated by inserting the equilibrium concentrations (or partial pressures) into the equilibrium constant expression.
  • 17. Example 2: Calculating the Value of the Equilibrium Constant 2 NO2 N2O4 What is the value of K when the concentration of NO2 (at equilibrium) is 0.0165 M and the concentration of N2O4 is 0.0417 M?
  • 18. Example 2 Solution: Calculating the Value of the Equilibrium Constant 2 NO2 N2O4 The equilibrium NO2 conc is 0.0165 M and N2O4 is 0.0417 M: ๐Š = ๐ ๐Ÿ ๐Ž ๐Ÿ’ ๐๐Ž ๐Ÿ ๐Ÿ = ๐ŸŽ.๐ŸŽ๐Ÿ’๐Ÿ๐Ÿ• ๐ŸŽ.๐ŸŽ๐Ÿ๐Ÿ”๐Ÿ“ ๐Ÿ = ๐Ÿ๐Ÿ“๐Ÿ‘ Remember, K is unitless!
  • 19. Example 3: Reversing the Reaction and the Equilibrium Constant Expression N2O4 2 NO2 The equilibrium NO2 conc is still 0.0165 M and N2O4 is 0.0417 M. What is the value for K?
  • 20. Example 3 Solution: Reversing the Reaction and the Equilibrium Constant Expression N2O4 2 NO2 The equilibrium NO2 conc is still 0.0165 M and N2O4 is 0.0417 Mโ€ฆ ๐Š = ๐๐Ž ๐Ÿ ๐Ÿ ๐ ๐Ÿ ๐Ž ๐Ÿ’ = ๐ŸŽ.๐ŸŽ๐Ÿ๐Ÿ”๐Ÿ“ ๐Ÿ ๐ŸŽ.๐ŸŽ๐Ÿ’๐Ÿ๐Ÿ• = ๐Ÿ”. ๐Ÿ“๐Ÿ‘ ร— ๐Ÿ๐ŸŽโˆ’๐Ÿ‘ This value is just K-1 for the previous reaction!
  • 21. Example 4: The Equilibrium Constant Value and Expression with Partial Pressures 2 NO2 N2O4 The equilibrium NO2 partial pressure is 1.26 atm and N2O4 is 0.199 atm: NOTE: These are not the same experimental conditions from the previous problem.
  • 22. Example 4 Solution: The Equilibrium Constant Expression with Partial Pressures 2 NO2 N2O4 The equilibrium NO2 partial pressure is 1.26 atm and N2O4 is 0.199 atm: ๐Š = ๐‘ท ๐‘ต ๐Ÿ ๐‘ถ ๐Ÿ’ ๐‘ท ๐‘ต๐‘ถ ๐Ÿ ๐Ÿ = ๐ŸŽ.๐Ÿ๐Ÿ—๐Ÿ— ๐Ÿ.๐Ÿ๐Ÿ” ๐Ÿ = ๐ŸŽ. ๐Ÿ๐Ÿ๐Ÿ“ NOTE: These are not the same experimental conditions from the previous problem.
  • 23. Next up, Properties of the Equilibrium Constant K (Pt 4)