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![Recall: Equilibrium Constant K
and Rate Constants
2 NO2 N2O4
k1[NO2]2 = k1[N2O4]
𝐊 =
𝐤 𝟏
𝐤−𝟏
=
𝐍 𝟐 𝐎 𝟒
𝐍𝐎 𝟐
𝟐
The equilibrium constant K is the ratio of
the forward and reverse rate constants.
k1
k1](https://image.slidesharecdn.com/unit15equilibriumpt3lomafinal-160601175726/75/Chem-2-Chemical-Equilibrium-III-The-Equilibrium-Constant-Expression-and-the-Law-of-Mass-Action-LOMA-3-2048.jpg)
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Chem 2 - Chemical Equilibrium III: The Equilibrium Constant Expression and the Law of Mass Action (LOMA)
The document discusses chemical equilibrium, focusing on the equilibrium constant expression and the law of mass action as demonstrated through various examples. It highlights that the equilibrium constant, K, is the ratio of the concentrations of products to reactants, remains constant under specific conditions, and is unitless due to standard concentration normalization. Numerous examples illustrate how to calculate K for different chemical reactions and the effects of reversing reactions on K's value.
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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 equilibrium… Macroscopic observableshave stopped changing. The forward and reverse reaction rates are equal.
- 3. Recall: Equilibrium ConstantK and Rate Constants 2 NO2 N2O4 k1[NO2]2 = k1[N2O4] 𝐊 = 𝐤 𝟏 𝐤−𝟏 = 𝐍 𝟐 𝐎 𝟒 𝐍𝐎 𝟐 𝟐 The equilibrium constant K is the ratio of the forward and reverse rate constants. k1 k1
- 4. The Equilibrium ConstantExpression 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 ofMass Action Consider the generalized reaction aA + bB ⇌ cC + dD Reactants A and B Products C and D coefficients
- 6. The Law ofMass Action For the reaction 𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃 The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is 𝐊 = 𝐂 𝐜 𝐃 𝐝 𝐀 𝐚 𝐁 𝐛 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
- 7. The Law ofMass Action What happens if we reverse the reaction? 𝐜𝐂 + 𝐝𝐃 ⇌ 𝐚𝐀 + 𝐛𝐁 The relationship between the value of the equilibrium constant K and the concentrations of reactants and products is still products over reactants! 𝐊 = 𝐀 𝐚 𝐁 𝐛 𝐂 𝐂 𝐃 𝐝 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
- 8. The Law ofMass Action 𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃 This relationship is true no matter the initial distribution (relative amounts) of reactants and products! 𝐊 = 𝐂 𝐜 𝐃 𝐝 𝐀 𝐚 𝐁 𝐛 𝐩𝐫𝐨𝐝𝐮𝐜𝐭𝐬 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬
- 9. 𝐏 𝐍𝐎 𝟐 𝐏𝐍 𝟐 𝐎 𝟒 “kinetics ” Remember… Once we reached the “equilibrium” part of the experiment, we ended up with 0.50 atm NO2 and 0.25 atm N2O4 … And it didn’t matter if we started with all products or all reactants! Time (s) P (atm) 0.75 0.50 0.25 1.0 𝐏 𝐍𝐎 𝟐 𝐏 𝐍 𝟐 𝐎 𝟒
- 10. Equilibrium Constant Kis Unitless Why is the value for K unitless? 𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃 Each concentration in the equilibrium constant expression is divided by a “standard concentration” of 1.0 M. 𝐊 = 𝐂 𝐜 𝟏. 𝟎 𝐃 𝐝 𝟏. 𝟎 𝐀 𝐚 𝟏. 𝟎 𝐁 𝐛 𝟏. 𝟎 All molarity units cancel out!
- 11. Equilibrium Constant Kis Unitless 𝐚𝐀 + 𝐛𝐁 ⇌ 𝐜𝐂 + 𝐝𝐃 When we divide by a standard concentration, we are left with “effective concentrations” or “activities” (a). 𝐊 = 𝒂 𝑪 𝐜 𝒂 𝑫 𝐝 𝒂 𝑨 𝐚 𝒂 𝑩 𝐛 “activity” can be though of as the “presence” of the reactant or product
- 12. Gases and theEquilibrium Constant Expression 𝐚𝐀 𝒈 + 𝐛𝐁 𝒈 ⇌ 𝐜𝐂 𝒈 + 𝐝𝐃(𝒈) The Law of Mass Action holds for gases in equilibrium, as well. 𝑷 𝑪 𝐜 represents the partial pressure of gas C, etc. 𝐊 = 𝑷 𝑪 𝐜 𝑷 𝑫 𝐝 𝑷 𝑨 𝐚 𝑷 𝑩 𝐛
- 13. Gases and theEquilibrium Constant Expression 𝐚𝐀 𝒈 + 𝐛𝐁 𝒈 ⇌ 𝐜𝐂 𝒈 + 𝐝𝐃(𝒈) Each partial pressure is divided by a standard pressure (1 atm), so the K for gaseous systems is also unitless! 𝐊 = 𝑷 𝑪 𝐜 𝑷 𝑫 𝐝 𝑷 𝑨 𝐚 𝑷 𝑩 𝐛 Partial pressures for gases are given in “activities”, just like solutions.
- 14. Example 1: TheEquilibrium 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: 𝐊 = 𝐍 𝟐 𝐎 𝟒 𝐍𝐎 𝟐 𝟐 The coefficient for N2O4 is 1
- 16. Calculating the Valueof 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: Calculatingthe 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: 𝐊 = 𝐍 𝟐 𝐎 𝟒 𝐍𝐎 𝟐 𝟐 = 𝟎.𝟎𝟒𝟏𝟕 𝟎.𝟎𝟏𝟔𝟓 𝟐 = 𝟏𝟓𝟑 Remember, K is unitless!
- 19. Example 3: Reversingthe 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 M… 𝐊 = 𝐍𝐎 𝟐 𝟐 𝐍 𝟐 𝐎 𝟒 = 𝟎.𝟎𝟏𝟔𝟓 𝟐 𝟎.𝟎𝟒𝟏𝟕 = 𝟔. 𝟓𝟑 × 𝟏𝟎−𝟑 This value is just K-1 for the previous reaction!
- 21. Example 4: TheEquilibrium 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: 𝐊 = 𝑷 𝑵 𝟐 𝑶 𝟒 𝑷 𝑵𝑶 𝟐 𝟐 = 𝟎.𝟏𝟗𝟗 𝟏.𝟐𝟔 𝟐 = 𝟎. 𝟏𝟐𝟓 NOTE: These are not the same experimental conditions from the previous problem.
- 23. Next up, Properties ofthe Equilibrium Constant K (Pt 4)