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2.03 Law of Mass Action for Combined Reactions

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Chimie avancée

Université du Kentucky

4.7 (430 notes) | 51K étudiants inscrits

A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

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4.7 (430 notes)

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KO

Jul 17, 2017

Excellent content and delivery!\n\nThis course has many well-explained examples and practice questions to reinforce the understanding of the taught concepts. Thank you!

CW

Sep 17, 2015

I used this course as a significant portion of my preparation for the CLEP Chemistry exam, and passed today with an A! Thank you!

À partir de la leçon

Chemical Equilibrium

This unit introduces the concept of chemical equilibrium and how it applies to many chemical reactions. The quantitative aspects of equilibrium are explored thoroughly through discussions of the law of mass action as well as the relationship between equilibrium constants with respect to concentrations and pressures of substances. Much of the discussion explores how to solve problems to find either the value of the equilibrium constant or the concentrations of substances at equilibrium. ICE (initial-change-equilibrium) tables are introduced as a problem-solving tool and multiple examples of their use are included. From a qualitative standpoint, Le Châtelier’s principle is used to explain how various factors affect the equilibrium constant of a reaction along with the concentrations of all species.

2.01 Dynamic Equilibrium6:28

2.02 Law of Mass Action7:34

2.03 Law of Mass Action for Combined Reactions4:36

2.04 Relationship Between Kc and Kp6:51

2.04a Relationship Between Kc and Kp2:34

2.05 Calculating the Equilibrium Constant13:09

2.05a Finding Kc5:04

2.06 Reaction Quotient5:55

2.06a Predicting Reaction Progress with the Reaction Quotient1:55

2.07 Calculating Equilibrium Concentrations12:54

2.07a Finding Equilibrium Concentrations, part 14:59

2.07b Finding Equilibrium Concentrations, part 23:20

2.07c Finding Equilibrium Concentrations, part 34:22

2.08 Le Chatelier's Principle Part A9:55

2.08a Le Chatelier's Principle Part B19:50

2.08b How Changes Shift the Equilibrium5:18

Enseigné par

Dr. Allison Soult
Lecturer
Dr. Kim Woodrum
Sr. Lecturer

Transcription

In this unit we will be looking at the law of mass action for combined reactions. Sometimes we know the equilibrium constant for two separate reactions but we need to find the equilibrium constant for the overall reaction. and we are going to talk about the mathematical manipulations we can use to figure out the K value or the equilibrium constant for the overall reaction. We are going to determine the affect of manipulating the equation on the equilibrium constant of a reaction. We are going to determine the affect of manipulating the We are going to determine the affect of manipulating the The first thing we can see is that when we reverse the reaction we inverse the equilibrium expression. So if we look at the equilibrium expression of the law of mass action for our first reaction. What we see is K for one equals C cubed over A time B squared. And we get this from the stoichiometry of our balance chemical equation. So we have the coefficients of 1 2 and 3. If I were to reverse the reaction I see a slight different K value and I see K minus 1 indicating a reverse reaction and I have A B squared over C cubed and I see K minus 1 indicating a reverse reaction and I have A B squared over C cubed Now mathematically this is just inverse for we we saw for K 1. and I have A B squared over C cubed Now mathematically this is just inverse for we we saw for K 1. So if I want to find the equilibrium constant for this reverse process all I have to do is take the inverse, or K to the minus one of the equilibrium constant for the first reaction. If I multiple the coefficients by some factor we actually raise the equilibrium constant to that same power. So for example, if we write the law of mass action for our first reaction here we have the same thing we did before K 1 = C Cubed over A B Squared. When I look at twice that reaction I say Two A plus 4 B yields 6 C, I have just multiplied everything by two. When I look at K 2 I get C to the sixth over A Square B to the fourth. Because I have different coefficients. I have to put those in their. But when I look at the relationship between K 2 and K 1 I see K 2 could also be written as B Cubed over A B Squared with the entire expression squared and that would be the same as K 1 squared So if I multiply a reaction by some factor, I could simply raise the K to that power remembering what we know about our mathematical relationship with powers that X squared squared is equal to X to the fourth X to the sixth is equal to X to the third to the second power [ X^3^2]. When we add reactions we have to take the product of those two K values so if I take A plus B yields C I can write my law of mass actions so my K 1 value there when I look at B plus D yields E I can write K 2 so E over B and D when I look at B plus D yields E I can write K 2 so E over B and D now if I add those two reactions I end up with A plus 2 B plus D as my reactants side and I end up with C and E on my product side so now if I write the K 3 if I write the equilibrium expression for that overall reaction I get C times E over A B squared D if I look at K 1 and K 2 I see this is simply the product if I look at K 1 and K 2 I see this is simply the product of those equilibrium expressions because I can break that down into each part and as a result I can find the because I can break that down into each part and as a result I can find the equilibrium constant of K 3 by simply multiply K 1 by K 2. So lets look at an example here how are K 1 and K 2 related to K 3 what is the relationship between these first two terms that will get me to my K 3 value. Hopefully you got B for the answer because when look at our reaction we see that reaction 3 is simply the sum of reaction one and reaction 2. Here we have A plus 2 A which gives us 3 A when have 2 B and 2 B we see we have a C on the product side but we also have a C on the reactant side, so they are not in the overall reaction and they cancel out. On the product side, we have 2 B and 2 D. So since my over reaction, the third reaction is simply the sum of reactions 1 and 2 the value of K 3 will be equal to K 1 times K 2. In the next unit we will look at the relationship between K P and K C K P is the equilibrium constant with respect to pressure and K C is the equilibrium constant with respect to their concentrations.

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