2.03 Law of Mass Action for Combined Reactions

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En provenance du cours de University of Kentucky

Advanced Chemistry

405 notes

University of Kentucky

Advanced Chemistry

405 notes

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.

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

Rencontrer les enseignants

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

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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