2.06 Reaction Quotient

Loading...

来自 University of Kentucky 的课程

高级化学

405 个评分

University of Kentucky

高级化学

405 个评分

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.

从本节课中

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

与讲师见面

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

In this unit we will learn to calculate the value of the reaction quotient

and determine what it mean for a reaction at equilibrium.

Our goal is to determine the direction of the reaction

by using the concentrations we are given initially

and how the value of the reaction quotient 'q' compares to 'k'.

In the previous examples we have worked

we have seen the reaction can only precede in one direction.

We either had the present of reactants

or the presents of products

and therefore the reaction had to go in a certain directionb

because we could not lose something we did not have.

However, we can also see that we have

reactions where have both initial concentrations of reactants and products.

We are going to have to use something called the reaction quotient

which is abbreviated 'q'

to determine if a system is at equilibrium

and if it is not to determine if a system is at equilibrium

and if it is not

how we know which direction it is going to go to get to equilibrium.

So when we calculate the value of

q we actually calculate it the exact same way we calculate k.

The reason for the different label is q we actually calculate it the exact same way we calculate k.

The reason for the different label is

because we do not know if we are at equilibrium

and so we cannot call it an equilibrium constant

so we call it q which stand for the reaction quotient.

Notice that for my reaction A going to B

we can write the reaction quotient, the concentration of B

over the concentration of A, or products over reactants.

I can then calculate the value of q

and compare it to the value of k I can then calculate the value of q

and compare it to the value of k

to determine which way the reaction will proceed. and compare it to the value of k

to determine which way the reaction will proceed.

The first example we will look at

is when q is less then k.

So here we have a value of q of 1.3

and k is 3.2.

What that tells us is that the value of

q needs to increase in order to reach equilibrium.

And for this value to increase

we need the concentration of b to increase.

And we need the value of A to decrease.

As a result that mean the reaction

will proceed in the forward direction.

We are consuming A and producing more B.

On the other side of the spectrum

when we look at a q value of 4.8

which is great then the k value of 3.2 for t a particular reaction when we look at a q value of 4.8

which is great then the k value of 3.2 for t a particular reaction

we see that we need to do just the opposite.

Our numerator, is too high

our denominator is too low

so what we need to happen now is for the value of

B, or the amount of B we have present to decrease

and the amount of A to increase.

And when that happens

we will see that ratio of concentration of B over the concentration of A

will also reduce

the value of q will reduce

and it will continue until we get to the value of k.

If we calculate the reaction

quotient, and we find that q is equal to k.

Then we know we are at equilibrium.

Now, remember we are dynamic equilibrium.

Reactants are still going to products

products are still going to reactants.

But the rate of those two reactions are exactly the same.

That is what we mean by that dynamic chemical equilibrium.

So lets look at an example

the reaction has an equilibrium constant of 0.078

at 100 degrees Celsius.

If the initial concentration of each of the three substances is

0.100 molar

which way will the reaction proceed to reach equilibrium.

Hopeful you saw, that reaction for each would proceed in the reverse direction.

We have to look at our reaction

which is a balanced equation that was given to us

we have to look at the value of k that was given

a 0.078

and we need to compare that to the value of q.

So what we are going to look at one the next slide and we need to compare that to the value of q.

So what we are going to look at one the next slide

is to see how we calculated that value of q.

So here I have summarized the information given in the problem.

The concentration of each of the three species

is 0.100 molar. The concentration of each of the three species

is 0.100 molar.

The k value is given as 0.078 at 100 degree Celsius.

Now what we need to calculate is q,

so I look at the concentration of the product

so the concentration of SO_2

times the concentration of Cl_2

and I need to divide that by

the concentration of SO_2-Cl_2

So here is my reaction quotient expression the concentration of SO_2-Cl_2

So here is my reaction quotient expression

I can now plug in the values that I know

which are going to be the same numbers for each concentration.

Because that is what was give in the problem, but it do not always have to be that way. which are going to be the same numbers for each concentration.

Because that is what was give in the problem, but it do not always have to be that way.

When I do this I see that my value of q is going to be equal to 0.100.

So now I have to look at the comparison between q and k.

And when I look at them I see that the value of

q, 0.100,

is greater then 0.078.

So I know that q is greater then k.

In order to reach equilibrium, I need the value of q to decrease

so I need q to decrease.

The way that the q decreases

is that the amount of my products decrease

and the amount of my reactant needs to increase.

So now, if this happens

the products decrease, the reactant increases So now, if this happens

the products decrease, the reactant increases

I am going to see that the value of q decreases the products decrease, the reactant increases

I am going to see that the value of q decreases

and what I see is that this will continue until q equals

k, at which point we will be at chemical equilibrium.

Now that we have figured out which way the reaction will proceed

we have to also figure out what equilibrium concentrations are Now that we have figured out which way the reaction will proceed

we have to also figure out what equilibrium concentrations are

we are going to go have to using an ICE table

but the problems are going to be a little bit more complicated

then what we saw in the previous section.

探索我们的目录

免费加入并获得个性化推荐、更新和优惠。

Coursera 致力于普及全世界最好的教育，它与全球一流大学和机构合作提供在线课程。

© 2019 Coursera Inc. 保留所有权利。

通过 App Store 下载

通过 Google Play 获取