2.07 Calculating Equilibrium Concentrations

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Do curso por University of Kentucky

Química Avançada

404 classificações

University of Kentucky

Química Avançada

404 classificações

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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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

Conheça os instrutores

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

In this unit we will learn how to calculate equilibrium concentraions.

Based on having some concentrations values and the value of k

we can use this information to calculate the equilibrium concentrations of our substances.

In order to find the equilibrium concentrations there are two things we can do

and the choice of those depends on what information we have. In order to find the equilibrium concentrations there are two things we can do

and the choice of those depends on what information we have.

If we know k and all but one of our

equilibrium concentrations or pressures

we can plug in our known values and solve for that one unknown.

Many of the problems we will look at

we will know the k value

and we will know the initial concentrations because we can

measure those when we do experiments

but we are going to have to calculate the equilibrium concentrations.

We will be using an ICE table that we introduced previously.

and by the time you are done with this unit

you should probably be dreaming on ICE tables. and by the time you are done with this unit

you should probably be dreaming on ICE tables.

If not, you probably have not done enough practice problems.

Here is an example of the first scenario.

We have a balanced chemical equation

of ammonia decomposing to nitrogen and hydrogen.

We have the value of k

notice that is has K_eq, that just stands for equilibrium.

We see a variety of subscripts when we look at K values.

In this case K_eq is the same as K_c.

We don't need to worry about the initial and change rows in our ICE

table because we know some of the equilibrium

concentrations and we are just solving for the one unknown.

So when we read through the problem

if the equilibrium concentrations of N_2 and H_2 So when we read through the problem

if the equilibrium concentrations of N_2 and H_2

are 0.500 molar if the equilibrium concentrations of N_2 and H_2

are 0.500 molar

and 0.100 molar respectively

what is the equilibrium concentration of NH_3?

Now, given that information

and knowing we can set up the k expression

we can actually solve for that unknown concentration.

Our k expression, the concentration of N_2

times the concentrations of H_2 cubed

divided by the concentrations of NH_3 squared

comes from our balanced chemical equation.

Remember it is products over reactants

and the coefficients become powers in the expression.

Now we can plug in the values that we know

and rearrange to solve for the NH_3.

We have NH_3 square equals 0.500 and rearrange to solve for the NH_3.

We have NH_3 square equals 0.500

times 0.100 cubed

divided by our k value of 3.81.

We can solve that to find NH_3 squared is

equal to 1.31 times 10 ^ -4.

Then the value of NH_3 as 0.0115 Molar.

If we were plug this value back into our k expression

with our concentrations of N_2 and H_2

we should find the value of k is 3.81.

Now lets look at a little bit more complicated situation.

Here we are given a balanced chemical equation

N_2 plus O_2 yield 2 NO.

We are given the k value 4.10 x 10 ^ -4 at 2000 Kelvin.

It does not really matter what the temperature is

what is important is that the temperature is constant through out the experiment.

I look at the information given in the problem.

0.500 moles of N_2 gas is mixed with

0.860 moles of O_2 gas

in a 2.00 liter tank at 2000 Kelvin.

What are the equilibrium concentrations of each substance?

So before I can do this problem

before I can even worry about the equilibrium concentrations

I need to know how to deal with the initial concentrations.

Note, that I am not given the concentration if N_2 and O_2 initially.

I am given the moles of N_2 and O_2

and I am given the volume.

So I can find the concentration in unit of molarity.

So for N_2 I take the 0.500 divided by 2

and get 0.250.

For Oxygen take 0.860 moles divided by 2

and get 0.430 molar.

I see there is no mention of the NO initially

so I can assume that value is 0.

Now, I have to look at the change row.

Remember that when we deal with the change row we have Now, I have to look at the change row.

Remember that when we deal with the change row we have

to go back to our balanced chemical equation

and look at the stoichiometry.

If I have look at the change in nitrogen and treat it as x

what I will see is that the change in nitrogen will be -x. If I have look at the change in nitrogen and treat it as x

what I will see is that the change in nitrogen will be -x.

The change in oxygen will be -x.

And the change of NO will be plus +2x.

Again, I am having to worry about whether I am gaining or losing

because I can only gain on the NO because I have nothing to lose there.

Now I have my change row

now I can set up my equilibrium row.

and it is simply the addition of

the initial row and the change row

so I get 0.250 - x for nitrogen

0.430 - x for oxygen

and 2x for my NO concentration. 0.430 - x for oxygen

and 2x for my NO concentration.

Now, I can take the k expression and my equilibrium row of values

and solve for the value of x

and therefore find the concentrations of

N_2, O_2, and NO at equilibrium.

So that is the next step, we have our data table

now I need to set up our k expression.

So k is going to equal to NO squared because I have NO as a product. now I need to set up our k expression.

So k is going to equal to NO squared because I have NO as a product.

2 is the coefficient of it, so that becomes the power.

N_2 is an reactant 2 is the coefficient of it, so that becomes the power.

N_2 is an reactant

and O_2 is a reactant so they both go into the denominator.

They both have coefficients of 1

so I do not have any powers associated with them.

Now, I have k equals NO squared

over N_2 times O_2

I can plug in the values I know over N_2 times O_2

I can plug in the values I know

and I find that I get 4.10 x 10 ^ -4

equals 2x squared

because I had 2x in my equilibrium row

I still have to remember to square that

because 2x represents the concentration.

The square is part of the law of mass action.

On the bottom, I have 0.250 - x The square is part of the law of mass action.

On the bottom, I have 0.250 - x

and 0.430 - x.

Now if I were to solve this

and rearrange to solve time for x

what I am going to find, is that I have a quadratic equation.

This makes the calculation a little more complicated. what I am going to find, is that I have a quadratic equation.

This makes the calculation a little more complicated.

However, in some cases

we can make a simplifying assumption.

That simplifying assumption is

if k is small That simplifying assumption is

if k is small

then we can assume x is much less then the original concentration.

So for example, if my concentration is 0.39

and I know the value of x is going to very small

then I can basically treat that number 0.39.

The value of x is not zero

but it is so small that it does not reflect

the value once I take significant figures into account.

So for example if the value of x was 0.0023

then what I would find is 0.39 - x

is equal to 0.3877

and I would still end up with same initial value of 0.39.

Note that k must have a small value.

After we do the calculation we will also look at

how we can test to make sure we made a valid assumption.

So why do we simplify, and when can we simplify?

Well, it is a small value of x.

The reaction favors the reactants.

That tells us, that if we start with reactants

and we have a small k value

very little of the product will actually be formed. and we have a small k value

very little of the product will actually be formed.

If very little of the product is formed

that means the value lost from the reactants side that change

row, the values with respect to x.

Will be small, and therefore

the concentration at equilibrium Will be small, and therefore

the concentration at equilibrium

will not be that different then the concentration initially.

Our check at the end

is to check to see if x is less Our check at the end

is to check to see if x is less

than 5% of the value from which it was subtracted.

If we look at the example we looked at on the previous slide

we had the value of 0.0023

and we were comparing that to 0.39.

We see that this number is much much less then 0.39

we also see that is less then 5%

of the value of .39

and so we have an acceptable approximation, we can use the simplifying assumption.

If the value is greater then 5%

we will have to redo our calculation If the value is greater then 5%

we will have to redo our calculation

using the quadratic equation.

So now we are going to show our law of mass action here.

We have our values substituted in that we have seen before.

Now we are going show how we use this simplifying assumption.

And the first thing I am going to do is say that

lets assume that x is much much less the 0.250.

Now by default if x much much less then 0.250

it must also be much much less then 0.430.

Making this simplifying assumption makes our problem much easier to solve.

We can avoid the quadratic equation

which greatly simplifies the mathematical calculations needed.

Because we have assumed that it is much much smaller

we actually can cancel out the minus x. [-x].

We assume that the value 0.250...

We assume that the value 0.250

- x is approximately equal to

0.250

and likewise for the 0.430.

Now we can rewrite our expression.

We still have the same value of k.

We still have the value of x on the top

because even though it is a small value it will still have a mathematically We still have the value of x on the top

because even though it is a small value it will still have a mathematically

significance when it is numerator multiplied by 2.

On the bottom, we have now simplified to having 0.250

and 0.430.

Now, we multiply 4.10 x 10 ^ -4

by both 0.250 and 0.430

and we get 4.41 x 10 ^ -5 by both 0.250 and 0.430

and we get 4.41 x 10 ^ -5

equals 2x

notice this is in brackets

squared. So everything inside those brackets needs to be squared. notice this is in brackets

squared. So everything inside those brackets needs to be squared.

I end up with x equals 3.32 x 10 ^ -3. squared. So everything inside those brackets needs to be squared.

I end up with x equals 3.32 x 10 ^ -3.

So I ended up with this being equal to 4 x squared. I end up with x equals 3.32 x 10 ^ -3.

So I ended up with this being equal to 4 x squared.

I divided both side by 4 So I ended up with this being equal to 4 x squared.

I divided both side by 4

and then I took the square root of the number

and got my x value to be 3.32 x 10 ^ -3.

We also have to say, is our assumption valid?

Was it reasonable

to make the approximation that x was much much less then 0.250?

We look at the value of x and divide

by value from which it was subtracted.

So 3.32 x 10 ^ -3

over .250 times 100

and we find that it is less then 5%.

It is only 1.33%.

So this lets us know that our assumption was valid.

We don't need to check the assumption for .430

because if it was less then 5% of .250

this number getting larger, our denominator getting larger.

is only going to be a smaller percentage.

Now we go back to our ice table

we see the values of our equilibrium concentrations

0.250 - x

0.430 - x, and 2x.

And we plug in our value of x

that we solved for in our previous slide

into these expressions

and find the equilibrium concentrations of

nitrogen, oxygen and NO.

So for nitrogen, 0.250 - x

is equal to 0.250 - 3.32 x 10 ^ -3 = .247.

For Oxygen we have .430 - x

and we end up with .427 molar. For Oxygen we have .430 - x

and we end up with .427 molar.

And for NO we we end up with 2x being equal to

6.54 x 10 ^ -3 molar.

Now, we have the equilibrium concentration for each of our substances.

We still have subtract x from our concentration of N_2 and O_2

even thought we made the simplifying assumption that they were much We still have subtract x from our concentration of N_2 and O_2

even thought we made the simplifying assumption that they were much

smaller then the value, in this case they still changed the value slightly.

not a lot, but it does change it, and we need to include it. smaller then the value, in this case they still changed the value slightly.

not a lot, but it does change it, and we need to include it.

But because it is such a small amount

the simplifying approximation made in the previous calculations is valid.

There are other examples of equilibrium problems as worked

examples that you can review to see some different example to see how these problems are worked.

We will not , however, do any problems involving the quadratic equation.

On all of our problems, we will be able to use the simplifying assumption. We will not , however, do any problems involving the quadratic equation.

On all of our problems, we will be able to use the simplifying assumption.

In the next unit we will look at LaSalle principle.

This idea helps us determine

how a system as equilibrium responds

when the equilibrium is disturbed.

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