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.

In this unit we are going to look at some additional examples

of how to calculate the equilibrium constant for our reaction.

In these examples we will be given information about the initial

and/or equilibrium concentrations.

We are going to show how we use the information

along with something known as an ICE table

to help us organize our data and solve for our equilibrium constant.

When we look at a reaction such as this one as we have

2 NO_2 in the gas phase

in equilibrium with N_2-O_4 also in the gas phase

We have a known K_c value here of

21 at 250 degrees Celsius.

If we change the temperature the equilibrium constant will change.

But if we are at the same temperature

what we find is regardless

of our initial concentrations

We see 4 different sets

we always end up with teh same K_c value. We see 4 different sets

we always end up with teh same K_c value.

Note that is initial concentrations

and the equilibrium concentrations

are all different from one another.

That when we get to equilibrium

what we are more worried about is the ratio

that defined by our K_c expression.

the concentration of the products

in this case divide by the

concentration of the NO_2 reactant

to the second power.

What we are going to use frequently in looking at equilibrium

problem is something known as an ICE table.

The ICE stands for Initial Change in Equilibrium.

This is way to sort the information that is given

and see what information we need to find out

and what we can assume based on the stoichiometry of the reaction.

Lets look at an example.

We have 2 SO_2 plus O_2 going to 2 SO_3

but have a balanced chemical equation

we have all our substances present in the gas phase

so in this case we could actually look at it in terms of the

pressures of the substances, or the concentrations.

This particular example goes through the

because the substances are all present in gases

we can look at either the K_c or the K_p using concentrations of pressures. because the substances are all present in gases

we can look at either the K_c or the K_p using concentrations of pressures.

In this problem we are given information about concentrations we can look at either the K_c or the K_p using concentrations of pressures.

In this problem we are given information about concentrations

and we are asked for the values of K_c

the first thing we want to do is read our problem and we are asked for the values of K_c

the first thing we want to do is read our problem

and see what information is given that we can put into our table.

So in our table we have 4 columns. We have one column and see what information is given that we can put into our table.

So in our table we have 4 columns. We have one column

for the I C E.

And then we have columns for each of the substances

in our chemical reaction.

So I start by seeing if the initial concentration

of SO_2 and O_2 are 0.100 molar.

So I am actually going to write that into my table.

I see that the equilibrium concentration of O_2

is 0.0750 molar. I see that the equilibrium concentration of O_2

is 0.0750 molar.

So now I have some information about my initial concentrations is 0.0750 molar.

So now I have some information about my initial concentrations

and one of my equilibrium concentrations. So now I have some information about my initial concentrations

and one of my equilibrium concentrations.

Know that of SO_3 we can assume starts at 0.

Now what we want to look at is the change row.

We get this information bu looking at the balance chemical equation.

What we can look at, is first I am going to look at oxygen because it has a coefficient of 1. We get this information bu looking at the balance chemical equation.

What we can look at, is first I am going to look at oxygen because it has a coefficient of 1.

I am going to say I lose oxygen

and I am know I am going to lose oxygen because I only have

reactants present, so the only way for this reaction

to proceed is towards the formation of SO_3.

When I say I am going to lose oxygen, I am going to say

X amount of oxygen, or X moles of oxygen.

When I look my balanced equation I see 2 in front of SO_2. X amount of oxygen, or X moles of oxygen.

When I look my balanced equation I see 2 in front of SO_2.

So I know the change in SO_2 is going to be 2 times

the amount of oxygen, because I have a coefficient of 2 for SO_2

and coefficient of 1 for oxygen.

I also know my change is going to be negative,

because just like with my oxygen I am going to lose SO_2 I also know my change is going to be negative,

because just like with my oxygen I am going to lose SO_2

and we are going to form SO_3.

We cannot lose any SO_3 because we do not start with any SO_3. and we are going to form SO_3.

We cannot lose any SO_3 because we do not start with any SO_3.

So when I look at the SO_3 it also has a 2 in front of it, so I know the change is going to be

2 x, or twice as much oxygen as I lose.

It is going to be a positive value because we are gaining SO_3.

Now I can write my equilibrium row

and put the values in that I know.

So I have 0.100 -2x and put the values in that I know.

So I have 0.100 -2x

in this case I already know the equilibrium concentration of oxygen.

So I do not need to put anything there.

And then I look at the SO_3

2x for the equilibrium concentration of SO_3.

But I need to find the value of X

because in order to find the value of K_c But I need to find the value of X

because in order to find the value of K_c

I need to know that value

to be able to plug in actual concentrations here for 2x and 2x.

But what I do know is that my equilibrium concentration of oxygen is

0.0750.

I know that is going to equal to the sum of the initial row

and the change row

so I have 0.100 - x.

I can rearrange.

I can actually solve for x, and what I end up with is X equals

0.0250.

Now, what this is going to let me do is plug this value for x

but into my equilibrium table, my ICE table Now, what this is going to let me do is plug this value for x

but into my equilibrium table, my ICE table

and solve for the actual concentration

of SO_2 and the actual concentration of SO_3 at equilibrium.

Once I know those actual concentrations I can plug them into my law of mass action.

and calculate the K_c value.

So here is my table, I have plugged these values in.

I have 0.10 - 2x

remembering that x had a value of 0.0250

so we end up with 0.05 for equilibrium concentrations for both

SO_2 and SO_3

and this is just how the numbers happened to work out.

There is nothing that requires that any of the concentration be equal to one another.

So now what is have are equilibrium concentrations for all three substances.

We can write the law of mass action.

So we have SO_3 squared

over SO_2 squared times O_2.

And again this comes the coefficients of my balanced chemical equation.

I can plug in the values I know

my concentrations I have in my equilibrium row of my ICE table.

Then I can solve the calculation

and find that the equilibrium constant is 13.3. Then I can solve the calculation

and find that the equilibrium constant is 13.3.

Now note, this is only true at this temperature.

If I change temperatures we are going to see a different value of K_c. Now note, this is only true at this temperature.

If I change temperatures we are going to see a different value of K_c.

If I were change my initial concentration of substances If I change temperatures we are going to see a different value of K_c.

If I were change my initial concentration of substances

I would still ultimately get to the same value of K_c

but what I would see is that my equilibrium concentrations

would be different from one another.

Now we are going to look at another example.

In this case what we are looking at are pressures instead of concentrations

but they way we approach the problem is no different. In this case what we are looking at are pressures instead of concentrations

but they way we approach the problem is no different.

We still have our ICE table

were we have out column I C E for Initial Change and Equilibrium

We still have a column for each of our three substances in our balanced chemical equation.

And we going to read through the problem and fill in what is given We still have a column for each of our three substances in our balanced chemical equation.

And we going to read through the problem and fill in what is given

so we can solve for the other rows and figure out what needs to go where.

So we have an initial pressure of ammonia

0.100 atmospheres.

And it is going to react, according to this reaction 0.100 atmospheres.

And it is going to react, according to this reaction

to NH_3 going to N_2 and 3 H_3 all in the gas phase.

At equilibrium the pressure of NH_3 is to NH_3 going to N_2 and 3 H_3 all in the gas phase.

At equilibrium the pressure of NH_3 is

0.80 atmospheres. At equilibrium the pressure of NH_3 is

0.80 atmospheres.

What are the initial pressures of N_2 and H_2?

Hopefully you realized that the pressures is going to be zero for both of them.

because there is no mention of having any N_2 and H_2 present Hopefully you realized that the pressures is going to be zero for both of them.

because there is no mention of having any N_2 and H_2 present

and if there is no mention of how much there is of something

initially, we can assume that the value is zero. and if there is no mention of how much there is of something

initially, we can assume that the value is zero.

So now we have our initial row complete

now we need to worry about the change row.

Remember to determine the terms we put in the change row for each substance

we have to refer back to our balanced chemical equation.

So when look at that equation, we see that B here is our correct answer

and what we look at is I look at 2 NH_3

that tells me the change is going to be 2x.

For N_2 because the coefficient is 1 the change will be x.

For H_2, the coefficient is 3 so I going to have 3x. For N_2 because the coefficient is 1 the change will be x.

For H_2, the coefficient is 3 so I going to have 3x.

I now need to figure, out what I am going to lose and what I am going to gain.

The easy way to look at this is I now need to figure, out what I am going to lose and what I am going to gain.

The easy way to look at this is

I have no pressure of Nitrogen or Hydrogen initially

which means I cannot lose them because i have nothing to lose.

Therefore I must be gaining nitrogen and gaining hydrogen.

If I am gaining products

I must be losing on the reactants side.

So I end up with -2x, x, and +3x for my change row. I must be losing on the reactants side.

So I end up with -2x, x, and +3x for my change row.

Now that we have both the initial and change rows

we can always use that information to find the values of X.

Because for NH_3 we know the initial pressure

and we know the equilibrium pressure and we know

the different in those two is equal to 2x.

So I can set up 1 - 2x

when we added my initial and change rows together

and it is going to be equal to 0.80.

I can then rearrange the equation

and solve for X, and find that the value of x

equal 0.10. and solve for X, and find that the value of x

equal 0.10.

I can then use this value of x equal 0.10.

I can then use this value of x

to find the value of each of the terms I can then use this value of x

to find the value of each of the terms

in the change row, as well as the equilibrium pressures of each substance. to find the value of each of the terms

in the change row, as well as the equilibrium pressures of each substance.

So now we need to look at what are the final pressures of N_2 and H_2.

Since the value of x is 0.10

we can say that x equal 0.10. Since the value of x is 0.10

we can say that x equal 0.10.

And we know that 3x will equal 0.30.

Since we started with no pressure of either nitrogen or hydrogen

the amount of change

will also equal the equilibrium

pressure of these two substances.

Now, that we the equilibrium pressures of each of our 3 substances

we can write the law of mass action

for this reaction, and solve for the value of K_p. we can write the law of mass action

for this reaction, and solve for the value of K_p.

So we know that when we are dealing with K_p we are going to be looking at pressure

so we have K_p equals

I have a 1 as a coefficient in front of my nitrogen

but I am not going to do anything to that value I just have the pressure of nitrogen

times

the pressure of hydrogen, and because there is coefficient of 3

I need to put that to the third power the pressure of hydrogen, and because there is coefficient of 3

I need to put that to the third power

over my reactant, NH_3 I need to put that to the third power

over my reactant, NH_3

so I have the pressure of NH_3.

And that is squared.

because I have a coefficient of 2 there.

Now using this information we can actually because I have a coefficient of 2 there.

Now using this information we can actually

calculate the K_p value for this reaction.

So what is the value K_p.

Well you should have gotten the answer D.

0.0042.

And we are going to look at on the next slide how we calculate that number.

Remember that I have to go back to my law of mass action

set it up for this balanced chemical equation and then Remember that I have to go back to my law of mass action

set it up for this balanced chemical equation and then

plug in the values that I know.

So if I do that, I know that I have K_p

equals the pressure of N_2

the pressure of H_2 cubed

over the pressure of NH_3 squared.

So I can look at my equilibrium pressure

and say the pressure of N_2

is 0.10.

The pressure of H_2 is 0.30.

But I put that to the third power. The pressure of H_2 is 0.30.

But I put that to the third power.

Over the pressure of the NH_3.

which is 0.80 Over the pressure of the NH_3.

which is 0.80

and that needs to be squared and when I do which is 0.80

and that needs to be squared and when I do

I get the value of 0.0042.

Now the value of this, and it is much much less then 1

tells me that my reactants are favored in this reaction.

We will form some products, it is an equilibrium process

but the amount of product formed is going to be very very small.

When we look at the amount of NH_3 we have left

compared to the amount of products we have we see that this is true.

We have .80 atmosphere of NH_3

only .1 and .3 atmospheres

of nitrogen and hydrogen.

In this examples we have looked at

we saw that the reaction could only proceed in one direction.

We had some reactants or some products present

and we knew we were going have to lose whatever we

had and gain the species we did not have.

However, not all reactions are that way.

Some cases we start with amounts of reactants and products However, not all reactions are that way.

Some cases we start with amounts of reactants and products

and before we can do any calculations

we determine equilibrium concentrations, or K_c or K_p values.

The first thing we have to figure out is which way the is the reaction going to precede

and to do that we use the reaction quotient.

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