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.

From the lesson

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 ChÃ¢telierâ€™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

Â

of how to calculate the equilibrium constant for our reaction.

Â

In these examples we will be given information about the initial

Â

and/or equilibrium concentrations.

Â

We are going to show how we use the information

Â

along with something known as an ICE table

Â

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

Â

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

Â

2 NO_2 in the gas phase

Â

in equilibrium with N_2-O_4 also in the gas phase

Â

We have a known K_c value here of

Â

21 at 250 degrees Celsius.

Â

If we change the temperature the equilibrium constant will change.

Â

But if we are at the same temperature

Â

what we find is regardless

Â

of our initial concentrations

Â

We see 4 different sets

Â

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

Â

we always end up with teh same K_c value.

Â

Note that is initial concentrations

Â

and the equilibrium concentrations

Â

are all different from one another.

Â

That when we get to equilibrium

Â

what we are more worried about is the ratio

Â

that defined by our K_c expression.

Â

the concentration of the products

Â

in this case divide by the

Â

concentration of the NO_2 reactant

Â

to the second power.

Â

What we are going to use frequently in looking at equilibrium

Â

problem is something known as an ICE table.

Â

The ICE stands for Initial Change in Equilibrium.

Â

This is way to sort the information that is given

Â

and see what information we need to find out

Â

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

Â

Lets look at an example.

Â

We have 2 SO_2 plus O_2 going to 2 SO_3

Â

but have a balanced chemical equation

Â

we have all our substances present in the gas phase

Â

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

Â

pressures of the substances, or the concentrations.

Â

This particular example goes through the

Â

because the substances are all present in gases

Â

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

Â

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

Â

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.

Â

In this problem we are given information about concentrations

Â

and we are asked for the values of K_c

Â

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

Â

the first thing we want to do is read our problem

Â

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

Â

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.

Â

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

Â

for the I C E.

Â

And then we have columns for each of the substances

Â

in our chemical reaction.

Â

So I start by seeing if the initial concentration

Â

of SO_2 and O_2 are 0.100 molar.

Â

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

Â

I see that the equilibrium concentration of O_2

Â

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

Â

is 0.0750 molar.

Â

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

Â

So now I have some information about my initial concentrations

Â

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

Â

and one of my equilibrium concentrations.

Â

Know that of SO_3 we can assume starts at 0.

Â

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

Â

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

Â

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.

Â

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

Â

I am going to say I lose oxygen

Â

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

Â

reactants present, so the only way for this reaction

Â

to proceed is towards the formation of SO_3.

Â

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

Â

X amount of oxygen, or X moles of oxygen.

Â

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

Â

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

Â

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

Â

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

Â

and coefficient of 1 for oxygen.

Â

I also know my change is going to be negative,

Â

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

Â

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

Â

and we are going to form SO_3.

Â

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

Â

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

Â

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

Â

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

Â

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

Â

Now I can write my equilibrium row

Â

and put the values in that I know.

Â

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

Â

So I have 0.100 -2x

Â

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

Â

So I do not need to put anything there.

Â

And then I look at the SO_3

Â

2x for the equilibrium concentration of SO_3.

Â

But I need to find the value of X

Â

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

Â

because in order to find the value of K_c

Â

I need to know that value

Â

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

Â

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

Â

0.0750.

Â

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

Â

and the change row

Â

so I have 0.100 - x.

Â

I can rearrange.

Â

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

Â

0.0250.

Â

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

Â

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

Â

but into my equilibrium table, my ICE table

Â

and solve for the actual concentration

Â

of SO_2 and the actual concentration of SO_3 at equilibrium.

Â

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

Â

and calculate the K_c value.

Â

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

Â

I have 0.10 - 2x

Â

remembering that x had a value of 0.0250

Â

so we end up with 0.05 for equilibrium concentrations for both

Â

SO_2 and SO_3

Â

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

Â

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

Â

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

Â

We can write the law of mass action.

Â

So we have SO_3 squared

Â

over SO_2 squared times O_2.

Â

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

Â

I can plug in the values I know

Â

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

Â

Then I can solve the calculation

Â

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

Â

and find that the equilibrium constant is 13.3.

Â

Now note, this is only true at this temperature.

Â

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

Â

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

Â

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

Â

If I were change my initial concentration of substances

Â

I would still ultimately get to the same value of K_c

Â

but what I would see is that my equilibrium concentrations

Â

would be different from one another.

Â

Now we are going to look at another example.

Â

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

Â

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

Â

but they way we approach the problem is no different.

Â

We still have our ICE table

Â

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

Â

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

Â

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.

Â

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

Â

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

Â

So we have an initial pressure of ammonia

Â

0.100 atmospheres.

Â

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

Â

And it is going to react, according to this reaction

Â

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

Â

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

Â

At equilibrium the pressure of NH_3 is

Â

0.80 atmospheres. At equilibrium the pressure of NH_3 is

Â

0.80 atmospheres.

Â

What are the initial pressures of N_2 and H_2?

Â

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

Â

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.

Â

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

Â

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

Â

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

Â

initially, we can assume that the value is zero.

Â

So now we have our initial row complete

Â

now we need to worry about the change row.

Â

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

Â

we have to refer back to our balanced chemical equation.

Â

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

Â

and what we look at is I look at 2 NH_3

Â

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

Â

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

Â

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.

Â

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

Â

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

Â

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.

Â

The easy way to look at this is

Â

I have no pressure of Nitrogen or Hydrogen initially

Â

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

Â

Therefore I must be gaining nitrogen and gaining hydrogen.

Â

If I am gaining products

Â

I must be losing on the reactants side.

Â

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

Â

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

Â

Now that we have both the initial and change rows

Â

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

Â

Because for NH_3 we know the initial pressure

Â

and we know the equilibrium pressure and we know

Â

the different in those two is equal to 2x.

Â

So I can set up 1 - 2x

Â

when we added my initial and change rows together

Â

and it is going to be equal to 0.80.

Â

I can then rearrange the equation

Â

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

Â

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

Â

equal 0.10.

Â

I can then use this value of x equal 0.10.

Â

I can then use this value of x

Â

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

Â

to find the value of each of the terms

Â

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

Â

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

Â

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

Â

Since the value of x is 0.10

Â

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

Â

we can say that x equal 0.10.

Â

And we know that 3x will equal 0.30.

Â

Since we started with no pressure of either nitrogen or hydrogen

Â

the amount of change

Â

will also equal the equilibrium

Â

pressure of these two substances.

Â

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

Â

we can write the law of mass action

Â

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

Â

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

Â

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

Â

so we have K_p equals

Â

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

Â

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

Â

times

Â

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

Â

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

Â

I need to put that to the third power

Â

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

Â

over my reactant, NH_3

Â

so I have the pressure of NH_3.

Â

And that is squared.

Â

because I have a coefficient of 2 there.

Â

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

Â

Now using this information we can actually

Â

calculate the K_p value for this reaction.

Â

So what is the value K_p.

Â

Well you should have gotten the answer D.

Â

0.0042.

Â

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

Â

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

Â

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

Â

set it up for this balanced chemical equation and then

Â

plug in the values that I know.

Â

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

Â

equals the pressure of N_2

Â

the pressure of H_2 cubed

Â

over the pressure of NH_3 squared.

Â

So I can look at my equilibrium pressure

Â

and say the pressure of N_2

Â

is 0.10.

Â

The pressure of H_2 is 0.30.

Â

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

Â

But I put that to the third power.

Â

Over the pressure of the NH_3.

Â

which is 0.80 Over the pressure of the NH_3.

Â

which is 0.80

Â

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

Â

and that needs to be squared and when I do

Â

I get the value of 0.0042.

Â

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

Â

tells me that my reactants are favored in this reaction.

Â

We will form some products, it is an equilibrium process

Â

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

Â

When we look at the amount of NH_3 we have left

Â

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

Â

We have .80 atmosphere of NH_3

Â

only .1 and .3 atmospheres

Â

of nitrogen and hydrogen.

Â

In this examples we have looked at

Â

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

Â

We had some reactants or some products present

Â

and we knew we were going have to lose whatever we

Â

had and gain the species we did not have.

Â

However, not all reactions are that way.

Â

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

Â

Some cases we start with amounts of reactants and products

Â

and before we can do any calculations

Â

we determine equilibrium concentrations, or K_c or K_p values.

Â

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

Â

and to do that we use the reaction quotient.

Â

Explore our Catalog

Join for free and get personalized recommendations, updates and offers.