0:17

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

Â 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

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

Â If we know k and all but one of our

Â equilibrium concentrations or pressures

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

Â Many of the problems we will look at

Â we will know the k value

Â and we will know the initial concentrations because we can

Â measure those when we do experiments

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

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

Â and by the time you are done with this unit

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

Â you should probably be dreaming on ICE tables.

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

Â 0:55

Here is an example of the first scenario.

Â We have a balanced chemical equation

Â of ammonia decomposing to nitrogen and hydrogen.

Â We have the value of k

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

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

Â In this case K_eq is the same as K_c.

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

Â table because we know some of the equilibrium

Â concentrations and we are just solving for the one unknown.

Â So when we read through the problem

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

Â if the equilibrium concentrations of N_2 and H_2

Â are 0.500 molar if the equilibrium concentrations of N_2 and H_2

Â are 0.500 molar

Â and 0.100 molar respectively

Â what is the equilibrium concentration of NH_3?

Â Now, given that information

Â and knowing we can set up the k expression

Â we can actually solve for that unknown concentration.

Â 1:43

Our k expression, the concentration of N_2

Â times the concentrations of H_2 cubed

Â divided by the concentrations of NH_3 squared

Â comes from our balanced chemical equation.

Â Remember it is products over reactants

Â and the coefficients become powers in the expression.

Â 2:03

Now we can plug in the values that we know

Â and rearrange to solve for the NH_3.

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

Â We have NH_3 square equals 0.500

Â times 0.100 cubed

Â divided by our k value of 3.81.

Â 2:19

We can solve that to find NH_3 squared is

Â equal to 1.31 times 10 ^ -4.

Â Then the value of NH_3 as 0.0115 Molar.

Â If we were plug this value back into our k expression

Â with our concentrations of N_2 and H_2

Â we should find the value of k is 3.81.

Â 2:43

Now lets look at a little bit more complicated situation.

Â Here we are given a balanced chemical equation

Â N_2 plus O_2 yield 2 NO.

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

Â It does not really matter what the temperature is

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

Â I look at the information given in the problem.

Â 0.500 moles of N_2 gas is mixed with

Â 0.860 moles of O_2 gas

Â in a 2.00 liter tank at 2000 Kelvin.

Â What are the equilibrium concentrations of each substance?

Â So before I can do this problem

Â before I can even worry about the equilibrium concentrations

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

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

Â I am given the moles of N_2 and O_2

Â and I am given the volume.

Â So I can find the concentration in unit of molarity.

Â So for N_2 I take the 0.500 divided by 2

Â 4:01

Now, I have to look at the change row.

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

Â Remember that when we deal with the change row we have

Â to go back to our balanced chemical equation

Â and look at the stoichiometry.

Â 4:11

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

Â 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

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

Â The change in oxygen will be -x.

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

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

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

Â 4:34

Now I have my change row

Â now I can set up my equilibrium row.

Â and it is simply the addition of

Â the initial row and the change row

Â so I get 0.250 - x for nitrogen

Â 0.430 - x for oxygen

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

Â and 2x for my NO concentration.

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

Â and solve for the value of x

Â and therefore find the concentrations of

Â N_2, O_2, and NO at equilibrium.

Â 5:06

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

Â now I need to set up our k expression.

Â 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.

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

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

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

Â N_2 is an reactant

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

Â They both have coefficients of 1

Â so I do not have any powers associated with them.

Â Now, I have k equals NO squared

Â over N_2 times O_2

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

Â I can plug in the values I know

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

Â equals 2x squared

Â because I had 2x in my equilibrium row

Â I still have to remember to square that

Â because 2x represents the concentration.

Â The square is part of the law of mass action.

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

Â On the bottom, I have 0.250 - x

Â and 0.430 - x.

Â Now if I were to solve this

Â and rearrange to solve time for x

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

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

Â This makes the calculation a little more complicated.

Â However, in some cases

Â we can make a simplifying assumption.

Â 6:16

That simplifying assumption is

Â if k is small That simplifying assumption is

Â if k is small

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

Â So for example, if my concentration is 0.39

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

Â then I can basically treat that number 0.39.

Â The value of x is not zero

Â but it is so small that it does not reflect

Â the value once I take significant figures into account.

Â So for example if the value of x was 0.0023

Â then what I would find is 0.39 - x

Â is equal to 0.3877

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

Â Note that k must have a small value.

Â After we do the calculation we will also look at

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

Â 7:09

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

Â Well, it is a small value of x.

Â The reaction favors the reactants.

Â That tells us, that if we start with reactants

Â and we have a small k value

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

Â very little of the product will actually be formed.

Â If very little of the product is formed

Â that means the value lost from the reactants side that change

Â row, the values with respect to x.

Â Will be small, and therefore

Â the concentration at equilibrium Will be small, and therefore

Â the concentration at equilibrium

Â will not be that different then the concentration initially.

Â Our check at the end

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

Â is to check to see if x is less

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

Â 7:48

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

Â we had the value of 0.0023

Â and we were comparing that to 0.39.

Â We see that this number is much much less then 0.39

Â we also see that is less then 5%

Â of the value of .39

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

Â 8:27

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

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

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

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

Â lets assume that x is much much less the 0.250.

Â 9:06

Because we have assumed that it is much much smaller

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

Â We assume that the value 0.250...

Â We assume that the value 0.250

Â - x is approximately equal to

Â 0.250

Â and likewise for the 0.430.

Â 9:27

Now we can rewrite our expression.

Â We still have the same value of k.

Â We still have the value of x on the top

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

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

Â significance when it is numerator multiplied by 2.

Â On the bottom, we have now simplified to having 0.250

Â and 0.430.

Â 9:49

Now, we multiply 4.10 x 10 ^ -4

Â by both 0.250 and 0.430

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

Â and we get 4.41 x 10 ^ -5

Â equals 2x

Â notice this is in brackets

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

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

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

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

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

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

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

Â I divided both side by 4

Â and then I took the square root of the number

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

Â 10:29

We also have to say, is our assumption valid?

Â Was it reasonable

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

Â We look at the value of x and divide

Â by value from which it was subtracted.

Â So 3.32 x 10 ^ -3

Â over .250 times 100

Â and we find that it is less then 5%.

Â It is only 1.33%.

Â So this lets us know that our assumption was valid.

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

Â because if it was less then 5% of .250

Â this number getting larger, our denominator getting larger.

Â is only going to be a smaller percentage.

Â 11:11

Now we go back to our ice table

Â we see the values of our equilibrium concentrations

Â 0.250 - x

Â 0.430 - x, and 2x.

Â And we plug in our value of x

Â that we solved for in our previous slide

Â into these expressions

Â and find the equilibrium concentrations of

Â nitrogen, oxygen and NO.

Â 11:32

So for nitrogen, 0.250 - x

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

Â For Oxygen we have .430 - x

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

Â and we end up with .427 molar.

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

Â 6.54 x 10 ^ -3 molar.

Â 11:58

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

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

Â 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

Â even thought we made the simplifying assumption that they were much

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

Â 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.

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

Â But because it is such a small amount

Â the simplifying approximation made in the previous calculations is valid.

Â 12:27

There are other examples of equilibrium problems as worked

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

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

Â 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.

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

Â