From your earlier experiences in chemistry you have learned to recognize that a chemical reaction can be represented by a chemical equation. A chemical equation contains the chemical formulas of reactants and products separated by an arrow. We then add coefficients to balance the chemical reaction. We in introductory chemistry have come to rely very heavily on the chemical equation as a representation of a chemical reaction. Some examples of chemical reactions that we have seen in class are;

Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq)

2H2(g) + O2(g) ---> 2H2O(g)

We will all admit that it is a little exciting to actually see some of these reactions occur. Another reaction that might be fun to watch can be expressed in the following equation;

2K(s) + Br2(l) ---> 2KBr(s)

These three reactions are examples of reactions that go too completion. A reaction that proceeds from left to right until one of the reactants is complete used up. It is very difficult for me to be able to show you the reverse reaction of any of these three systems.

In Chapter 17 we begin to look at chemical reactions in more detail. To afford this type of scrutiny we will be studying a particular type of reaction referred to as an equilibrium reaction. The term equilibrium is not unfamiliar to us because we used it in Chapter 13 and 14 to describe vapor pressure. The equilibrium vapor pressure for a liquid was attained when the rate of condensation of a liquid was equal to the rate of evaporation of the liquid. This equilibrium vapor pressure was dependent on the temperature of the system. We will study the idea of chemical equilibrium in more detail in this chapter and relate it to chemical systems. Let's begin by looking at an example of a chemical reaction that we know to exhibit some of the observable behavior that we associate with chemical equilibrium. I'm not providing you with a definition of chemical equilibrium just yet, rather I would like to show you some examples of chemical systems which exhibit the behavior that we associate with chemical equilibria.

The reaction we will study is;

Fe3+(aq) + SCN-(aq) ---> FeSCN2+(aq)

View the movie on the right and answer the following three questions:

1. Identify the chemical species present in the solution poured into the beaker on the left in the movie. Answer

2. Identify the chemical species present in the solution poured into the beaker on the right in the movie. Answer

3. What evidence is there for a chemical reaction when the two solutions are mixed? Answer

In the reaction the color of each species is;

Fe3+(aq) + SCN-(aq) ---> FeSCN2+(aq)

yellowish

clear

reddish-brown

 

In the movie equal samples of the reaction mixture are added to two petri dishes.

To the solution in the petri dish on the left several drops of Fe3+ are added.

Describe what you observe when the Fe3+ is added to the solution in the petri dish.

What chemical species must have been formed to account for your observation?

What chemical species did the Fe3+, that was added to the solution in the petri dish, react with to produce the change you observed?

Answer

 

 

In this movie equal samples of the reaction mixture were added to two petri dishes. To the petri dish on the left several drops of Fe3+ have been added.

To the solution in the petri dish on the right several drops of SCN- are added.

Describe what you observe when the SCN- is added to the solution in the petri dish.

What chemical species must have been formed to account for your observation?

What chemical species did the SCN-, that was added to the solution in the petri dish, react with to produce the change you observed?

Answer

 

What is really interesting about adding Fe3+(aq) to the petri dish on the left and SCN-(aq) to the petri dish on the right are the results we observe. Adding some Fe3+(aq) and observing the solution get darker implies that there must be some unreacted SCN-(aq) in the petri dish on the left. So when the original sample of Fe3+(aq) and SCN-(aq) were mixed there is unreacted SCN-(aq) available to react with any additional Fe3+(aq) .

But wait, when some SCN-(aq) is added to petri dish on the right the color gets darker implying there is unreacted Fe3+(aq) in that solution. So when the original sample of Fe3+(aq) and SCN-(aq) were mixed there is unreacted Fe3+(aq) available to react with any additional SCN-(aq).

How can that be!!! How can it be possible to mix the original samples of Fe3+(aq) and SCN-(aq) split the resulting mixture and find there is still some unreacted Fe3+(aq) and SCN-(aq) in both solutions?!

The only way this could happen is if when the original solutions of Fe3+(aq) and SCN-(aq) were mixed, some product, FeSCN2+(aq) was formed, but the reaction did not completely use up either of the reactants. Some Fe3+(aq) and SCN-(aq) remained unreacted. So when the sample was split into two beakers the addition of more reactant resulted in more reaction.

What is even more interesting is what happens when F-(aq) is added to a fresh sample of FeSCN2+(aq). Before describing what is observed it is important to know that mixing F-(aq) with Fe3+(aq) produces the colorless compound FeF63-. So adding F-(aq) will react with any Fe3+(aq) in the solution.

Describe what you observe when the F- is added to the solution in the beaker.

Explain why the solution changes color.

Answer

 

 

 

 

Amazing, it is as though the reaction

Fe3+(aq) + SCN-(aq) ---> FeSCN2+(aq)

can also go backwards;

FeSCN2+(aq) ---> Fe3+(aq) + SCN-(aq)

This is unlike any reaction we have investigated before.

Actually this is not that unique, and we describe such a reaction as an equilibrium reaction. The equation is written;

Fe3+(aq) + SCN-(aq) FeSCN2+(aq)

Where the symbol represents a reaction that is capable of going in both directions.

Here is the clip from our discussion in class about this experiment.

Now we are ready to investigate an equilibrium reaction from a submicroscopic, or particulate level view.

The reaction we will study is;

A(g) + BC(g) ---> AB(g) + C(g)

We have a reaction mixture which contains 3 mol of A(g) and 3 mol of BC(g).

 

 

 

A(g) + BC(g) ---> AB(g) + C(g)

We have a reaction mixture which contains 3 mol of A(g) and 3 mol of BC(g).

The question I have is, after the reaction begins can you predict the amount of AB that will be formed and the amounts of AB(g) + C(g) that will form after the reaction has progressed for a while?

 

 

 

Lets watch what happens when the reaction

A(g) + BC(g) ---> AB(g) + C(g)

occurs.

 

 

The reaction between A(g) + BC(g) is

A(g) + BC(g) ---> AB(g) + C(g)

We can set-up an important table that summarizes the initial and final conditions we observe in the reaction. Initially we had 3.0 mols of A(g) and 3.0 mol of BC(g). (View the discuss for experiment #1 from lecture.)

Experiment #1

[A]

[BC]

[BC]

[C]

Initial

3.0

3.0

0

0

Change

 

 

 

 

Equilibrium

 

 

 

 

The reaction was started and allowed to proceed until it was 'over'. We observed the reaction actually never stopped, but it did get to a point where it did not change much.

Experiment #1

[A]

[BC]

[BC]

[C]

Initial

3.0

3.0

0

0

Change

 

 

 

 

Equilibrium

0.6

0.6

2.4

2.4

Now we can complete the Change row of information in the table. In the case of A the number can be calculated by subtracting the initial amount from the final amount

Experiment #1

[A]

[BC]

[BC]

[C]

Initial

3.0

3.0

0

0

Change

-2.4

-2.4

+2.4

+2.4

Equilibrium

0.6

0.6

2.4

2.4

Notice the numbers in the change row follows the stoichiometry of the reaction, that is, 2.4 mol A reacted and 2.4 mol BC reacted and 2.4 mol AB and 2.4 mol of C formed.

So what about a new experiment?

Experiment #2

[A]

[BC]

[BC]

[C]

Initial

4.0

4.0

0

0

Change

 

 

 

 

Equilibrium

 

 

 

 

What would you predict the equilibrium amounts of A, BC, AB and C will be? Answer (View the discuss for experiment #2 from lecture.)

In the next experiment we halve (based on experiment #2) the initial amounts of A and BC.

Experiment #3

[A]

[BC]

[BC]

[C]

Initial

2.0

2.0

0

0

Change

 

 

 

 

Equilibrium

 

 

 

 

What would you predict the equilibrium amounts of A, BC, AB and C will be? Answer (View the discuss for experiment #3 from lecture.)

In experiment #4 we'll start with some BC and C

Experiment #4

[A]

[BC]

[BC]

[C]

Initial

0

0

3.0

3.0

Change

 

 

 

 

Equilibrium

 

 

 

 

What would you predict the equilibrium amounts of A, BC, AB and C will be? Answer (Hint: look at experiment #1) (View the discuss for experiment #4 from lecture.)

In experiment #5 we'll start with some BC and C

Experiment #5

[A]

[BC]

[BC]

[C]

Initial

0

0

4.0

4.0

Change

 

 

 

 

Equilibrium

 

 

 

 

What would you predict the equilibrium amounts of A, BC, AB and C will be? Answer (Hint: look at experiment #2)

In experiment #6 we'll try something complete different. In this case we know the initial amounts of A, BC, AB and C, AND one of the equilibrium amounts.

Experiment #6

[A]

[BC]

[BC]

[C]

Initial

4.0

3.2

0

0

Change

 

 

 

 

Equilibrium

 

 

 

2.8

What do you predict the equilibrium amounts of A, BC, AB and C will be? Answer

In experiment #7 we'll try a real tough question. In this case we know the initial amounts of A, BC, AB and C.

Experiment #7

[A]

[BC]

[BC]

[C]

Initial

4.0

2.6

1.0

2.0

Change

 

 

 

 

Equilibrium

 

 

 

 

What would you predict the equilibrium amounts of A, BC, AB and C will be?

This is a more difficult question. Let consider some possible guesses for the final equilibrium amounts of A, BC, AB and C to see if we can determine what the answer is.

Look at each of the guesses below to see whether it is a reasonable guess or if it is an unreasonable guess. How do we tell whether the guess is reasonable or not? The first check is to see if the guess follows the stoichiometry of the reaction.

Guesses

Guess

A

BC

BC

C

Reasonable (Y/N)

1

3.0

2.0

2.0

3.0

(Yes/No)

2

2.5

1.1

2.5

3.5

(Yes/No)

3

2.2

.8

2.8

3.8

(Yes/No)

4

1.85

0.45

3.15

4.15

(Yes/No)

5

4.2

2.8

1.2

2.2

(Yes/No)

6

4.5

3.1

.5

1.5

(Yes/No)

Click here to continue.