CH-180
Equilibrium and Rates
One of the most important goals of science is to learn enough about a natural system to
allow us to control that system. If we can control the system, we can cause it to behave in a way
that benefits us. Medicine is an excellent example of an entire discipline built on the idea that the
behavior of a system (an ill patient) can be altered, to bring about a desirable outcome (healing).
The ability to control the progress and outcome of a chemical reaction gives the chemist the means
to more economically produce valuable chemicals (drugs, plastics, ceramics, etc.) and to reduce
production of harmful chemicals (e.g., acid rain).
Just as medicine's success depends on a thorough understanding of how the human body works, the
chemist's success in controlling reactions depends on a thorough understanding of how chemical
reactions work. In this lab, we look at two main factors that influence the progress of a chemical
reaction: equilibrium and the rate of reaction.
Equilibrium
For some chemical reactions, a percent yield of 100% is not achieved, regardless of how
carefully we set up and perform the reaction. This is sometimes due to side reactions, which are
alternative reactions, which also use up the reactants that we intend to be used in the main reaction.
Often, however, percentage yields are not 100% because the reactants and products can exist in a
state of equilibrium, i.e., the chemical reaction occurs in both the forward and reverse directions,
and the forward and reverse reactions can both occur at the same rate. The double arrow indicates
that equilibrium is established:
A + BC + D
Equilibrium exists whenever the forward and reverse reactions occur at the same rate. It is possible,
however, to change the reaction conditions so that the reaction is temporarily out of equilibrium.
The reaction responds to the change by producing either more products (forward reaction) or more
reactants (reverse reaction) until a new equilibrium is established.
The French chemist Le Châtelier studied equilibrium reactions in the 1880's. From his
research, he was able to conclude that equilibrium systems always respond to a disturbance in a way
that minimizes the disturbance. This means that ...
1. If extra reactants are added, or if some of the products are removed from an equilibrium system,
the system responds by producing more products, until a new equilibrium is established. In other
words, the forward direction of the reaction is favored until the new equilibrium is reached.
2. If extra products are added, or if some of the reactants are removed from an equilibrium system,
the system responds by producing more reactants, until a new equilibrium is established. In other
words, the reverse direction of the reaction is favored until the new equilibrium is reached.
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Let's now summarize Le Châtelier's Principle in a table. For equilibrium system A + B  C + D,
Disturbance
add extra A or B
remove some C or D
add extra C or D
remove some A or B
Response
more C and D are produced
more C and D are produced
more A and B are produced
more A and B are produced
Reaction Direction Favored
forward ()
forward ()
reverse ()
reverse ()
We can generalize Le Châtelier's Principle to account for other disturbances as well, such as
changes in pressure and temperature.
For this experiment, you will work with three equilibrium systems:
1. FeCl3(aq) + 3NH4SCN(aq)
yellow
colorless
Fe(SCN)3(aq) + 3NH4Cl(aq)
deep red
colorless
Here we see a reaction involving the polyatomic ion thiocyanate, SCN-.
2. SbCl3(aq) + H2O(l)
colorless colorless
SbOCl(s) + 2HCl(aq)
white ppt colorless
The SbOCl precipitate in this reaction is called antimony oxychloride.
3. H Phen(aq)
colorless
H+(aq) + Phen-(aq)
colorless red
HPhen is our abbreviation for phenolphthalein, a complicated organic molecule. The H appearing
at the beginning of "HPhen" tells us that phenolphthalein is really an acid. As an acid,
phenolphthalein can undergo neutralization if a base is present. For example:
HPhen(aq) + NaOH(aq)  Na+(aq) + Phen-(aq) + H2O(l)
Thus Phen- (phenolphthalein’s red form) exists if a base is present. If an acid is present, the Phenform can react with the acid to form the colorless HPhen form of phenolphthalein. For example:
Phen-(aq) + HCl(aq)  HPhen(aq) + Cl-(aq)
Because phenolphthalein's color differs in acidic and basic solutions, we can use it as an indicator to
tell us whether a solution is acidic or basic. We used indicators in one of our previous experiments.
Do you remember what indicators we used?
Rates of Reaction
Let's now expand our power over chemical reaction systems by learning how bond type,
concentration, temperature, and catalysts control the rate at which a chemical reaction occurs. Le
Châtelier's Principle allows us to maximize or minimize the amount of product formed in a
reaction. But Le Châtelier's Principle says nothing about how long it takes to form the product.
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What good would it do to maximize the yield of a compound if it took months - or thousands of
years - for the reaction to produce that maximum yield? Time is an important factor in any
chemical reaction, especially for reactions which benefit us economically: By speeding up reaction
rates, winemakers can produce their wine in a shorter period of time, thereby cutting the time that
they must wait to realize profits from sales; by slowing reaction rates, grocers can transport fresh
fruits and vegetables from warm climates to cold climates with less fear of spoilage; pharmaceutical
manufacturers can ensure that their medicines remain potent, even after months on the shelf.
While it may be the chemist who is most interested in chemical kinetics (the study of the
rates of chemical reactions) people in all walks of life can find situations in which understanding
how to control the rate of a reaction results in better health, higher income, or safer living.
Bond type, concentration, temperature, and catalysts are all major factors influencing
reaction rates, but there are many other influencing factors as well, such as particle size,
atmospheric pressure, and degree of mixing (stirring.) Moreover, we should remember that the
effects of bond type, concentration, temperature, and catalysts differ for every chemical reaction;
indeed, rate studies are more difficult to understand than are equilibrium studies, because factors
affecting rate create a greater variety of effects than do factors affecting equilibrium. Nevertheless,
we can get a general idea of how rate is affected by considering - on the molecular level - how each
of our four main factors act on reacting molecules:
A. Bond Type. In general, the reaction rate is slower for covalently-bonded reactants than for
ionic-bonded reactants. Why? Consider the following: when covalently-bonded compounds react,
bonds must be broken to create product molecules. Broken bonds violate the octet rule for at least
two atoms in the reactant molecules. But, atoms are reluctant to violate the octet rule, so the
reaction will occur only slowly. Also, covalent bonds have a specific location within the molecule,
so that only those reactants that collide in just the right way are able to react:
O
CH3
H3 C
H
H
H
but
H
N:
H
O
N:
H
yields
H3 C
CH3
H3 C
CH3
+ H2O
N:
H
Of course, the reactants may bump around until they get into the right orientation to react, but this
bumping and jostling takes time, slowing the rate of reaction.
Ionic compounds, on the other hand, consist of charged atoms (ions) that typically continue to obey
the octet rule even during reaction. Take, for example, the ionic reaction,
Na2SO4 + BaCl2  BaSO4 + 2NaCl
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A check of the charges in all four compounds shows that all ions retained their original charges.
Also, ions show their electrical charges equally in all directions, so there is no preferred orientation
for reaction:
charge is observed
Ba2+
to be the same
from all directions
B. Concentration. The rate of reaction usually increases when the concentration of reactants
increases. An increase in reactants’ concentration means that more molecules will be present in the
reaction vessel to collide; collisions of reactant molecules lead to chemical reaction. This is really a
matter of probability. Examples of concentration effects are easy to find; here are two examples:
1. Iron oxidizes much more rapidly in 100% oxygen than in air (which is only 20% oxygen):
Fe + air (20% O2)  Fe2O3
slow rusting over several days
Fe + O2 (pure oxygen)  Fe2O3 Fe rapidly burns with orange flame
2. Fats dissolve faster in concentrated NaOH solution than in dilute NaOH solution. This
knowledge helps when treating a clogged drain: drain clogs are often caused by grease buildup in
the plumbing. Products like Drano or Liquid Plumr contain NaOH in high concentration to
dissolve the fatty grease. Directions on the containers of both products tell the user to run only a
little water after treatment. The manufacturers know that running a lot of water after treatment
would dilute the NaOH so much that the drain would either remain clogged, or clear very slowly:
Fatty acid + NaOH (conc.)  soluble salt of fatty acid (very fast)
Fatty acid + NaOH (dilute)  soluble salt of fatty acid (slow)
C. Temperature. The reaction rate usually increases when the temperature of the reaction
mixture is raised. To understand how temperature increases reaction rate, we must look at the
energy involved in a reaction. The graph gives a
reaction’s "energy history." Two features are important:
First, the energy levels of the reactants and products: the
products are at lower energy than the reactants, indicating
that the atoms involved in the reaction are more stable
configured as product molecules than they are as reactant
molecules. Second, in order to get from the reactants to
products, the atoms must gain energy temporarily (this is
signified by the big hump in the graph). It is this second
feature that is influenced by temperature increases.
If we increase the reaction mixture’s temperature, more energy is available (as heat) for the
reactant molecules to take up in order to progress beyond the "energy hump" and form products. If
more energy is available, a greater number of reactant molecules can make it "over the hump" in a
given period of time. For example:
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1. Water is produced by the reaction of hydrogen and oxygen.
room
2H2(g) + O2(g)  2H2O(g)
(billions of years)
temp.
spark
2H2(g) + O2(g)  2H2O(g)
(explosion!)
or flame
2. Food spoils readily at higher temperatures.
Room Temp.
spoils in 2 days
hamburger
spoils in a week
Refrigerator
D. Catalyst. The reaction rate increases when a
catalyst is present. Neither a product nor a reactant, a
catalyst is present in a reaction for the solely to
increase the reaction rate. A catalyst increases
reaction rate by providing an energy "shortcut" - a way
of going from reactants to products requiring less
energy than would be required without the catalyst.
How catalysts provide this shortcut varies from catalyst to catalyst, and is the subject of much
research. In living systems, enzymes often function as catalysts:
C6H12O6 + 6O2  6CO2 + 6H2O (many years)
glucose
enzyme
C6H12O6 + 6O2  6CO2 + 6H2O (a few hours)
Although we present equilibrium and rates as different ways to control a chemical reaction,
the distinction is more conceptual than actual. In fact, equilibrium can be readily explained on the
basis of the rates of the forward and reverse reactions that together make up an equilibrium system.
We know we can increase the rate of a reaction by increasing the concentration of one (or more) of
the reactants. This is precisely what happens when we disturb equilibrium! If we increase the
concentration of "A" in a system, we will raise the rate of the forward reaction:
A+BC+D
However, the concentration of "C" and "D" did not change immediately after we increased the
concentration of "A", so the rate of the reverse reaction did not change:
C+DA+B
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Consequently, "A" and "B" molecules are converted to "C" and "D" faster than "C" and
"D" are converted to "A" and "B". We experience a buildup of "C" and "D"; the equilibrium
shifted right, just as Le Châtelier predicted!
Equilibrium is eventually re-established: as we convert "A" and "B" to "C" and "D", we
increase the concentration of "C" and "D", and reduce the concentration of "A" and "B". Can you
guess the result? The rate of the forward reaction (A + B  C + D) goes down, and the rate of the
reverse reaction (C + D  A + B) goes up, until they become equal once more, and equilibrium is
restored. The main thing to remember is that equilibrium is restored in a system that now has
more "C" and "D" present than it had before we added the extra "A" and "B".
Procedure
I. Equilibrium
1. FeCl3(aq) + 3NH4SCN(aq)
Fe(SCN)3(aq) + 3NH4Cl(aq)
a. Observe the color of the reactants before they are mixed, and record your observations.
b. Place 50 mL distilled H2O into a clean 150 mL beaker. Add 1 mL each of 0.1 M FeCl3 and 0.1
M NH4SCN. Stir the solution with a glass stir rod. Record observations of the solution's color.
c. Transfer 5 mL of the solution into each of four 13  100 mm test tubes. Keep one of the test
tubes (#1) for the purpose of comparing the color in the other three tubes with the original solution.
d. To test tube #2, add 1 mL of 0.1 M FeCl3 solution. Record your observations.
e. To test tube #3, add 1 mL of 0.1 M NH4SCN solution. Record your observations.
f. To test tube #4, add 1 mL of 0.1 M NH4Cl solution. Again, record your observations.
2. SbCl3(aq) + H2O(l)
SbOCl(s) + 2HCl(aq)
Using the work that you did in the previous section as a guide (not necessarily for the amounts of
each solution), design and carry out an experiment that clearly demonstrates control of this system’s
equilibrium. In other words, you should be able to use your procedure to push the reaction towards
products (SbOCl – a white precipitate – and HCl), and then, by performing another step in your
procedure, push the reaction back toward the reactants (SbCl3 solution and H2O). To get started on
your design, you should include the following:
 The reaction should be carried out in a clean beaker.
 Use 2 mL of 0.1 M SbCl3 to start.
 Use 6M HCl as your source of hydrochloric acid.
 Be as precise as possible when recording your observations.
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3. HPhen(aq)
+
-
H (aq) + Phen (aq)
Again, design and carry out an experiment that shows how changes in the amount of hydrogen ion
shifts the equilibrium. Think about how you can change (both increase and decrease) the [H+].
You will use no more than 3-5 drops of phenolphthalein solution in a test tube to get an observable
effect.
II. Rates of Reaction
A. Bond Type. Reactions performed are:
6FeSO4 + 2KMnO4 + 4H2SO4
3H2C2O4 + 2KMnO4 + H2SO4
3Fe2(SO4)3 + 2MnO2(s) + K2SO4 + 4H2O (ionic)
6CO2 + 2MnO2(s) + K2SO4 + 4H2O
(covalent)
1. Place 2 mL of distilled water and ten drops of 3 M H2SO4 in each of three clean test tubes,
labeled as numbers 1, 2, and 3.
2. Into test tube (tt) #1, add nothing. Add a few crystals of iron (II) sulfate (FeSO4) to tt #2, and a
few crystals of oxalic acid (H2C2O4) to tt #3.
3. Shake tt #2 and tt #3 until all solids are dissolved.
4. Add two drops of 0.02 M KMnO4 solution to each test tube and mix thoroughly. Contrast the
rate of color disappearance in test tubes 2 and 3 against test tube 1. Record your conclusions.
B. Concentration. We’ll perform three reactions in quick succession:
slow
3HSO3-(aq)
IO3-(aq)
+
hydrogen sulfite iodate
 3SO42-(aq) + 3H+(aq) + I- (rate-determining step1)
sulfate
fast
IO3-(aq)
+ 5I (aq) + 6H (aq)  3I2(s) + 3H2O(l)
-
+
(excess)
fast
I2(s) + starch  starch-I2 complex (dark blue)
1. Thoroughly clean two graduated cylinders and a 150 mL beaker. Your success here depends on
clean glassware! Load one cylinder only with A; the other only with B.
2. Measure 10.0 mL of potassium iodate solution (contains IO3-; labeled as solution A) into a
cylinder.
We observe changes in this step’s rate. The other two steps are fast; they are needed only to create the
dark blue color that indicates the completion of the slow, rate-determining step.
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3. Measure 10.0 mL of sodium hydrogen sulfite solution (contains
into the second cylinder.
HSO3-;
labeled as solution B)
4. Simultaneously pour solutions A and B into a 150 mL beaker containing 50 mL of distilled
water. Stir the mixture for only 2 or 3 sec with a stirring rod. Use a watch with a second hand to
record the number of seconds that elapse from the moment of mixing to the appearance of a blue
color. Record your result.
5. Repeat the previous procedures, but this time use a potassium iodate solution made by diluting
5.0 mL of solution A with 5.0 mL of distilled water (concentration of solution B remains the same,
so use 10.0 mL of B). Record your result.
C. Temperature. Reactions performed are the same ones used in part B. Again, make sure the
glassware is CLEAN!
1. Measure 10.0 mL of solution A into a cylinder.
2. Measure 10.0 mL of solution B into the other cylinder.
3. Simultaneously pour solutions A and B into a 150-mL beaker containing 50 mL of distilled
water that has been carefully heated on a hotplate to a temperature between 33 and 35 °C.
(Check with your thermometer.) Record the time required for the blue color to appear.
4. Repeat the preceding procedures, but this time, initially cool the 50 mL of water to 10 °C by
using an ice bath. Record the time required for the blue color to appear.
D. Catalyst.
Reactions performed are:
2H2O2  2H2O + O2(g)
(no catalyst)
FeCl3
2H2O2  2H2O + O2(g)
(catalyst)
1. Measure 5 mL of 3% hydrogen peroxide into two separate test tubes.
2. Add five drops of 0.1 M iron (III) chloride (FeCl3) solution to one of the test tubes and five drops
of water to another test tube.
3. Observe the evolution of oxygen in each test tube.
III. End of Lab Cleanup
We have two duties to complete before we leave the lab:
1. Clean any equipment and glassware around your work area, and
2. Restock the lab equipment drawers that are assigned to you by the instructor on the day
of the lab.
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Name ________________________
Equilibrium and Rates Report
1. Observations of FeCl3(aq) + 3NH4SCN(aq)
Color of FeCl3
Fe(SCN)3(aq) + 3NH4Cl(aq)
Color of NH4SCN _____________________
Color of solution when FeCl3 and NH4SCN are mixed (test tube #1) _____________________
Color after more FeCl3 is added (tt #2) _____________________
Color after more NH4SCN is added (tt #3) _____________________
Color after NH4Cl is added (tt #4) _____________________
Are your observations consistent with Le Châtelier's Principle?
Explain how you reached your answer:
2. Observations of SbCl3(aq) + H2O(l)
SbOCl(s) + 2HCl(aq)
Color of SbCl3 solution (in bottle) _____________________
Describe the procedure that you used:
YES
NO (circle one)
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Observations:
3. Observations for HPhen(aq)
H+(aq) + Phen-(aq)
Color of phenolphthalein (in bottle) _____________________
Describe the procedure that you used:
Observations:
4. Questions. By analogy with your observations in this experiment, predict which direction,
forward or reverse, is favored when each of the following reactions undergo the specified
disturbances:
a. CaF2(s)  Ca2+(aq) + 2F-(aq)
disturbance
addition of Ca2+
addition of NaF
favored direction (  or  )
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b. 2K2CrO4(aq) + H2SO4(aq)  K2Cr2O7(aq) + H2O(l) + K2SO4(aq)
disturbance
addition of K2CrO4
removal of H2O
addition of KOH
addition of K2Cr2O7
favored direction (  or  )
c. CaF2(s)  Ca2+(aq) + 2F-(aq) (i)
HF(aq) + H2O  H3O+(aq) + F-(aq) (ii)
disturbance
addition of HF
addition of any base
removal of Fremoval of HF
favored direction (  or  )
Note: in (c), we have two equilibria in one sample. In other words, all the chemical species shown
in equations (i) and (ii) are present in a single solution. The question asks you to predict how the
specified disturbances in equation (ii) will affect equation (i). Note also that hydronium ion, H3O+,
will react with any base added to the solution.
d. Hemoglobin in red blood cells carries O2 as the blood circulates from the lungs to the
cells. We can represent the uptake of O2 in the following way, where "Hem" represents a
hemoglobin molecule:
Hem + O2  HemO2
Physiologists know that the concentration of oxygen is much higher in the lungs than in
the body's cells. Use this information and the equilibrium concept to explain how
hemoglobin can take up O2 in the lungs and release it in the body's cells.
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II. Rates of Reaction
Bond Type: Circle the type of bond -- ionic or covalent -- which reacted fastest.
Concentration and Temperature: Summarize your conclusions regarding the effect of
concentration and temperature on reaction rate, using the data recorded in the table
below.
Concentration:
Temperature:
Present your data in this table:
Section
B
B
C
C
Volume of A (mL)
Volume of B (mL)
Temp. (°C)
Time of Reaction (sec)
Catalyst: Which decomposition of hydrogen peroxide (with or without the catalyst; circle one)
reacted fastest? Describe the observations that led you to your conclusion.