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Name:______________________________
Date:_____________
CH105
Pre-Lab 7: Chemical Equilibrium
Directions: Read the Goals, Background, Safety, and Procedure sections for this
experiment, then answer the following questions in the space provided. For short
answer questions, write complete sentences and provide a reason for the answer. For
calculation questions, show all work and report answers in a box with the appropriate
significant figures and units. Pencil is acceptable for this assignment.
1. Write the forward reaction for the following equilibrium equation.
Fe3+ + NCS- →
← FeNCS2+
2. Write the reverse reaction for the following equilibrium equation.
Fe3+ + NCS- →
← FeNCS2+
3. Write the Keq expression for the following equilibrium equation.
3 N2H4 (g) + 4 ClF3 (g) →
← 12 HF (g) + 3 N2 (g) + 2 Cl2 (g)
4. The equilibrium constant for the following reaction is 0.212 at 100°C.
N2O4 (g) →
← 2 NO2 (g)
Is the forward or reverse reaction favored at 100°C?____________________________
Explain.
5. If you cool the equilibrium system from #4 and the K value
decreases, is the forward reaction endothermic or exothermic?_________________
Explain.
1
PCC
Experiment 7:
Chemical Equilibrium
Goals: • Represent chemical equilibrium with models.
• Determine an equilibrium constant using models.
• Observe Le Chatelier’s Principle for the iron-thiocyanate and
acetic acid-sodium acetate systems.
• Based on observations, determine whether the forward reaction
is endothermic or exothermic for the iron-thiocyanate
equilibrium system.
Purpose: Explore chemical equilibrium through models and chemical
reactions. Visualize Le Chatelier’s Principle with chemical systems.
Background:
Dynamic Chemical Equilibrium Defined
Until this point in your chemistry career, almost all of the reactions you
have studied involved one strong arrow pointing to the right (→). This
arrow showed you that reactions began with reactants and ended with
products.
While this way of representing chemical equations is useful to
understanding the concepts of reactants and products, it is slightly
deceptive. In reality, not all chemical reactions proceed only in a forward
direction. Some chemical reactions can reverse and turn products back
into reactants.
For instance, imagine you are making a sandwich in your kitchen. You
take 2 pieces of bread for every slice of cheese and make a cheese
sandwich.
2 slices of bread + 1 slice of cheese → 1 cheese sandwich
Logically, your reactants (bread and cheese) can become a sandwich, if
you provide the reaction energy. This reaction represents the forward
reaction.
This is not the only direction the reaction can proceed, though. Imagine
you have a cheese sandwich, but you want to eat only the bread. You
could break down your cheese sandwich into 2 slices of bread and 1 slice
of cheese.
1 cheese sandwich → 2 slices of bread + 1 slice of cheese
This second reaction would be considered the reverse reaction because
you took a product (cheese sandwich) and turned it back into reactants.
2
PCC
Chemical Equilibrium • 3
Standing in your kitchen, you could create an equilibrium system by
making cheese sandwiches and breaking them back down into bread and
cheese.
2 slices of bread + 1 slice of cheese →
← 1 cheese sandwich
Many chemical systems establish chemical equilibrium in a similar
manner: reactants create products which in turn create reactants. On a
macroscopic scale, it appears that no reaction is occurring; however, on
the molecular level, the process is occurring with reactants creating
products as fast as products are creating reactants.
In your kitchen, this equilibrium would make it appear as if the total
number of sandwiches, slices of bread, and slices of cheese were not
changing (macroscopic); however, you would still be breaking apart
sandwiches as fast as you were making sandwiches (molecular).
This constant action on the molecular level is the reason chemical
equilibrium is frequently referred to as dynamic chemical equilibrium.
Chemical Equilibrium and Concentration
In most chemical systems, the concentration of reactants and products
are not equal at equilibrium. What is equal at equilibrium is the rate (or
speed) of the forward and reverse reactions.
In your kitchen, you might have 25 cheese sandwiches made and only 2
slices of bread and 1 slice of cheese. As long as the rate of sandwich
making and sandwich separation are equal, you will have 25 cheese
sandwiches, 2 slices of bread, and 1 slice of cheese. The concentrations
of reactants and products do not have to be equal at equilibrium.
Consequently, chemists try to account for these differences in
concentration by writing equilibrium constant expressions. These
expressions help chemists to identify whether the forward or reverse
reactions are favored from a concentration standpoint. Consider the
general equation below:
aA +bB →
← cC + dD
A chemist would look at that equation and write an equilibrium constant
[C]c  [D]d
expression (Keq) that looks like: K eq 
[A]a  [B]b
If the value of Keq is greater than 1, then the products (produced by the
forward reaction) are favored. If the value of Keq is less than 1, then the
reactants (produced by the reverse reaction) are favored.
PCC
4 • Experiment 7
Let’s return to our cheese sandwich example. To write the Keq
expression, we must look at our equation.
2 slices of bread + 1 slice of cheese →
← 1 cheese sandwich
Here, the Keq expression would be:
K eq 
[cheese sandwich]1
[bread]2  [cheese]1
When the coefficient in an equilibrium expression is 1, it is not usually
written as an exponent. Any number to the first power is equal to itself.
The Keq expression could be rewritten as:
Keq 
[cheese sandwich]
[bread]2  [cheese]
The value of the Keq expression could be calculated from the equilibrium
concentration values for bread, cheese, and cheese sandwiches. Earlier,
the equilibrium concentration of cheese sandwiches was said to be 25,
while there were 2 slices of bread and 1 slice of cheese. Plugging these
numbers into the Keq expression, we find the value of Keq is 6.25.
K eq 
[25]
 6.25
[2]2  [1]
Because the Keq value is greater than 1, the products (or forward
reaction) is favored. The equilibrium is said to be “towards the right.”
We already suspected that because we knew the number of sandwiches
was greater than the amount of bread and cheese; however, the Keq value
confirmed it. This is extremely important at times that we do not know
the concentrations of reactants or products, but we do know the Keq
value.
When the equilibrium constant expression is written for the ionization
reaction between a weak acid and water, it is called a Ka. Acetic acid,
HC2H3O2, is a weak acid. When placed in water, the following
equilibrium reaction occurs:
HC2H3O2 (aq) + H2O (l) →
← C2H3O2 (aq) + H3O+ (aq)
The concentration of water will not change significantly because it is the
solvent for this reaction. Liquid and solid concentrations are not
included in equilibrium constant expressions; only gas and aqueous
phases, which can change concentration, are included. As long as some
liquid or solid is present, the reaction will occur.
Chemical Equilibrium • 5
Consequently, the Ka expression for acetic acid is:
Ka 
[C 2 H 3O -2 ]  [H 3O  ]
[HC 2 H 3O 2 ]
Several strong acids exist. Among these is hydrochloric acid. When
placed in water, hydrochloric acid completely ionizes as shown in the
following reaction:
HCl (aq) + H2O (l) → Cl- (aq) + H3O+ (aq)
Notice the arrow in this reaction is pointing only to the right. This
reaction is not an equilibrium system. The reaction is said to go to
completion. Consequently, the number of chloride ions and hydronium
ions formed by this reaction are equal to the number HCl molecules
dissolved in water. That is not the case for weak acids. For weak acids,
the number of anions and hydronium ions produced from each molecule
of weak acid has to be determined from the pH of the solution, known Ka
value, or percent ionization data.
Le Chatelier’s Principle
Once a chemical system establishes equilibrium, it will resist changes
introduced as stresses to the chemical system. These stresses can
include changes in concentration, pressure, volume, heat, temperature,
or presence of a catalyst.
Changes in Concentration
Whatever stress is introduced, the system will work to counteract the
stress and reestablish equilibrium. Consequently, if more reactant or
product is added to a chemical system at equilibrium, the chemical
system will shift to reestablish equilibrium by using part of the added
substance.
The shift in the favored reaction happens because the value of Keq at a
given temperature is a constant. In order to maintain the Keq for the
temperature, the chemical system has to make products (when extra
reactants are present or some products are removed) and use products
(when extra products are present or some reactants are removed). Table
1 summarizes the various concentration stresses a chemical system at
equilibrium can experience as well as the responses of the system in
order to maintain Keq at a given temperature.
PCC
6 • Experiment 7
Table 1.
Summary of Concentration Stresses, Responses, and Keq Effects.
Stress
Response
Keq
increase in a reactant
use reactants to produce products
does not change
as long as
decrease in a reactant
use products to make reactants
temperature is
increase in a product
use products to make reactants
constant
decrease in a product
use reactants to produce products
This process is impossible to observe if the reactants and products all
look alike; however, when the reactants and products have different
colors, the shift of the chemical system can be seen with the naked eye.
Changes in Volume and Pressure
Volume and pressure changes can be treated in much the same way as
changes in concentration when the equilibrium system involves gases.
As long as the system maintains constant temperature, the Keq value will
not change. Consequently, as the volume of a gaseous mixture that was
at equilibrium is increased, the chemical system will shift to minimize
this increase in volume by increasing the pressure.
Remember, volume and pressure are inversely related. When the volume
of a system increases, the pressure is lowered. The chemical system
increases pressure by shifting to the side of the reaction with more moles
of gas. As the number of gas molecules increase, the pressure increases
because these newly formed molecules will run into the walls of the
container more frequently than the lower number of molecules did.
Similarly, when the volume of a chemical system at equilibrium is
decreased, the equilibrium will shift to the side of the reaction with fewer
moles of gas in order to lower the pressure. Table 2 summarizes the
results of volume and pressure stresses on a chemical system at
equilibrium.
Table 2.
Summary of Volume and Pressure Stresses, Responses, and Keq Effects.
Stress
Response
Keq
volume increase
does not change
shift to side with more moles of gas
as long as
pressure decrease
temperature
is
volume decrease
shift to side with fewer moles of gas
constant
pressure increase
Changes in Heat
Heat can be treated as a reactant if the chemical system is endothermic
or a product if the chemical system is exothermic. An addition of heat
will shift the equilibrium in the direction that the heat is used. Heat is
not accounted for in the Keq expression. Consequently, the Keq value will
actually change with as the amount of heat changes. As the reaction
Chemical Equilibrium • 7
compensates for the change in heat, the concentrations of reactants and
products will change. This change will minimize the effect of changing
the heat of the system; however, because heat is not part of the Keq
equation, the Keq value will change to reflect the new concentrations of
the reactants and products.
(Before reading the next section on temperature changes, remember,
heat and temperature are not the same thing. A substance at its boiling
point or freezing point (constant temperatures) can experience a change
in heat without experiencing a change in temperature. Most equilibrium
systems are not studied at their boiling point or freezing point, though.
Consequently, a change in heat generally corresponds to a change in
temperature and vice versa.
Also note, heat can be written into an equation as a reactant or a
product, as appropriate. Occasionally, the heat for the reaction is
written quanitatively to the right of the equation as a H value. If the H
value is positive, the forward reaction is endothermic. If the H value is
negative, the forward reaction is exothermic.)
Changes in Temperature
The temperature of a reaction will affect the Keq value. As the
temperature is increased, the reaction that counteracts that effect
(endothermic) will become dominant. As the temperature is decreased,
the reaction that counteracts that effect (exothermic) will become
dominant. Consequently, by observing a chemical system at equilibrium
at two different temperatures, the forward reaction can be identified as
endothermic or exothermic depending on the favored conditions.
For instance, the following equilibrium is established when cobalt (II)
chloride, CoCl2, is dissolved in hydrochloric acid, HCl (aq):
→
Co(H2O)2+
6 (aq) + 4 Cl (aq) ← CoCl4 (aq) + 6 H2O (l)
pale pink
deep blue
The Co(H2O)2+
6 and CoCl4 (aq) are called complex ions. They have unique
colors that allow the forward and reverse directions of a reaction to be
distinguished from one another.
At room temperature, the equilibrium condition for this reaction looks
violet because both ions exist in solution. When the reaction is heated
and the temperature increases, more CoCl-4 (aq) ions form and the
reaction looks more blue. When the reaction is cooled and the
temperature is lowered, more Co(H2O)2+
6 (aq) ions form and the reaction
looks pink.
PCC
8 • Experiment 7
If this chemical system at equilibrium is working to counteract the stress
of increased heat by producing more CoCl-4 (aq) ions and counteract the
stress of decreased heat by producing more Co(H2O)2+
6 (aq) ions, heat
must be a reactant.
→
Co(H2O)2+
6 (aq) + 4 Cl (aq) + heat ← CoCl4 (aq) + 6 H2O (l)
As this reaction at equilibrium is heated by an outside source, the
chemical system works to reduce the heat by using it and produces more
of the product ions as a result. As the reaction at equilibrium is cooled
by an outside source, the chemical system works to increase the heat
and makes more reactant ions as a result. Consequently, we are able to
deduce that the forward reaction is endothermic (uses heat), while the
reverse reaction is exothermic (produces heat).
Table 3 summarizes the relationship between heat and temperature
stresses on an equilibrium system and the possible responses of the
system depending on whether the forward reaction is endothermic or
exothermic.
Table 3.
Summary of Heat/Temperature Stresses, Possible Responses of the Equilibrium
System Based on Nature of the Forward Reaction, and Resulting Change in Keq.
Stress
Possible Responses
Forward Reaction is…
Keq
make more products
endothermic
↑
heat/raise temperature
make more reactants
exothermic
↓
make more products
exothermic
↑
cool/lower temperature
make more reactants
endothermic
↓
Presence of a Catalyst
When a catalyst is added to a reaction, it does not actually react. It
merely lowers the activation energy of the reaction. That allows both the
forward and reverse reactions to occur more quickly. The concentration
of reactants and products do not change when a catalyst is added.
Consequently, the equilibrium does not shift when a catalyst is added.
On the macroscopic scale, no visible changes occur to a system at
equilibrium to which a catalyst has been added. On the molecular scale,
the rate of changing between reactants and products is faster.
To go back to the cheese sandwich analogy, a catalyst would enable you
to take apart and reassemble the cheese sandwiches faster. The number
of cheese sandwiches, slices of bread, and slices of cheese would not
change.
Chemical Equilibrium • 9
Reactions in today’s experiment
While examining Le Chatelier’s Principle, you will make the ironthiocyanate equilibrium system by reacting 1 M Fe(NO3)3 (aq) and 1 M
NH4NCS (aq).
The net ionic equation for this reaction is shown below.
Fe3+ (aq) + NCS- (aq) →
← FeNCS2+ (aq)
yellow
colorless
red
Depending on the Keq value for this reaction at room temperature, this
equilibrium system may be yellow (Keq less than 1; reactants favored), red
(Keq greater than 1; products favored), or orange (Keq approximately equal
to 1; reactants and products both present).
You will keep some of this solution as a color reference as you perform
several mini-experiments on this equilibrium system. As you perform
the experiments, watch to see if the color changes to more yellow (reverse
reaction favored and more reactants produced) or more red (forward
reaction favored and more products produced).
In preparation for the lab, you might want to predict which type of color
change you would expect from the following reactions:
If you add Fe(NO3)3 (aq), you will essentially add more Fe3+ to the system
at equilibrium. What type of stress is that? If you said, “It’s a
concentration stress, I am adding more reactant ions.” You would be
correct. What effect will that have on the chemical system? Would you
expect the solution to become more yellow or more red? Will the
equilibrium shift to the right (make more products) or to the left (make
more reactants)?
If you add 1 M NH4NCS, you will essentially add more NCS- (aq) to the
system at equilibrium. What type of stress is that? What effect will that
have on the chemical system? Would you expect the solution to become
more yellow or more red? Will the equilibrium shift to the right or to the
left?
If you add 0.1 M SnCl2 (aq), the tin (II) ions will reduce iron (III) ions to
iron (II) ions as seen in the equation below:
Sn2+ (aq) + 2 Fe3+ (aq) →
← 2 Fe2+ (aq) + Sn4+ (aq)
You will essentially remove some Fe3+ from the system at equilibrium.
What type of stress is that? What effect will that have on the chemical
system? Would you expect the solution to become more yellow or more
red? Will the equilibrium shift to the right or to the left?
PCC
10 • Experiment 7
If you add 0.1 M AgNO3 (aq), the silver ions will react with the
thiocyanate ions to give a white precipitate of silver thiocyanate as seen
in the equation below:
Ag+ (aq) + NCS- (aq) →
← AgNCS (s)
You will essentially remove some NCS- from the system at equilibrium.
What type of stress is that? What effect will that have on the chemical
system? Would you expect the solution to become more yellow or more
red? Will the equilibrium shift to the right or to the left?
If you add 0.1 M Na2HPO4 (aq), the hydrogen phosphate ions form a
complex ion with iron (III) ions as shown in the equation below:
+
Fe3+ (aq) + HPO24 (aq) →
← FeHPO4 (aq)
How will that affect the system at equilibrium? What type of stress is
that? Would you expect the solution to become more yellow or more red?
Will the equilibrium shift to the right or to the left?
If you add 1 M NH3 (aq), it will react as a base to form a precipitate or a
colloidal suspension of iron (III) hydroxide when mixed with iron (III) ions
as shown in the equations below:
+
NH3 (aq) + H2O (l) →
← NH4 (aq) + OH (aq)
Fe3+ (aq) + 3 OH (aq) →
← Fe(OH)3 (s)
How will that affect the system at equilibrium? What type of stress is
that? What effect will that have on the chemical system? Would you
expect the solution to become more yellow or more red? Will the
equilibrium shift to the right or to the left?
Procedure:
Modeling Equilibrium Constants
Equilibrium #1: Consider the equation A →
← B. For this Keq = [B]/[A] and
the value of Keq is 8.0 (or 8:1).
1. Obtain 2 petri dishes and 27 BB’s. One Petri dish will represent A,
the reactant side of the reaction, and the other will represent B, the
product side of the reaction.
2. Place all 27 BB’s on the reactant side (in A).
3. Move the BB’s from the reactant side to the product side until the Keq
expression as a value of 8.0. (Hint: Move one BB to petri dish B. What’s the
ratio of B:A? If it is not 8:1, then move another BB and recheck. Keep going until the
ratio is 8:1.)
Chemical Equilibrium • 11
4. At this point the system has reached equilibrium. How many BB’s are
on the reactant side at equilibrium? the product side? Record these
values on your data sheet.
5. Add 18 new BBs to the product side. Find the new equilibrium values
for A and B and record them on your data sheet. Remember, Keq =
8.0, regardless of how many total BB’s you have, the system will shift
the BB’s (with your help) to maintain the Keq value.
Equilibrium #2: Consider the equation C →
← D. For this Keq = [D]/[C] and
the value of Keq is 1/15 (or 0.067).
1. Begin with 32 BB’s on the product side. Move the BB’s to establish
equilibrium, then record the number of reactant and product BB’s on
your data sheet.
2. Start again with all 32 BB’s on the reactant side. Move the BB’s to
establish equilibrium, then record the number of reactant and
product BB’s on your data sheet.
3. Keq for A →
← B was 120 times larger than Keq for C →
← D. What does the
size of the Keq value have to do with the amount of reactants present
at equilibrium?
Equilibrium #3: Sometimes two equilibrium systems share reactants or
products. These equilibria will affect each other. They
cannot be written as one equation because they are
separate processes that happen to overlap.
Consider the equation E →
← F. For this Keq = [F]/[E] and
the value of Keq is 1/4 (or 0.25). Also, consider the
equation F →
← G. For this equation, Keq = [G]/[F] and the
value of Keq is 5.0.
1. Obtain 3 petri dishes to represent E, F, and G positions. Begin with
30 BB’s all representing E molecules. Move the BB’s from E to F and
from F to G to establish both equilibria. (Hint: Move one BB to petri dish F.
What’s the ratio of F:E? If it is not 1:4, then move another BB and recheck. Keep
going until the ratio is 1:4. Keep in mind, you will also need to have enough BB’s in
petri dish G that the ratio of G:F is 5:1.)
2. Record the values of E, F, and G at equilibrium on your data sheet.
3. Add 10 new BB’s to the F dish. Find the new equilibrium values of E,
F, and G. Record these values on your data sheet.
PCC
12 • Experiment 7
4. Do the two Keq values allow you to predict which compounds are most
abundant at equilibrium? Explain.
Le Chatelier’s Principle
Iron-thiocyanate system
1. Add 1 drop each of 1 M Fe(NO3)3 and 1 M NH4NCS to 25 mL of
distilled water. Mix well.
Fe3+ + NCS- →
← FeNCS2+
yellow colorless
red
2. Add a few drops of this solution to each of seven wells in a spot plate.
One well will serve as a color standard against which to judge color
changes in the other wells. The other six wells will be used to
introduce an external stress to this equilibrium system.
3. Add one drop of 1 M Fe(NO3)3 to one of the wells, mix, and observe.
Record your observations on the data sheet.
4. Add one drop 1 M NH4NCS to a second well, mix, and observe.
Record your observations on the data sheet.
5. Add one drop 0.1 M SnCl2 to a third well, mix, and observe. Record
your observations on the data sheet. Tin (II) ions reduce iron (III) ions
to iron (II) ions as seen in the equation below:
Sn2+ + 2 Fe3+ →
← 2 Fe2+ + Sn4+
6. Add one drop 0.1 M AgNO3 to a fourth well, mix, and observe. Record
your observations on the data sheet. Silver ions react with the
thiocyanate ions to give a white precipitate of silver thiocyanate as
seen in the equation below:
Ag+ + NCS- →
← AgNCS (s)
7. Add one drop 0.1 M Na2HPO4 to a fifth well, mix, and observe. Record
your observations on the data sheet. Hydrogen phosphate ions form a
complex ion with iron (III) ions as shown in the equation below:
+
Fe3+ + HPO24 →
← FeHPO4
Chemical Equilibrium • 13
8. Add one drop of 1 M NH3 (probably labeled NH4OH) to a sixth well,
mix, and observe. Record your observations on the data sheet. Any
base will form a precipitate or a colloidal suspension of iron (III)
hydroxide when mixed with iron (III) ions as shown in the equations
below:
+
NH3 + H2O →
← NH4 + OH
Fe3+ + 3 OH →
← Fe(OH)3 (s)
9. Pour 4-5 mL of the iron-thiocyanate solution made in step 1 into two
test tubes. Set one tube aside as a color standard against which to
judge color changes in the other tube.
10. Gently warm the other tube in a hot water bath on a hot plate. Do
not boil the solution. Observe and record your observations on your
data sheet.
11. After you have determined which way the reaction shifted, cool the
tube in a beaker of ice water. Record your observations on your data
sheet. Do these results indicate that the forward reaction is
exothermic or endothermic?
12. Make sure the reaction is reversible. Place the test tube in the warm
water bath again. Record your observations on your data sheet.
13. Place the test tube in the ice water bath again. Record your
observations on your data sheet.
14. Pour all of the solutions from this part of the experiment into the
waste container labeled “used iron-thiocyanate solutions.”
PCC
Name:______________________________
Date:_____________
CH105/Sp06
Data Sheet 7: Chemical Equilibrium
Directions: Record the data as it is collected onto this sheet in BLUE or BLACK ink. Do
not use white out. Correct mistakes by making a single line through the error and
writing the new information above or beside the mistake.
Modeling
A
B
Total
27 BB on A side
initially
18 new BB added
to B side
27
45
C
D
Total
32 BB on D side
initially
32 BB on C side
initially
32
32
Keq for A →
← B was 120 times larger than Keq for C →
← D. What does the size of
the Keq value have to do with the amount of reactants present at equilibrium?
E
F
G
30 BB on E
side initially
10 new BB
added to F
Total
30
40
Do the two Keq values allow you to predict which compounds are most
abundant at equilibrium? Explain.
14
PCC
Experiment 7
Data Sheet
Iron-thiocyanate system
Fe3+ + NCS- →
← FeNCS2+
yellow colorless
red

If you add a “stress” and the solution turns more yellow than the standard,
the reaction shifted to the left.

If you add a “stress” and the solution turns more red than the standard,
then the reaction shifted to the right.

If you add a “stress” and the solution looks the same, then no shift
occurred.
Stress
Color/Observations
Equilibrium Shift (Left or Right?)
Color/Observations
Equilibrium Shift (Left or Right?)
none
+ 1 drop Fe(NO3)3
+ 1 drop NH4NCS
+ 1 drop SnCl2
+ 1 drop AgNO3
+ 1 drop Na2HPO4
+ 1 drop NH3
Stress
heat
cool
heat
cool
Is the forward reaction endothermic or exothermic?____________________________
Explain.
15
PCC
Name:______________________________
Date:_____________
CH105/Sp06
Post Lab 7: Chemical Equilibrium
Directions: Answer the following questions in the space provided. For short answer
questions, write complete sentences and provide a reason for the answer. For
calculation questions, show all work and report answers in a box with the appropriate
significant figures and units. Pencil is acceptable for this assignment.
1. The following reaction was examined at 250°C:
PCl5 (g) →
← PCl3 (g) + Cl2 (g)
At equilibrium, [PCl5] = 4.2 x 10-5 M, [PCl3] = 1.3 x 10-2 M, and
[Cl2] = 3.9 x 10-3 M. Calculate the equilibrium constant for the reaction.
2. Is the forward or reverse reaction favored in question 1?____________________
Explain.
3. If the concentration of Cl2 (g) is increased
at 250°C, will the concentration of PCl3 (g)
increase, decrease, or stay the same?______________________________________
Explain.
4. If the concentration of Cl2 (g) is increased
at 250°C, will the concentration of PCl5 (g)
increase, decrease, or stay the same?______________________________________
Explain.
5. The equilibrium constant for the following reaction is 0.212 at 100°C.
N2O4 (g) →
← 2 NO2 (g)
H = 57.2 kJ*
*Did you see the H value information on page 7 of the lab?
As the temperature is lowered, what will the
equilibrium constant increase, decrease, or stay the same?______________
Explain.
16
PCC