EQUILIBRIUM
My Goals for this Lesson:

Explain the concept of dynamic equilibrium.
I’m preparing to understand and to be able to explain the concept of dynamic equilibrium.
Introduction
Fill in the blanks using the interactive component on the Introduction page.
Imagine that there are two island communities connected to each other by a bridge. One island has a
population of 100,000, and the other has a population of 10,000. How would the flow of traffic on the
bridge affect the populations on each island?
If the bridge only allowed one-way traffic, how would that affect the populations on each island?
The population on the island from which the traffic was leaving would
on the other island would
while the population
.
What if the bridge allows for two-way traffic, but there is only one lane heading east and three lanes
heading west?
The island on the east side of the bridge would be
while the island on the west side of the bridge would be
. This means that the population in the east would
would
more occupants than it would be
,
more occupants than it would be
as the population in the west
.
How would the populations on each island be affected if the traffic flow on the bridge was exactly the
same in both directions?
Even though the identities of the cars and people would be constantly changing as people moved across
the bridge in either direction, the total population on each island would stay
This illustrates a concept in chemistry that we call
.
. As you learn about
,
it will be important to remember that the rates, or speeds, of opposite reactions are equal when a
system is at
. Just as the traffic did not stop on the bridge, reactions do not stop when
they reach
, but the equal rates provide us with steady amounts on either side of the
reaction.
Equilibrium
Fill in this table with your own words using the interactive section from the Lesson.
When nitrogen dioxide gas (NO2) is sealed into an evacuated glass container at
room temperature, you can observe the initial dark brown color begin to fade as
the brown nitrogen dioxide is
to colorless dinitrogen tetroxide
(N2O4).
If only the
reaction were to occur, we would expect the sample to
eventually be completely
as all of the NO2 is converted to N2O4.
However, even over a long period of time, you will see that the gas inside the
container
becomes completely colorless. This is because the
reaction (2NO2 → N2O4) does not go to
.
The gas mixture of NO2 and N2O4 inside the container eventually reaches a
constant
color that does not change at a constant temperature.
This constant color shows us that the concentrations of NO2 and N2O4 are no
longer
completion.
, which seems to indicate that the forward reaction has stopped before reaching
If you could observe the reaction at the
stopped. The
level, you would see that the reaction has not
reaction did not fully complete because the
was also occurring. When the forward and the reverse reactions occur at the
reaction (N2O4 → 2NO2)
rate or speed, the
concentration of the reactants and products have stopped changing.
At the beginning of this reaction, only the
reaction was occurring because only nitrogen dioxide
was present in the container. As the nitrogen dioxide particles collided together and reacted, its
concentration
as it was consumed in the reaction to form the product. Remember that as
the concentration of reactants decrease during a reaction, the rate of the reaction also decreases.
At the same time, the concentration of the dinitrogen tetroxide
as it was being formed by
the forward reaction. As more and more dinitrogen tetroxide particles were produced, the collisions
between these particles increased. Remember that an increase in concentration increases the probability
of successful collisions, increasing the rate of the reaction. This means that the rate of the reverse
reaction (N2O4 → 2NO2) is increasing as the rate of the forward reaction is decreasing.
Eventually, the rates of the forward and the reverse reactions will be
. This
does not mean that the concentrations of the reactants and the products will be
equal with each other, but that each of the concentrations will remain constant once
the rates of the forward and the reverse reactions are the same. Once the forward
and the reverse reactions are occurring at the same rate, the system is said to be in
a state of dynamic equilibrium.
Dynamic equilibrium is a state of
reactions
to occur, but the
in which the forward and the reverse
of the forward and the reverse
reactions are equal. All reactions have the potential to reach equilibrium if they
occur in a
system.
Example - Treadmill
We can compare an equilibrium system to a
. If you have ever walked on a
treadmill, you know that you are walking forward as the belt of the treadmill is moving in
the
direction.
If you walk or run
than the rate of the treadmill’s belt, you will move
on
the treadmill.
If you are moving forward at a speed that is
moving
than the belt’s speed, you will find yourself
(and possibly falling off the back of the treadmill).
If you are able to walk forward at the
rate that the treadmill’s belt is
moving backward, you will find yourself walking in place. This does not mean that you or the
belt has stopped moving; it just means that you are moving in
directions at
rates. This is true of a chemical system at equilibrium. When the forward and the
reverse reactions are occurring at the
products will
the
rate, the concentrations of the reactants and the
to remain constant. This does not mean that the reactions have stopped, but that
of the forward and the reverse reactions are happening at the
speed.
Phase Equilibrium
In your own words, tell how phase changes also undergo dynamic equilibrium.
In your own words, tell how water sealed in an evacuated flask can demonstrate
dynamic equilibrium.
Graphs
Use the lesson to describe the parts of each graph and how they relate to dynamic equilibrium.
Equilibrium Constant
Science is based largely on experimental investigations, and the description of
equilibrium systems is no different. In 1864, two Norwegian chemists, Cato
Guldberg and Peter Waage, used data and observations from numerous reaction
systems to propose the law of mass action to describe the equilibrium condition.
Let’s examine the basic concepts of the law of mass action.
Consider the following general reversible reaction. In this example, w moles of A react with x moles of B
to produce y moles of C and z moles of D in the forward reaction (w, x, y, and z are coefficients, A and B
are reactants in the forward reaction, and C and D are products of the forward reaction).
wA + xB  yC + zD
At equilibrium, the concentrations of each substance in the system can be plugged into the equation
below. The law of mass action proposes that, for a reaction at a given temperature, this equation can be
used to relate the equilibrium concentrations of the substances in the system to the reaction’s
equilibrium constant (K).
K = [C]y x [D]z
[A]w x [B]x
Remember that brackets are used to represent concentration in molarity (M).
The value of a system's equilibrium constant (K) is constant at a given temperature, regardless of the
initial amounts of the reactants and the products introduced into the system. Although there is only one
value for the equilibrium constant for a system at a given temperature, there are infinite combinations
of equilibrium concentrations of the reactants and the products within the system.
Use the interactive section on Properties to complete this chart.
Each set of equilibrium concentrations is called an
position reached by a system depends on the
. The specific equilibrium
concentrations of each substance and can be
predicted based on the value of the equilibrium constant (K).
Equilibrium constants and the law of mass action can be used to
and make
the behavior of a variety of equilibrium systems. We can use the values of K to
about
the progress
of different reactions.
Because the law of mass action provides an equation where the concentration of products is in the
numerator and the concentration of reactants is in the denominator, it follows that the
the value of the equilibrium constant is related to the position of the equilibrium.
As you can see from the law of mass action's equation, the greater the value of
K, the farther to the right the system will proceed before reaching equilibrium.
A system with an equilibrium constant value greater than one.
According to the law of mass action:
If K > 1, that means [C]Y × [D]Z > [A]W × [B]X.
This means that in a system where K is greater than one, you will expect the forward reaction to
proceed farther to the right before the system reaches equilibrium, giving
more products than reactants when the system reaches equilibrium.
A system with an equilibrium constant value less than one.
A very low value for K (a decimal value less than one, because K cannot be negative)
means that the equilibrium system lies to the left. The forward reaction will not
proceed very far before equilibrium is established, and the system will have more
reactants than products at equilibrium.
According to the law of mass action:
If K < 1, that means [C]Y × [D]Z ‹ [A]W × [B]X.
This means that in a system where K is less than one, you will expect the forward reaction to not
proceed very far to the right before the system reaches equilibrium, giving more reactants than
products when the system reaches equilibrium.
It is important to note that the value of K indicates how far the forward reaction proceeds before
reaching equilibrium. However, the value of K does not indicate the
amount of time it will take for the system to reach equilibrium.
Be sure to do the “Let’s Review” section on the Lesson page.
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