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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry > Unit 7

Lesson 1: Introduction to equilibrium# Dynamic equilibrium

Many physical and chemical processes are reversible. A reversible process is said to be in dynamic equilibrium when the forward and reverse processes occur at the same rate, resulting in no observable change in the system. Once dynamic equilibrium is established, the concentrations or partial pressures of all species involved in the process remain constant. Created by Jay.

## Want to join the conversation?

- Why would the reaction be at equilibrium in 0.2M and 0.6M? Shouldn't it be at equilibrium in 0.25M and 0.5M, because we need the number of moles of X2 be twice the number of 2X?(2 votes)
- Equilibrium means that the forward and reverse reaction rates are equal, not that the product and reactant concentrations be equal. We can have wildly different concentrations for both and a reaction can be at equilibrium. At equilibrium the reaction reaches a point where the concentrations of the chemicals do not appreciably change anymore because products are being formed at the same rate reactants are.

Hope that helps.(4 votes)

- Are X_2 and X actual gases? If so, which gases are they?(1 vote)
- So those aren't any specific gases. Using X2 and X is just a general way to write the equation which could use any elements we wanted. Think of X being used here as a variable in the same way that it's used in algebra.

Hope that helps.(2 votes)

## Video transcript

- [Instructor] To illustrate
the concept of equilibrium, let's say that we have a beaker and we put some water into our beaker. And also we make sure that
our beaker has a lid on it. Some of those water molecules
are going to evaporate and turn into a gas. And eventually once we
have enough gaseous water, some of the gaseous water
is going to condense and turn back into liquid water. To represent these two processes, we can show liquid water on the left and gaseous water on the right. So in the forward process, liquid water turns into gaseous water. And this forward arrow here represents the process of vaporization. And when gaseous water turns
back into liquid water, that's represented by this
arrow here on the bottom, and so that's the process of condensation. Since we start with liquid water, at first, the rate of vaporization is greater than the rate of condensation. But eventually we reach
a point where the rate of vaporization is equal to
the rate of condensation. And when that happens, if
you're turning liquid water into gaseous water at the same rate, you're turning gaseous water
back into liquid water, the number of water molecules
in a liquid and gaseous state would remain constant. So when the rate of vaporization
is equal to the rate of condensation, we've reached
a state of equilibrium. And this is a dynamic equilibrium
because if we zoom in and we look at this, water
molecules are being converted from the liquid state to the
gaseous state all the time and molecules are going from
the gaseous state back to the liquid state all the time. However, since the rates are equal, the number of molecules in the liquid and gaseous state remained constant. And if we look at it from a
macroscopic point of view, the level of water wouldn't change at all. Now let's apply this concept
of dynamic equilibrium to a hypothetical chemical reaction. In our hypothetical reaction
X2 which is a reddish brown gas, decomposes into its
individual atoms to form 2X and individual atoms are colorless. So in the forward reaction,
we're going from X2 to 2X. So X2 is decomposing to 2X. And in the reverse
reaction, the two atoms of X are combining together to form X2. When we have a forward
reaction and a reverse reaction by convention, we say,
what's on the left side, are the reactants, and
what's on the right side, are the products. And by using these terms,
we can avoid confusion. Let's say that we start our
reaction with only reactants. So only X2 is present in
this first container here. And there are five particles of X2. If every particle represents 0.1 moles, since we have five particles of X2, we have 0.5 moles of X2. And let's say, this is
a one liter container. 0.5 divided by one would be 0.5 molar. So the initial concentration
of gas, X2 is 0.5 molar. And since we don't have any of the X, there are no white dots
in this box, right? The initial concentration
of X would be zero molar. So let me just write that in here, 0M. Next we wait 10 seconds. So we start off at time's
equal to zero seconds, and now we're at time's
equal to 10 seconds. And now we can see, there are three particles of X2 in our box. And so that would be 0.3M, so let's go ahead and write 0.3M in for our concentration. And now we have some particles of X. There are one, two, three, four particles. And once again, if each
particle represents 0.1 moles that's 0.4 moles of X
divided by one or 0.4M. We wait another 10 seconds, so when time is equal to 20 seconds, now there are two particles of X2 and one, two, three, four,
five, six particles of X. So now the concentrations are
0.2M for X2 and 0.6M for X. We wait another 10 seconds
for a total of 30 seconds. And there are still two particles of X2 and six particles of X. And so the concentrations
after 30 seconds, the concentration of X2
is 0.2M and of X is 0.6M. Notice how the concentration
of X2 went from 0.5M to 0.3M to 0.2, and then it
was also 0.2 after 30 seconds. So it became constant when
time is equal to 20 seconds. The concentration of X went
from zero to 0.4 to 0.6 and then it was also 0.6 after 30 seconds. So the concentrations became
constant when time is equal to 20 seconds, which means the
reaction reached equilibrium after 20 seconds. So at time is equal to zero
was not at equilibrium. When time is equal to 10 seconds,
it was not at equilibrium. Only when time was equal to 20 seconds, did it reach equilibrium. And that equilibrium, the
rate of the forward reaction is equal to the rate of
the reverse reaction. And if that's true, then
X2 is being turned into 2X at the same rate that 2X is
being turned back into X2. And if those rates are equal,
the concentrations of X2 and X at equilibrium
would remain constant. We can see the same concept, if you look at a graph of
concentration versus time. Concentration of X2 starts
at 0.5M when time is equal to zero seconds,
and then drops to 0.3M after 10 seconds. And after 20 seconds, it's at 0.2M and then stays constant after that. For the concentration of
X, we start out at 0M, we increase 2.4 after 10 seconds, then we're at 0.6 and
then we are constant. So if you think about a line, if we just draw a dash
line here at 20 seconds. That's the dividing line
between on the left, where we're not at equilibrium. And so the concentrations
are always changing. And then to the right of that dotted line, we are at equilibrium where the concentrations remain constant. So the equilibrium concentration
of X2 gas is equal to 0.2M and the equilibrium concentration of X gas is equal to 0.6M. Finally, let's use these
particular diagrams to think about what we would see at a macroscopic level as the reaction proceeds to equilibrium. And the first particulate diagram we see only red particles. So only our reactants X2 are
present in the beginning. However, as time goes on,
the number of red particles decreases from five in the
first particular diagram to three in the second to two and then the number stays at
two because remember we reach equilibrium after 20 seconds. So what we would see from a
macroscopic point of view, is we'd start out with a
reaction vessel that is a darker brown red, and then it would
be a lighter brown red. And then finally, when
we reach equilibrium and even lighter brown red, and it would stay that
same light brown red, because we've reached equilibrium
and the concentrations of reactants and products
remain constant at equilibrium. Even though our reactants are
turning into our products. Our products are turning back
into our reactants at the same rate and therefore the concentrations of both reactants and
products are constant.