Dynamic equilibrium | Equilibrium | AP Chemistry | Khan Academy

To illustrate the concept of equilibrium, let's say that we have a beaker and we put some water into our beaker. We also 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. 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. 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. 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 remain constant.

If we look at this 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, where individual atoms are colorless.

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.

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.

Let's say this is a 1-liter container. 0.5 divided by 1 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: zero molar.

Next, we wait 10 seconds. So we start off at time equals zero seconds, and now we're at time equals 10 seconds. Now we can see there are three particles of X₂ in our box, and so that would be 0.3 molar. So let's go ahead and write 0.3 molar in for our concentration.

And now we have some particles of X. There are one, two, three, four particles. Once again, if each particle represents 0.1 moles, that's 0.4 moles of X divided by 1, or 0.4 molar.

We wait another 10 seconds. So when time is equal to 20 seconds, now there are two particles of X₂ and six particles of X. So now the concentrations are 0.2 molar for X₂ and 0.6 molar 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. So the concentrations after 30 seconds, the concentration of X₂ is 0.2 molar and of X is 0.6 molar.

Notice how the concentration of X₂ went from 0.5 molar to 0.3 molar 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 0 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 equal to 0, it was not at equilibrium. When time equals 10 seconds, it was not at equilibrium. Only when time was equal to 20 seconds did it reach equilibrium, and at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction.

If that's true, then X₂ is being turned into 2X at the same rate that 2X is being turned back into X₂. If those rates are equal, the concentrations of X₂ and X at equilibrium would remain constant.

We can see the same concept if we look at a graph of concentration versus time. The concentration of X₂ starts at 0.5 molar when time is equal to 0 seconds and then drops to 0.3 molar after 10 seconds. After 20 seconds, it's at 0.2 molar and then stays constant after that.

For the concentration of X, we start out at 0 molar, we increase to 0.4 after 10 seconds, then we're at 0.6, and then we are constant.

So if we think about a line, if we just draw a dashed 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 0.2 molar, and the equilibrium concentration of X gas is equal to 0.6 molar.

Finally, let's use these particulate diagrams to think about what we would see at a macroscopic level as the reaction proceeds to equilibrium. In the first particular 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.

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, an even lighter brown-red.

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.