Direction of reversible reactions | Equilibrium | AP Chemistry | Khan Academy

As an example of a reversible reaction, let's look at the hypothetical reaction where diatomic gas X₂ turns into its individual atoms, X. It would turn into two of them, so X₂ goes to 2X. The forward reaction is X₂ turning into 2X, and the reverse reaction is 2X combining to form X₂.

Let's say the X₂ is a reddish-brown gas. If we assume that both the forward and the reverse reactions are elementary reactions, we can actually write the rate law from the balanced equation. So for the forward reaction, let's go ahead and write the rate of the forward reaction is equal to the rate constant for the forward reaction, which we will symbolize as k with the subscript f times the concentration of—if we're going in the forward direction, the reactant would be X₂—so times the concentration of X₂.

Since we have a coefficient of 1 in front of X₂ for this elementary reaction, this would be raised to the first power. Next, we can write the rate law for the reverse reaction. The rate of the reverse reaction is equal to the rate constant, and we'll put in a subscript r here, so that's the rate constant for the reverse reaction. In the reverse reaction, 2X combines to form X₂, so this would be times the concentration of X, and since we have a 2 as our coefficient, we need to raise the concentration of X to the second power.

Next, let's look at these particulate diagrams and think about what happens for the forward reaction. So we start at time is equal to zero, and we start with only X₂. Here are five particles of X₂. If we wait ten seconds, now we've gone from five particles of X₂ to only three particles of X₂. Overall, two of those particles of X₂ have turned into X, and now there are four particles of X in this second particulate diagram.

We wait another 10 seconds for a total of time is equal to 20 seconds, and we've gone from three particles of X₂ to only two particles of X₂, and we've increased in the particles of X, so now we're up to six particles of X. The concentration of X₂ has decreased; we went from five particles of X₂ to three particles of X₂ to only two particles of X₂.

If we look at the rate law for the forward reaction, the rate of the forward reaction is proportional to the concentration of X₂. So, if the concentration of X₂ decreases, the rate of the forward reaction also decreases. We can see the same concept if we look at a graph of rate versus time. If we look at this line right here, we're starting at a certain rate for the forward reaction, and as the concentration of X₂ decreases, we can see the rate of the reaction decrease.

The rate of the reaction stops decreasing when we get to time is equal to 20 seconds. Next, let's think about the rate of the reverse reaction. Well, when time is equal to zero, the rate of the reverse reaction is zero, and that's because when we start out, we have only X₂; we don't have any X present, so the reverse reaction doesn't happen. But as soon as some of that X₂ turns into X, it's possible for the reverse reaction to happen.

As we increase in the amount of X, we look at our rate law here for the reverse reaction; as we increase in the concentration of X, the rate of the reverse reaction should increase as well. That's why we see the rate of the reverse reaction increase as time increases. So, as the forward reaction is happening, the reverse reaction is also occurring at the same time.

However, we don't really see that when we look at our particulate diagrams. In our particulate diagrams, we see a net conversion of X₂ into 2X. For example, looking from the first particulate diagram to the second, we see that two particles of X₂ have turned into four particles of X, and going from the second diagram to the third diagram, we see that another particle of X₂ has turned into 2X. Therefore, we have six particles of X at time is equal to 20 seconds.

Since we see a net conversion of reactants to products in our particulate diagram, the rate of the forward reaction must be greater than the rate of the reverse reaction, and we can see that. So before time is equal to 20 seconds here, if we look at our rates, let's just pick, for example, time is equal to 10 seconds; for the forward reaction, there's a higher rate than for the reverse reaction.

At time is equal to 20 seconds, the rate of the forward reaction becomes equal to the rate of the reverse reaction. Here’s the line on our graph where the rates become equal, and also notice the rates become constant at this point. When the rate of the forward reaction is equal to the rate of the reverse reaction, the reaction has reached equilibrium. So to the right of the dotted line, the reaction is at equilibrium, and to the left of the dotted line, the reaction is not at equilibrium.

Since the rate of the forward reaction is equal to the rate of the reverse reaction at equilibrium, X₂ is turning into 2X at the same rate that 2X is turning back into X₂. Therefore, the concentrations of X₂ and X at equilibrium remain constant, and we can see that when we look at the particulate diagrams for time is equal to 20 seconds and time is equal to 30 seconds.

Both of these particulate diagrams have two X₂ particles and six X particles. Let's look at a summary of what the rates of the forward and reverse reaction mean in terms of reactants and products. If the rate of the forward reaction is greater than the rate of the reverse reaction, that means there's a net conversion of reactants to products. Therefore, over time, the amount of reactants would decrease, and the amount of products would increase.

Eventually, the rate of the forward reaction becomes equal to the rate of the reverse reaction, and that means the reaction is at equilibrium, and there's no net change in the amounts of reactants or products. Reactants are turning into products at the same rate that products are turning back into reactants. Finally, if the rate of the reverse reaction is greater than the rate of the forward reaction, there's a net conversion of products to reactants.

So products are turning into reactants faster than reactants are turning into products. In the example that we looked at, the rate of the forward reaction was greater than the rate of the reverse reaction, and eventually, the rates became equal and the reaction reached equilibrium. If we had looked at an example of this, this third case here, where the rate of the reverse reaction is greater than the rate of the forward reaction, eventually the two rates would become equal and this reaction would reach equilibrium.