So far, we've assumed that every reaction we're given goes to completion, meaning that if we're given the stoichiometrically perfect amount of reactants, all of the reactions will react to form products.
For example, the Haber process:
`N_(2(g)) +3H_(2(g)) rArr 2NH_(3(g))`
From our understanding of chemical equations so far, the equation above states that `1 "mol" N_(2(g))` and `3 "mol" H_(2(g))` will react to form `2 "mol" NH_(3(g))` . This statement, in fact, is still true! For every `1 "mol" N_(2(g))` and `3 "mol" H_(2(g))` that react, `2 "mol" NH_(3(g))` will be formed.
The subtlety lies in the fact that not all of the `N_(2(g))` and `H_(2(g))` will react. Only a fraction of the reactants will react to form `NH_(3(g))`. Additionally, consider the following reaction:
`2NH_(3(g)) rArr N_(2(g)) +3H_(2(g))`
Once the `N_(2(g))` and `H_(2(g))` react to form `NH_(3(g))`, what's to prevent the `NH_(3(g))` from dissociating back to `N_(2(g))` and `H_(2(g))`?
It turns out that this is exactly what happens! Once the `N_(2(g))` and `H_(2(g))` react to form `NH_(3(g))`, the `NH_(3(g))` will dissociate to form back the original reactants. Thus, both equations are valid in a sense.
You can think of chemical equations as a back and forth, where the reactants will convert to products, products back into reactants, reactants back to products, ad nauseum. Eventually, there will reach a point in which the rate of `"reactants" rArr "products"` will be equal to the rate of `"products" rArr "reactants"` . On a macroscopic level, it would seem that there was no change occuring at all! This is called a dynamic equilibrium and is the point in which we say the reaction is at equilibrium.
Equilibrium occurs when the rate of reactants converting to products is equivalent to the rate of products converting to reactants.
For reactions in equilibrium, we're going to write the equation with a ↔ instead of a `rArr` in order to reflect that the reactants and products are continuously converting back and forth.
There are still reactions that react to completion, however. For these reactions, we'll continue using `rArr` in order to denote that the reaction is going to completion. For example, the dissociation of a strong acid is always to completion:
`HBr rArr H^+ + Cl^-`
This doesn't mean that the reaction isn't in equilibrium - just that the equilibrium position involves just the product. We'll explore this in the next post.
1. Chemical equations usually don't involve reactions going to completion. The products will react to form reactants the same way that reactions react to form the products.
2. The equilibrium position is defined as the position in which the rates of reactants to products and products to reactants is equal.
3. At equilibrium, there is no visible change.
4. The ↔ denotes a reaction at equilibrium. `rArr` denotes a reaction going to completion.
1. LA Smog and air quality
LA is well known for having poor air quality. In many pictures, the air is a noticeably brown/grey. In fact, anywhere where the air quality is poor, the air looks distinctly brown. Why is that?
It turns out that the culprit is the following reaction:
`N_2O_(4(g)) ↔ 2NO_(2(g))`
`N_2O_(4(g))` is a colorless gas whereas `NO_(2(g))` is a very dark red gas. The reason that the air in LA is brown is because of the equilibrium position of this reaction. The air isn't fully clear because there is some `NO_(2(g))`, but not fully dark red because of the `N_2O_(4(g))`.
Interestingly enough, it turns out that the equilibrium can shift more towards `NO_(2(g))` at sufficient temperatures. This means that, as the temperature increases, the air becomes browner. We haven't gone over how the equilibrium position can change yet, but this is just something to keep in mind.