Chem – Le Chatelier’s Principle: Volume and Pressure

How do you relate volume and pressure to Le Chatelier’s Principle?

The next hurdle we have to overcome is explaining pressure and volume. Remember we are not going to think about them in the same way as concentration and temperature. Also when we talk about pressure and volume we will not say that we increase the pressure or volume of the reactants side or the products side. This is chemical reaction, both the reactants and products, are in one container. Whatever we do to the container we do to both sides of the chemical equation.

200kJ + N2(g) + 3 H2(g) <—-> 2 NH3(g)

For pressure and volume the focus will be on what the total moles are present on each side of the chemical equation. In my example, on the left side of the chemical equation we have 1 mole of N2 and 3 moles of H2. Notice I just took the numbers of the coefficients and used them for the number of my moles. So if I have 1 mole and 3 moles that adds up to a total of 4 moles on the reactants side. Notice I do not care exactly what chemical each of the moles are. I am treating all moles as the same here. In my example, on the right side of the chemical equation I have 2 moles of NH3. So I have 2 moles total on the products side.

Moles are no different than any other object or group. If I were to say to you that I had boxes of all the same size and shape, and then I asked you which one takes up more space, 4 boxes or 2 boxes? You would say 4 boxes of course. So if I ask you which one takes up more space 4 moles or 2 moles? You would say 4 moles of course. Therefore, in my example, the reactants side takes up more space. Remember we are treating all moles as the same regardless of what chemical they come from.

200kJ + N2(g) + 3 H2(g) <—-> 2 NH3(g)

So now we can see that it is all an issue of space. Lets try to talk about volume first since it is a little easier than pressure. If we increase the volume of the container, in my example above, then we are going to allow for more space. If we allow for more space the equilibrium will cause a shift toward the reactants (left), because that side takes up more space. If we decrease the volume of the chemical equation then the equilibrium will cause a shift toward the products (right) because the products take up less space.

Remember that the states of solids and liquids don’t count for anything having to do with an equilibrium.

Examples: Which way will the equilibrium shift when we change the volume as mentioned?

Decrease volume

2 C4H10(l) + 13 O2(g) <—-> 8 CO2(g) + 10 H2O(g)

Decrease volume

6 Ag(s) + Ca3(PO4)2(s) <—-> 3 Ca(aq) + 2 Ag3PO4(s)

Answer: Shift Left (solids count as zero moles)

Just like volume pressure is all about the space that the reactants versus products take up. However, just like in the gas law lesson pressure is inverse (reverse) to the relationship of volume. That means pressure does the opposite of what volume does. I like to think of pressure as a squeeze. If you imagine a balloon between your hands, increasing the pressure on that balloon would be to try to squeeze your hands together. What happens to the space in that balloon? It goes down right? So if we increase the pressure we decrease the volume. Likewise, if I start pulling my hands apart from the now squished balloon I decrease the pressure. What happens to the space? The space increases right? So if we decrease the pressure we increase the volume. Let us take a look at my example problem below.

200kJ + N2(g) + 3 H2(g) <—-> 2 NH3(g)

Which side has more moles total? The left side (1 + 3). Which side has less moles total? The right side (2). So if I increase the pressure on the container that this reaction is in, the equilibrium shifts to the right because the space in the container is now smaller. However, if I decrease the pressure on the container that this reaction is in, the equilibrium shifts to the left because the space in the container is now more or larger.

Examples: Which way will the equilibrium shift when we change the pressure as mentioned?

Increase pressure

2 N2O3(g) <——> 2 N2(g) + 3 O2(g)

Decrease pressure

Ba(OH)2(aq) + 2 HCL(aq) <—-> 2 H2O(l) + BaCl2(aq)

Answer: Shift Left (remember don’t count the moles of water because it is liquid)

PRACTICE PROBLEMS: Which way will the equilibrium shift when we change the volume or pressure as mentioned?

Decrease pressure

6 C4(s) + 12 O2(g) + 24 H2(g) <—-> 4 C6H12O6(s)

Increase volume

C4(s) + 4 O2(g) <—-> 4 CO2(g)

Answer: No shift (same moles on both sides when you count solid as zero)

Decrease volume

H2SO4(aq) <—–> H2(aq) + S(s) + 2 O2(aq)

Decrease pressure

2 Fe3+(aq) + 3 CO32-(aq) <—-> Fe2(CO3)3(s)

Shift Left

Increase volume

6 C4(s) + 12 O2(g) + 24 H2(g) <—-> 4 C6H12O6(s)

Increase pressure

H2SO4(aq) <—–> H2(aq) + S(s) + 2 O2(aq)  