Chemistry » Chemical Equilibrium » Le Chatelier's Principle

# Effect of Pressure on Equilibrium

## The effect of pressure on equilibrium

If the pressure of a gaseous reaction mixture is changed the equilibrium will shift to minimise that change.

• If the pressure is increased the equilibrium will shift to favour a decrease in pressure.

• If the pressure is decreased the equilibrium will shift to favour an increase in pressure.

When the volume of a system is decreased (and the temperature is constant), the pressure will increase. There are more collisions with the walls of the container. If there are fewer gas molecules there will be fewer collisions, and therefore lower pressure. The equilibrium will shift in a direction that reduces the number of gas molecules so that the pressure is also reduced. So, to predict in which direction the equilibrium will shift to change pressure you need to $$\color{darkgreen}{\text{look at the number of gas molecules in the balanced}}$$ $$\color{darkgreen}{\text{reactions}}$$.

#### Tip:

Recall that:

$\text{p} \propto \dfrac{\text{T}}{\text{V}}$

That is, if the temperature remains constant, and the volume is increased, the pressure will decrease. The figure below shows a decrease in pressure by an increase in the volume, and an increase in pressure by a decrease in the volume.

For example, the equation for the reaction between nitrogen and hydrogen is shown below:

$$\color{blue}{\text{N}_{2}\text{(g)}} + \color{blue}{\text{3H}_{2}\text{(g)}} \leftrightharpoons \color{red}{\text{2NH}_{3}\text{(g)}}$$

The ratio in the balanced equation is $$1:3:2$$. That is, for every $$\text{1}$$ $$\text{molecule}$$ of $$\color{blue}{\text{N}_{2}\text{ gas}}$$ there are $$\text{3}$$ $$\text{molecules}$$ of $$\color{blue}{\text{H}_{2}\text{ gas}}$$ and $$\text{2}$$ $$\text{molecules}$$ of $$\color{red}{\text{NH}_{3}\text{ gas}}$$ (from the balanced equation). Therefore the ratio is $$\color{blue}{\textbf{4 molecules of reactant gas}}$$ to $$\color{red}{\textbf{2 molecules of product gas}}$$.

• An increase in pressure will:

• Favour the reaction that decreases the number of gas molecules.

• There are $$\color{red}{\textbf{fewer molecules of product gas}}$$ than reactant gas, so the forward reaction is favoured.

• The equilibrium will shift to the right and the yield of $$\color{red}{\textbf{NH}_{3}}$$ will increase.

• A decrease in pressure will:

• Favour the reaction that increases the number of gas molecules.

• There are $$\color{blue}{\textbf{more molecules of reactant gas}}$$, so the reverse reaction is favoured.

• The equilibrium will shift to the left and the yield of $$\color{red}{\textbf{NH}_{3}}$$ will decrease.

Consider illustration in the figure below. (a) A decrease in the pressure of this reaction favours the reverse reaction (more gas molecules), the equilibrium shifts to the left. (b) An increase in the pressure of this reaction favours the forward reaction (fewer gas molecules), the equilibrium shifts to the right.

The figure above shows how changing the pressure of a system results in a shift in the equilibrium to counter that change. In the original system there are 12 molecules in total: $${\color{skyblue}{6{\text {H}}_{2}}} + {\color{blue}{2{\text{N}}_{2}}} \leftrightharpoons {\color{red}{4{\text{NH}}_{3}}}$$

If you decrease the pressure (shown by an increase in volume), the equilibrium will shift to increase the number of gas molecules. That shift is to the left and the number of $$\text{H}_{2}$$ and $$\text{N}_{2}$$ molecules will increase while the number of $$\text{NH}_{3}$$ molecules will decrease: If you increase the pressure (shown by a decrease in volume), the equilibrium will shift to decrease the number of gas molecules. That shift is to the right and the number of $$\text{H}_{2}$$ and $$\text{N}_{2}$$ molecules will decrease while the number of $$\text{NH}_{3}$$ molecules will increase: Note that the total number of nitrogen and hydrogen atoms remains the same in all three situations. Equations (a) and (b) are not balanced equations.

Another example is the reaction between sulfur dioxide and oxygen:

$$\color{blue}{\text{2SO}_{2}\text{(g)}} + \color{blue}{\text{O}_{2}\text{(g)}} \leftrightharpoons \color{red}{\text{2SO}_{3}\text{(g)}}$$

In this reaction, $$\color{red}{\textbf{two molecules of product gas}}$$ are formed for every $$\color{blue}{\textbf{three molecules of}}$$ $$\color{blue}{\textbf{reactant gas}}$$.

• An increase in pressure will:

• Favour the reaction that decreases the number of gas molecules.

• There are $$\color{red}{\textbf{fewer molecules of product gas}}$$ than reactant gas, so the forward reaction is favoured.

• The equilibrium will shift to the right and the yield of $$\color{red}{\textbf{SO}_{3}}$$ will increase.

• A decrease in pressure will:

• Favour the reaction that increases the number of gas molecules.

• There are $$\color{blue}{\textbf{more molecules of reactant gas}}$$ than product gas, so the reverse reaction is favoured.

• The equilibrium will shift to the left and the yield of $$\color{red}{\textbf{SO}_{3}}$$ will decrease.