Chemical Equilibrium

Up until 1803, scientists believed that all reactions were irreversible. Say you want to cook an egg. You place it in a pan of boiling water and leave it to simmer for a few minutes. When you return, the previously-runny egg has turned solid and is ready to eat. But cooling the egg - say, by freezing it - won’t turn it back into its raw, liquid form. Scientists thought that this was the case for all reactions.

However, in 1803, Claude Louis Berthollet observed salt crystals forming at the edge of a salt lake in Egypt. He noticed that it was the reverse of a common reaction, in which sodium carbonate and calcium chloride reacted to produce sodium chloride and calcium carbonate. He hypothesised that some reactions could indeed go backwards. These are known as reversible reactions. If you leave a reversible reaction in a sealed container, it will eventually form something known as a state of equilibrium.

• This article is about chemical equilibrium.
• We’ll start by explaining what reversible reactions are and exploring the different types of chemical equilibria.
• We’ll then cover Le Châtelier’s principle and the factors that affect equilibrium.
• We will then look at examples of how reversible reactions are used in industry.
• Finally, we'll discuss equilibrium constants, with a particular focus on Kc and Kp.

Reversible reactions and chemical equilibrium

Many reactions are irreversible. You put the reactants together, supply them with enough energy and provide just the right conditions, and they react to form new products. If you mix these products together, nothing will happen - there will be no further reaction. Think of it like driving down a one-way street.

But some reactions are reversible. This means that under slightly different conditions, the products of the reaction can react again to reform the original reactants. In this case, the street is two-way - you can drive down it from either direction.

We show reversible reactions using half arrows: ⇋. When writing reversible reactions, we say that the reaction going from left to right, that is to say from the reactants to the products, is the forward reaction. The reaction going from right to left, from products to reactants, is the backwards reaction. This makes it easier to distinguish between the two reactions. However, you could just as easily write the equation the other way round! Take a look at the example below:

A + B ⇋ C + D

Going from left to right, $\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$. This is the forward reaction.

Going from right to left, $\mathrm{C}+\mathrm{D}\to \mathrm{A}+\mathrm{B}$. This is the backward reaction.

But we could also swap the equation round:

C + D ⇋ A + B

Now $\mathrm{C}+\mathrm{D}\to \mathrm{A}+\mathrm{B}$is the forward reaction and $\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$ is the backward reaction!

If left alone in a sealed system, reversible reactions reach a state of dynamic chemical equilibrium. We often just call this an equilibrium for short - other chemists will know what you are talking about.

We defined chemical equilibrium at the start of the article. A dynamic chemical equilibrium has two defining features:

• The rates of the forward and backward reactions are equal.

• The concentrations of reactants and products remain the same.

Let's go back to our example:

At a state of dynamic equilibrium, A and B react to form C and D. At the same time, C and D react to form A and B. However much C and D we produce is used up to remake A and B; the same amount of A and B is then reused to make C and D once more. The process is continuous. Overall, the concentrations of A, B, C, and D remain constant. This reversible reaction has reached dynamic equilibrium.

Types of chemical equilibrium

There are two different types of chemical equilibrium you should know about.

• Homogeneous equilibrium

• Heterogeneous equilibrium

Homogeneous equilibrium

The word homogeneous comes from the Greek words homos, meaning 'the same', and genos, meaning 'race' or 'type'. In a homogenous equilibrium, all the species present are in the same state. For example, they might all be liquid, aqueous, or gaseous.

An example is the Haber process. This is used to make ammonia. Nitrogen and hydrogen gases react to produce ammonia, which is also in the gaseous state:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

Heterogeneous equilibrium

Heterogeneous is also based on the Greek language, but this time it comes from the word heteros, meaning 'other'. In a heterogeneous equilibrium, the species present are in multiple different states.

An example is the decomposition of solid calcium carbonate. This decomposes into calcium oxide, another solid, and carbon dioxide, a gas:

${\mathrm{CaCO}}_3\left(\mathrm{s}\right)\rightleftharpoons \mathrm{CaO}\left(\mathrm{s}\right)+{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Chemical equilibrium and Le Châtelier's principle

Le Châtelier was a French chemist most famous for his work on chemical equilibria. He proposed a principle to explain how systems in dynamic equilibrium respond to changing conditions.

OK - how about that in plain English, please!

We know that if you take any reversible reaction and leave it in a sealed container for long enough, it will reach dynamic chemical equilibrium. The rates of the forward and backward reactions are the same, and the concentrations of products and reactants remain constant. However, Le Châtelier stated that if we change the conditions inside of the container, we can change the rates of the two reactions. For example, we could increase the temperature, and this may favour the forward reaction. Or we could increase the pressure, and this may favour the backward reaction. This is called shifting the position of the equilibrium. If we shift the equilibrium to the right, we say the equilibrium favours the forward reaction. If we shift it to the left, we say that it favours the backward reaction.

However, the change in rate isn't random. The equilibrium system always tries to reduce the impact of the change in conditions.

• Increasing the temperature favours the endothermic reaction. This is because the endothermic reaction takes in excess heat.

• Increasing the pressure favours the reaction that produces the fewest moles of gas. This is because all gases take up equal volume at the same temperature and pressure, and having fewer molecules of gas in a container reduces the pressure.

• Increasing the concentration of one of the reactants favours the forward reaction. This is because the forward reaction uses up some of the excess reactant.

• Adding a catalyst doesn't change the position of the equilibrium. This is because catalysts speed up the overall rate of reaction - they don't favour a particular reaction.

Le Châtelier's principle is useful because it allows us to influence the yield of a reversible reaction. Let's look at some real-life examples.

Examples of chemical equilibrium

There are many examples of systems at equilibrium. We're going to focus on three in particular:

• Methanol production

• Ethanol production

• Ammonia production

But before we dive into these processes, you need to understand compromise conditions.

Take our general reaction involving A, B, C, and D again. We want to maximise our yield of C and D. Let's say that the forward reaction is exothermic. According to Le Châtelier's principle, this means that lowering the temperature increases the rate of the forward reaction - the system will favour the exothermic reaction in order to try and produce extra heat. This will therefore increase our yield of C and D.

However, reducing the temperature slows down the overall rate of reaction and therefore reduces our yield. Whilst a low temperature might produce a lot of C and D, high temperature results in an overall faster rate of reaction. A medium temperature is used instead. This takes both yield and rate of reaction into consideration and in fact gives us more of C and D than a low temperature - simply because the rate of reaction is higher. This is an example of a compromise condition.

Now let's look specifically at the examples mentioned above.

Methanol production

Here's the equation for the production of methanol:

$\mathrm{CO}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{CH}}_3\mathrm{OH}\left(\mathrm{g}\right)\mathrm{\Delta H}=-91\mathrm{kJ}{\mathrm{mol}}^{-1}$

Note the following conditions:

• The forward reaction is exothermic. This means a lower temperature favours the forward reaction and increases the yield of methanol. However, a low temperature slows the rate of reaction and so a compromise temperature of 500 K is used.

• The forward reaction produces fewer moles of gas. This means that a higher pressure favours the forward reaction and increases the yield of methanol. However, maintaining a high pressure is expensive, so a compromise pressure of 10,000 kPa is used.

• We use a copper catalyst to increase the overall rate of reaction.

Ethanol production

The conditions for industrial ethanol production are very similar to those for methanol production. Here's the equation and the conditions:

${\mathrm{C}}_2{\mathrm{H}}_4\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\left(\mathrm{g}\right)\mathrm{\Delta H}=-46\mathrm{kJ}{\mathrm{mol}}^{-1}$

• The forward reaction is exothermic. This means a lower temperature favours the forward reaction and increases the yield of ethanol. However, a low temperature slows the rate of reaction and so a compromise temperature of 570 K is used.

• The forward reaction produces fewer moles of gas. This means that a higher pressure favours the forward reaction and increases the yield of ethanol. However, maintaining a high pressure is expensive and so a compromise pressure of 6,500 kPa is used.

• We use a phosphoric acid catalyst to increase the overall rate of reaction.

• Adding an excess of steam shifts the equilibrium to the right and increases the yield of ethanol. However, too much steam dilutes the catalyst and so slows down the rate of reaction. Instead, the ethanol is removed as it is formed, decreasing its concentration and therefore favouring the forward reaction.

Ammonia production

Ammonia is produced industrially using something called the Haber process. Again, its conditions follow the same principles as methanol and ethanol production. Here's the equation:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\mathrm{\Delta H}=-92\mathrm{kJ}{\mathrm{mol}}^{-1}$

Note the following:

• The forward reaction is exothermic. This means a lower temperature favours the forward reaction and increases the yield of ammonia. However, a low temperature slows the rate of reaction and so a compromise temperature of 670 K is used.

• The forward reaction produces fewer moles of gas. This means that a higher pressure favours the forward reaction and increases the yield of ammonia. However, maintaining a high pressure is expensive and so a compromise pressure of 20,000 kPa is used.

• We use an iron catalyst to increase the overall rate of reaction.

• The ammonia is removed as it is formed, decreasing its concentration and therefore favouring the forward reaction.

Summary

Here's a handy table to help you compare the three processes:

A table comparing methanol, ethanol and ammonia production. Anna Brewer, StudySmarter Originals

Equilibrium constants

Finally, let's take a look at equilibrium constants.

Here's something important to note: for a given equilibrium reaction at a certain temperature, equilibrium constants are always the same. Take our general reaction involving A, B, C and D again. No matter how much of A and B we start with, provided we keep the temperature the same, we'll always end up with the same equilibrium constant. This means that'll we'll always end up with the same ratio of C and D to A and B. It works the other way too - even if we start with just C and D, and no A or B, we'll end up with the same equilibrium constant.

Types of equilibrium constant

There are a few different types of equilibrium constant:

• Kc
• Kp
• Kw
• Ka
• Kb

Kc

Kc is an equilibrium constant involving concentration. You work with Kc when you have equilibria containing gaseous or aqueous species.

Here's the equation for Kc. It might look a little complicated - but don't worry, we'll talk through it in just a second:

For the reaction $\mathrm{aA}+\mathrm{bB}\rightleftharpoons \mathrm{cC}+\mathrm{dD}$, $\mathrm{Kc}=\frac{{{\left[\mathrm{C}\right]}_{\mathrm{eqm}}}^{\mathrm{c}}{{\left[\mathrm{D}\right]}_{\mathrm{eqm}}}^{\mathrm{d}}}{{{\left[\mathrm{A}\right]}_{\mathrm{eqm}}}^{\mathrm{a}}{{\left[\mathrm{B}\right]}_{\mathrm{eqm}}}^{\mathrm{b}}}$

What does this all mean? Well, square brackets represent concentration, so [A]_eqm^a means the concentration of A at equilibrium, raised to the power of a. What is a? Take a look at the general equation. a is the molar ratio of A. So if we have two moles of A in the equation, and the equilibrium concentration of A is 0.5 mol dm^-3, [A]_eqm^a = 0.5².

To work out Kc, we find a similar value for each of our products and multiply them together. We then find similar values for each of our reactants and multiply them together. We then divide the product value by the reactant value to find Kc.

Here's an example.

Using the equilibrium concentrations given to us, the equation for Kc looks like this:

Calculating Kc. Anna Brewer, StudySmarter Originals

Kp

Kp is very similar to Kc. However, instead of equilibrium concentrations, it uses equilibrium partial pressures.

To calculate the partial pressure of a gas, you need to know its molar fraction. You can find this by dividing the number of moles of the gas at equilibrium by the total number of all the moles of gas in the system. You then multiply the molar fraction by the total pressure of the system to find the gas's partial pressure. Once you have done this with all of the gases present, you can calculate Kp.

Kp takes the following equation:

For the reaction ${\mathrm{A}}^{\mathrm{a}}\left(\mathrm{g}\right)+{\mathrm{B}}^{\mathrm{b}}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{C}}^{\mathrm{c}}\left(\mathrm{g}\right)+{\mathrm{D}}^{\mathrm{d}}\left(\mathrm{g}\right)$, $\mathrm{Kp}=\frac{{p_{\mathrm{C}}}^{\mathrm{c}}\mathrm{x}{p_{\mathrm{D}}}^{\mathrm{d}}}{{p_{\mathrm{A}}}^{\mathrm{a}}\mathrm{x}{p_{\mathrm{B}}}^{\mathrm{b}}}$

Here, ${p_{\mathrm{A}}}^{\mathrm{a}}$ represents the partial pressure of gas A at equilibrium - for ease, we've left the eqm sign out.

Here's how you'd go about working out Kp:

Calculating Kp. Anna Brewer, StudySmarter Originals

Kw, Ka, and Kb

Kw, Ka, and Kb are equilibrium constants that involve the dissociation of molecules into ions in solution. The molecules only partially dissociate, meaning that the system forms a dynamic equilibrium.

• Kw looks at the dissociation of water molecules into ${\mathrm{H}}^{+}$ and ${\mathrm{OH}}^{-}$ ions.

• Ka looks at the dissociation of weak acid molecules into ${\mathrm{H}}^{+}$ and ${\mathrm{A}}^{-}$ ions.

• Kb looks at the dissociation of weak base molecules into ${\mathrm{BH}}^{+}$ and ${\mathrm{OH}}^{-}$ ions.

Their equations are derived from Kc. However, we ignore some of the terms in the equations. This is because they are so large that they are practically constant and overpower the other values.

For water:

$\mathrm{Kw}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\sout{\left[{\mathrm{H}}_2\mathrm{O}\right]}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]$

As before, square brackets represent concentration. Remember to use concentration at equilibrium - we've left the eqm symbol out to simplify the equation. Here, you should also leave out the concentration of water. Because water only partially dissociates into ions, the concentration of water is a very large value and so is almost constant, so we can just ignore it.

For weak acids:

$\mathrm{Ka}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$

And finally, for weak bases:

$\mathrm{Kb}=\frac{\left[{\mathrm{B}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[\mathrm{BOH}\right]}$