Chemical Equilibrium

Before 1803, scientists believed that all reactions were irreversible. For example, if you boiled an egg, you couldn't turn it back into a runny liquid by freezing it. But in 1803, a scientist named Claude Louis Berthollet observed something interesting: salt crystals forming at the edge of a salt lake in Egypt. He realized that some reactions could actually go backwards. These are called reversible reactions. When a reversible reaction is left alone in a sealed container, it eventually reaches a state of chemical equilibrium, where the rates of the forward and backward reactions are equal and the concentrations of reactants and products don't change. This is also called dynamic equilibrium.

In this article, we'll explore chemical equilibrium. We'll start by explaining what reversible reactions are and look at the different types of chemical equilibria. Then, we'll talk about Le Châtelier's principle and the factors that affect equilibrium. After that, we'll discuss examples of how reversible reactions are used in industry. Finally, we'll focus on equilibrium constants, specifically Kc and Kp.

Reversible reactions and chemical equilibrium

Reactions can either be irreversible or reversible. In an irreversible reaction, the reactants react to form new products that cannot react again to form the original reactants. But in a reversible reaction, the products can react again to reform the original reactants under slightly different conditions. This is like a two-way street where you can drive from either end.

We use half arrows (⇋) to show reversible reactions. The reaction going from reactants to products is called the forward reaction, while the reaction going from products to reactants is the backward reaction. But you can also swap the equation around.

In a sealed system, reversible reactions reach a state of dynamic chemical equilibrium. This is where the rates of the forward and backward reactions are equal, and the concentrations of reactants and products remain the same.

Changes in conditions such as temperature, pressure, or concentration can shift the equilibrium position. Le Châtelier's principle helps us to predict how the equilibrium position will shift. For example, increasing the temperature favours the endothermic reaction, while increasing the pressure favours the reaction that produces fewer moles of gas.

Understanding these principles helps us to influence the yield of a reversible reaction. In the next section, we'll explore 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.

Compromise conditions are conditions that don’t necessarily give the greatest yield of the product, but are the most economical when it comes to balancing factors like cost and rate of reaction.

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

The equation for the production of methanol is:

CO(g) + 2H2(g) ⇋ CH3OH(g)

The forward reaction is exothermic, meaning that a lower temperature would increase the yield of methanol. However, a low temperature would also slow down the overall rate of reaction. Therefore, a compromise temperature of 500 K is used.

The forward reaction produces fewer moles of gas, so a higher pressure would also increase the yield of methanol. However, maintaining a high pressure is expensive. Therefore, a compromise pressure of 10,000 kPa is used.

A copper catalyst is used to increase the overall rate of reaction.

These compromise conditions help to balance factors like yield and cost in the production of methanol.

Ethanol production

The equation for industrial ethanol production is:

C2H5OH(g) ⇋ C2H4(g) + H2O(g)

Similar to methanol production, the forward reaction is exothermic, meaning that a lower temperature would increase the yield of ethanol. However, a compromise temperature of 570 K is used to balance yield and rate of reaction.

The forward reaction produces fewer moles of gas, so a higher pressure would increase the yield of ethanol. However, a compromise pressure of 6,500 kPa is used because maintaining a high pressure is expensive.

A phosphoric acid catalyst is used to increase the overall rate of reaction.

To increase the yield of ethanol, excess steam is added to shift the equilibrium to the right. However, too much steam would dilute the catalyst and slow down the rate of reaction. Instead, the ethanol is removed as it is formed, decreasing its concentration and therefore favouring the forward reaction.

These compromise conditions help to balance factors like yield, rate of reaction, and cost in the production of ethanol.

Ammonia production

The Haber process is an artificial nitrogen fixation process used in the industrial production of ammonia. The reaction combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic. The conditions used in the Haber process follow the same principles as methanol and ethanol production. The forward reaction is exothermic, so a lower temperature favours the forward reaction and increases the yield of ammonia. However, a low temperature slows the rate of reaction, so a compromise temperature of 670 K is used.The forward reaction produces fewer moles of gas, so a higher pressure favours the forward reaction and increases the yield of ammonia. However, maintaining a high pressure is expensive, so a compromise pressure of 20,000 kPa is used.An iron catalyst is used 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

Equilibrium constants are important in understanding the behavior of chemical reactions at equilibrium. They are values that compare the amount of products and reactants at equilibrium, and are always constant for a given reaction at a specific temperature. This means that the ratio of products to reactants will always be the same, regardless of the initial concentration or pressure of the reactants.

While equilibrium constants are unaffected by changes in concentration and pressure, they can be affected by changes in temperature. This is because temperature affects the rate of the forward and reverse reactions, which in turn affects the equilibrium constant. For exothermic reactions, increasing the temperature will shift the equilibrium towards the reactants, while for endothermic reactions, increasing the temperature will shift the equilibrium towards the products.

Understanding equilibrium constants is crucial in the design and optimization of chemical reactions, as it allows us to predict the behavior of the reaction at equilibrium and make adjustments to optimize the yield of desired products.

Types of equilibrium constant

For the reaction E(aq) + 2F(aq) ⇌ 2G(aq), the equation for Kc is:

Kc = ([G]eq2) / ([E]eq1 × [F]eq2)

Using the equilibrium concentrations given to us, we have:

Kc = (0.42) / (0.2 × 0.32)
Kc = 4.2

So the equilibrium constant Kc for this reaction is 4.2. This means that at equilibrium, the concentration of the product G is 4.2 times higher than the concentration of the reactants E and F combined.

Calculating Kc
Calculating Kc

For the reaction H(g) + I(g) ⇌ J(g) + K(g), the equation for Kp is:

Kp = (pJe × pKf) / (pHg × pIh)

To calculate the partial pressure of each gas, we first need to determine the molar fraction of each gas at equilibrium. We have 1.5 moles of H and I, 3 moles of J, and 2 moles of K, for a total of 8 moles. Therefore:

  • The molar fraction of H is 1.5/8 = 0.1875
  • The molar fraction of I is also 0.1875
  • The molar fraction of J is 3/8 = 0.375
  • The molar fraction of K is 2/8 = 0.25

To find the partial pressure of each gas, we multiply its molar fraction by the total pressure of the system:

  • pH = (0.1875)(400 kPa) = 75 kPa
  • pI = (0.1875)(400 kPa) = 75 kPa
  • pJ = (0.375)(400 kPa) = 150 kPa
  • pK = (0.25)(400 kPa) = 100 kPa

Now we can plug these values into the Kp equation:

Kp = (150)(100) / (75)(75)
Kp = 2

So the equilibrium constant Kp for this reaction is 2. This means that at equilibrium, the partial pressure of the products J and K combined is twice as high as the partial pressure of the reactants H and I combined.

Calculating Kp
Calculating Kp

Thank you for the detailed explanation! As an AI language model, I don't have the capability to explore those articles. However, I appreciate the information you provided on the different types of equilibrium constants and their equations. It's important to understand the differences between Kc, Kp, Kw, Ka, and Kb, as they apply to different types of reactions and systems. By knowing how to calculate these equilibrium constants, we can better understand and predict the behavior of chemical reactions.

Great summary! Understanding chemical equilibrium is crucial in chemistry as it helps us predict and control the behavior of chemical reactions. By knowing the different types of equilibrium and equilibrium constants, we can optimize reactions for desired outcomes, such as in industrial processes. Additionally, knowing how to calculate equilibrium constants and how they are affected by temperature can help us understand the underlying principles of chemical reactions.

Chemical Equilibrium

What does equilibrium mean in chemistry?

In chemistry, equilibrium describes the state of a reversible reaction where the rates of the forward and backward reactions are equal and the concentrations of the products and the reactants stay the same.

What is the difference between equilibrium and equilibria? 

Equilibrium is singular whereas equilibria is plural.

What is the formula for equilibria?

You can use the equilibrium constant, Kc, to represent an equilibrium reaction.  To calculate the equilibrium constant, raise the equilibrium concentration of each of your products to the molar ratio of that product given in the equation and multiply these terms together. Do the same with the equilibrium concentrations of the reactants. To calculate Kc, divide the product value by the reactant value.

What is the purpose of chemical equilibrium?

In biology, equilibria play an important role in maintaining optimum bodily functions. For example, your blood constantly maintains a dynamic equilibrium, which keeps its pH stable. In chemistry, we can use equilibria for industrial purposes. At equilibrium, the concentrations of products and reactants in a reaction don't change. However, we can influence the position of the equilibrium to change these concentrations. This is useful in industry as it can help us improve the yield of a reversible reaction. This is one example of the purpose of chemical equilibrium.

Describe how a reaction reaches equilibrium. 

Take a reversible reaction with the reactants A and B and the products C and D. Let's say that you mix some of A and B together in a sealed container. A and B will react rapidly to form some of C and D. As more and more of C and D are formed, A and B's rate of reaction will slow. At the same time, some of C and D will start reacting to reform A and B. Eventually, the rate of the forward reaction (the reaction between A and B) equals the rate of the backward reaction (the reaction between C and D). At this point, the concentrations of A, B, C and D remain constant. The system has reached a dynamic equilibrium.

Quiz questions showing the correct answer and a leaderboard with friends.

Create chemistry notes and questions for free

96% of learners report doubling their learning speed with Shiken

Join Shiken for free

Try Shiken Premium for free

Start creating interactive learning content in minutes with Shiken. 96% of learners report 2x faster learning.
Try Shiken for free
Free 7 day trial
Cancel anytime
30k+ learners globally
Shiken UI showing questions and overall results.