When a chemical reaction is at equilibrium the concentration of each reactant and the concentration of?

Overview of Week 2 Inquiry Question – What factors affect equilibrium and how?

Learning Objective #1 – Le Chatelier’s Principle

Learning Objective #2 – Disturbance factors that affect equilibrium position

Learning Objective #3 – Equilibrium progress over time using Collision Theory

Learning Objective #4 – Activation Energy and Equilibrium Position

Learning Objective #5 – Heat of Formation and Equilibrium Position

NEW HSC Chemistry Syllabus Notes Video – Factors that affect Equilibrium 

Week 2 Homework Set – Essential for Band 5

Week 2 Curveball Question – Moving from Band 5 to Band 6

Week 3 Extension Questions – Regularly posted exam questions & solutions

Solutions to Week 2 Questions

Overview of Week 2 Inquiry Question

In this week, we will go over Le Chatelier’s Principle and Collision Theory to answer this week’s Inquiry Question of “What affects equilibrium and how?”

Le Chatelier’s principle will help you predict the direction (left or right) an equilibrium reaction will proceed due to a change in temperature, concentration of substances involved in the equilibrium reaction, pressure and/or volume! 

Recall those experiments you did in week 1 to test whether a reaction is a reversible (i.e. a dynamic equilibrium reaction).

For example: Fe3+ (aq) + SCN– (aq) ⇌ FeSCN2+ (aq)

Since the above reaction is reversible,  i.e. Fe3+ and SCN– is converted from FeSCN2+ but how? Well, this can be explained by Le Chatelier’s Principle which will be the focus of this week’s notes and syllabus inquiry question.

After exploring Le Chatelier’s Principle, we will apply collision theory as another way to explain shifts in an equilibrium reaction’s position. 

Recall that at the end of last week’s notes, we applied collision theory to one-way reactions. In this week’s notes, we will apply collision theory to equilibrium (two-way) reactions.

Lastly, we will explore how activation energy and heat of reaction will affect the position of equilibrium and thus the concentration of reactants and products.

Without further ado, let’s get into exploring how Le Chatelier’s Principle can help us predict the direction of which an equilibrium reaction will proceed and the factors which can affect the equilibrium position in a closed system!

Learning Objective #1: Investigate the effects of temperature, concentration, volume and/or pressure on a system at equilibrium and explain how Le Chatelier's principle can be used to predict such effects.

Le Chatelier’s Principle 

The definition of Le Chatelier’s Principle is as follows:

In the event that there is a disturbance in the closed system affecting its equilibrium position, the system will counteract this change by shifting its equilibrium position to minimises the effect of the disturbance. 

This change or disturbance exerted on the closed system may be caused by a change in temperature, concentration of reactants or products, pressure or volume. 

It is important to note that Le Chatelier’s Principle only applies to equilibrium reactions that occur in closed systems and NOT open systems. 

Let’s see Le Chatelier’s Principle in action by exploring an example for each of the disturbance factors that we have just listed above and their effect on the equilibrium position.

Potential Equilibrium Disturbance #1: Change in Temperature

To illustrate Le Chatelier’s Principle in action during the event of a temperature change in the system, let’s use the following dynamic equilibrium reaction as an example:

N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

The above reaction is called the Haber Process and occurs in pressurised system like most gas reactions. The forward reaction (i.e. hydrogen reacting with nitrogen to produce ammonia) is exothermic. 

Because of this, we know that heat is released into the surrounding environment when ammonia is formed.

Here is a trick that will reduce your mental workload in general when solving equilibrium reaction problems: 

If the forward reaction is exothermic, always write ‘+ Heat’ on the right side of the equilibrium equation. 

If the forward reaction is endothermic, always write ‘+Heat’ to the left side of the equilibrium equation.

The logic behind why you write '+ Heat' on one side of the chemical is as follows:

Let’s have a look at exothermic reactions first.

For Exothermic Reactions: The system’s enthalpy (total internal energy) decreases but the surrounding environment’s enthalpy increases. 

So for exothermic reactions, the change in system’s enthalpy is negative and change in surrounding environment’s enthalpy is positive. 

Collectively, this mean that ammonia AND heat is released into the environment as products which is why we write ‘+ Heat’ on the right side of the chemical reaction for forward reaction that are exothermic.

Vice versa, the reason why you write ‘+ Heat’ when, in short, the net change in internal heat energy (enthalpy) of the system is positive when the reaction proceeds. 

So, we write ‘+ Heat’ on the left side of the chemical equation if the forward reaction is endothermic. 

In general:

IF the equilibrium position shifts to right in favour of the forward reaction, this will mean that the concentration of products will increase and the concentration of reactants will decrease. Yes, this will mean that the rate of the forward reaction will increase. When equilibrium is re-established again, the rate of the forward reaction will be equal to the rate of the reverse reaction.

IF the equilibrium position shifts to the left in favour of the reverse reaction, this will mean that the concentration of reactants will increase and the concentration of products will decrease. Yes, this will mean that the rate of the reverse reaction will increase. When equilibrium is re-established again, the rate of the reverse reaction will be equal to the rate of the forward reaction. 

Let’s get back on our adventure in exploring how Le Chatelier’s principle may apply to the Haber Process equation. 

Side Note: The Haber Process was ‘randomly’ chosen as our example as it was common equilibrium reaction used in the old syllabus.

So how will the concentration of products and reactants change when the temperature of the system (such as the pressurised vessel) is increased? How about when the temperature of the system was lowered? 

We can use Le Chatelier’s principle to predict the direction of which the equilibrium position will shift either in favour of the forward reaction or reverse reaction. Let’s see how. 

For temperature as a disturbance factor on equilibrium, we will explain it in two ways.

Scientifically speaking, the second way is more accurate. So I recommend you using the second method in your response to exam questions. 

However, the first method of explanation may help aid your visualisation of how the equilibrium position  shifts its position when there is a change in the system’s temperature.

First way of explanation – LESS ACCURATE from thermodynamic standpoint

Well, let’s first suppose the change in temperature is positive, i.e. temperature in the system has increased. 

This could be due to you heating up the system in any way of form. For example: Heating up the container which has the Haber process equilibrium reaction is taking place inside. So, the gas molecules will gain kinetic energy.

N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

Since we have “+ Heat” on the right hand side of the chemical equation, what you should imagine is that increasing temperature will create greater disturbance on the right hand side of the equation than on the left hand side. 

This means that, according Le Chatelier’s Principle,  the system will shift its equilibrium position to the left in order to counteract and minimise the disturbance or change (i.e. increase in temperature). 

This would mean that species on the right side of the reaction (i.e. NH3 molecules) will react to form more species on the left side of the chemical equation (i.e. N2 and H2). 

The concentration of ammonia will therefore fall and the concentration of H2 and N2 will increase. Note that the change in the concentration of each species will be proportional/correspond to their molar ratios in the chemical equation. 

Now, let’s see the opposite scenario. Let’s suppose we have a decrease in the system’s temperature. 

Maybe you put the container, which has the Haber Process is taking place inside, in a cold ice bath. 

You can imagine this by thinking that ‘+ Heat’ is being taken away from right hand side of the equation. 

This therefore leaves you with just 

N2(g) + 3H2(g) ⇌ 2NH3(g) 

INSTEAD OF: N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

This creates a disturbance in the system. So, according to Le Chatelier’s Principle, the equilibrium system will shift to the right in order produce more heat to counteract and minimise the change (i.e. the decrease in temperature) 

As a result of this shift in equilibrium position, the concentration of ammonia will increase and the concentration of H2 and N2 will decrease. 

Again, the change in the concentration of the species will be proportional to their molar ratios in the Haber Process chemical equation.

Second way of explanation – MORE ACCURATE, from a thermodynamic standpoint

What will happen to the equilibrium position when temperature of the system increased?

Since the reverse reaction of the Haber Process is endothermic, the equilibrium position will shift to the left hand side, in favour of the endothermic reaction, to consume the excess heat inputted into the system to counteract and minimise the increase in temperature.

What will happen to the equilibrium position when the temperature of the system decreases?

Since the forward reaction of the Haber Process is exothermic, the equilibrium position will shift to the right hand side, in favour of the exothermic reaction, to produce more heat to counteract and minimise the drop in temperature.

• NOTE: The concentration of ammonia can be written as [NH3] where the brackets means concentration.

So, what do I mean by the change in the concentration of the species will be in proportion to their molar ratios in the chemical equation?

Well, if the [NH3] decreases by 2 moles per litres (2M), then 1 mole of [N2] and 3 moles of [H2] will be formed. 

This is because the molar ratio of N2:H2:NH3 is 1:3:2 in the Haber Process chemical equation.

NOTE: The unit for concentration is molarity (M) which is moles per litre. 

Potential Equilibrium Disturbance #2: Change in Concentration!

Let’s continue using the Haber Process to illustrate how the Le Chatelier’s Principle can be used to explain how an equilibrium system will shift its position when there is a change in the concentration of one of its reactant or products!

N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

If we increase the concentration of ammonia by pumping more ammonia into the closed system, we will predict that the equilibrium reaction to the left! 

“Why is that?”, you may ask. 

Well, according to Le Chatelier’s principle, the disturbance experienced by the system will be an increase in the [NH3]. In order to counteract and minimise this disturbance, the system will shift its equilibrium position to the left to decrease the [NH3] which means that the [N2] and [H2] will increase. 

As explained before, the increase in nitrogen’s and hydrogen’s molarity will depend on mole ratio. 

• NOTE: At the end of this week’s notes there will be homework questions that requires you to determine exactly how much reactants and products increase or decrease. So, give them a go and check your answers after.

Similarly, if you pump N2 or H2 into the system, it will shift the equilibrium position to the right as per Le Chatelier’s Principle to counteract and minimise the the increase in nitrogen and hydrogen gas concentrations. This will therefore result in an increase in [NH3] to re-establish equilibrium. 

NOTE: Suppose there is a dynamic equilibrium that has a solid either as a reactant or product. Increasing the concentration of the solid WILL NOT affect equilibrium position as their concentration is constant when temperature remains constant. In terms of concentration remains constant, we will formally explore this in more detail in Week 3 and Week 4 notes.

NOTE: If pure liquids in dynamic equilibrium, i.e. they are acting as a solvent for the reaction, also do not affect equilibrium position. We will formally introduce and explore this idea about pure liquids in Week 4 notes.

Potential Equilibrium Disturbance #3: Change in Pressure!

NOTE: Changes in pressure only affects gas species in a chemical equilibrium. 

Changes in the system’s pressure will have no effect on solids and liquids. 

A system would shift its equilibrium position to counteract and minimise a disturbance such as a change in pressure. 

It is important to compare the moles of gases on each side of the chemical reaction in order to predict the direction which the equilibrium will shift its position when system’s pressure changes.

So, for the Haber Process: N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

There are 4 moles of gas on the left hand side of the chemical equation and 2 moles on the right.

This means the nitrogen and hydrogen gases collectively exerts more pressure than the ammonia molecules on the system. 

This means if there is an increase in pressure, the system will shift its equilibrium position to the side with the least moles of gas, i.e. to the right, to counteract and minimise the increase in pressure. 

• NOTE: To be more detailed, you could explain in terms of concentration whereby increasing in pressure would affect the reactants (nitrogen and hydrogen) more than the products (ammonia) as there are greater number of moles of gas. Therefore the increase in reactants’ concentration would be greater than ammonia’s concentration. So, the system will shift to the right to minimise the overall increase in concentration as per Le Chatelier’s Principle to re-establish equilibrium.

Vice versa, if there is a decrease in pressure in the system, the equilibrium will shift to the side with more moles of gas, i.e. to the left, to counteract and minimise the decrease in pressure. 

Potential Equilibrium Disturbance #4: Change in Volume!

NOTE: Change in volume only affects gas species in a chemical equilibrium. 

Changes in the system’s volume will have zero effect on solids and liquids for HSC purposes.

Like with changes in pressure questions, we need to compare the moles of gases on each side of the chemical reaction in order to predict the direction which the equilibrium will shift its position when system’s volume changes.

So using the Haber Process as our example again: N2(g) + 3H2(g) ⇌ 2NH3(g) + Heat

There are 4 moles of gas on the left hand side and 2 moles on the right hand side of the chemical equation.

If the volume of the system increases, it means there is decrease in pressure. This is because there is more room for the particles to move about in the system. You can think about this such that the particles will spend less time colliding with the walls of the container as the distance between the walls are greater (due to increased volume of container).

• Alternatively, to be more detailed: You can explain using the concentration of gases. That is, as the system volume increases, the concentration of gases on the left hand side of the reaction (nitrogen and hydrogen) will be affected more greater than the gas molecules (ammonia) on the right hand side of the chemical reaction. This is because there are more moles of gas on the left hand side of the reaction as per chemical formula, 2:1 to be precise. This means that the decrease concentration of gases on the left hand side of the reaction will be greater than on the gas molecules on the right hand side.

So, if the volume of the system increases (i.e. a decrease in pressure), the equilibrium will shift to the side of the chemical equilibrium with more moles of gas to counteract and minimise the decrease in pressure, as per Le Chatelier’s Principle to restore equilibrium. That is, the equilibrium position will shift to the left. 

By effect, a shift in equilibrium position to the left will increase the [N2] and [H2] would result in more moles of gas in the system. This will increase the system’s pressure and re-establish equilibrium.

This also means that the [NH3] will decrease in proportion to the molar ratio with [N2] and [H2] species as the equilibrium shifts the left. 

Vice versa, when the volume of the system decreases the system will shift its equilibrium position to the side with least moles of gas to minimise the increase in pressure and gas concentration. So, the [NH3] will increase and [N2] and [H2] will decrease.

Inverse Relationship between Pressure and Volume as per Boyle’s Law

There is an inverse relationship between Pressure and Volume in terms of shifting the equilibrium position!

• Increasing the system’s pressure will have the same effect on shifting the equilibrium position as decreasing volume of the system.

• Decreasing the system’s pressure will have the same effect on shifting the equilibrium position as increasing volume of the system.

Learning Objective #2 - Explain the overall observations about equilibrium in terms of the Collision Theory

Recall that collision theory has two components from Week 1’s Notes.

The first criteria is that the collision energy of the reacting species must be greater than or equal to the activation energy for the chemical reaction to occur.

The second criteria that must be satisfied is that the reacting species must collide at an effective orientation. 

In Week 1’s of HSC Chemistry Syllabus Notes, you have learnt about how the collision theory is used to explain how temperature, pressure, volume and concentration affect rate of reaction. 

In this learning objective, we will elaborating on what we learnt in Week 1. More specifically, we will be exploring how the concentration of reactants and products changes from the start to end of a chemical equilibrium reaction (“overall observations about equilibrium”).

Then, we will explain such changes in terms of the collision theory instead of using Le Chatelier’s Principle.

Overall Observations of Equilibrium