HSC Chemistry Syllabus Notes

Module 5 / Inquiry Question 2

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

In the graph above, we started off the chemical equilibrium reaction with no ammonia, i.e. [NH3] = 0 M.

So what happened to the concentration of ammonia over time? Well, the nitrogen gas reacted with hydrogen gas to form ammonia. So, over time, the [NH3] increased. This should be obvious.

You can also see that the change in concentration varies as time of reaction increased.

In fact, the rate of reaction can be calculated as concentration over time.

This means that the slope at any given time in the above concentration vs time graph is actually the rate of reaction at that given time.

As you can see from the graph, the rate of reaction changes as time of reaction increases.

Eventually, change in concentration is CONSTANT for all three reacting species and this means that the system is at equilibrium as the change in concentration of all species are constant. That is, the rate of the forward reaction is equal to the rate of the reverse reaction.

The slope is zero when equilibrium is established as you can see on the graph! What does that remind you of?
Yes! The concept of dynamic equilibrium! We have touched on this in the beginning of Week 1’s Notes.

But, before the Haber Process has reached equilibrium, the rate of reaction (concentration/time) was not equal as the slope of each species is changing as it proceeds to equilibrium as shown in the graph.

How can we explain this change in slope as the system proceeds to equilibrium?

Le Chatelier’s Principle isn’t really helpful here. This is because, although Le Chatelier can help us predict the direction which the equilibrium system will shift towards, it explain the shift at the molecular level.

So what can we use to explain the differing rate of reaction? We can use collision theory!

At time = 0, the nitrogen and hydrogen molecules react with each other to form ammonia.
Over time, the [NH3] increases and the [N2] and [H2] decreases.

You can see that the rate of formation of ammonia decreases over time by measuring the slope on the graph. Similarly, the rate of consumption for nitrogen and hydrogen also decreases over time. Why? Collision Theory!

Well, as more ammonia is being formed, more hydrogen and nitrogen molecules are being used up.

This means there is less hydrogen and nitrogen molecules to react with each other and form ammonia. This, however, does not mean that the probability of effective orientation between N2 and H2 have decreased nor have their collision energy has decreased.

It just means, for a given period of time, the amount of hydrogen and nitrogen molecules being converted into ammonia has decreased. This is because you have fewer reactants that can react with other in the system.

NOTE: Later in the course, you will understand that varying concentration can change the effective orientation of reactants. This would then affect the number of successful collisions occurring. Here, we are assuming that the probability of effective orientation and collision energy does not change with varying concentration. Therefore, any changes to the rate of reaction is due to the frequency of collision being decreased due to lower concentration of reactant and so the number of successful collisions is affected.

ANALOGY: Say you have a robot that is capable of making 5 pizzas an hour. Initially, you put enough ingredients for the robot to make 5 pizza per hour neatly on the table.

However, by midnight, a strange wind entered the house. The wind scattered the ingredients across the house. You need to personally find and collect the ingredients and give them to the robot in order for it to make 5 pizza an hour.

The random locations of the ingredients that was scattered by the wind means that you only give the robot enough ingredients for it to make 2 (or any number less than 5) pizza per hour as you need time to find the ingredients. So, the rate of production has decreased over time.

However, it is important to note that the decrease in the rate of production does not mean the robot will make the pizza different as the manufacturing process (reaction) is the same. That is, effective orientation & collision energy required by the reactants to make the product is still the same as the manufacturing process (reaction) is still the same.

In our case, it is just that there is fewer reactants (available ingredients) available for reaction to form ammonia (to provide the robot to make pizza).

Before the system has established or reached equilibrium

In terms of collision theory, if the rate of reaction (slope) is declining for the forward reaction, it would mean that the number of successful collision between reactants are decreasing (as less of them are available in the system as per analogy). Vice versa, if the rate of reaction is increasing, it means that there are more successful collisions.

When system has reached or equilibrium

Back to the graph, when the equilibrium is reached, the rate of reaction is constant. The rate of forward and reverse reactions are equal. This is because, when equilibrium is established, as x amount of ammonia is produced, x amount of ammonia is being also being converted into nitrogen and hydrogen – Rate of forward and reverse reactions are equal and constant at equilibrium.

In terms of collision theory, the number of successful collision between the reactants to produce products and between products to produce reactants are equal at equilibrium.

This would mean that the rate of forward and reverse reaction is equal at equilibrium.

Relating changes in concentration to collision theory

Suppose you have the following equilibrium established: A(g) + B(g) <-> C(g) + D(g)

By increasing concentration of substance A, there will be more of substance A with the effective collision orientation when colliding and reacting with substance B.

As per collision theory, this would mean that more substances A and B will satisfy the collision theory and be converted to substances C and D.

This would mean there will be more successful collisions between species undergoing the forward reaction compared to species undergoing to the reverse reaction

This would mean that the forward rate of reaction for the above equilibrium will be favoured, i.e. the forward rate of reaction will be greater than the reverse rate of reaction.

This would mean that the equilibrium position will shift to the right so the [C] and [D] would increase.

Please NOTE: Of course, if you increase concentration, the frequency of collision also increased as reacting species spend more time interacting with each other. However, more frequent collisions do NOT mean that the reacting substances will have the effective orientation. By increasing the frequency of collisions between reacting substances with the effective collision orientation, the rate of reaction would increase because there is more frequent collision with increased effective collision orientation.

Notice that this is explaining the shift in equilibrium position using collision theory. You can also explain this shift in equilibrium position via Le Chatelier’s Principle.

Vice versa, if you decrease the concentration of A, there will be fewer substance A with effective collision orientation when it collides and react with substance B.

As per collision theory, this would that less substances A and B will satisfy the collision theory and be converted to substance C and D.

This would mean that, if you decrease the concentration of A, the reverse rate of reaction will be greater than the forward rate of reaction.

This would mean that the equilibrium position will shift to the left so that the [A] and [B] will increase.

NOTE: The reason to why increasing surface area of reacting species will increase rate of reaction is that because there will be increased number of reacting species reacting with effective collision orientation.

Relating changes in temperature to collision theory

If you increase temperature, the frequency of collision increased as reacting species spend more time interacting with each other. However, more frequent collisions do NOT increase the amount of reacting species that satisfy the collision theory (similar to the case of changing concentration above) as it is insignificant compared to increasing system’s temperature.

If the system’s temperature is increased then the amount of reacting species that have collision energy equal or greater the activation energy of the reaction would increase.

The increase in temperature will have greater effect for reacting species that are proceeding to undergo the endothermic reaction of the equilibrium than the reacting species undergoing the exothermic reaction.

This can be explained using Arrhenius’s equation in determining the rate constant (k) for a reaction which we have learnt in Preliminary HSC Chemistry.

Recall that in equilibrium reactions, if one direction is endothermic then the reverse reaction must therefore be exothermic.

As temperature increases, the increase in the amount of reacting species undergoing the endothermic reaction that will have their collision energy greater or equal to the activation energy will be greater than the increased amount of reacting species proceeding the exothermic reaction.

We can explain this using the Arrhenius equation. The term ‘k’ is the rate constant which directly affect the rate of reaction of the forward and reverse reaction.

Both the forward and reverse reaction have their own unique rate constant (k).

In endothermic reactions, the activation energy barrier (Ea) as seen in the fraction will be higher than the activation energy barrier in the exothermic reaction.

This would mean that an increase in temperature will have a greater effect on the amount of reacting species undergoing the endothermic reaction than those species undergoing to the exothermic reaction.

NOTE: You can test this by plugging in some random numbers into the Arrhenius equation and adjusting the activation energy to see the effect on ‘k’ which is directly related to rate of the forward and reverse reactions.

So, as temperature increases, the equilibrium position will shift in favour of the endothermic reaction as, compared to the exothermic reaction, there will be a higher increase in the species that will have greater or equal collision energy than the activation energy.

Learning Objective #3 - Examine how activation energy and heat of reaction affect the position of equilibrium

What is activation energy?

Recall from Preliminary HSC Chemistry that activation energy is the energy required form the necessary chemical bond(s) to convert reactants into products (or vice versa). That is, energy required to reach the transition state.

When the reacting species collide, their kinetic energy will be converted into collision energy.

Sufficient collision energy must be present to break the chemical bonds required for the reactants to be converted into products (or vice versa) in order for a chemical reaction to take place.

If the reaction only requires one chemical bond to be broken, then activation energy is equal to the bond energy for the chemical bond that is required to be broken.

Usually, however, more than one chemical bond is required to be broken so activation energy does not equal to the bond energy.

So what is relationship between activation energy and equilibrium?

Recall the Haber Process equilibrium graph in the previous learning objective. For an equilibrium chemical reaction to reach equilibrium, it requires some time.

Equilibrium reactions with smaller activation energies will reach the state of chemical equilibrium in less time than reactions requiring higher activation energies.

So why lower activation energy allows equilibrium to be reached faster?

Lower activation energy means that the minimum collision energy required has been lowered. This means that the number of reactants (or products as dynamic equilibrium is two way reaction) that satisfy the minimum collision energy required as per collision theory has been increased.

Therefore the rate of successful collision reactions have been increased which equates to an increase in the rate of reaction.

If you increase the rate of reaction, the state of equilibrium can be attained with less time.

In an equilibrium graph like the on the Haber Process that we have shown in the previous learning objective, you can imagine the slopes of the N2, H2 and NH3 curves are becoming steeper due to increased rate of reaction.

An alternative reaction pathway with a lower activation energy can be achieved using a catalyst so that more reacting species will have collision energy that is equal or greater than the activation energy of the new reaction pathway that is made available by the catalyst!

Catalysts are substances that participate in the equilibrium chemical reaction but are NOT consumed in the overall reaction.

What does this mean? Well, this means you can still reuse the catalyst after putting it in the equilibrium reaction as a reactant as it will reappear as a product at the end and not be consumed. The catalyst equally speeds up the rate of the forward and reverse reactions so that equilibrium position is reached in less time due to lowering the activation energy required for the bonds.

NOTE: As explained in Week 1’s notes, catalysts do not affect the position of the equilibrium. Instead, only less time is required to reach the point of equilibrium by increasing the number of successful collisions between reactants and between products by the same amount.

Thus, the forward and reverse rate of reactions increase by the same amount.

The example below involves a catalyst participating in a three-step one-way reaction:

Step 1: A + B –> C

Step 2: C + D –> E

Step 3: E + A -> D + G

Overall Reaction: 2A + B -> G

The catalyst is the reactant that are used to speed up the reaction but is NOT consumed in the OVERALL reaction.

Hence, A and B and G cannot be catalysts.

In our example, D is the catalyst. This is because it appeared as a reactant in Step 2 but reappeared as (formed) as a product in Step 3, i.e. not consumed.

Fun Fact: Species E is called an intermediate substance. This is because it initially appeared as a product and is consumed as a reactant in Step 3 of the reaction to make more the final product, G.

The diagram below illustrates how a catalyst provides an alternative (different) pathway, with a lower activation energy, for the reaction X + Y –> Z.

The red line illustrates that alternative reaction pathway has lower activation energy provided by a catalyst compared to the original reaction pathway, which is illustrated in black, without a catalyst.

The fluctuation of the activation energy in the reaction alternative pathway (red line) provided by catalyst is NOT required for HSC Chemistry.

However, if you are curious and wish to learn more about the fluctuation curve, you can research about the intermediate compound formation theory that is used to explain catalysts.

The change in enthalpy of a chemical equilibrium reaction will determine whether a chemical reaction is exothermic or endothermic as mentioned earlier.

Suppose we have the following chemical equilibrium reaction:

A + B ⇌ C + D ; where ΔHforward reaction is negative.

This means the forward reaction (A + B -> C + D) is exothermic and the reverse reaction (D + C -> A + B) is endothermic.

The heat of reaction will have affect on the equilibrium position as per Le Chatelier’s Principle.

We have already explored this concept under the ‘Potential Equilibrium Disturbance #1: Change in Temperature’ section earlier in this week’s notes.

Week 2 Homework Set (Essential for Band 5)

Question 1: Explain how Le Chatelier’s Principle can be used to predict how the equilibrium position will shift for heating cobalt (II) chloride hydrate [4 marks]

Question 2: Explain how Le Chatelier’s Principle can be used to predict how equilibrium will shift for the interaction between nitrogen dioxide and dinitrogen tetroxide. Suppose that, for this question, the disturbance in the equilibrium system is change in temperature, pressure, volume and concentration of the reactants. ALL FOUR! [6 marks]

Question 3: Explain how Le Chatelier’s Principle can be used to predict how equilibrium position will shift when the concentrations of ions are varied when the Fe3+ + SCN– ⇌ FeSCN2+ reaction is at equilibrium. [4 marks]

Question 4: Define Le Chatelier’s Principle [1 mark]

Question 5: Suppose you are given this equilibrium reaction: 2A + 3B ⇌ 2C + 6D.The change in enthalpy for the forward reaction is 170kJ. What is the change in enthalpy for the reverse reaction? What principle supports this? [2 marks]

Question 6: How does a catalyst affect the equilibrium position of a chemical reaction? [2 marks]

Question 7: Refer the diagram below to answer the following questions.

How does the rate of reaction change over time?
Explain what has happened to the system between t=2 and t=3 minutes?
Construct a diagram to illustrate how calcium carbonate’s mass change over time
The decomposition of calcium carbonate is an endothermic reaction. If there is a decrease in the system’s temperature (i.e. the furnace’s temperature has been lowered), how will the [CO2] change and how the rate of forward reaction be affected?

Question 8: Distinguish between activation energy and collision energy [2 marks]

Question 9: Explain the inverse relationship between volume and pressure in terms of shifting the equilibrium position a chemical reaction [4 marks]

Question 10: Describe what is meant by equilibrium position [2 marks]

Week 2 Curveball Questions

Curveball Question 1: Distinguish between a catalyst and an intermediate molecule in a chemical reaction? [2 marks]

Curveball Question 2: Suppose you are given this equilibrium reaction: 2A + 3B ⇌ 2C + 6D. It is safe to say that at equilibrium, there will always be twice as much [D] than [B]? Why or Why not? [2 marks]

Curveball Question 3: Suppose you are given the following reaction in a closed system [4 marks]

N2O4 (g) ⇌ 2NO2 (g) where the change in enthalpy for the forward reaction is positive.

Draw an equilibrium graph of concentration against time for the following change:

Suppose at t=0 minutes, there is 6M of N2O4 and 2M of NO2 at equilibrium.

At t=4 minutes, there is a disturbance in the system. The concentration of both dinitrogen tetroxide and nitrogen dioxide were increased as a chemist reduced the volume of the closed system by lowering a piston.

Curveball Question 4: Suppose you are given the following reaction in a closed vessel [8 marks]

N2 (g) + 3H2 (g) ⇌ 2NH3 (g)

The forward reaction is exothermic.

Draw an equilibrium graph of concentration against time for the following changes.

• At t=0, the above reaction is at equilibrium.

• At t=1, you added more nitrogen gas into the vessel

• At t=2, you increased the pressure of the vessel

• At t=3, you increased the temperature of the vessel

Solutions to Week 2 Homework Set

Solution to Question #1:

Le Chatelier Principle helps predict the shift in equilibrium position by stating that in the event that there is a disturbance in the system, the system will counteract this change by shifting its position to minimise the effect of the disturbance. The forward reaction of the cobalt (II) chloride hydrate and cobalt chloride dehydrate equilibrium is endothermic.

Co(Cl)2•6H2O (s) <-> CoCl2 (aq) + H2O(g)

By increasing temperature of the system, the equilibrium position will shift to the right, in favour of forward endothermic reaction to consume the excess heat. This would mean that the concentration of cobalt (II) chloride dehydrate will increase and the concentration of cobalt (II) hexahydrate will decrease.

Solution to Question #2:

The equilibrium between nitrogen dioxide and dinitrogen tetroxide can be expressed as the following:

2NO2(g) <-> N2O4(g)

By increasing the concentration of nitrogen dioxide, the equilibrium position will shift to the right to decrease the [NO2] which increases the [N2O4] and vice versa.

If the temperature of the system increased, the equilibrium position will shift to the left, in favour of the reverse endothermic reaction as per Le Chatelier’s principle to consume the excess heat. If the temperature of the system decreased, the equilibrium position will shift to the right, in favour of the forward exothermic reaction as per Le Chatelier’s principle to release heat and minimise the drop in temperature.

By increasing the system’s pressure, the equilibrium position will shift to the side of the chemical equation with the least moles of gas as per Le Chatelier’s principle to restore the system’s overall gas pressure. Since the molar ratio of gases between NO2 and N2O4 is 2:1, the equilibrium will shift to right, increasing the [N2O4] and decreasing [NO2]. Vice versa for decreasing in system’s pressure.

By increasing the system’s volume, the equilibrium position will shift to the side of the chemical equation with more moles of gas. This is due to Le Chatelier’s principle to minimise and restore the overall gas concentration through the system. Since the molar ratio of gases between NO2 and N2O4 is 2:1, the equilibrium will shift to left, increasing the [NO2] and decreasing [N2O4]. Vice versa for decreasing in system’s volume.

Solution to Question #3:

By increasing the concentration of Fe3+ ions, the equilibrium position will shift to the right as per Le Chatelier’s principle to minimise the increase in [Fe3+]. This would mean that the [FeSCN2+] will increase and [SCN–] will decrease. The same effect and reason will apply to increasing the concentration of [SCN–]. If the concentration of FeSCN2+ ions was increased, the equilibrium position will shift to the left to minimise the increase in [FeSCN2+] as per Le chatelier’s principle. This will result in a increase in [Fe3+] and [SCN–] and decrease in [FeSCN2+]. It is important to note that although [FeSCN2+] will decrease due to shift in equilibrium position as per Le Chatelier’s principle, the final concentration of FeSCN2+, after the shift, will be greater than the initial concentration of [FeSCN2+]. This is because le chatelier’s principle will not completely eliminate all the extra moles of FeSCN2+ added into the system.

Solution to Question #4:

Solution to Question #5:

The change in enthalpy for the reverse reaction is -170 kJ. The law of conservation of energy supports this.

Solution to Question #6:

A catalyst will not change the equilibrium position for an equilibrium reaction, i.e. the concentration of products and reactants. A catalyst increases both the forward and reverse rate of reaction so that the time taken to reach equilibrium is lowered.

Extra notes below: (not part of marking criteria for 2 marks – if the question is 4 marks, perhaps you can include the following).

Catalyst speeds up reaction by providing an alternative pathway of with lower activation energy required for the reactants interact and form products (and vice versa for equilibrium reactions). Since the alternative pathway has a lower alternative energy which means more species (reactant and/or products) will have enough collision energy that’s equal or greater than the new (lower) activation energy. this means that more species are satisfies the collision theory and thus increasing the rate of reaction.

Solution to Question #7:

The rate of reaction increases from t=0 minutes to t=2 minutes. From t=2 minutes to t=3 minutes, the rate of reaction is constant, i.e. the rate of the forward reaction is equal to the rate of the reverse reaction.

From t = 2 minutes to t = 3minutes, equilibrium has been established for the reaction (decomposition of calcium carbonate) in the system. The rate of the forward reaction is equal to the rate of the reverse reaction.

In the event of a decrease in temperature in the system, as per Le Chatelier’s Principle, the equilibrium position will shift to the left in favour of the exothermic reaction to release heat. This would minimise the drop in temperature and the forward (endothermic reaction) will decrease until equilibrium has been re-established.

The concentration of carbon dioxide will decrease as the reverse reaction is favoured.

Do notice in the graph that the rate at which the mass of calcium carbonate is decreasing over time. Eventually, at equilibrium, there is the decrease of calcium carbonate’s mass is zero, i.e. the change in calcium carbonate’s mass is constant. This is because at equilibrium, the rate of the forward and reverse reaction is constant. As calcium carbonate is decomposed into calcium oxide and carbon dioxide, equal moles of calcium carbonate is formed AT THE SAME TIME AT EQUILIBRIUM (from the reaction of calcium oxide and carbon dioxide).

Solution to Question #8:

Activation energy is the energy required to break and form chemical bonds to convert reactants into products. Collision energy is the energy converted from kinetic energy when reacting species collide.

Solution to Question #9:

Changes to a system’s volume and pressure will affect the concentration of gases at equilibrium in the system. By increasing the system’s volume, the equilibrium position will shift to the side of the chemical equation with more moles of gas. This is due to Le Chatelier’s principle to minimise and restore the overall gas concentration through the system. Vice versa, by decreasing the system’s volume, the equilibrium position will shift to the side of the chemical equation with less moles of gas as per Le Chatelier’s principle to restore the system’s overall gas concentration.

By increasing the system’s pressure, the equilibrium position will shift to the side of the chemical equation with the least moles of gas as per Le Chatelier’s principle to restore the system’s overall gas pressure. Vice versa, by decreasing the system’s pressure the equilibrium position will shift to the side of the chemical with more moles of gas as per Le Chatelier’s principle to restore the system’s overall gas pressure.

Hence, there is an inverse relationship between pressure and volume. This is illustrated in as Boyles’ Law: P1V1 = P2V2 or P is proportional to 1/V at constant temperature (Year 11 syllabus).

Solution to Question #10:

Equilibrium position is the concentration of reactant(s) and product(s) at a particular point in time in an equilibrium reaction. Generally, the position at which the concentration of reactants and products has no tendency change is referred to as the equilibrium position in an equilibrium reaction.

Solutions to Week 2 Curveball Questions

Solution to Curveball Question #1:

A catalyst generally appears or is added as a reactant in a chemical reaction but is not consumed, i.e. it will re-appear as a product. An intermediate is first appears as a product (rather than as a reactant) of a chemical reaction but is consumed in the later stages of the chemical reaction as a reactant.

Solution to Curveball Question #2:

The concentration ratio between reactant and products at equilibrium is determined by the equilibrium constant. As there are many equilibrium positions (concentration of reactant and product at a given point in time) that can satisfy the equilibrium constant, the amount of [D] to [B] may not necessarily twice as much. However, it could be an possibility. You will learn about equilibrium constant in the next week’s notes. This was just an open-eye question to get your awareness.

Solution to Curveball Question #3:

The decrease concentration of nitrogen due to the effect of Le Chatelier’s principle to counteract and minimise the disturbance, in this case the decrease in volume, will not drop lower than the initial concentration of nitrogen when the equilibrium reaction was initially at equilibrium.