M5 EAR 1-2

IQ1. What happens when chemical reactions do not go to completion?

IQ1. SYLLABUS POINTS

5.1.1 Investigating reversible reactions.

5.1.2 Static and dynamic equilibrium; open and closed systems.

5.1.3 Entropy and enthalpy in non-equilibrium systems.

5.1.4 Collision theory and reaction rate in equilibrium reactions.

5.1.1 conduct practical investigations to analyse the reversibility of chemical reactions, for example:

cobalt(II) chloride hydrated and dehydrated
iron(III) nitrate and potassium thiocyanate
burning magnesium
burning steel wool

· conduct investigations to collect valid and reliable primary and secondary data and information CH12-3· communicate scientific understanding using suitable language and terminology for a specific audience or purpose CH12-7

So far we have considered reactions as proceeding to completion, eg

CaCO_3(s) → CaO_(s) + CO_2(g)

but many products may react to reform the original substances. This is called a reversible reaction. This is a common feature of many physical processes.

Consider the following chemical reaction:

NO_2(g) ⇌ N₂O_4(g)

Each of the gases, NO_2(g) and N₂O_4(g), has a distinctive appearance. At 298 K (25 degrees C) the mixture is yellow, and there is more of the product.

Both reactions continue to occur, but the rate of the forward reaction equals the rate of the reverse reaction. The reaction is said to be in a state of equilibrium.

If we increase the temperature, the colour darkens, due to the presence of more reactant, while at lower temperatures a lighter colour indicates more product.

For gases, the pressure can also affect the equilibrium situation.

At equilibrium, although the process is dynamic (forward and reverse reactions both occurring) at the molecular level, the overall concentrations of the substances remain the same (rate of forward reaction equals the rate of the reaction that reverses it).

A significant number of chemical reactions are equilibrium reactions, which can also be referred to as reversible or non-spontaneous. In these reactions, the forward reaction does not go to completion; instead, a certain quantity of the reactants remains.

At a molecular level, the reverse reaction (from products to reactants) is also occurring. This means that the reactants are never fully used up: they are being re-formed in the reverse reaction.

If the rates of the forward reaction and reverse reaction are not the same (which can easily be the case), then the amounts of the reactants and products will be changing. When the reaction rates become the same, then the concentrations will become constant. At this point the system is in equilibrium. More specifically, dynamic equilibrium because individual molecules are still being converted between the reactants and products (but at the same rate).

The rate of a reaction is proportional to the concentration of the reactants. That means that if more of the reactants are present, the reaction will occur more quickly. As the reaction proceeds in the forwards direction, the reactants will be used up and the rate of the forward reaction will slow. The rate of the reverse reaction is proportional to the concentration of the products of the original reaction. Therefore, as this reaction proceeds and more products are produced, the rate of the reverse reaction will increase. We can visualise this process with the graph below to realise that an equilibrium point will always be established in a system where a reverse reaction is possible.

However, the location of that equilibrium point depends on the particular reaction. In some cases, we say that an equilibrium lies to the right, which means that there are more of the products than the reactants at equilibrium. An equilibrium which lies to the left has more of the reactants than the products at the equilibrium point. In these cases the reaction will only proceed a small amount before equilibrium is reached.

Example:

H₂O _(l)+ CO_{2 (g)} ⇌ H₂CO_{3 (aq)}

The first thing to notice is the bi-directional arrow (double arrow), used in chemical reactions to indicate that an equilibrium is established. Under standard conditions (100kPa of pressure, 25 degrees Celcius), the equilibrium for this reaction lies to the left. This means that only a small amount of the CO₂ present in a system will be dissolved in water. Once we reach an equilibrium, both the forward reaction, where CO₂ dissolves in water, and the reverse reaction, where it leaves the solution are occurring. However, the rate of these reactions is the same, so the relative amounts of dissolved and undissolved CO₂ will not change.

12C5 2a NO2 MP4.mp4

ATAR Notes Key Points:

→ should be used to represent irreversible chemical reactions
⇌ should be used to represent reversible chemical reactions (Silove, 2018, p. 2)
Anhydrous means substances that do not contain water molecules
-hydrate means substances that contain some number of water molecules. The prefix determines how many there are "mono-” = one, “di-” = two, “tri-” = three, “tetra-” = four, “penta-” = five, “hexa-” = six, “hepta-” = seven, “octa-” = eight, “nona-” = nine, and “deca” = ten.

Copper sulfate pentahydrate contains 5 water molecules surrounding each copper cation and the sulfate anion combination.

VIEW videos

EAR#1 https://www.youtube.com/watch?v=13oUEFo6IXw&index=5&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [5.10]

ACTIVITY 5.1.1A

Reversible and Irreversible Reactions

ACTIVITY 5.1.1B

This will be completed as a demonstration due to the nature of the cobalt chloride.

VIEW video https://www.youtube.com/watch?v=GS9kIj9n-BU [1.38mins]

Activity 5.1.1C

Try the pHet simulation: https://phet.colorado.edu/en/simulation/reversible-reactions

Task 5.1.1A

1. Some ionic substances exist in both a hydrated and anhydrous form. Use an example to explain the difference between the terms anhydrous and hydrated as they apply to crystals.

2. Use the following examples to help you distinguish between reactions which are reversible and those which proceed to completion:

a) Cobalt(II) chloride – hydrated and dehydrated

b) Iron(III) nitrate and potassium thiocyanate

c) Burning magnesium

d) Burning steel wool

3. The following word equations represent reversible processes. Write a balanced chemical equation for each of the following (Schell & Hogan, 2018, p. 5):

a) Sulfur dioxide and oxygen produces sulfur trioxide

b) Nitrogen and hydrogen produces ammonia

c) Nitrogen dioxide produces dinitrogen tetroxide

d) Phosphorus pentachloride produces phosphorus trichloride and chlorine gas

e) Hydrogen and iodine produce hydrogen iodide

Answers:

Q2 Reversible; a) and b). Irreversible: c) and d)

Q3 too hard to get subscripts and double arrows, so imagine they are there!

a) 2SO2 + O2 === 2SO3 (all gases)

b) N2 + 3H2 === 2NH3 (all gases)

c) 2NO2 === N2O4 (both gases)

d) PCl5 === PCl3 + Cl2 (solid, liquid, gas)

e) H2 + I2 === 2HI (all gases)

5.1.2 model static and dynamic equilibrium, and analyse the differences between open and closed systems

· develop and evaluate questions and hypotheses for scientific investigation CH12‑1· conduct investigations to collect valid and reliable primary and secondary data and information CH12-3· communicate scientific understanding using suitable language and terminology for a specific audience or purpose CH12-7

Open and Closed Systems:

Consider the three types of systems: open, closed and isolated [Open, closed and isolated systems (Woollett, Birtles, & McInnes, 2018, p. 502)]

In a closed system, the products of a reaction remain in the system. Under certain circumstances, they may react to reform the reactant(s).

Reactions should not be though of as one substance completely changing into another. They should be visualised as having two components acting in opposite directions, initially at different rates.

A reaction rate determines how fast a reaction proceeds, and is mathematically defined as the:

change in concentration of a species delta [ ]

---------------------------------------------------------------- = ------------------------

change in time delta t

An example of a fast reaction would be fire burning (combustion with oxygen) and an example of a slow reaction is iron rusting (oxidation of a metal). Reactions do not only proceed forward; reactions often occur in a mixture of forward and reverse, the more dominant being the observed course of the reaction.

When the system reaches a point where the two components are acting at equal rates; the system has reached equilibrium.

Rate of the forward reaction (reactants → products) = Rate of the reverse reaction (products → reactants)

Static and Dynamic Equilibrium:

There are two different types of equilibrium: static equilibrium and dynamic equilibrium.

Static equilibrium

Static equilibrium occurs when all particles in the reaction are at rest and there is no motion between reactants and products. (Static equilibrium can also be seen as a steady-state system in a physics-based view.) Dynamic forces are not acting on the potential energies of the reverse and forward reactions.

For static equilibrium, the rates of forwards and backwards reactions are zero.

An example of static equilibrium is graphite turning into diamond. This reaction is considered at static equilibrium after it occurs because there are no more forces acting upon the reactants (graphite) and products (diamond).

C(graphite)→C(diamond)

Consider the diagrams below showing the general reaction: C→D

NOTE: Static equilibrium does not necessarily mean there are no moles on the reactant side after the reaction has occurred. This is only one example of static equilibrium.

Dynamic equilibrium

At dynamic equilibrium, reactants are converted to products and products are converted to reactants at an equal and constant rate.

Reactions do not necessarily—and most often do not—end up with equal concentrations. Equilibrium is the state of equal and opposite rates, not equal concentrations.

TASK 5.1.2A

View the models below and make sure you understand how each relates to equilibrium.

1. How are models helpful?

2. What limitations do they have?

STATIC chemical equilibrium is only achieved when the system:

is closed. (Matter will NOT be lost or gained.)
involves constant (=unchanging) macroscopic properties (eg colour, pressure, volume, electrical conductivity, pH), although changes are occurring at the molecular level.
is static, the rate of forward reaction = rate of reverse reaction = zero

A completion reaction involving a limiting agent could be regarded as a static equilibrium. No further reaction occurs and there are no macroscopic changes occurring to a system. In static equilibrium, the rates of forwards and backwards reactions are zero. A reaction at static equilibrium is irreversible.

DYNAMIC chemical equilibrium is only achieved when the system:

is closed. (Matter will NOT be lost or gained.)
involves constant (=unchanging) macroscopic properties (eg colour, pressure, volume, electrical conductivity, pH), although changes are occurring at the molecular level.
is dynamic, the rate of the forward reaction = the rate of the reverse direction.

An equilibrium process such as the NO2 / N2O4 equilibrium is regarded as a dynamic equilibrium: while there appear to be no macroscopic changes taking place, at the particle level both the forward and reverse reactions are occurring, but at the same rate. In dynamic equilibrium, rates of forwards and backwards reactions are equal. A reaction at dynamic equilibrium can be reversible (ie can proceed in the direction of products to reactants).

VIEW Videos:

EAR#2 https://www.youtube.com/watch?v=HFpkOarB6Zc&index=4&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.32]
HSC Chem Dynamic Equilibrium Model https://www.youtube.com/watch?v=zwakUht8590&t=35s [6.11]

ACTIVITY 5.1.2A

Complete modelling exercise ref https://youtu.be/fVeYoclfBN4 or https://www.youtube.com/watch?v=C5jDmG4nVV8 [9.39 mins]

1. Partner A fill Cylinder A to about 2/3 full, Partner B leave Cylinder B empty.

2. Each partner push the straw to the bottom of the cylinder, put a finger tightly on the top of the straw or straws to trap the water, and then transfer to the other cylinder. "But there's no water to move out of B!" I know. Do it anyway!

3. Once the water is transferred, put the straws back into the cylinder, one straw in A and two into B. Re-measure water and record it. (May have to estimate at the start).

4. Continue until there is no change. Call teacher to check.

"Why do you think you're at equilibrium?"

"The amounts aren't changing any more."

"Are they the same?"

"No, I thought they would be, but they're not."

"So what is the same?" "Why aren't the amounts changing even though you're moving water?"

"The amounts moving back and forth are the same."

Optional extra: start with equal amounts.

ACTIVITY 5.1.2B

In pairs, taking turns to be the calculator and the checker, complete the following:

12C5 2b Equilibrium Model WS.pdf

TASK 5.1.2C

1. Use examples to help you explain the difference between a closed system and an open system.

2. What are the three important features of an equilibrium system?

3. Soft drinks contain dissolved carbon dioxide which bubbles out of the solution when the pressure is reduced according to the following equation.

CO_2(aq) ⇌ CO_2(g)

Apply your understanding of equilibrium to what happens when a soft drink bottle is opened and then sealed.

4. Use the diagrams below to help you explain the concept of equilibrium as it applies to chemical processes.

This Photo by Unknown Author is licensed under CC BY-SA (Hegarty, Year 12 Chemistry Student Workbook Book 1, 2018, p. 5)

Answers:

Q3 In outline, opening the bottle reduces pressure. Favours the reaction where more moles of gas are present (ie forward reaction), so CO2 will bubble out of the liquid. Some gas will escape, so the soft drink system will have a lower CO2 concentration. When re-sealed, pressure is increased, favours formation of fewer moles of gas ie reverse reaction, some gas will go back into solution, but the concentration will be less.

REVIEW 5.1.2

Visit https://www.chemguide.co.uk/physical/equilibria/introduction.html for a model and explanation of equilibrium. (Ignore the name of the iron oxide for the moment.)
Make sure you can define: open/closed/isolated systems, homogeneous/heterogeneous systems, reversible/irreversible reactions, static/dynamic equilibria, reaction rate,

5.1.3 analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:

a) combustion reactions

· select and process appropriate qualitative and quantitative data and information using a range of appropriate media CH12-4

We have previously discussed the concepts of enthalpy and entropy in relation to spontaneity of chemical reactions. What relevance do they have to reversible reactions?

Theoretically if the system is closed, then sufficient activation energy could drive the products to reform the reactants.

In practice, many chemical reactions take place in an open system where energy and matter are lost from the system. Once this occurs, this mass cannot be recovered and this may result in a process being irreversible. One good example is a combustion reaction. A combustion reaction is a non-equilibrium system.

An interesting fact about combustion reactions is the products are generally independent of the fuel and only dependent on the concentration of oxygen.

If we were to carry out a series of combustion reactions involving the fuels ethanol, methane, octane and glucose (in the presence of sufficient oxygen for complete combustion), the products of all four reactions would be carbon dioxide and water.

If a combustion reaction was carried out in a closed container, would we expect water and carbon dioxide to reform the fuel and oxygen?

Both CO₂ and H₂O are stable, to form each one is exothermic and increases the entropy of the system, Gibbs Free energy is high, and so combustion is a non-equilibrium system.

Let's use octane. If we look at spontaneity using Gibbs Free energy at 100^oC, and we know the ΔH^o for the combustion of octane (ΔH = -5114 kJ.mol^-1) and the ΔS^o (+384 J.mol^-1.K^-1) then we can work out the ΔG^o value.

Do this, don't forget to change the units for T and ΔS^o (see diagram below).

A negative value for ΔG indicates a spontaneous reaction. Energy is released as the reaction proceeds until all the fuel or oxygen is consumed (Chan, et al., 2019, p. 65).

VIEW Videos:

EAR #3 https://www.youtube.com/watch?v=aV_-5R3A52M&index=3&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz

TASK 1.3.1

1. Write an equation for the complete combustion of methane and use it, and/or the values given in the table below, to help you determine whether this reaction is spontaneous at room temperature (SLC). (Davis, Disney, & Smith, 2018, p. 26)

2. Given the enthalpy of formation of solid glucose is -1273 kJ.mol^-1 and its entropy is +212 J.mol^-1.K^-1, use any relevant values from Question 1 to determine the spontaneity of respiration at SLC.

5.1.3 analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:

b) photosynthesis

· select and process appropriate qualitative and quantitative data and information using a range of appropriate media CH12-4

Combustion is an exothermic reaction which increases the entropy of the system. It is logical to think of combustion reactions as spontaneous and irreversible.

By contrast, photosynthesis is an endothermic process which is not spontaneous. It requires an input of energy, and a catalyst to lower the activation energy in order to proceed. Photosynthesis is chemically the reverse of respiration. Respiration is a form of combustion of glucose and is a spontaneous reaction.

Rather than a simple reaction, photosynthesis is a complex series of reactions which take place in the chloroplasts of plants. Sunlight powers photosystems in the thylakoid membranes of the chloroplasts to break down water molecules. This involves a series of electron transfer pathways. We do not expect the specifics of photosynthesis to be examined as the thermodynamic analysis is very challenging. The diagram below (Sato, 2012) may provide a sense of the complexity of this process and some keys component terms for those who wish to dig deeper, or for the Biologists.

It is sufficient to identify that photosynthesis includes a number of steps which involve both spontaneous and non-spontaneous processes. However we classify photosynthesis as a non-spontaneous, non-equilibrium process.

VIEW"

PPT Photosynthesis

12C5 1.1c Photosynthesis PPT.pptx

VIEW Videos:

EAR#4 https://www.youtube.com/watch?v=5qMMQEjUIxo&index=2&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [6.37]
Light and dark reaction https://www.youtube.com/watch?v=v-G-d27C1TU [4.37]

TASK 5.1.3A

1. Write the overall equation for photosynthesis. Include the enthalpy value (+2805 kJ.mol^-1).

2. If the ΔS for photosynthesis is -212 J.mol^-1.K^-1, use the enthalpy value from question 1 to show that photosynthesis is non-spontaneous at SLC.

3. State whether the following statements are true or false. If false, identify how you could turn them into true statements. (Chan, et al., 2019, p. 69)

A) Photosynthesis may be considered to be a non-equilibrium system overall.

B) The change in entropy, ΔS, for photosynthesis is positive.

C) The change in enthalpy, ΔH, for photosynthesis is positive.

D) Spontaneous reactions drive photosynthesis, increasing the entropy of the universe overall.

E) The net products of photosynthesis are glucose and oxygen.

ANSWERS:

Q1 6CO2 + 6H20 + 2805 kJ.mol-1 --------- C6H12O6 + 6O2 (g, l, aq, g)

Q2.

Q3 B is the only incorrect statement. From the equation for photosynthesis, there are 6 molecules of gas on each side of the equation, but there is a decrease in the number of particles in the system. Hence ΔS will be negative. (Another way of looking at this is that the production of a large product molecule from many smaller reactant molecules is creating greater order.)

1.4 investigate the relationship between collision theory and reaction rate in order to analyse chemical equilibrium reactions

· select and process appropriate qualitative and quantitative data and information using a range of appropriate media CH12-4· analyse and evaluate primary and secondary data and information CH12‑5 · solve scientific problems using primary and secondary data, critical thinking skills and scientific processes CH12‑6

During the Year 11 course, we applied the concept of collision theory to explain reaction rates in chemical reactions. How can we apply this to equilibrium systems?

Collision Theory: “for a reaction to occur, the particles must collide with sufficient energy to break the bonds and have the appropriate orientation to allow the new bonds to form”. (Davis, Disney, & Smith, 2018, p. 32)

We can use the energy profile graph (below, left) to analyse chemical equilibrium reactions too. When analysing the energy profile diagram there are several points to note:

The ΔH value is the same whether we are considering X → Y or Y → X, only the sign is different (opposite).
If the forward reaction is endothermic, the reverse reaction will be exothermic.
The activation energy E_A will not be the same for the forward and reverse reactions and will be higher for the endothermic reaction than the exothermic reaction.
The catalyst reduces the activation energy for both the forward and reverse reactions. This means the system will reach equilibrium more quickly. It will not change:

- ΔH

- amount of reactant at equilibrium

- amount of product at equilibrium

Only if the activation energy for both forward and reverse reactions is sufficiently low (or overcome) will both reactions occur simultaneously.

If we analyse the diagram (below, right Chan, et al., 2019, p. 60), we see that initially there is 1 mole of nitrogen and 3 moles of hydrogen. The product of this reaction, ammonia, is not present at the start. There is a high concentration of nitrogen and hydrogen molecules and these collide frequently (more collisions) with sufficient energy to break the bonds and form ammonia (more successful collisions). However as the reaction proceeds, the concentrations of hydrogen and nitrogen decrease (reducing the numbers of collisions, and therefore the forward reaction rate) as the concentration of ammonia increases. This increases the likelihood of two ammonia molecules colliding with sufficient energy to reform the reactants.

The forward reaction rate is slowing as the reverse reaction rate increases, until they reach a point where the reaction rates are equal. Here there is no net change in the concentration of the substances present.This is equilibrium.

VIEW video:

EAR#5 https://www.youtube.com/watch?v=S-0n7juijS0&index=1&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.05]

TASK 5.1.4A

1. Write a balanced equation to represent the synthesis of ammonia.

2. Look at the graph below. What information in this graph tells you that this is an equilibrium reaction?

3. Use the graph to work out the number of moles of each species at equilibrium.

4. Work out the change in the number of moles of each species as the reaction proceeds from the start to the point where equilibrium is reached.

ANSWERS:

1. N2 + 3H2 ===== 2NH3 (all gases)

both reactant and product species are present (has not gone to completion ie reversible reaction)
constant macroscopic properties, ie no change to the concentrations of any of the species; therefore rate forward = rate reverse

3. Each square on y-axis is 0.33 moles/L, so NH3 is 0.5, N2 is 0.7, H2 is 2.1 (need 3 times as much H2 as N2, look at the equation)

4. NH3 is 0 + 0.5 = 0.5 mol/L; N2 is 1.0 - 0.7 = 0.3 mol/L; H2 is 3 - 2.1 = 0.9 mol/L

IQ2 What factors affect equilibrium and how?

IQ2. SYLLABUS POINTS

5.2.1 Factors that affect equilibrium and Le Châtelier.

5.2.2 Equilibrium observations and collision theory.

5.2.3 Activation energy, heat of reaction and equilibrium position.

5.2.1 investigate the effect of change on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects.

Le Chatelier’s Principle

Le Chatelier’s Principle applies to systems which are in equilibrium.

When a system in equilibrium is subjected to a change in concentration, temperature, external pressure, or some other factor which disturbs the equilibrium, the system reacts to counteract that change.

Le Chatelier’s Principle can be used to predict how a system in equilibrium will react to a change.

Le Chatelier's way of saying this makes it sound as if the system thinks, which it doesn't, so I prefer to say

"...the response of the system is such that the change is counteracted."

OVERVIEW

VIEW:

Equilibrium and Le Chatelier's Principle PPT
Which way will the equilibrium shift? (Le Chatelier) https://www.youtube.com/watch?v=BPDkl92NCUs [8.30 mins]

12C5 1.0 Equilibrium PPT.ppt

5.2.1 investigate the effect of change on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects.

a) How will a change in TEMPERATURE affect an equilibrium reaction?

KEY POINTS:

Endothermic reactions absorb heat to move forward
Exothermic reactions release heat to move forward
An increase in temperature will favour the endothermic reaction

The position of equilibrium changes for a change in temperature. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change made.

If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature. It will do that by favouring the reaction which absorbs heat.

H₂ + I₂ ↔ 2 HI Enthalpy of reaction (Delta H) = -10.4 kJ/mol

In the equilibrium above, that will be the reverse reaction because the forward reaction is exothermic.

So, according to Le Chatelier's Principle the position of equilibrium will move to the left. Less hydrogen iodide will be formed, and the equilibrium mixture will contain more unreacted hydrogen and iodine.

PRAC 5.2.1A Cobalt chloride and temperature change

Consider the equilibrium system below:

A12C5 2a Cobalt Reaction.pdf

PRAC 5.2.1B NO₂ / N₂O₄ and temperature change

In a chemical reaction, equilibrium is the state in which the rate of the forward reaction is equal to the rate of the reverse reaction. A system will remain in equilibrium unless it is stressed or disturbed. Le Chatelier’s Principle states that “when a stress is placed on a system at equilibrium, the system will shift to offset the stress applied”.

Equilibrium shifts AWAY from what is ADDED and TOWARDS what is TAKEN

Changes in concentration, temperature or pressure can stress an equilibrium system and may cause a shift in the position of equilibrium.

The reaction being studied is:

2NO₂(g) ↔ N₂O₄(g) ΔH = -58.0 kJ/mol N₂O₄

Brown Colourless

This reaction is exothermic in the forward reaction (produces N₂O₄), and endothermic in the reverse reaction (produces NO₂). This is because N₂O₄ is more stable than NO₂ (the bonds in the N₂O₄ are stronger than the bonds in the NO₂).

NO₂ gas exists as a mixture of NO₂ and N₂O₄ at equilibrium (giving it a yellow-brown color).

In the forward reaction that releases heat, two NO₂ molecules dimerize or combine to form N₂O₄. In the reverse reaction that requires heat, N₂O₄ decomposes into NO₂ as shown in the diagram in the box below:

Because the reaction as written is exothermic (heat is produced in the forward reaction) we can expect that an increase in temperature in this system will cause the reaction to reverse (favour the endothermic reaction) to use the excess heat. Thus, a decrease in temperature will cause a forward shift to produce heat.

Procedure (Demo) Temperature:

1. Place the three NO₂/N₂O₄ tubes in beakers of cold, room temperature and hot water, respectively. The depth of the colour shown in the three vials is a direct indication of the extent of reaction.

2. Compare the colour of the gases in each vial and record observations. Infer what substance is in each of the three vials based on the colour.

3. Place the tube in hot water into the ice water. ‘Think visually’ about the changes that are occurring as the colour changes occur. What process is being favoured in terms of what molecules are doing at a faster rate? Deduce an equilibrium equation based on the changes. Consider what effects the addition and removal of heat have: two molecules of NO₂(g) forming one molecule of N₂O₄(g).

4. What changes have occurred in the hot water? Why ? Explain in terms of Le Chatelier’s Principle.

5. Repeat by placing the tube in the boiling water back into the cold water.

6. What changes have occurred? ‘Think visually’ about the changes that are occurring as the colour changes occur. Why? Explain in terms of Le Chatelier’s Principle.

7. Removed from the hot and cold water and allow to reach room temperature. ‘Think visually’ about the changes that are occurring as the colour changes occur. Why? Explain in terms of Le Chatelier’s Principle.

8. VIEW video https://www.youtube.com/watch?v=aOuqVrp7zkc&feature=youtu.be (see below)

Count the number of NO₂ and N₂O₄ molecules at the beginning and at the end to notice that the number of NO₂ and N₂O₄ molecules are constant, even though some NO₂ molecules become N₂O₄ and vice versa. At equilibrium, reactions are still going on (as in the simulation) but the ratio of N₂O₄ to NO₂ molecules remains constant.

Results:

Consider the following reaction:

2NO₂(g) ↔ N₂O₄(g) ΔH = -58.0 kJ/mol N₂O₄

Chemical equilibrium is the point at which the rate of the forward reaction is equal to the rate of the reverse reaction. At this point, the concentrations of all species are constant because these processes are occurring at the same rate. At equilibrium, as much N₂O₄ reacts to form NO₂ as NO₂ reacts to re-form N₂O₄, however, the number of molecules in each phase are NOT equal. Be sure you understand what this means.

According to Le Chatelier’s Principle, a stress placed on a system at equilibrium will cause the equilibrium to shift to counteract the stress. For example, a temperature increase in the above reaction will favour the reverse reaction to use the excess heat and form brown NO₂ gas. A temperature decrease in the above reaction favours the forward reaction to produce heat and form colourless N₂O₄ gas.

Review NO₂ / N₂O₄ Prac understandings:

VIEW videos

EAR#6 Effect of Temperature on an Equilibrium System https://www.youtube.com/watch?v=7r4jb1-QNq4&index=8&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.17 mins]
NO2 / N2O4 and temperature

12C5 2a NO2 MP4.mp4

TASK 5.2.1A

1. (2007, Q28e)

The equation below represents a system in equilibrium:

N_2(g) + 3H_2(g) ↔ 2NH_3(g) ΔH = -92.4 kJ

Predict the effect of decreasing the temperature on the above equilibrium and explain your reasoning using:

a) Le Chatelier's Principle

b) Collision theory

ANSWERS:

a) Decreasing the temperature will slow both forward and reverse reactions, but it will do so more for the endothermic reaction, which is the reverse reaction in this equilibrium system. The exothermic reaction, the forward reaction, will be favoured, so there will be more product formed.

b) Decreasing the temperature will decrease the number of collisions, as well as the number of particles with E greater than or equal to activation E. The reaction in which there are fewer successful collisions is the endothermic reaction, so these will be reduced to a greater degree. The forward reaction will have more successful collisions, and so more product will be formed.

5.2.1 investigate the effect of change on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects.

b) How will a change in PRESSURE affect an equilibrium reaction?

Pressure affects the concentration of the particles in a gas. The same types of changes can be observed for the changes to concentrations.

eg. I_2(g) ⇌ I_2(s)

purple shiny grey-black

Increasing vapour pressure of gas leads to formation of more grey-black crystals. The equilibrium has shifted to the right as a result of more collisions between I₂ gas molecules hence an increase in the rate of the forward reaction.

Adding more solid iodine has no perceivable effect, no increase in purple gas, no increase in pressure. There is no change in the system this time. More solid takes up more space, so its concentration has not changed. The maximum vapour pressure cannot be exceeded unless there is a change in temperature.

Increasing the volume of the vessel increases the quantity of purple gas. Some of the crystals sublime. The equilibrium shifts to the left. The gas molecules collide less often in a larger space hence the rate of the forward reaction decreases. The rate of the reverse reaction has not changed but it is now greater than that of the forward reaction, so more iodine gas is formed.

Suppose that the equilibrium system given below is in a cylinder fitted with a movable piston. The increase in pressure favours the reverse reaction.

CH4(g) + H2O(g) ⇌ CO2(g) + 3H2(g) (endothermic)

VOLUME

Changing the volume and maintaining the same pressure does not alter the equilibrium established by the system.
Decreasing the volume increases the pressure (and therefore concentration), while increasing the volume decreases the pressure (and therefore concentration).
It is possible to change the pressure of the system without changing its volume. The total pressure of the system can be increased by adding an inert gas.
Addition of inert gas at constant volume increases the total gas pressure and decreases the mole fractions of the gaseous species. However, the partial pressure of each gas does not change. As such, the presence of an inert gas does not affect the equilibrium condition.

KEY POINTS:

PRESSURE

changes in pressure do not affect the concentration of the reacting species in condensed phases (e.g. solids and liquids) because solids and liquids are incompressible (can't be compressed).
concentrations of gases are greatly affected by changes in pressure
- INCREASE in PRESSURE [is an INCREASE in CONCENTRATION of the gas] favours the reaction that decreases the total number of moles of gases. In the example given below, the increase in pressure favours the reverse reaction.
- DECREASE in PRESSURE favours the reaction that increases the total number of moles of gases.

ANOTHER EXPLANATION:

This only applies to systems involving at least one gas.

The position of equilibrium may be changed if there is a change in the pressure of a system. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change made.

That means that if you increase the pressure, the position of equilibrium will move to decrease the pressure again - if that is possible. It can do this by favouring the reaction which produces the fewer molecules. If there are the same number of molecules on each side of the equation, then a change of pressure makes no difference to the position of equilibrium.

Explanation

Where there are different numbers of molecules on each side of the equation

Let's look at the same equilibrium we've used before. This one would be affected by pressure because there are 3 molecules on the left but only 2 on the right. An increase in pressure would move the position of equilibrium to the right.

Because this is an all-gas equilibriium, it is much easier to use K_p:

Where there are the same numbers of molecules on each side of the equation

In this case, the position of equilibrium isn't affected by a change of pressure.

A SPECIAL CASE - addition of an inert gas

The addition of an inert gas can affect the equilbrium, but only if the volume is allowed to change.

Addition of an inert gas at constant volume:

When an inert gas is added to the equilibrium at constant volume, the total gas pressure will increase. But the actual concentrations of the products and reactants (i.e. ratio of their moles to the volume of the container) will not change. So when an inert gas is added to the system in equilibrium at constant volume there will be no effect on the equilibrium.

Addition of an inert gas at constant pressure:

When an inert gas is added to the equilibrium at constant pressure, then the total volume will increase. The concentration of reactants and products will decrease. So the equilibrium will shift towards an increase in number of moles of gases.

Eg:

2NH3(g) ↽−−⇀ N2(g) + 3H2(g)

The addition of an inert gas at constant pressure to the above reaction will shift the equilibrium towards the forward direction because the number of moles of products is more than the number of moles of the reactants.

Suppose there are two groups of people who don't like each other. A fight starts. Now you and your friends step into the mix to try to talk them into stopping the fight.Would there still be a fight? Yes, but they've spread out because of you and your friends taking up space, and it's harder for them to reach each other through your group. So there is less association (fighting groups) and more dissociation (individuals spread out). In our equilibrium, there are now more individual particles = more moles of gas = forward reaction is easier, reverse is harder.

PRAC 5.2.1C (Demo) Pressure change in NO₂ / N₂O₄ equilibrium:

The PRESSURE in the equilibrium can be changed by pushing the syringe in and out. SO the syringe is pushed in and out and the change in colour of the gas observed (this is qualitative analysis) showing a shift in equilibrium when the pressure changed.

At high pressure (ie syringe pushed in) the equilibrium shifted to the side with less moles - the N₂O₄ side - which meant the gas became a lighter brown colour as more N₂O₄ was made. With lower pressures it shifts the opposite way.

VIEW Video:

EAR#8 https://www.youtube.com/watch?v=8adN4rKGNTk&index=6&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [6.31 mins]

TASK 5.2.1B

1. (07, Q28e)

The equation below represents a system in equilibrium:

N_2(g) + 3H_2(g) ↔ 2NH_3(g) ΔH = -92.4 kJ

Predict the effect of changing each of the following factor on the above equilibrium. Explain your reasoning.

a) increasing the pressure

b) increasing the volume

5.2.1 investigate the effect of change on a system at equilibrium and explain how Le Chatelier’s principle can be used to predict such effects.

c) How will a change in CONCENTRATION affect an equilibrium reaction?

KEY POINTS:

CONCENTRATION

The position of equilibrium is changed if you change the concentration of something present in the mixture. According to Le Chatelier's Principle, the position of equilibrium moves in such a way as to tend to undo the change made.

Suppose you have an equilibrium established between four substances A, B, C and D.

According to Le Chatelier's Principle, if you decrease the concentration of C, for example, the position of equilibrium will move to the right to increase the concentration again.

Changing the concentration of either reactants or products can disturb the equilibrium condition.

INCREASING the concentration of the REACTANTS shifts the equilibrium to favour formation of products, to the RIGHT.
INCREASING the concentration of the PRODUCTS shifts the equilibrium favour formation of reactants, to the LEFT. DECREASING the concentration of the REACTANTS will have the same effect.

Suppose that the equilibrium system given below is in a cylinder fitted with a movable piston. Increasing the amount of hydrogen gas favors equilibrium to form CH4 and H2O. Decreasing the concentration of CH4 favors equilibrium to the left, thereby producing more CH4 and H2O.

CH4(g) + H2O(g) ⇌ CO2(g) + 3H2(g) (endothermic)

And as an extra:

The position of equilibrium is not changed if you add (or change) a catalyst.

A catalyst brings a system to reach equilibrium faster.

A catalyst speeds up both the forward and back reactions by exactly the same amount. Dynamic equilibrium is established when the rates of the forward and back reactions become equal. If a catalyst speeds up both reactions to the same extent, then they will remain equal without any shift in position of equilibrium.

PRAC 5.2.1D Iron / thiocyanate and concentration change

TX2 PA 5.1 P27

When potassium thiocyanate [KSCN] is mixed with iron(III) nitrate [Fe(NO3)3] in solution, an equilibrium mixture of Fe3+, SCN– and the complex ion FeSCN2+ is formed (equation 1). The solution also contains the spectator ions K+ and NO3 – .

The relative amounts of the ions participating in the reaction can be judged from the solution colour: in neutral to slightly acidic solutions, Fe3+ is light yellow, SCN– is colorless, and FeSCN2+ is red.

If the solution is initially reddish, and the equilibrium shifts to the right (more FeSCN2+), the solution becomes darker red, while if the equilibrium shifts to the left (less FeSCN2+), the solution becomes lighter red or straw yellow.

You will add various reagents to this reaction at equilibrium to see if/how those reagents shift the equilibrium position of the reaction using the colour of the resulting solution. In the reaction between the iron(III) ion and the thiocyanate ion:

Fe3+ + SCN– FeSCN2+ (1)

yellow colorless blood red

When solutions containing Fe³⁺ ion and thiocyanate ion are mixed, the deep red thiocyanatoiron (III) ion ([FeSCN]²⁺) is formed. As a result of the reaction, the starting concentrations of Fe³⁺ and SCN⁻ will decrease. For every mole of [FeSCN]²⁺ that is formed, one mole of Fe³⁺ and one mole of SCN⁻ will react.

The equilibrium constant expression (K_c) for this reaction is formulated as follows:

Kc=[FeSCN2+] / [Fe3+][SCN−]

Square brackets are used to indicate concentration in moles/litre, mol/L or molarity (M).

The value of K_c is constant at a given temperature. At a given temperature, the same value of the K_c will be obtained no matter what initial amounts of Fe³⁺ and SCN⁻ were used.

To find K_cfor this reaction experimentally, spectrophotometry can be used to measure the appearance of red color—indicating forming the [FeSCN]²⁺ ion. The amount of light absorbed by the red complex is measured at 447 nm, the wavelength at which the complex most strongly absorbs. The absorbance (A) of the complex is proportional to its concentration (M) and can be measured directly on the spectrophotometer:

A=kM

The Beer-Lambert Law relates the amount of light being absorbed to the concentration of the substance absorbing the light and the path length through which the light passes:

A=ϵbc

In this equation, the measured absorbance (A) is related to the molar absorptivity constant (ε epsilon), the path length (b), and the molar concentration (c) of the absorbing species. The equation shows how the concentration is directly proportional to absorbance.

PRAC 5.2.1E Cobalt Chloride and concentration change

TX2 PA 5.2 P30

A12C5 2a Cobalt Reaction.pdf

PRAC 5.2.1F

Aim: To investigate chromate/dichromate equilibrium

Equipment:

3 test tubes in rack
5 plastic pipettes
10mL measuring cylinder
0.1M HCl 10 mLs in small beaker
0.1M NaOH 10 mLs in small beaker
0.1M 5mLs barium nitrate in small beaker
0.1M potassium dichromate 5mL in each of 2 test tubes

Risk Assessment:

Complete a risk assessment

Procedure:

add NaOH dropwise to the dichromate in the two test tubes until the colour changes.

2. to the first test tube, add HCl dropwise until the colour changes.

3. to the second add barium nitrate dropwise until colour changes.

When we increase the concentration of reactants (or products) the number of collisions will increase, thus increasing the reaction rate in a given direction, in this case forwards (or backwards).

eg. 2CrO₄^2-_(aq) + 2H⁺_(aq) ⇌ Cr₂O₇^2-_(aq) + H₂O_(l)

yellow orange

Increasing [H⁺] produces a more orange system. The equilibrium has shifted to the right. More collisions between the H⁺ ions and CrO₄^2- ions increases the rate of the forward reaction.

Adding OH^- ions produces a more yellow system. The OH^- ions neutralise the H⁺ ions and the equilibrium shifts to the left as the number of collisions between the H⁺ ions and CrO₄^2- ions decreases.

VIEW Video:

EAR#7 https://www.youtube.com/watch?v=XtwGgD4Oyj0&index=7&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [5.46 mins]

12C5 2.1c LeChatgraphs WS.pdf

TASK 5.2.1C

1. (07, Q28e)

The equation below represents a system in equilibrium:

N_2(g) + 3H_2(g) ↔ 2NH_3(g) ΔH = -92.4 kJ

Predict the effect on the above equilibrium of a decrease in the concentration of ammonia by liquefying it. Explain your reasoning.

2. (10, Q18) Chromate and dichromate ions form an equilibrium according to the following equation.

2CrO₄²⁻_(aq) + 2H⁺_(aq) ⇌ Cr₂O₇²⁻_(aq) + H₂O_(l)

Which solution would increase the concentration of the chromate ion (CrO₄²⁻) when added to the equilibrium mixture?

(A) Sodium nitrate (B) Sodium chloride

3. (09, Q23, 6 marks)

The graph shows the variation in concentration of reactant and products as a function of time for the following system.

Identify and explain each of the changes in conditions that have shaped the curves during the time the system was observed.

REVIEW 5.2.1

VIEW Videos:

Le Chatelier's Principle of Chemical Equilibrium - Basic Introduction https://www.youtube.com/watch?v=wjcPEHhiik8 [10.58 mins]

REVIEW QUESTIONS 5.2.1

5.2.2 explain the overall observations about equilibrium in terms of the collision theory

Recall the three important features of equilibrium systems.

Equilibrium is achieved when the system is closed. (Matter will neither be lost or gained.)
Constant macroscopic properties are an indicator of dynamic equilibrium. (includes colour, pressure, volume, electrical conductivity.)
Chemical equilibria are dynamic. (the rate of the forward reaction = the rate of the reverse direction).

Various factors affect both the position of equilibrium and how the system can be changed. Factors which affect a system in equilibrium include:

Temperature
Gas pressure (if one or more substances exist in their gaseous form)
Concentration

Note while both the presence of a catalyst and the surface area of one or more reactants can change the rate of the reaction, neither can shift the position of the equilibrium.

We can use collision theory to help explain why the three factors can shift the equilibrium.

The more reactant particles there are, the more collisions occur, the more product particles are made. However, the collision theory model is only useful in reactions of the form A + B ↔ C. It adequately explains the effect on temperature and concentrations. Yet concentrations of reactants and temperature changes cannot alone be the determining factor of reaction rate.

Temperature: Increasing heat energy increases kinetic energy of the particles. This increases their velocity. This also means they are more likely to have sufficient energy to overcome the activation energy needed for the reaction to occur. This increases the frequency of collisions and improves the chances of successful reactions when a collision occurs.

Pressure: Increasing pressure for gases involves increasing the density or decreasing the volume. Pushing the same number of particles into a smaller space will increase the likelihood of collisions hence increase the rate of reaction.

Concentration: Increasing the concentration of a substance increases the numbers of particles and hence there is a higher likelihood of reactant particles colliding to form one or more products, or vice versa.

It is also worth noting that the stereochemistry (the arrangement in space, the geometry) requirement of the reaction has an affect on collision success, especially in biochemical processes. When HCl approaches ethene, only collision #1 will produce a reaction. Can you think about why this might be the case?

Read notes on Reaction Rates to review

12C5 Collision Theory.doc

VIEW Videos:

EAR#9 https://www.youtube.com/watch?v=fDRBflUoBgI&index=10&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.06 mins]
Collision Theory and Equilibrium https://www.youtube.com/watch?v=wT1yXC2yJfs [7.07 mins]

TASK 5.2.2A

1. VIEW video and explain changes in terms of collision theory

https://www.youtube.com/watch?v=zVZXq64HSV4 [1.08 mins]

2. Write a balanced equation for the equilibrium system shown below and continue the graphs to show how the system would adjust to this change in pressure. Explain what you have drawn. (Chan, et al., 2019, p. 100)

TASK 5.2.2B

1. Look at the table on the right below and then use the table on the left to assist. Instead of justifying the changes on the basis of Le Chatelier's Principle, provide a thermodynamic or kinetic explanation for why the system will shift as indicated. The first one has been completed for you.

5.2.3 examine how activation energy and heat of reaction affect the position of equilibrium

Recall that Activation Energy (E_A) is the energy needed for a chemical reaction to take place. Specifically it is the energy required to form the activated complex and ensure that the products form as a result.

A high activation energy for a reaction can mean very few particles have sufficient energy to react. In any gas, the particles will have a wide range of energies. This can be shown on a graph known as the Maxwell-Boltzmann Distribution (see below).

One simple way to increase the energy of the particles so they approach the energy needed to react is to add heat energy. The rate of reaction increases when temperature is increased as both the collision energy (main effect) and collision frequency (secondary effect) increase. However, the increase in collision energy is more important as it accounts for about 95% of the increase in reaction rate for a given reaction.

“In terms of collision theory, a change in temperature has a greater impact on the rate of a reaction with a large E_A than one with a small E_A”. (Hegarty, 2018, p. 34)

“On average, a 10^oC temperature rise approximately doubles the rate of reaction. The higher the activation energy, the greater the effect an increase in temperature will have.”

More information can be found at http://www.a-levelchemistry.co.uk/uploads/9/0/4/5/90457821/2.2_notes.doc

Looking at the image below, imagine the products sitting above the letter D and the reactants sitting below the letter B. It is much easier for the reactants to climb up the top of hill C then it is for the products to climb up the hill. Once a reactant climbs the hill and slides down to products it will not be able to go back to reactants by climbing up the hill unless we give it more energy by increasing temperature. Raising the temperature gives more of the products enough energy to climb up the hill and go to reactants.

So, the exothermic reaction is favoured at low temperatures and the endothermic at high temperatures.

VIEW Videos

EAR#10 https://www.youtube.com/watch?v=3woQ-1P29KA&index=9&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.01 mins]

TASK 5.2.3A

1. Using activation energy, explain why a decrease in temperature always shifts the equilibrium to favour the exothermic reaction.

2. Use the Maxwell-Boltzmann Distribution to explain two different ways of increasing the number of particles which have sufficient energy to react.

REVIEW 5.2.3

VIEW videos:

Le Chatelier's Principle https://www.khanacademy.org/science/chemistry/chemical-equilibrium/factors-that-affect-chemical-equilibrium/v/le-chatelier-s-principle [14.42 mins]
Equilibrium: Crash Course Chemistry https://www.youtube.com/watch?v=g5wNg_dKsYY [10.56 mins]

REVIEW TASK 5.2.3B

Complete HSC Qs. Answers supplied are from the HSC Marking Centre. They are sample answers, and correct, but not the BEST answers.

12C5 MDQ2 HSC Qs.docx

12C5 MDQ2 HSC ANS.docx

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