DOR3

MDQ3 How can enthalpy and entropy be used to explain reaction spontaneity?

Text Reference: Chapter 16

3.1 analyse the differences between entropy and enthalpy

The Laws of Thermodynamics

Thermodynamics is the study of energy. This usually focuses on heat, temperature and (another physics concept we have not explored yet) work.

First Law: The Law of conservation of energy.

Energy can be neither created or destroyed, only transformed from one form of energy into another. The total amount of energy that exists is a constant (Rosenberg, et al., 2013).

We have focussed on the exchange of energy between the chemical system and its surroundings. The key forms of energy are the chemical potential energy stored in chemical bonds, the kinetic energy of the particles and the heat energy associated with the moving particles and/or the breaking and making of bonds.

If pressure is constant, any change in heat associated with the chemical reaction is equal to the change in enthalpy, which is specifically the difference in energy used to break and make chemical bonds.

Second Law: The Law of entropy.

The sum of the entropy change of the system and surroundings for any spontaneous process is always greater than zero. The total entropy of the universe increases in any spontaneous process (Brown, et al., 2010).

To understand the second law of thermodynamics, we need to understand entropy. Entropy has the symbol S and (like enthalpy) it is easier to measure changes in entropy rather than absolute values for a system. So we will look at ΔS.

As for enthaply change, change in entropy can be found using the following equation:

Δ_universeS = Δ_sysS + Δ_surrS (>0) or ΔS_(reaction) = ΔS_(products) - ΔS_(reactants)

Don't need this one yet

Third Law: The law of equivalence.

The entropy of a pure crystalline substance at absolute zero is zero (Brown, et al., 2010, p. 164).

As we remove as much energy as possible from a system and the temperature drops, so the particles will lose kinetic energy. Theoretically, absolute zero (0 K) is the temperature at which the particles have zero kinetic energy and hence have an entropy of zero. If we could reach this theoretical temperature, the particles would be in the most ordered state possible.

VIEW :

Entropy PPT

VIEW videos:

DOR#11 Entropy and enthalpy [6.48 mins] https://www.youtube.com/watch?v=Dxb_WopoLec&index=13&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8
Embrace the chaos: crash course chemistry #20 [13.41] https://www.youtube.com/watch?v=ZsY4WcQOrfk
Entropy Explained [5.20] http://digg.com/video/entropy-explainer

11C4 3a Entropy PPT 2.pptx

Entropy, ExplainedWhy does ice melt but not un-melt? Why does your coffee cool down but never heat up by itself? Why does air escape a tire never to return again — without, at least, some human intervention? Enter entropy.

3.2 use modelling to illustrate entropy changes in reactions

We can look at the entropy in systems using the concepts of macrostate and microstate.

Microstate is a specific way in which we can arrange the energy of the system. Each atom within a system is a microstate of that system.

Macrostates are comprised of many microstates which determine the overall properties of the system including temperature, pressure and volume.

This can be most easily demonstrated using the analogy of a coin toss.

If 3 coins are tossed:

macrostates: microstates

3 Heads HHH

2 Heads, 1 Tail HTH, THH, HHT

1 Head, 2 Tails HTT, THT, TTH

3 Tails TTT

Entropy may then be thought of as the probablility of each outcome. The more likely outcomes are random configurations rather than ordered outcomes. Hence the chances of all 3 coins being identical is only 2/8 or ¼ and this decreases as we add more coins. Hence disordered arrangements are much easier to achieve than ordered ones. http://www.chem1.com/acad/webtext/thermeq/TE1.html#SEC (Lower, 2018)

Entropy increases as the distribution of particles becomes more random. We can model this by thinking about the changes in state from a solid to a liquid to a gas. If we were to plot entropy vs temperature, we might have a graph which looks similar to the one below.

The behaviour of the individual particles provides a clue about their macroscopic state eg solid, liquid or gas, but the microscopic level could reveal differences in kinetic energy between atoms or molecules. This randomness is one reason why not every molecule of water changes from liquid to gas at exactly 100°C.

We need to remember that in chemical systems:

There are an extremely large number of particles (think Avogadro’s number)
Once a change begins, it proceeds spontaneously
Thermal energy is constantly exchanged between particles in the system and between the system and its surroundings.
Thermal energy spreads rapidly and randomly throughout energetically accessible microstates within the system

EXTRA INFORMATION:

http://www.chem1.com/acad/webtext/thermeq/TE1.html#SEC

VIEW videos:

DoR#12 Modelling entropy [6.50] https://www.youtube.com/watch?v=8WiHvQJGaHg&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=12
Illustrating Entropy [4.17] https://www.youtube.com/watch?v=P9eksvw2e6w
Entropy [5.17] https://www.youtube.com/watch?v=o_PmWaW-JNE

HOMEWORK TASK 3.2.2

Complete Tx1 16.1 Key Qs p508

3.3 predict entropy changes from balanced chemical reactions to classify as increasing or decreasing entropy

We can use our knowledge of chemical reactions to predict the changes in entropy for a particular reaction. In order to do this, we need to observe the following generalisations:

As temperature increases, entropy increases; higher KE leads to an increase in possible arrangements of the particles
Substances increase in entropy as they change from solid (highly regular, ordered structure) through liquid (less ordered, some freedom of movement, but contained in volume) to gas (disordered, highly irregular, random motion)
Adding particles to a system creates more arrangements, so increasing entropy.
Dissolving (dissolution) gives the particles greater arrangements, so this also increases entropy

(Woollett, et al., 2018, pp. 505-506)

An example from your Pearson text is below:

EXTRA FOR EXPERTS: https://chem.libretexts.org/Textbook_Maps/General_Chemistry_Textbook_Maps/Map%3A_Chemistry%3A_The_Central_Science_(Brown_et_al.)/19%3A_Chemical_Thermodynamics/19.4%3A_Entropy_Changes_in_Chemical_Reactions

VIEW Videos:

DoR#13 Predicting entropy changes [5.50] https://www.youtube.com/watch?v=9bj24hD76bw&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=11
Predict the entropy change for a given reaction or process [HL IB Chemistry15:20] https://www.youtube.com/watch?v=ro3AnXdajAM
How to calculate change in entropy [5.53] https://www.youtube.com/watch?v=XMtnseGcnVc

TASK 3.3.1

1. Predict whether the entropy of the system increases in the following chemical reaction: (Woollett, et al., 2018, p. 507)

CaO_(s) + H₂O_(l) → Ca(OH)_2(s)

2. Predict the sign of ΔS for each of the following chemical processes. No numbers are necessary, only a +ve or -ve. (Rosenberg, et al., 2013, p. 262)

a) Hard-boiling an egg

b) C_{(s, graphite)} → C_{(s, diamond)}

c) Br_2(l) → Br_2(g)

d) O_2(g) → 2O_(g)

e) N_{2(g, 10 atm pressure)} → N_{2(g, 1 atm pressure)}

f) N_2(g) + 3H_2(g) → 2NH_3(g)

g) C_(s) + H₂O_(g) → CO_(g) + H_2(g)

h) Synthesis of water

3. Explain why the melting of a solid will increase entropy while the formation of a precipitate in a chemical reaction can decrease entropy of the system.

TASK 3.3.2

Complete online quiz:

General: http://chemunder.chemistry.ohio-state.edu/under/chemed/qbank/quiz/bank15.htm

GO TO Qualitative Prediction of Entropy of Reaction Quiz

TASK 3.3.3

complete Tx2 WS 4.6 Order and disorder p168

HOMEWORK TASK 3.3.4

complete Tx1 16.2 Review, Key Qs p514

3 .4 explain reaction spontaneity using terminology, including:

a. Gibbs Free Energy

Spontaneous reactions can:

increase or decrease entropy

and/or

absorb or release energy.

This means in order to classify a reaction as occurring spontaneously, we need to take into account both enthalpy and entropy.

These two concepts are brought together in a function known as the Gibbs Free Energy. Mathematically:

G = H - TS

where T is the absolute temperature (Kelvin) of the system. We need to look at the Gibbs Free Energy change if we are to analyse a system and determine whether it is spontaneous. Since we cannot directly measure H and S, but only the changes in them, we cannot directly measure G, just the change ∆G.

ΔG_sys = ΔH_sys - TΔS_sys

A spontaneous process is one that increases entropy.

For a spontaneous process (∆G<0) at constant T and P, ∆G is the maximum useful work obtainable from a system (-w)

For a non-spontaneous process (∆G>0) at constant T and P, ∆G is the minimum work that must be done to the system to make the process take place.

The effects of the change in enthalpy ΔH and entropy ΔS on a reaction are combined to give the change in Gibbs Free Energy ΔG at constant temperature and pressure.

∆G_sys = ∆H_sys - T∆S_sys

ΔG > 0 Reaction is not spontaneous

ΔG = 0 System is at equilibrium

ΔG < 0 Reaction is spontaneous

(Hurst 2018 pp57,58)

VIEW videos:

DoR#14 Gibbs free energy [6.03] https://www.youtube.com/watch?v=VPmVEX4xswA&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=15
Gibbs Free Energy, Entropy, and Enthalpy [5.32] https://www.youtube.com/watch?v=XvuRJuXykyw
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy [8.11] https://youtu.be/8N1BxHgsoOw
Gibbs Free Energy [12.59] https://www.youtube.com/watch?v=DPjMPeU5OeM

Complete Tutorial Gibbs Free Energy: Definition & Significance http://study.com/academy/lesson/gibbs-free-energy-definition-significance.html

Gibbs Free Energy: Definition & Significance - Video & Lesson Transcript | Study.comHave you ever wondered why some things just happen spontaneously? Have you ever asked yourself why salt dissolves in water? In this lesson, you...

3 .4 explain reaction spontaneity using terminology, including:

b. enthalpy

Spontaneous reactions are reactions which progress towards completion without any outside intervention and which proceed in a particular direction. A gas escaping from a gas tap will diffuse around the laboratory, however gas distributed around a laboratory will not spontaneously diffuse back into the gas tap. Iron nails may rust, but rusty nails will not return to pure iron in the solid state.

Temperature is a factor affecting spontaneity. A chemical or physical process which may not be spontaneous at cool temperatures may become spontaneous once its temperature reaches a critical level.

We can identify and analyse four types of reactions when seeking to classify them as spontaneous.

Type 1 reactions: these react as soon as they are mixed, eg Mg plus HCl.
Type 2 reactions: do not react at room temperature, but will react if provided with a spark or a small amount of heat, Mg plus O₂.
Type 3 reactions: do not react unless we continuously heat them. They do not generate sufficient heat energy as they progress, so must be constantly supplied with heat, eg decomposition of CuCO₃.
Type 4 reactions: do not occur even when very high temperatures are maintained, eg very high temperatures will boil water, but not decompose it. (Smith & Davis, 2017, p. 377)

It is easy to see why type 1 is classified as spontaneous and type 4 as non-spontaneous. Can you explain why type 2 is spontaneous but type 3 is not?

VIEW videos:

DoR#15 Reaction spontaneity [7.00] https://www.youtube.com/watch?v=cN-d3BNcx7Q&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=14

TASK 3.4.2

1. Which of the following statements is true concerning a chemical reaction?

a. A catalyst reduces the enthalpy change for the reaction

b. Endothermic reactions have lower activation energies than exothermic reactions

c. The ΔH value for the forward reaction is negative the ΔH value of the reverse reaction

d. Exothermic reactions are always spontaneous

2. Given the following heats of formation data:

Mg Δ_fH = 0 kJ.mol^-1

O₂ Δ_fH = 0 kJ.mol^-1

MgO Δ_fH = −601 kJ.mol^-1

What is the change in enthalpy for the following reaction?

2Mg_(s) + O_2(g) → 2MgO_(s)

a. 1202 kJ.mol^-1

b. −601 kJ.mol^-1

c. −1202 kJ.mol^-1

d. 601 kJ.mol^-1

3 .4 explain reaction spontaneity using terminology, including:

c. entropy

Exothermic processes are processes which release energy from the system to its surroundings. When ΔH is negative, reactions will generally be spontaneous as this is the favourable enthalpy change for spontaneity. If accompanied by an increase in entropy, such a process will always be spontaneous. Where an exothermic process is accompanied by a decrease in entropy, the reaction will be spontaneous at low temperatures and may be non-spontaneous when the temperature becomes too high (exceeds a threshhold level).

Endothermic processes are processes which absorb energy from the surroundings. When ΔH is positive, reactions will generally be non-spontaneous as this is the non-favourable enthalpy change for spontaneity. If accompanied by a decrease in entropy, such a process will always be non-spontaneous. Where an endothermic process is accompanied by an increase in entropy, the reaction will be spontaneous if temperatures are low and if the entropy change is sufficiently high.

VIEW videos:

DoR#16 Spontaneity and entropy [5.44] https://www.youtube.com/watch?v=ycGlszgIIqM&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=18

TASK 3.4.2

Work through tutorial https://www.ck12.org/book/CK-12-Chemistry-Second-Edition/r6/section/22.4/

3.5 solve problems using standard references and (Gibbs free energy formula) to classify reactions as spontaneous or non-spontaneous

The sign of ΔG provides information about the nature of the reaction for a reversible reaction. Not all reactions go from reactants to products. Some reach a point where there is no further change in the quantities of reactants and products, despite all species being present. One example is the equilibrium established between nitrogen dioxide and dinitrogen tetroxide.

N₂O_4(g) ⇌ 2NO_2(g)

Whether we start with nitrogen dioxide or dinitrogen tetroxide, we will end up with the same proportions of each. We describe such reactions as reaching equilibrium. When a reaction is at equilibrium, ΔG = 0 .

In such reactions, the equilibrium can be reached from either the left (reactants) or the right (products) but will always occur when the Gibbs Free Energy is at its mimimum. We will look at equilibrium reactions in greater detail in the HSC Course.

We can also use the formula to calculate spontaneity.

Interpreting the Gibbs Free Energy change needs to be understood in the context of the reaction, after considering both the enthalpy change and the entropy change. It is also important to consider standard conditions and non-standard conditions.

If at a particular temperature ΔG^o is negative, then under standard conditions the reaction proceeds in the forward direction at that temperature.

If at a particular temperature ΔG is positive, then the reaction proceeds in the reverse direction at that temperature. (Smith & Davis, 2017, p. 376)

When ΔG < 0, the process is exergonic and will proceed spontaneously in the forward direction to form more products.

When ΔG > 0, the process is endergonic and not spontaneous in the forward direction. Instead, it will proceed spontaneously in the reverse direction to make more starting materials.

When ΔG = 0, the system is in equilibrium and the concentrations of the products and reactants will remain constant.

NOTE: Chemists normally measure energy (both enthalpy and Gibbs free energy) in kJ mol^-1 (kilojoules per mole) but measure entropy in J K^-1 mol^-1 (joules per kelvin per mole). So it is necessary to convert the units - usually by dividing the entropy values by 1000 (X 10^-3)so that they are measured in kJ K^-1 mol^-1.

STEPS:

Define Gibbs Free energy using equation and identify all values in the equation.

Use the standard references to substitute values into the equation.

Calculate the change in entropy for reactions using ΔS^o = ∑S^o (products) - ∑S^o (reactants)

Calculate the change in enthalpy for reactions using ΔH^o = ∑H^o (products) - ∑H^o (reactants)

NESA chemistry data sheet (approved) and standard enthalpies and standard entropies of common compounds: http://www.mrbigler.com/misc/energy-of-formation.PDF

VIEW Videos:

DoR#17 Problems involving Gibbs Free Energy [5.44] https://www.youtube.com/watch?v=qHmw0Q0cO-w&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=17

Worked Example

Calculate delta H_f^o and delta S_f^o for the following reaction and decide in which direction each of these factors will drive the reaction.

N₂(g) + 3 H₂(g) 2 NH₃(g)

Solution

Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information:

Compound H_f^o(kJ/mol) S_f^o(J/mol-K)

N₂(g) 0 191.61

H₂(g) 0 130.68

NH₃(g) -46.11 192.45

The reaction is exothermic ( H_f^o < 0), which means that the enthalpy of reaction favors the products of the reaction:

delta H^o = H_f^o(products) - H_f^o(reactants)

= [2 mol NH₃ x 46.11 kJ/mol] - [1 mol N₂ x 0 kJ/mol + 3 mol H₂ x 0 kJ/mol]

= -92.22 kJ

The entropy of reaction is unfavorable, however, because there is a significant increase in the order of the system, when N₂ and H₂ combine to form NH₃.

delta S^o = S^o(products) - S^o(reactants)

= [2 mol NH₃ x 192.45 J/mol-K] - [1 mol N₂ x 191.61 J/mol-K + 3 mol H₂ x 130.68 J/mol-K]

= -198.75 J/K

TASK 3.5.1

1. Use the data in the table below to answer the questions .

(a) The following equation shows one of the reactions that can occur in the extraction of iron.

Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)

(i) Calculate the standard enthalpy change and the standard entropy change for this reaction.

(ii) Explain why this reaction is feasible at all temperatures.

(b) The reaction shown by the following equation can also occur in the extraction of iron.

Fe₂O_3(s) + 3C_(s) → 2Fe_(s) + 3CO_(g) ΔH = +492.7 kJ mol^–1

The standard entropy change, ΔS , for this reaction is +542.6 J K^–1 mol^–1

Use this information to calculate the temperature at which this reaction becomes feasible, assuming that ΔH and ΔS do not vary with temperature.

(c) Calculate the temperature at which the standard free-energy change, ΔG has the same value for the reactions in parts (a) and (b).

TASK 3.5.2

work through tutorials

- Gibbs Free Energy, Entropy, and Enthalpy [5.33]

https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/gibbs-free-energy-example

- Gibbs free energy and spontaneity[17.39]

https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/a/gibbs-free-energy-and-spontaneity

Gibbs free energy exampleDetermining if a reaction is spontaneous by calculating the change in Gibbs free energy. Also calculates the change in entropy using table of standard entropies.

PRACTICAL

complete Webquest / Practical http://highschoolenergy.acs.org/content/hsef/en/energy-theories/entropy-enthalpy/_jcr_content/toparticleparsys/columnsbootstrap/column1/acscontainer/containerPar/download/file.res/Teacher's_Key.pdf

TASK 3.5.3

1. The enthalpy and entropy change of a reaction are −29.3 kJ.mol^-1 and −2.54 J.mol^-1.K^-1 respectively at 25°C. What is the free energy change in kJ.mol^-1?

2. The enthalpy and entropy change of a reaction are −34.7 kJ.mol^-1 and −6.95 J.mol^-1.K^-1 respectively at 25°C. What is the free energy change in kJ.mol^-1?

3. The enthalpy and entropy change of a reaction are −6.1 kJ.mol^-1 and +38.9 J.mol^-1.K^-1 respectively at 25°C. What is the free energy change in kJ.mol^-1?

These questions can also be attempted and marked online here: http://chemunder.chemistry.ohio-state.edu/shell-cgi/world/genquiz.pl

3.6 predict the effect of temperature changes on spontaneity

There is one more variation on the situations we have explored above. There are occasions when both ΔS° and ΔH° have the same sign. In such cases, temperature becomes the determinant of the spontaneity of the reaction. We can rearrange the Gibbs Free Energy equation to calculate the minimum temperature required for a reaction to become spontaneous.

If we look at the Haber process:

N_2(g) + 3H_2(g) ® 2NH_3(g)

And are given the following values: ΔS° = −198 J.mol^-1.K^-1 and ΔH° = −91.8 kJ.mol^-1

Then we can calculate the minimum temperature required for this reaction to become spontaneous (the point of equilibrium).

ΔG^o = ΔH^o - TΔS^o

0 = −91.8 – T x –(198/1000)

91.8 = 0.198T

T = = 464 K

Hence the minimum temperature required for the Haber Process to be spontaneous is 464 K or (464 – 273 = 191°C).

(Schell & Hogan, 2018, p. 137)

PRACTICAL:

observe aluminium placed into three beakers of different temperature dilute HCl and record reaction rate

VIEW videos:

DoR#18 Temperature change and spontaneity [7.49] https://www.youtube.com/watch?v=428RP9enuj4&list=PLeFSFSJ9WqSA24lCCivXWB_Fruf_Jh6a8&index=16
Predict the effect of a change in temperature on the spontaneity of a reaction [5.29] https://youtu.be/dMoN__BiwNY

TASK 3.6.1

1. For a particular reaction, ΔH = 53 kJ and ΔS = 115 J/K

a) Is this reaction spontaneous at 25°C?

b) Is this reaction spontaneous at 250°C?

c) At what temperature does ΔG = 0 (the system is in equilibrium)? (Hegarty, 2018, p. 80)

TASK 3.6.2

complete Worksheet 4.7 To be or not to be? Tx2 p169

HOMEWORK TASK 3.6.3

Complete 16.3 Review, Key Qs p519
COMPLETE UNDER TEST CONDITIONS then review answers Chapter 16 review p520-1

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