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MDQ3 How can enthalpy and entropy be used to explain reaction spontaneity?
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
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.
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.
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:
ΔuniverseS = ΔsysS + ΔsurrS (>0) or ΔS(reaction) = ΔS(products) - ΔS(reactants)
Don't need this one yet
Third Law: The law of equivalence.
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 PPT:
Entropy
11C4 3a Entropy PPT 2.pptxVIEW 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
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 a larger number of arrangements is far more probable than a smaller number of arrangements. 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 macrostate eg solid, liquid or gas, but the microstate 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:
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:
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) Br2(l) → Br2(g)
d) O2(g) → 2O(g)
e) N2(g, 10 atm pressure) → N2(g, 1 atm pressure)
f) N2(g) + 3H2(g) → 2NH3(g)
g) C(s) + H2O(g) → CO(g) + H2(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.
ANSWERS
Decreases, fewer microstates l --> s
a) Decreases, fewer microstates l --> s
b) Decreases, fewer microstates with increased bonding C-C
c) Increases l --> g
d) Increases 1 mole --> 2 moles
e) Increases, lower pressure, more positions
f) Decreases 4 moles gas to 2 moles gas
g) Increases s --> g
h) decreases 3 moles --> 2 moles, g --> l
The melting of a solid increases entropy because the particles change from a fixed, ordered arrangement to a more disordered liquid state. In the liquid, particles have more freedom of movement and there are many more possible arrangements of particles and energy distributions (=more microstates), so entropy increases.
The formation of a precipitate can decrease entropy because dissolved ions that were freely moving in solution become arranged into an ordered solid lattice. This reduces the number of possible particle arrangements and energy distributions (fewer microstates), so the entropy of the system decreases.
TASK 3.3.2
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
Chemical reactions can:
absorb or release energy.
and/or
increase or decrease entropy
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. Gibbs Free Energy, put simply, is the free energy available in a system to do work - carry out a reaction.
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.
ΔGsys = ΔHsys - TΔSsys
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.
∆Gsys = ∆Hsys - T∆Ssys
ΔG > 0 Reaction is not spontaneous
ΔG = 0 System is at equilibrium
ΔG < 0 Reaction is spontaneous
(Hurst 2018 pp57,58)
For calculating ΔG∘, ΔS∘ must be converted from J mol−1 K−1 to kJ mol−1 K−1 because ΔH∘ is usually given in kJ mol−1.
The equation ΔG = ΔH − TΔS
Units must match:
ΔH∘ → kJ mol−1
T → (C) --> K
ΔS∘ → (J) --> kJ mol−1 K−1
ΔG∘ = kJ mol−1 - ( K x kJ mol−1 K-1)
= kJ mol−1 - kJ mol−1
= kJ mol−1
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
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 ΔfH = 0 kJ.mol-1
O2 ΔfH = 0 kJ.mol-1
MgO ΔfH = −601 kJ.mol-1
What is the change in enthalpy for the following reaction?
2Mg(s) + O2(g) → 2MgO(s)
a. 1202 kJ.mol-1
b. −601 kJ.mol-1
c. −1202 kJ.mol-1
d. 601 kJ.mol-1
ANSWERS
1
a. False – A catalyst lowers activation energy, not the enthalpy change (ΔH).
b. False – Activation energy is independent of whether a reaction is endothermic or exothermic.
c. True – ΔH (forward) = −ΔH (reverse).
d. False – Exothermic reactions are not always spontaneous (spontaneity depends on ΔG, not just ΔH).
2.
ΔH = ∑ΔHf(products) − ∑ΔHf(reactants)
Products:
2 × (−601) = −1202 kJ
Reactants:
Mg = 0, O₂ = 0 ∑ΔHf(reactants) = 0
ΔH = (−1202) − (0) = −1202 kJ
The standard enthalpy of formation (ΔfH) is defined as the enthalpy change when 1 mole of a compound is formed from its elements in their standard states
Mg(s) is already magnesium in its standard state
O₂(g) is already oxygen in its standard state
They are not forming from anything else, so ΔfH = 0 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.
N2O4(g) ⇌ 2NO2(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 ΔGo 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 ΔSo = ∑So (products) - ∑So (reactants)
Calculate the change in enthalpy for reactions using ΔHo = ∑Ho (products) - ∑Ho (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 Hfo and delta Sfo 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 Hfo(kJ/mol) Sfo(J/mol-K)
N2(g) 0 191.61
H2(g) 0 130.68
NH3(g) -46.11 192.45
The reaction is exothermic ( Hfo < 0), which means that the enthalpy of reaction favors the products of the reaction:
delta Ho = Hfo(products) - Hfo(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 So = So(products) - So(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.
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(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.
Fe2O3(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]
- Gibbs free energy and spontaneity[17.39]
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:
N2(g) + 3H2(g) ® 2NH3(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).
ΔGo = ΔHo - TΔSo
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
Check-ins:
Check-ins:
Why is the synthesis of water spontaneous?
Why is the synthesis of water spontaneous?
2H2(g) + O2(g) → 2H2O(l)
1. Change in entropy (ΔS)
1. Change in entropy (ΔS)
3 moles reactants (higher # arrangements) to 2 moles product (lower # arrangements) = entropy decreases
Products: gas (higher # arrangements) to liquid (lower # arrangements) = entropy decreases
ΔS < 0 (negative) The entropy decreases because gaseous reactants form a liquid product, reducing the number of accessible microstates and the degree of disorder.
2. Change in Enthalpy
2. Change in Enthalpy
ΔH is negative (exothermic)
3. Spontaneity
ΔH is very large and negative (exothermic) with strong O–H bonds formed = favours spontaneity
ΔS is negative but smaller in magnitude, does not favour spontaneity
From ΔG = ΔH − TΔS, spontaneity will depend on temperature:
Reaction is still spontaneous at standard conditions because ΔH dominates.
ΔG < 0
NOTE:
NOTE:
Spontaneous does not necessarily mean fast.
The reaction needs activation energy to start, even though ΔG < 0
Why don't all the liquid water paticles become a gas at the same time when water is boiled and reaches 100C?
At 100 °C, water molecules do not all have the same energy (Maxwell-Boltzmann distribution). Instead, the total energy of the liquid is distributed among molecules in a large number of possible microstates (higher entropy).
Some microstates contain molecules with enough energy to overcome intermolecular attractive forces and enter the gas phase. When these high-energy particles escape from the liquid, they remove more than the average kinetic energy. This leaves the remaining liquid with a slightly lower average energy. However, heat is continuously supplied at the boiling point. Through collisions, this energy is rapidly redistributed among particles, creating a Maxwell–Boltzmann distribution with a new high-energy tail.
This continual redistribution produces new high-energy arrangements, allowing more molecules to escape.
Vaporisation occurs progressively because molecules do not all occupy the same energy microstate at the same time; they continuously move between possible microstates, with the system favouring the higher number of microstates available in the gas phase.
11C Entropy Gibbs Qs AS.pdfGoogle Sites
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