19-givengenchem.bu.edu/ch131-fall-2020/_lectures/19-given.pdf · Lecture 19 CH131 Fall 2020...

Lecture 19 CH131 Fall 2020 11/24/2020 9:33 AM

Copyright © 2020 Dan Dill [email protected] 1

TP Which of the following is the correct expression for ∆𝑆° of the reactionC 𝑠 O 𝑔 → CO 𝑔

1. ∆𝑆° C, 𝑠 ∆𝑆° O ,𝑔 ∆𝑆° CO,𝑔

2. ∆𝑆° C, 𝑠 ∆𝑆° O ,𝑔 ∆𝑆° CO,𝑔

3. ∆𝑆° CO,𝑔4. ∆𝑆° CO,𝑔

5. 𝑆° C, 𝑠 𝑆° O ,𝑔 𝑆° CO,𝑔

6. 𝑆° C, 𝑠 𝑆° O ,𝑔 𝑆° CO,𝑔

7. 𝑆° CO,𝑔8. 𝑆° CO,𝑔

Copyright © 2020 Dan Dill [email protected] 19 CH131 Fall 2020

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Lecture 19 CH131 Fall 2020Tuesday, November 24, 2020

• Complete: Absolute entropy 𝑆°• Entropy change of reaction ∆𝑆°• Free energy change, Δ𝐺• Effect of temperature on spontaneity

Next lecture: Complete: Effect of temperature on spontaneity; Begin ch14: Chemical equilibrium

How to determine ΔSorxn ?

The essential starting point is that at 0 K, for each substance,

𝑊 1 and so 𝑆 0.

This is known as the third law of thermodynamics.


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Starting with 𝑆 0 at 𝑇 0 , adding a little heat 𝑑𝑞 , entropy 𝑑𝑆

~will be added, raising 𝑇 a little bit, say to 1 K.

Adding a little more heat 𝑑𝑞 , entropy 𝑑𝑆

~will be added, raising 𝑇 another little bit, say to 2 K.


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1 2

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Continuing in this way, up to a final temperature 𝑇, the sum of all of the small additions 𝑑𝑆 are 𝑆 for the substance at 𝑇,

𝑆° 𝑇 𝑑𝑆 𝑑𝑆 ⋯ 𝑑𝑆

These total values are called absolute entropies 𝑆° at temperature 𝑇.


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Find S at a particular temperatureMake a sketch of how you expect the entropy of water to change with temperature, starting from 𝑆 0 at 𝑇 0 K and ending at the entropy at𝑇 400 K.

Lecture 19 CH131 Fall 2020

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Copyright © 2020 Dan Dill [email protected]

Find S at a particular temperatureLecture 19 CH131 Fall 2020

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Find S at a particular temperatureAbsolute entropies typically are tabulatedat 298 K.

Note, these are entropies, not entropy changes.


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How to get entropy change of reaction, ΔSorxn ?

1. Get absolute entropies 𝑆°

2. ∆𝑆° 𝑆° 𝑆°


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TP Which of the following is the correct expression for ∆𝑆° of the reactionC 𝑠 O 𝑔 → CO 𝑔

1. ∆𝑆° C, 𝑠 ∆𝑆° O ,𝑔 ∆𝑆° CO,𝑔

2. ∆𝑆° C, 𝑠 ∆𝑆° O ,𝑔 ∆𝑆° CO,𝑔

3. ∆𝑆° CO,𝑔4. ∆𝑆° CO,𝑔

5. 𝑆° C, 𝑠 𝑆° O ,𝑔 𝑆° CO,𝑔

6. 𝑆° C, 𝑠 𝑆° O ,𝑔 𝑆° CO,𝑔

7. 𝑆° CO,𝑔8. 𝑆° CO,𝑔


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Entropy of reaction, ΔSorxn

∆𝑆° 𝑆° 𝑆°

Rules of thumb:• If more gas moles formed, ∆𝑆° large and positive• If more gas moles consumed, ∆𝑆° large and negative• If gas moles unchanged, ∆𝑆° small but positive or negative


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ΔSorxn = So(products) − So(reactants)

2 Zn 𝑠 O2 𝑔 2 ZnO 𝑠

∆𝑆° 2 43.7 2 41.6 205.0 200.8 J/K

Δ𝑛 1, so ∆𝑆° is large and negative

N2 𝑔 O2 𝑔 2 NO 𝑔

∆𝑆° 2 210.8 191.6 205.0 25 J/K

Δ𝑛 0, so ∆𝑆° is small


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An alternative criterion of spontaneityQuantifying spontaneity, we have learned so far:

𝑊 → 𝑆 𝑘 ln 𝑊 → ∆𝑆 𝑘 ln

∆𝑆 ∆𝑆 ∆𝑆

∆𝑆 ∆

∆𝑆 ∆ ∆𝑆


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An alternative criterion of spontaneityFor phase transitions:

∆𝑆 ∆ ∆𝑆

∆ ∆


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An alternative criterion of spontaneityFor chemical reactions:

∆𝑆 ∆ ∆𝑆

∆ ∆𝑆


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TP While ∆𝑆 takes into account entropy change in both the surroundings and the system, the expression ∆𝑆 ∆ ∆𝑆 means that for chemical reactions, …

1. the entropy change of the surroundings is not important2. then entropy change of the surroundings still plays a role3. only exothermic reactions are spontaneous4. Further information needed


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An alternative criterion of spontaneityWhile the expression

∆𝑆 ∆ ∆𝑆

has only properties of the system, always remember that ∆ is the entropy change of the surroundings.

This shows that spontaneity is favored by exothermic reactions ∆𝐻 0 because they increase the entropy of the surroundings.

However, whether a reaction is spontaneous ∆𝑆 0 also depends on ∆𝑆 .


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Free energy change, ∆𝐺To emphasize that we only have to know the system quantities ∆𝐻 and ∆𝑆 , the free energy change of reaction is defined as

𝑇∆𝑆 ∆𝐺 ∆𝐻 𝑇∆𝑆

The components of Δ𝐺 are usually written with “rxn” or “sys” omitted …

Δ𝐺 ∆𝐻 𝑇∆𝑆

It turns out that free energy change is the work than can be done on the surroundings excluding any pressure volume work , and so tells us how much work we can get from a chemical transformation.


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System-only spontaneity measureΔ𝐺 depends only on system quantities, but it is equivalent to ∆𝑆 .

If Δ𝐺 0, then spontaneous ∆𝑆 0 and process provides work

If Δ𝐺 0, then equilibrium ∆𝑆 0 and no work is involved

If Δ𝐺 0, then non-spontaneous ∆𝑆 0 and process requires work


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Problem 7e and 8e 13.31 Lecture 19 CH131 Fall 2020

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Effect of temperature on spontaneity


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Atomic or molecular oxygen?At what temperature will oxygen spontaneously decompose, O2 𝑔 2 O 𝑔 ?• Δ𝐻f O, 𝑔 249.2 kJ/mol• 𝑆o O, 𝑔 161.1 J/ K mol• 𝑆o O2, 𝑔 205.0 J/ K mol

How to proceed?

At 300 K,Δ𝐺 2 249.2 300 K 10 3 2 161.1 205.0 463 kJSince 0, not spontaneous, so mostly molecules at 300 K


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Atomic or molecular oxygen?At what temperature will oxygen spontaneously decompose, O2 𝑔 2 O 𝑔 ?

At 300 K,

Δ𝐺 2 249.2 300 K 10 3 2 161.1 205.0 463 kJ

At what 𝑇 will decomposition become spontaneous?

ΔG 0 2 249.2 𝑇 10 3 2 ∙ 161.1 205.0

𝑇 2 ∙ 249.2 103/ 2 161.1 205.0 4253 K

So, for 𝑇 above 4253 K, mostly atoms


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