Ppt04 (PS8), Thermodynamics energy changes (heat flows) Very global (“the universe”) Applies to all fields of science Led to creation of heat engines/refrigerators/air.

Ppt04 (PS8), Thermodynamics• energy changes (heat flows)• Very global (“the universe”)• Applies to all fields of science• Led to creation of heat

engines/refrigerators/air conditioners• In chemistry, focus is on physical and

chemical changes• How can we use “spontaneous processes” to

“do work”?• We can use tabulated data to make

predictions (and even calculate K’s for possible processes!)

1Ppt08

“Spontaneous” “fast” !!

• Spontaneity refers to “directionality”– Does it occur (on its own) in the forward [or stated]

direction, or does it not?• Thermodynamics will answer / address this question• Does not depend on how the process is carried out.

• Fast refers to rate at which it occurs (in the spontaneous direction)

– How fast will it occur?• Kinetics will answer / address this question• Does depend on how the process occurs (“pathway”;

transition state, activation energy, etc.)

• Recall that we should be spontaneously combusting right now!!

2Ppt08

Figure 17.4 Thermodynamics domain vs. Kinetics domain

“Pathway dependent”

THERMODYNAMICSInitial and final states,

spontaneity “Path independent”

3Ppt08

Reminder: 1st Law of Thermodynamics

• The total amount of energy in the universe never changes (it’s constant)

Euniv= 0 Esys = -Esurr

• “If energy leaves the system, it must go to the surroundings, and vice versa”

• Has nothing to do with “spontaneity”– 1st Law is consistent with a casserole dish

coming out of the oven colder than when it was put in, as long as the oven would get hotter

Ppt08 4

Spontaneity and the 2nd Law

• A spontaneous process is one that occurs (at a specified set of conditions) “without external intervention”.– Gases expand into the volume of their

containers– Ice will melt at 10°C– Water will freeze at -10°C– A chemical reaction will occur in the forward

direction if Q < K

Ppt08 5

Spontaneity and the 2nd Law (continued)

• At the same conditions, the reverse process of a spontaneous one is not spontaneous– Gases won’t spontaneously “contract” into a

smaller volume (perfume molecules going back into the bottle after opened?)

– Ice will not freeze at 10°C– Water will not melt at -10°C– A chemical reaction will not occur in the

reverse direction if Q < KPpt08 6

Spontaneity and the 2nd Law (continued)

• What determines whether a given process is spontaneous or not?– This question is not addressed by the 1st Law;

it is addressed by the 2nd Law.

• 2nd Law:

For a spontaneous process to occur, the entropy (S) of the universe must increase.

Suniv > 0 for any spontaneous process

Ppt08 7

**Suniv Ssys!** Be careful!

• The universe is made up of two “parts”—system and surroundings:

ΔSuniv = ΔSsys + ΔSsurr

• It doesn’t ultimately matter if ΔSsys is positive, or if ΔSsurr is positive. The key is whether ΔSuniv is positive! (next slide).

Ppt08 8

Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl)

This turns out to be very temperature dependent (See Section 17.4 in Tro). We will come back to this issue later.

See next slide

plus equals

ΔSsys + ΔSsurr = ΔSuniv

9Ppt08

Bar plots to visualize idea on prior slide(system and surroundings both contribute to Suniv!)

NOTE: These two examples both have Ssys < 0 and Ssurr > 0, but all possible variations are possible!

ΔSsys + ΔSsurr = ΔSuniv

|ΔSsurr| > |ΔSsys| |ΔSsys| > |ΔSsurr|

10Ppt08

What is entropy?

• Hard thing to define conceptually!– Something like “energy dispersal”--not necessarily “spatial”

dispersal, but energy dispersed over the various motions of atoms & molecules. Reflects quality or usefulness of energy.

• Let’s start with individual substances– Look at what affects the amount of entropy in a sample of a

single substance.

• Then we’ll look at mixtures of substances (in a reactive system, e.g.)– When a change occurs in the system, Ssys changes because

there are new substances present (or new physical states)• We’ll consider the surroundings last

– Just a “bunch of substances not doing anything” (Prof. Mines’ view)

11Ppt08

Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 17.6)

• Entropy of a sample of a substance increases with:

• Recall that we’ll typically only specify substances that undergo some change to be part of the system.

– Energy added (T increase or phase change)• Solid < Liquid << Gases

– Volume (gas or solution species)– Complexity of basic unit of substance– Number of moles of the substance

• Entropy, like energy, is a “per mole” kind of quantity

12Ppt08

Differences in entropy of physical states (assume a given T)

13Ppt08

Fig. 17.5 Entropy of a substance increases with T, and depends (significantly) on state (s, l, g)

14Ppt08

One “way” to arrange the energy (one [micro]state)

Two “ways” to arrange the energy (two [micro]states)

E = 5 J E = 5 J E = 5 J

Both systems have 4 J of energy, but the entropy of System B is greater because it has two ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see Slide 17)!

Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E

System A (4 J) System B (4 J)

15Ppt08

What if an energy state at 5 J was present? Would the # of microstates change in either system?

No, because there isn’t enough energy (in either system) to access that 5 J state. But…

What if 2-3 more units of energy were added (to make a total of 6-7 J)?

Two “ways” to arrange the energy

(vs. one before); two[micro]states

Three “ways” to arrange the energy

(vs two before); three [micro]states

E = 5 J

In both cases, adding energy (raising T) leads to increased S

Nanoscopic Interpretation of Entropy: # of ways to arrange (disperse?) E

System A (6-7 J)

System B (6-7 J)

Both systems have 6-7 J of energy, but the entropy of System B is (still) greater because it has three ways to “arrange” the energy. A greater # of “accessible” (i.e., “low”) energy levels leads to greater entropy (see next slide)!

16Ppt08

E

E = 5 J

More energy levels in gas!

17Ppt08

Patterns for Entropies of Substances (see handout/outline for details; Tro does this later; Section 17.6)

• Entropy of a sample of a substance increases with:

– Energy added (T increase or phase change)

• Solid < Liquid << Gases

– Volume (gas or solution species)

– Complexity of basic unit of substance

– Number of moles of the substance• Entropy, like energy, is a “per mole” kind of quantity

18Ppt08

Adding units of energy

Adding more accessible energy

states

(NOTE: Appendix has lots more data. Use appendix for PS8!)

19Ppt08

(From another text [McMurry])

20Ppt08

Predicting whether processes are “entropy increasing” or “decreasing” (in the system)

(the most dominant effect in most cases is the number of moles of

gases made and lost [ngas], not the complexity of the substance[s]):

• C(s) + 2 H2(g) CH4(g)

• N2(g) + 3 H2 (g) 2 NH3(g) [see next slide]

• 2 CO(g) + O2(g) 2 CO2(g)

• CaCO3(s) CaO(s) + CO2(g)

• 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)

21Ppt08

Can quantify the change in S (at a given T) using standard entropies of substances

• C(s) + 2 H2(g) CH4(g) Srxn° = 186.2 – (2.4 + 2 x 130.6)

2.4 2 x 130.6 186.2 = -77.4 J/K (per mol of C reacted)

(consistent with prediction, S decreases)

– Although CH4 is more complex than H2, it doesn’t have twice the

entropy (per mole) as H2, so the dominant effect here (as usual) is the

change in the number of moles of gases on reaction, ngas)

• N2(g) + 3 H2 (g) 2 NH3(g) Srxn° = 2 x 192.3 – (191.5 + 3 x 130.6)

191.5 3 x 130.6 2 x192.3 = -198.7 J/K (per mol of N2)

– Again, consistent with prediction (S decreases). Though NH3 more

complex, only 2 moles of it form compared to 4 moles of gas “lost”).

22Ppt08

The entropy of a perfect crystal at 0 K is zero (3rd Law).

Tro, Fig. 17.8Zumdahl, Fig 16.5

23Ppt08

W = “the number of microstates” (ways to arrange energy)

Entropy increases for a substance as energy is added to it (either T increases, or a phase change occurs)

Curve differs for different substances.

This is how standard entropies of substances are determined (at, say, 298 K)

24Ppt08

Relating Entropy of the Surroundings to a Property of the System--What is “free

energy” (G)?

• See Section IV of Handout Outline

25Ppt08

© 2011 Pearson Education, Inc.

EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S

SOLUTION

Consider the reaction for the decomposition of carbon tetrachloride gas:

(a) Calculate G at 25 C and determine whether the reaction is spontaneous.

(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any) the reaction becomes spontaneous.

The reaction is not spontaneous.

At the specific conditions / concentrations of the reaction system!

Consider only part (a) for now.

26Ppt08

(a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules).

RECALL: Interplay of ΔSsys and ΔSsurr in Determining the Sign of ΔSuniv (Table 16.3, Zumdahl)

This turns out to be very temperature dependent (See Section 17.4 in Tro). We will come back to this issue later.

See next slide

plus equals

ΔSsys + ΔSsurr = ΔSuniv

27Ppt08

Ppt08 28

Revisiting Earlier Example

If both of these plots represent the same exact process under the same exact conditions EXCEPT THAT ONE IS AT A HIGHER TEMPERATURE, then which plot represents the higher T and which the lower T? How do you know?

What would the bar plot look like if the T went to infinity? To zero?

HHint: S

Tsys

surr

29Ppt08

(see next slides →)

As T → , |Ssurr| → 0!(so Suniv → Ssys)

HS

Tsys

surr

T increasing T~

30Ppt08

As T → 0, |Ssurr| → !(so Suniv → Ssurr)

T decreasingT very small

HS

Tsys

surr

(Y-axis scale much bigger (zoomed out); Ssys same as on right.)

31Ppt08

Summary, T-dependence of Suniv

• At very high T, Suniv approaches Ssys

– Because Ssurr becomes tiny (and Ssys remains essentially unchanged)

• At very low T, Suniv approaches Ssurr

– Because Ssurr becomes huge, while Ssys remains essentially unchanged

32

NOTE: Many people end up taking a different approach to this “T-dependence” issue: They look at Gsys rather than Suniv. We’ll explore this next!

**A more detailed summary table of this will be discussed a bit later.

Ppt08

Thus: As T goes to…

33

…infinity, G →

G = H –TS

…zero, G → H (Ssurr dominates)

-TS (Ssys dominates)

Ppt08

Tro’s “version” of this

34

Thus: As T goes to…

…infinity, G →

G = H –TS

…zero, G → H -TS

-TS H

Gsys “view”

NOTE: You could also look at each of these from a Suniv “view”!Ppt08

“Global” (entropy) view (Review of earlier slide)

35

Thus: As T goes to……infinity, Suniv →

Suniv = Ssys + Ssurr; Ssurr =

…zero, Suniv → Ssurr Ssys

Ssys Ssurr

Suniv “view”

surrq

T

Ssurr Ssys

Ssurr > 0)

Ssurr > 0)

Ssurr < 0)

Ssurr < 0)

Ssys < 0)

Ssys < 0)

Ssys > 0)

Ssys > 0)

Ppt08

36

Table 16.4 (Zumdahl) Results of the Calculation of ΔSuniv and ΔG° for the Process H2O(s) H2O(l) at

-10°C, 0°C, and 10°C

T-Dependence Explored Further

Ppt08

© 2011 Pearson Education, Inc.

FOR PRACTICE 17.3Consider the reaction:

C2H4(g) + H2(g) C2H6(g) ∆H = –137.5 kJ; ∆S = –120.5 J/K

Calculate G at 25 C and determine whether the reaction is spontaneous. Does G become more negative or more positive as the temperature increases?

EXAMPLE 17.3 Computing Gibbs Free Energy Changes and Predicting Spontaneity from H and S

(b) Since S is positive, G will become more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 17.9 to find the temperature at which G changes from positive to negative (set G = 0 and solve for T). The reaction is spontaneous above this temperature.

(a) Use Equation 17.9 to calculate G from the given values of H and S. The temperature must be in kelvins. Be sure to express both H and S in the same units (usually joules).

SOLUTION

Consider the reaction for the decomposition of carbon tetrachloride gas:

(a) Calculate G at 25 C and determine whether the reaction is spontaneous.(b) If the reaction is not spontaneous at 25 C, determine at what temperature (if any)

the reaction becomes spontaneous.

The reaction is not spontaneous.

At the specific conditions / concentrations of the reaction system!

Now consider part (b)

37Ppt08

Relationship of G to Q (and G)

• See Outline (and board)

• ΔG = ΔG° + RT ln Q

38Ppt08

Relationship of G to K

• See Outline—Derive from prior slide’s relationship!

39Ppt08

Table 16.6(Zumdahl) Qualitative Relationship Between the Change in Standard Free Energy and the

Equilibrium Constant for a Given Reaction

40

G = -RT lnK

Ppt08

41Ppt08

From another text

These data provide a more direct way to calculate the G for a chemical reaction equation (i.e., you don’t need to

calculate H and S if you know Gf values!

42Ppt08

Fig. 17.10 (Tro). Why “free” energy?

43

H = -74.6 kJS = -80.8 J/KG = -50.5 kJ

Ppt08

Table 16.1 The Microstates That Give a Particular Arrangement (State)

Ppt08 44

Table 16.2 Probability of Finding All the Molecules in the Left Bulb as a Function of

the Total Number of Molecules

Ppt08 45