chapter 16 spontaneity, entropy and free energy. contents l spontaneous process and entropy l...
Post on 22-Dec-2015
228 views
TRANSCRIPT
![Page 1: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/1.jpg)
Chapter 16
Spontaneity, Entropy and Free energy
![Page 2: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/2.jpg)
Contents Spontaneous Process and Entropy Entropy and the second law of thermodynamics The effect of temperature on spontaneity Free energy Entropy changes in chemical reactions: The
Third Law of Thermodynamics Free energy and chemical reactions The dependence of free energy on pressure Free energy and equilibrium Free energy and work
![Page 3: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/3.jpg)
16.1 Spontaneous Process and Entropy A process that will occur without outside
intervention. Thermodynamics can’t determine how fast the
process is (may be fast or slow). Kinetics tell us that the rate of reaction depends
on: activation energy; temperature, concentration and catalysts; i.e, it depends on the pathway of process.
Thermodynamics compares initial and final states and does not require knowledge of the pathway.
Kinetics describes pathway between reactants and products.
We need both thermodynamics and kinetics to describe a reaction completely.
![Page 4: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/4.jpg)
Spontaneity process that will occur without outside intervention. Rolling a ball down a hill (gravity); steel rusting (?);
wood burning (exothermic process); transfer of heat from hot to cold (exothermic); freezing of water (exothermic); Melting of ice (endothermic!!); etc.
What common characteristic derives all those processes to be spontaneous?
It is an increase in a property called Entropy, S.
Thus all spontaneous processes occur as a result of an increase of the S of the universe.
![Page 5: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/5.jpg)
What is Entropy?
Entropy, S, is measure of randomness, or disorder in the system. For example molecular randomness.
Naturally, things move from order to disorder! (from lower S to higher S.)
![Page 6: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/6.jpg)
EntropyS defined in terms of probability.Substances take the most likely
arrangement (that have more probabilities).
The most likely arrangement is the most randomrandom.
Can we calculate the number of arrangements for a system?
![Page 7: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/7.jpg)
Entropy and Gases Gas placed in one bulb of a
container will spontaneously expand to fill the entire vessel (vessel of two bulbs) evenly.
Probabilities of finding equal number of molecules in each bulb are huge.
Finding molecules in one bulb is highly improbable; thus the process does not occur spontaneously.
![Page 8: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/8.jpg)
Entropy and state of matter Gases completely fill their chamber
because there are many more ways to do that than to leave half empty.
Ssolid <Sliquid <<Sgas In solids, molecules are very close and thus
they have relatively few positions available to them
Entropy also describes the number of possible positions of a molecule
There are many more ways for the molecules to be arranged as a liquid than a solid.
Gases have a huge number of positions possible.
![Page 9: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/9.jpg)
Example
Which has higher positional S?– Solid CO2 or gaseous CO2?– N2 gas at 1 atm or N2 gas at 0.01 atm?
Predict the sign of S– Solid sugar is added to water
• + because larger volume– Iodine vapor condenses on cold surface to
form crystals
• - because gs, less volume
![Page 10: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/10.jpg)
Positional EntropyPositional Entropy
Entropy also describes the number of
possible positions of a molecule A gas expands into a vacuum because
the expanded state has the highest
positional probability of states available
to the system; i.e., the highest positional
entropy Probability depends on the number of
configurations in space ( positional
microstates)
![Page 11: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/11.jpg)
Solutions form because there are many more possible arrangements of dissolved pieces than if they stay separate; there is an increase in entropy Generally, in any spontaneous process, there is always an increase in the entropy of the universe This is the second law of thermodynamicssecond law of thermodynamics;
Suniv = Ssys + Ssurr
IfSuniv is positive the process is spontaneous. IfSuniv is negative the process is spontaneous in the opposite direction. Suniv > 0 for any spontaneous processFirst law: The energy of a universe is constantFirst law: The energy of a universe is constant
16.2 Entropy and the second law of thermodynamics
![Page 12: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/12.jpg)
For exothermic processes Ssurr is positive For endothermic processes Ssurr is negative Consider this process (endothermic)
H2O(l)H2O(g) Ssys is positive (change from L to G) Ssurr is negative (random motion of atoms in the
surrounding decreases) Which one will control the process: Ssys or Ssurr? Suniv;
If the ∆SIf the ∆Ssyssys and ∆S and ∆Ssurrsurr have different signs, the have different signs, the temperature determines the ∆Stemperature determines the ∆Sunivuniv
16.3 The effect of temperature on spontaneity
![Page 13: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/13.jpg)
Determining ∆Ssurr Sign of ∆Ssurr
– depends on direction of heat flow
– ∆Ssurr + for exothermic reactions
– ∆Ssurr - for endothermic reactions Magnitude of ∆Ssurr
– Depends on temperature– Heat flow = ∆H at constant P– Very small at high T, increases as T
decreases
T
HS reactionsurr
![Page 14: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/14.jpg)
SsysSsurr Suniv Spontaneous?
+ + +
- --
+ - ?
+- ?
No
Yes
Yes at Low temp.
Yes at High temp.
Entropy of Sys and Ssurr
in determining the sign of Suniv
Ssurr = -H/T
![Page 15: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/15.jpg)
Example A process has a H of +22 kJ and a S of -13
J/K. At which temperatures is the process spontaneous?
– if there is no subscript, S = Ssys
Suniv > 0 to be spontaneous Ssys + Ssurr > 0
KTK
J
T
JT
J
K
JT
HS reactionsys
1700
13000,22
0000,22
13
0
![Page 16: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/16.jpg)
Example
For methanol, the enthalpy of vaporization is 71.8 kJ/mol and the entropy of vaporization is 213 J/K. What is the normal boiling point of methanol?
Ssys= 213 J/K and H = 71.8 kJ/molK = 71,800 J/molK
– at the boiling point, the vaporization begins to be spontaneous
Suniv = 0 to be at bp Ssys + Ssurr = 0
CKTK
J
T
JT
J
K
J
T
HS reactionsys
64337213800,71
0800,71
2130
![Page 17: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/17.jpg)
16.4 Free Energy, G G: Gibbs free energy G: helps the determination of the T dependence
on spontaneity G ≡ H - TS definition of G ∆G ≡ ∆ H - T ∆ S for constant T
All quantities refer to the system. When no subscript the quantity refers to
the system
T-
S T -
T-
H
T-
G
syssurr S S
T
G
![Page 18: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/18.jpg)
If G is negative at constant T and P, the process is spontaneous.
The process is spontaneous in the direction G decreases
- G means + Suniv
So what sign would the ∆G of reaction with a + ∆Suniv have?
– Negative
syssurruniv S S T
GS
at constant T and P
![Page 19: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/19.jpg)
Example
Is the following reaction spontaneous at -10°C in
the forward direction? H2O(s) H2O(l)
– where ∆H°=6.03x103 J/mol and ∆S°=22.1 J/mol.K
mol
J102.2G
)molK
J22.1( 63K)2( - )
mol
J106.03( G
2
3
∆Go = ∆ Ho - T ∆ So
![Page 20: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/20.jpg)
J/molK -0.82.12 9.22 Suniv
Another way to check for spontaneity of a reaction?
– Check to see if the ∆Suniv is positive
– How can we solve for ∆Ssurr?– use ∆H° and T
At -10°C, is it spontaneous?
molKJsurr /9.22263K
J/mol1003.6 S
3
NO!- the reverse is spontaneous
T
HS reactionsurr
syssurruniv S SS
![Page 21: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/21.jpg)
Example 2 At what T is this reaction
spontaneous at 1 atm
Of pressure?
– Br2(l) Br2(g)– ∆Ho=31.0 kJ/mol, – ∆So=93.0 J/molK
K 333 )
molKJ
0.39(
)mol
J100.13
T
ΔS
ΔH T
ST- H ΔG
0 G point, boiling At the
3
o
o
oo
o
o
![Page 22: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/22.jpg)
G=H-TS
HS Spontaneous?
+ - At all Temperatures
+ + Spontaneous at high temperatures
- - Spontaneous at low temperatures
+- Not spontaneous at any temperature,
Reverse is spontaneous
![Page 23: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/23.jpg)
16.5 Entropy changes in chemical reactionsThe Third Law of Thermodynamics
16.5 Entropy changes in chemical reactionsThe Third Law of Thermodynamics
So far we dealt with physical changesIn a chemical reaction, when the number of
gaseous molecules increases, the positional
disorder will increase and consequently S would
be Positive and Visa Versa Consider the following examples:
N2(g) + 3H2(g) 2NH3(g)
4NH3(g) + 5O2 (g) 4NO(g) + 6H2O(g)
CaCO3(s) CaO (s) + CO2(g)
![Page 24: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/24.jpg)
The changes in enthalpy determines the exothermicity and endothermicity at constant P
The changes in entropy determines the
spontaneity at constant P and T Are there values given for S? The entropy of a pure crystal at 0 K is 0.
– Molecular motion is almost zero and only one arrangement is possible.
Thus, the entropy of a perfect crystal is zero; this statement is called the Third Law of Thermodynamics
![Page 25: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/25.jpg)
At T >0, some changes in order will occur, consequently, S >0
This value gives us a starting point. Standard Entropies Sº ( at 298 K and 1 atm) of
substances are listed. Sº can then be determined for products and
reactants
Sreaction = npS(products) nrS(reactants)
Entropy is a state function of the system (It is not pathway dependent)
![Page 26: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/26.jpg)
Entropy is an extensive property. It depends on the amount of substance present
Number of moles of reactants and products must be taken into account
What is the expected S (0, 0r +ve or –ve) for the following reaction?
Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)
More complex molecules possess higher Sº
![Page 27: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/27.jpg)
Example
Find the ∆S° at 25°C for:
– 2NiS(s) + 3O2(g) 2SO2(g) + 2NiO(s)
53 205 248 38
K
J
molK
Jmol
molK
Jmol
molK
Jmol
molK
JmolSreaction
149)205(3)53(2
)38(2)248(2
![Page 28: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/28.jpg)
16.6 Free Energy and Chemical Reactions
Gº is needed when dealing with chemical reactions
Gº = standard free energy change. Free energy change that will occur if
reactants in their standard state turn to products in their standard state.
N2(g) + 3H2(g) 2NH3(g) Gº= -33.3 kJ
Gº can’t be measured directly, but can be calculated from other measurements.
Gº=Hº-TSº
![Page 29: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/29.jpg)
Free Energy in Reactions
There are tables of Gºf . The standard free energy of formation for
any element in its standard state is 0. Why Gº is useful? To compare relative tendencies of
reactions to occur. The more –ve the value of Gº, the further
a reaction will go to the right to reach equilibrium
![Page 30: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/30.jpg)
How do we calculate Gº?There are three Ways to Calculate ∆G°
Use equation: ∆G° ≡ ∆ H ° - T ∆ S ° Use Hess’s Law
– Rearrange equations to get the given equation
– Add ∆G° values for the equations Use ∆Gf° : standard free energy of formation
)tan()( tsreacfrproductsfpreaction GnGnG
![Page 31: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/31.jpg)
1. Use the equation: Gº=Hº-TSº This equation is applicable for
reactions taking place at constant temperature
![Page 32: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/32.jpg)
![Page 33: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/33.jpg)
![Page 34: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/34.jpg)
![Page 35: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/35.jpg)
2. Claculation of Gº using Hess’s law
Free energy is a state function just like enethalpy, i.e., it depends upon the pathway of the reaction
It can be found same as H using Hess’s law
![Page 36: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/36.jpg)
![Page 37: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/37.jpg)
3. Calculation of Gº using the standard free energies of formation Gºf
Gºf of a substance is – the change in free energy that
accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states
Gº for a specific reaction is obtained from the equation:
G = npGf(products) nrGf(reactants)
![Page 38: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/38.jpg)
There are tables of Gºf . The standard free energy of
formation for any element in its standard state is 0.
Remember also that the number of moles of each reactant and product must be used in calculating Gº for a reaction
![Page 39: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/39.jpg)
![Page 40: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/40.jpg)
Exercise Is the following reaction spontaneous under the
standard conditions? :
C2H4(g) + H2O(l) C2H5OH(l)– Find Gº for the reaction using Gºf values
from the table– If Gº is –Ve, the process is spontaneous and
the formation of ethanol is favorable– Of course to judge whether the reaction is
feasible or not, we have to study its kinetics to know whether the reaction is fast or slow.
– If it is slow adding a catalyst may be considered
– Also, remember that Gº depends on temperature: Go=Ho-TSo
![Page 41: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/41.jpg)
16.7 The Dependence of Free Energy on Pressure
At a large volume the gas has many more positions available for its molecules than at low volumes
S large V > S small V S low P > S high P Since S depends on P then G will depend on P.
It can be shown that: G = Go + RTln(P) (Go = free energy at 1 atm)
![Page 42: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/42.jpg)
By proper derivation we got: G = Gº +RTln(Q)
– where Q is the reaction quotient (P of the products /P of the reactants).
![Page 43: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/43.jpg)
![Page 44: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/44.jpg)
![Page 45: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/45.jpg)
16.8 Free energy and equilibrium
According to thermodynamics the equilibrium occurs at the lowest value of free energy available
At equilibrium G = 0, Q = K G = Gº +RT ln(Q) Thus, Gº = -RT lnK
![Page 46: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/46.jpg)
Gº K
=0 =1
<0 >0
>0 <0
Gº = -RT lnK
![Page 47: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/47.jpg)
Temperature dependence of K
Gº= -RT lnK = Hº - TSº ln(K) = - Hº/RT + Sº/RPlot ln(K) VS 1/TA straight line gives a:
Slope = Hº/R
Intercept = Sº/R
![Page 48: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/48.jpg)
![Page 49: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/49.jpg)
![Page 50: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/50.jpg)
Example
![Page 51: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/51.jpg)
![Page 52: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/52.jpg)
![Page 53: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/53.jpg)
16.9 Free energy and Work
Free energy is that energy free to do work. The maximum amount of work possible at a
given temperature and pressure. Never really achieved because some of the
free energy is changed to heat during a change, so it can’t be used to do work.
![Page 54: Chapter 16 Spontaneity, Entropy and Free energy. Contents l Spontaneous Process and Entropy l Entropy and the second law of thermodynamics l The effect](https://reader035.vdocument.in/reader035/viewer/2022081513/56649d775503460f94a59522/html5/thumbnails/54.jpg)
Reversible v. Irreversible Processes
Reversible: The universe is exactly the same as it was before the cyclic process.
Irreversible: The universe is different after the cyclic process.
All real processes are irreversible -- (some work is changed to heat)