Lecture 09 Powerpt Slides

7/29/2019 Lecture 09 Powerpt Slides

1/21

Lecture 8 Summary

Ssys SsurrSuniv

sys < 0

Ssys

SsurrSuniv

sys < 0

Ssys

Ssurr

Suniv

sys > 0

Spontaneous, exothermic

Spontaneous, exothermic

Spontaneous, endothermic


2/21

Lecture 9: Gibbs Free Energy G

Reading: Zumdahl 10.7, 10.9

Outline

Defining the Gibbs Free Energy (G)

Calculating G (several ways to)

Pictorial Representation ofG


3/21

Defining G

Recall, the second law of thermodynamics:

Suniv = Stotal = Ssys + Ssurr

Also recall:

Ssurr = -Hsys/T

Substituting,

Stotal = Ssys + Ssurr-Hsys/T

-TStotal = -TSsys+Hsys

Multipling all by (-T) gives


4/21

We then define:

G = -TStotal

Substituting

G = H - TS

G = -TSsys+Hsys

G = The Gibbs Free Energy @const P

Giving finally


5/21

G and Spontaneous Processes

Recall from the second law the conditions of spontaneity:

Three possibilities:

IfSuniv> 0..process is spontaneous

IfSuniv< 0..process is spontaneous in oppositedirection.

IfSuniv= 0.equilibrium

In our derivation ofG, we divided by -T;

therefore, the direction of the inequality

relative to entropy is now reversed.


6/21

Three possibilities in terms ofS:

IfSuniv> 0..process is spontaneous.

IfSuniv< 0..process is spontaneous in oppositedirection.

IfSuniv= 0. system is in equilibrium.

Three possibilities in terms ofG:

IfG < 0..process is spontaneous.

IfG > 0..process is spontaneous in opposite

direction.

IfG = 0.system is in equilibrium


7/21

Spontaneous Processes: temperature dependence

Note that G is composed of both H and S termsG = H - TS

IfH < 0 and S > 0.spontaneous at all T

A reaction is spontaneous ifG < 0. So,

IfH > 0 and S < 0.not spontaneous at any T

IfH < 0 and S < 0. becomes spontaneous at low T

IfH > 0 and S > 0.becomes spontaneous at high T


8/21

Example: At what T is the following reaction

spontaneous?

Br2(l) Br2(g)

where H = 30.91 kJ/mol, S = 93.2 J/mol.K

G = H - TS


9/21

Try 298 K just to see result at standard conditions

G = H - TS

G = 30.91 kJ/mol

- (298K)(93.2 J/mol.K)

G = (30.91 - 27.78) kJ/mol= 3.13 kJ/mol > 0

Not spontaneous at 298 K


10/21

At what T then does the process become spontaneous?

G = H - TS

T = /S

T = (30.91 kJ/mol) /(93.2 J/mol.K)

= 0

Just like our previous calculation

T = 331.65 K


11/21

Calculating G

In our previous example, we needed to determine

Hrxn and Srxn separately to determine Grxn

But G is a state function; therefore, we can

use known G to determine Grxn using:

Grxn = Gprod. - Greact.


12/21

Standard G of Formation: Gf

Like Hf and S, the standard Gibbs freeenergy of formation Gf is defined as thechange in free energy that accompanies theformation of 1 mole of that substance for its

constituent elements with all reactants andproducts in their standard state.

As forHf , Gf

= 0 for an element in its standard

state:

Example: Gf (O2(g)) = 0


13/21

Example

Determine the Grxn for the following:

C2H4(g) + H2O(l) C2H5OH(l)

Tabulated Gffrom Appendix 4:

Gf(C2H5OH(l)) = -175 kJ/mol

Gf(C2H4(g)) = 68 kJ/mol

Gf(H2O (l)) = -237 kJ/mol


14/21

Using these values:

C2H4(g) + H2O(l) C2H5OH(l)

Grxn = Gf(C2H5OH(l)) - Gf(C2H4(g))

-Gf(H2O (l))

Grxn = Gprod. - Greact.

Grxn = -175 kJ - 68 kJ -(-237 kJ)

Grxn = -6 kJ < 0 ; therefore, spontaneous


15/21

More G Calculations

Similar to H, one can use the G for

various reactions to determine G for the

reaction of interest (a Hess Law forG)

Example:

C(s, diamond) + O2(g) CO2(g) G = -397 kJ

C(s, graphite) + O2(g) CO2(g) G = -394 kJ


16/21

C(s, diamond) + O2(g) CO2(g) G = -397 kJ

C(s, graphite) + O2(g) CO2(g) G = -394 kJ

CO2(g) C(s, graphite) + O2(g) G = +394 kJ

C(s, diamond) C(s, graphite) G = -3 kJ

Grxn< 0..rxn is spontaneous


17/21

Grxn Reaction Rate

Although Grxn can be used to predict if areaction will be spontaneous as written, it

does not tell us how fast a reaction will

proceed. Example:

C(s, diamond) + O2(g) CO2(g)

Grxn = -397 kJ

But diamonds are forever.


18/21

Example Problem

Is the following reaction spontaneous under standard conditions?

4KClO3(s) 3KClO4(s)KCl(s)

Hf(kJ/mol) S (J/mol.K)

KClO3(s) -397.7 143.1

KClO4(s) -432.8 151.0

KCl (s) -436.7 82.6


19/21

Example Problem Solution

Calulating Hrxn

4KClO3(s) 3KClO4(s)KCl(s)

Hrxn = 3Hf KClO4( ) Hf KCl( )- 4Hf KClO3( ) = 3(-432.8kJ) (-436.7kJ)- 4(397.7kJ)

= -144 kJ

Calulating Srxn

Srxn = 3S KClO4( ) S KCl( )-4S KClO3( )

= 3(151.0JK) (82.6J

K)- 4(143.1J

K)

= -36.8JK


20/21

Example Problem Solution

Calulating Grxn

Grxn = Hrxn -TSrxn

= -144 kJ- 298K( ) -38.6JK( )1kJ

1000J

= -133kJ

Grxn < 0 ;therefore, reaction is spontaneous.


21/21

Example Problem Continued

For what temperatures will this reaction be spontaneous?

Answer: For T in which Grxn < 0.

Grxn

= Hrxn

-TSrxn

0= Hrxn -TSrxnH

rxn

Srxn=

-133kJ

-38.6JK( ) 1kJ1000J

= 3446K=T

Spontaneous as long as T < 3446 K.

Documents

Lecture 09 Powerpt Slides