Ch19 Thermo 2 Kotz

7/29/2019 Ch19 Thermo 2 Kotz

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Gibbs Free Energy, GGibbs Free Energy, G

Multiply through by -TMultiply through by -T-TS-TSunivuniv = H= Hsyssys - TS- TSsyssys

-TS-TSunivuniv = change in Gibbs free energy= change in Gibbs free energy

for the system = Gfor the system = Gsystemsystem

UnderUnderstandard conditionsstandard conditions

GGoosyssys = H= Hoo

syssys - TS- TSoo

syssys

Suniv =Hsys

T+ Ssys

SSunivuniv = S= Ssurrsurr + S+ Ssyssys


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GGoo = H= Hoo - TS- TSoo

GibbsGibbs free energyfree en

ergy change =change =total energy change for systemtotal energy change for system

- energy lost in disordering the system- energy lost in disordering the system

If reaction isIf reaction is exothermic (negative Hexothermic (negative H oo)) (energy dispersed)(energy dispersed)

and entropy increases (positive Sand entropy increases (positive S oo))

(matter dispersed)(matter dispersed) thenthen GG oomust bemust be NEGATIVENEGATIVE

reaction is spontaneous (and product-

favored).


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Gibbs free energy change =Gibbs free energy change =

total energy change for systemtotal energy change for system

- energy lost in disordering the system- energy lost in disordering the system

If reaction isIf reaction is

endothermic (positive Hendothermic (positive H oo))

and entropy decreases (negative S

and entropy decreases (negative S oo

)) thenthen GG oo must bemust be POSITIVEPOSITIVE

reaction isreaction is not spontaneousnot spontaneous (and is(and is reactant-reactant-favoredfavored).).


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HHoo SSoo GGoo ReactionReaction

exo()exo() increase(+)increase(+) Prod-favoredProd-favored

endo(+)endo(+) decrease(-)decrease(-) ++ React-favoredReact-favored

exo()exo() decrease(-)decrease(-) ?? T dependentT dependent

endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent


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Two methods of calculating GTwo methods of calculating Goo

a)a) Determine HDetermine Hoorxnrxn and Sand Soorxnrxn and useand use

GIbbs equation.GIbbs equation.

b)b) Use tabulated values ofUse tabulated values of

free energiesfree energies

of formation, Gof formation, Gffoo..

GGoorxnrxn == GGffoo (products) -(products) - GGff

oo (reactants)(reactants)




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Free Energies of FormationFree Energies of Formation

Note that GNote that Gfffor an element = 0for an element = 0


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Calculating GCalculating GoorxnrxnCombustion of acetyleneCombustion of acetylene

CC22HH22(g) + 5/2 O(g) + 5/2 O22(g) --> 2 CO(g) --> 2 CO22(g) + H(g) + H22O(g)O(g)

Use enthalpies of formation to calculateUse enthalpies of formation to calculate

HHoorxnrxn

= -1238 kJ= -1238 kJ

Use standard molar entropies to calculateUse standard molar entropies to calculate

SSoorxnrxn = -97.4 J/K or -0.0974 kJ/K= -97.4 J/K or -0.0974 kJ/K

GG

oo

rxnrxn = -1238 kJ - (298 K)(-0.0974 J/K)= -1238 kJ - (298 K)(-0.0974 J/K)= -1209 kJ= -1209 kJ

Reaction isReaction is product-favoredproduct-favored in spite of negativein spite of negativeSSoorxnrxn..

Reaction isReaction is enthal drivenenthal driven


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Calculating GCalculating Goorxnrxn

Is the dissolution of ammonium nitrate product-Is the dissolution of ammonium nitrate product-favored?favored?

If so, is it enthalpy- or entropy-driven?If so, is it enthalpy- or entropy-driven?

NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)


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From tables of thermodynamic data we findFrom tables of thermodynamic data we find

HHoorxnrxn = +25.7 kJ= +25.7 kJ

SSoorxnrxn = +108.7 J/K or +0.1087 kJ/K= +108.7 J/K or +0.1087 kJ/K

GGoorxnrxn = +25.7 kJ - (298 K)(+0.1087 J/K)= +25.7 kJ - (298 K)(+0.1087 J/K)

= -6.7 kJ= -6.7 kJ

Reaction isReaction is product-favoredproduct-favored in spite of negativein spite of negativeHHoorxnrxn..

Reaction isReaction is entropy drivenentropy driven

NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)


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Two methods of calculating GTwo methods of calculating Goo

a)a) Determine HDetermine Hoorxnrxn and Sand Soorxnrxn and useand use

GIbbs equation.GIbbs equation.

b)b) Use tabulated values ofUse tabulated values of

free energiesfree energies

of formation, Gof formation, Gffoo..


oo (reactants)(reactants)GGoo

rxnrxn == GGffoo (products) -(products) - GGff



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Combustion of carbonCombustion of carbon

C(graphite) + OC(graphite) + O22(g) --> CO(g) --> CO22(g)(g)

GGoorxnrxn = G= Gffoo(CO(CO22) - [G) - [Gffoo(graph) + G(graph) + Gffoo(O(O22)])]

GGoorxnrxn = -394.4 kJ - [ 0 + 0]= -394.4 kJ - [ 0 + 0]

Note that free energy of formation of an elementNote that free energy of formation of an element

in its standard state is 0.in its standard state is 0.GGoorxnrxn = -394.4 kJ= -394.4 kJ

Reaction isReaction is product-favoredproduct-favored as expected.as expected.

GGoorxnrxn == GGffoo (products) -(products) - GGffoo (reactants)(reactants)GGoo

rxnrxn == GGffoo

(products) -(products) - GGffoo

(reactants)(reactants)


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Free Energy and TemperatureFree Energy and Temperature

2 Fe2 Fe22OO33(s) + 3 C(s) ---> 4 Fe(s) + 3 CO(s) + 3 C(s) ---> 4 Fe(s) + 3 CO22(g)(g)HHoorxnrxn = +467.9 kJ= +467.9 kJ SS

oorxnrxn = +560.3 J/K= +560.3 J/K

GGoorxnrxn = +300.8 kJ= +300.8 kJ

Reaction isReaction is reactant-favoredreactant-favored at 298 Kat 298 K

At what T does GAt what T does Goorxnrxn just change from beingjust change from being

(+) to being (-)?(+) to being (-)?When GWhen Goorxnrxn = 0 = H= 0 = H

oorxnrxn - TS- TS

oorxnrxn

T =HrxnSrxn

=467.9 kJ

0.5603 kJ/K= 835.1 K


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More thermo?More thermo? You betcha!You betcha!


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FACT: GFACT: Goorxnrxn is the change in free energyis the change in free energy

when pure reactants convert COMPLETELYwhen pure reactants convert COMPLETELY

to pure products.to pure products.

FACT: Product-favored systems haveFACT: Product-favored systems have

KKeqeq > 1.> 1.

Therefore, both GTherefore, both Grxnrxn and Kand Keqeq are related toare related toreaction favorability.reaction favorability.

Thermodynamics and KThermodynamics and Keqeq


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KKeqeq is related to reaction favorability and sois related to reaction favorability and soto Gto Goorxnrxn..

The larger the value of K the more negativeThe larger the value of K the more negativethe value of Gthe value of Goorxnrxn

GGoorxnrxn = - RT lnK= - RT lnK

where R = 8.31 J/Kmolwhere R = 8.31 J/Kmol



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Calculate K for the reactionCalculate K for the reaction

NN22OO44 --->2 NO--->2 NO22 GGoo

rxnrxn = +4.8 kJ= +4.8 kJGGoorxnrxn = +4800 J = - (8.31 J/K)(298 K) ln K= +4800 J = - (8.31 J/K)(298 K) ln K

GGoorxnrxn = - RT lnK= - RT lnK

lnK = -4800 J

(8.31 J/K)(298K)

= - 1.94


K = 0.14K = 0.14

When GWhen Goorxnrxn > 0, then K < 1> 0, then K < 1


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1717G, G, and KG, G, and Keqeq

G is change in free energy at non-G is change in free energy at non-standard conditions.standard conditions.

G is related to G G is related to G

G = G + RT ln Q

G = G + RT ln Q

where Q = reaction quotientwhere Q = reaction quotient

When Q < K or Q > K, reaction isWhen Q < K or Q > K, reaction is

spontaneous.spontaneous.

When Q = K reaction is at equilibriumWhen Q = K reaction is at equilibrium

When G = 0 reaction is at equilibriumWhen G = 0 reaction is at equilibrium

Therefore, G = - RT ln K Therefore, G = - RT ln K


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G, G, and KG, G, and Keqeq

Figure 19.10Figure 19.10


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Product favoredProduct favored

reactionreaction

GGoo and K > 1and K > 1

In this case GIn this case Grxnrxn isis

< G< Goorxnrxn , so state with, so state with

both reactants andboth reactants and

products present isproducts present is

MORE STABLE thanMORE STABLE than

complete conversion.complete conversion.



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Product-favoredProduct-favoredreaction.reaction.

2 NO2 NO22 ---> N---> N22OO44

GGoorxnrxn = 4.8 kJ= 4.8 kJ

Here GHere Grxnrxn is less thanis less than

GGoorxnrxn , so the state, so the state

with both reactantswith both reactants

and productsand productspresent is morepresent is morestable than completestable than completeconversion.conversion.



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Reactant-favoredReactant-favoredreaction.reaction.

NN22OO44 --->2 NO--->2 NO22

GGoorxnrxn = +4.8 kJ= +4.8 kJ

Here GHere Goorxnrxn is greateris greater

than Gthan Grxnrxn , so the, so the

state with bothstate with both

reactants andreactants andproducts present isproducts present is

more stable thanmore stable than

complete conversion.complete conversion.



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KKeqeq is related to reaction favorability.is related to reaction favorability.

When GWhen Goorxnrxn < 0, reaction moves< 0, reaction moves

energetically downhillenergetically downhill

GGoorxnrxn is the change in free energy whenis the change in free energy when

reactants convert COMPLETELY toreactants convert COMPLETELY to

products.products.


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Ch19 Thermo 2 Kotz