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25-4-2011

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At what temperature will the reaction below starts At what temperature will the reaction below starts to be non-spontaneous:to be non-spontaneous:

NN2 2 (g) + 3H(g) + 3H22(g) (g) 2NH 2NH33 (g) (g)

H° - T H° - T S° = S° = G°G°- 92 kJ -198.5 J/K (- 92 kJ -198.5 J/K (S is -ve)S is -ve)@ T Low @ T Low H° dominatesH° dominates G° = -92000 –(-198.5) = -91802J G° = -92000 –(-198.5) = -91802J Spontaneous Spontaneous@ T High@ T High --TTS° dominatesS° dominates G° (+) G° (+) Non-spontaneous Non-spontaneous

To go from spontaneous to non-spontaneous, To go from spontaneous to non-spontaneous, G° = 0G° = 0T = T = - 92kJ- 92kJ = 463.5 K = 463.5 K

-0.198 kJ / K-0.198 kJ / K

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CaCO3 (s) CaO (s) + CO2 (g)

H0 = 177.8 kJ/mol

S0 = 160.5 J/K·mol

G0 = H0 – TS0

At 25 oC, G0 = 177.8–(298*160.5*10-3) = 130.0 kJ/mol,Suggesting a non-spontaneous process at this temperature.

G0 = 0 at 1108 K or 835 oC, where the reaction becomes possible

At what temperature will the reaction below starts to At what temperature will the reaction below starts to be spontaneous:be spontaneous:

Equilibrium Pressure of CO2

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The Significance of the Sign and Magnitude of Go

Go is a large, negative quantity and equilibrium is very far to the right (towards products)

Gprod << Greac

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The Significance of the Sign and Magnitude of Go

Go is a large, positive quantity and equilibrium is very far to the left (towards reactants)

Gprod >> Greac

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The Significance of the Sign and Magnitude of Go

The equilibrium lies more toward the center of the reaction profile

Gprod Greac

When Q is smaller than K (Q<Keq ) this means that reactants have a large value. This shifts the equilibrium towards product formation, and vice versa.

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Q= [NH3]2

]H2[3]N2[

When Q is smaller than K this means that reactants have a large value. This shifts the equilibrium towards product formation, and vice versa.

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Phase Transition

Phase transitions can be looked at as an equilibrium process, in which both phases will be present. Therefore,

G = 0 and:

0 = H – TS

S = H/T

We can then write:

Sfusion = Hfusion/Tmp and Svap = Hvap/Tbp

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The heat of fusion and vaporization of benzene are 10.9kJ/mol and 31.0kJ/mol, respectively. Calculate the entropy changes from solid to liquid and liquid to vapor. At 1 atm, benzene melts at 5.5 oC and boils at 80.1 oC.

Sfusion = Hfusion/Tmp (solid liquid)

Sfusion = (10.9*103J/mol)/(273 + 5.5)K

Sfusion = 39.1 J/K mol

Svap = Hvap/Tbp (liquid gas)

Svap = (31.0*103J/mol)/(273 + 80.1)K

Svap = 87.8 J/K mol

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Recall that G and K (equilibrium constant) apply to standard conditions

Recall that G and Q (equilibrium quotient) apply to any conditions.

It is useful to determine whether substances under any conditions (non-standard conditions) will react:

QRTGG ln

Gibbs Free Energy and the Equilibrium Constant

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Calculate G at 298K for the reaction below, where 1.0 atm N2(g), 3.0 atm H2(g) and 1.0 atm NH3(g) are present in the mixture. Go at 298K = -33.3 kJ

N2(g) + 3H2(g) 2NH3(g)Solution

Q = P2NH3/(PN2*P3

H2)Q = 1.02 / (1.0 * 3.03)Q = 3.7 x 10-2

G = Go + RT ln QG = -33.3 kJ + 8.314 J/K * 298K * ln (3.7 x 10-2)G = -33.3 kJ - 8.17 kJ = -41.5 kJ

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Is the reaction 2 SO2 (g) + O2 (g) 2 SO3 (g) expected to be spontaneous at 25 oC and 1 atm. Gf

o (SO2 (g)) = -300 kJ/mol and Gfo

(SO3 (g)) = -370 kJ/mol.

First, you should remember that Gfo (O2 (g)) =

0 since the element in its standard state has both Gf

o and Hfo = 0

Go = 2 Gfo (SO3 (g)) – { 2 Gf

o (SO2 (g)) + Gfo (O2 (g))}

Go = 2*(-370 kJ/mol) – { 2*(-300 kJ/mol) + 0 }= -140 kJ/mol

The value of Go is very negative suggesting that the reaction is far to the right. Thus the reaction is spontaneous

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Free Energy and Equilibrium

Under non-standard conditions, we need to use G instead of G°.

Q is the reaction quotiant.

Note: at equilibrium: G = 0.away from equil, sign of G tells which way rxn goes

spontaneously.

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At equilibrium, Q = K and G = 0, so

We can conclude: If G < 0, then K > 1 (products are

favored). If G = 0, then K = 1 (equilibriunm state). If G > 0, then K < 1 (reactants are

favored).

eq

eq

KRTG

KRTG

QRTGG

ln

ln0

ln

Gibbs Free Energy and the Equilibrium Constant

This equation is one of the most important relationships in thermodynamics because it allows us to calculate K from standard free energy changes.

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Calculate the equilibrium constant, Kp, at 25 oC, for the reaction:

2H2O (l) 2H2 (g) + O2 (g)

Where Gof (H2O(l)) = 237.2 k/mol, Go

f (H2(g)) = 0.0 k/mol Go

f (O2(g)) = 0.0 k/mol

Gorxn = {(2*0.0 + 1*0.0)} –(- 2*237.2) =474 kJ

Gorxn = -RTlnKp

474.4*103 J/mol = - 8.314 J/K mol*298K lnKp

Kp = 7*10-84 since Kp is very small it is not wise to use partial pressures to calculate it.

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G Equations

G°

n G°prod - n G°

reaction

n G°prod - n G°

reaction

H° - S°H° - S°

- RT ln Keq

G - RT ln QG - RT ln Q

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Coupled Reactions

A thermodynamically unfavored reaction can be made to proceed in the forward direction by coupling to a thermodynamically favored one:

Example:Zn(s) Zn(s) + S(s) has a Go

rxn = 198 kJ/molHowever, the reaction:S(s) + O2(g) SO2(g) has a Go

rxn = -300 kJ/mol

Coupling the two reactions gives:ZnS(s) + O2(g) Zn(s) + SO2(g) where Go

rxn = (- 300 –(+198)} = -112 kJ/mol, which is favorable

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Selected Problems

1, 4, 5, 7, 8, 10-14, 17, 19-25, 27, 31