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HYPOTHESIS: if !S is positive, the reaction will be spontaneous.%

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'.,-$#-(&$[^%Spontaneity depends on two thermodynamic functions: !H

Exothermic reactions (!H <0) are usually spontaneous. !S

Spontaneous if %!Ssystem + !Ssurroundings > 0

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Criteria for spontaneity: on'['$(B%\%i%3o:X-&F%%%',%

!G < 0 Spontaneous !G > 0 Not spontaneous !G = 0 System at equilibrium.

Thermodynamic function that determines the amount of energy available to do work: called FREE ENERGY

%%!G = !H " T!S

T is absolute temperature (in K). !G refers to a reaction at constant T and P.

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!Gf° is the :3829849%m411%1214n7%5m%m54=83;52!!Gf° Free energy change in forming one mole of a compound from its elements each in their standard states.

Units: !Gf°: kJ/mole !Hf°: kJ/mole S°: J/mole-K

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!G°rxn = #!Gf°(prods) ' #!Gf°(reactants)

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!H° and !S° do not vary much with T, BUT !G° does!!!

For other temperatures: !G°rxn " !H°rxn # T!S°rxn

Use tables of %!Hf° and S° to find !H°rxn and !S°rxn

and plug in to $ !G°rxn at a specific T. %

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Use !G = !H " T!S !H and !S do not change substantially with temperature,

but %G does.

Be able to qualitatively predict the EFFECT OF TEMPERATURE ON FREE ENERGY%

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Q. Is this reaction spontaneous at 298 K? At 1100 oC? CaCO3(s) !CaO(s) + CO2(g) So 92.88 39.75 213.6 (J/mol K)

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!"#$%"#..(-'%YX)&-G%."#'(%$)#-'&+,-'<%•  During phase changes temperature is a constant. (Heat

transfer is reversible: qrev) •  !S for a phase transition is predictable: for example it is

always positive at melting or boiling. (Disorder increases.) What about !G = ??? at equilibrium

Entropy and heat: !Strans = qrev = !Htrans T T

e.g., for melting ice, !Sfus = !Hfus/273 K (endothermic) " " " + + m.p.= 0°C

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Calculate the boiling point of bromine.

!Hovap = 31.0 (kJ/mol)

!Sovap = 92.9 (J/mol-K)

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So far we have considered only standard conditions ($Ho, $So, $Go) and how to estimate $Go at different T.

What happens under other conditions?

!G = !G° + RT ln Q !G = !G° + 2.303 RT log10Q aA + bB cC + dD

REACTION QUOTIENT !

Q = C[ ]c D[ ]d

A[ ]a B[ ]b

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!G = !G° + RT lnQ

How do concentrations affect Q and !G?

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Is this consistent with LeChatelier’s Principle?

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Q = NH 3[ ]2

N2[ ] H2[ ]3N2 (g) + 3 H2(g) NH3 (g)

AgCl(s) Ag+(aq) + Cl#(aq)

1) Is this spontaneous under standard conditions?

!G°rxn = +55.6 kJ/mol 2) What if [Ag+] = [Cl"] = 1 & 10"10 M at 298K?

!Grxn = !G° + RT ln Q

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AgCl(s) Ag+(aq) + Cl#(aq) 3) What if [Ag+] = [Cl#] = 1.34 % 10#5 M at 298K?

!G = !G° + RT ln Q

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!G = !G° + RT ln Q

At equilibrium, !G = Q =

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What is the solubility product constant Ksp of AgBr at 25oC? AgBr(s) Ag+(aq) + Br"(aq)

!Gf

o Ag+ (aq) 77.1 kJ/mol !Gf

o Br# (aq) #104 kJ/mol !Gf

o AgBr (s) #96.9 kJ/mol

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At equilibrium point, !G = 0 for interconverting

reactants M products

Note:

This does NOT mean !Go = 0

For a spontaneous process, !G = Wmax = The maximum work that can be obtained from a process at constant T and P. For a non-spontaneous process, !G = Wmin = The minimum work that must be done to make a process go at constant T and P.

What is “free” about Free Energy? !G is how much work can be done by a process and is the portion of the energy change of a spontaneous reaction that is free to do useful work. But in actuality, not all of this “free energy” will be able to be used to do work. Some the energy enters the environment as heat. =>%?,@#-% A"(B%CCD% D]%

RELATIONSHIP BETWEEN !G AND WORK

Calculate the maximum energy available from the oxidation of octane at 25 oC and 1 atm.

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Calculate the work done in a combustion reaction.

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1)  Combustion (burning) = exo 2)  Breaking multiple bonds = endo 3)  Forming bonds = exo 4)  Phase changes s M ! M g = endo 5)  Opposite processes (g #% ! M s) = exo 6)  Measured T #% OR heat given off = exo 7)  Measured T $% OR heat added = endo

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!S is positive (disorder increases) when: 1)  # molecules of gas #% during a reaction 2)  Phase changes%s M% ! M% g 3)  Solid dissolves in solution 4)  # particles #% during a reaction 5)  # atoms in a molecule is larger 6)  T increases 7)  V increases

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