General InformationGeneral Information

"Thermodynamics""Thermodynamics", is the study of , is the study of the energy changes or transfers the energy changes or transfers accompanying physical and accompanying physical and chemical processeschemical processes

helps ushelps us predict whether a rxn will predict whether a rxn will occur under certain conditionsoccur under certain conditions

DO NOT ASSUME THAT ALL DO NOT ASSUME THAT ALL REACTIONS HAPPEN REACTIONS HAPPEN AUTOMATICALLY!AUTOMATICALLY!

SpontaneitySpontaneity

""spontaneousspontaneous" - a process can " - a process can occur (not necessarily fast)occur (not necessarily fast)

""non-spontaneousnon-spontaneous" - a process " - a process cannot occurcannot occur

General InformationGeneral Information many spontaneous rxns many spontaneous rxns

release heat energy = release heat energy = "exothermic";"exothermic";

products have less E than products have less E than reactants ("downhill") reactants ("downhill") (see diagram)(see diagram)

some rxns absorb energy some rxns absorb energy = "endothermic" = = "endothermic" = products have more products have more energy than the reactants energy than the reactants ("uphill") (see diagram)("uphill") (see diagram)

EndothermicEndothermic

ExothermicExothermic Ea = Activation Energy

a = Ea

b = Ea for reverse reactionc = enthalpy change

General InformationGeneral Information not all exothermic rxns are not all exothermic rxns are

spontaneous; not all spontaneous; not all spontaneous rxns are spontaneous rxns are exothermic, exothermic, so some other factor so some other factor must influence spontaneitymust influence spontaneity

Four Laws of Four Laws of ThermodynamicsThermodynamics

Zeroth Zeroth

law:law:If TIf T11 = T = T22 and T and T22 = T = T33, then T, then T11 = = TT33

First First

law:law:Conservation of energyConservation of energy

Second Second

law:law:You can't get something from You can't get something from nothing; in fact, you can't even nothing; in fact, you can't even break evenbreak even

Third Third Law:Law:

When there is no motion & When there is no motion & perfect arrangement, perfect arrangement, there is there is no disorder, such as at no disorder, such as at absolute zeroabsolute zero

EnthalpyEnthalpy

HH H H = heat content = heat content HHf f = heat of formation= heat of formation

HHrxn rxn = heat of reaction (which is the = heat of reaction (which is the sum of sum of HHff))

HHoo = the "= the "oo" means that conditions " means that conditions are at standard, which for are at standard, which for thermodynamics is at 25thermodynamics is at 25ooC and C and 101.3 kPa101.3 kPa

Hess’ Law Hess’ Law HHrxnrxn = [= [nnHHff ]]productsproducts - [ - [nnHHff ]]reactantsreactants Consult a Thermodynamics chart; be Consult a Thermodynamics chart; be careful of careful of

the units and phasethe units and phase (s, cr, l, aq, g) (s, cr, l, aq, g) ie: 2NOie: 2NO(g)(g) + O + O2(g)2(g) 2NO 2NO2(g)2(g)

2(+90.25kJ/mol) 0 2(+33.18kJ/mol)2(+90.25kJ/mol) 0 2(+33.18kJ/mol) 2(-90.25kJ/mol) + 2(+33.18kJ/mol)2(-90.25kJ/mol) + 2(+33.18kJ/mol) -180.5kJ/mol + 66.36kJ/mol -180.5kJ/mol + 66.36kJ/mol

= -114.1kJ/mol = = -114.1kJ/mol = HHrxnrxn

And since this is a negative value, it is an EXOTHERMIC REACTION

Why is this zero?

The sum of…

Coefficients from the BCE

Sign of Sign of H H

--HH ExothermicExothermic Favors Favors spontaneitspontaneityy

++HH EndothermEndothermicic

Does not Does not favor favor spontaneitspontaneityy

EntropyEntropy

SS S S = a measure of the amount of = a measure of the amount of

disorder in a system (randomness)disorder in a system (randomness) it is natural for things to it is natural for things to become more become more

disordereddisordered, and takes energy to make , and takes energy to make things ordered againthings ordered again

SSrxnrxn = [= [nnSS ] ]productsproducts - [ - [nnSS ] ]reactantsreactants

solids have solids have low low SS; ; gases have gases have high high SS

EntropyEntropy

Sign of Sign of S S

++SS More More disorderdisorder

Favors Favors spontaneitspontaneityy

--SS More More order order (less (less disorder)disorder)

Does not Does not favor favor spontaneitspontaneityy

Entropy ProblemEntropy Problem Again,Again,SSrxnrxn = [= [nnSS ] ]productsproducts - -

[[nnSS ] ]reactantsreactants

NN22HH4(l)4(l) + 2H + 2H22OO2(l)2(l) N N2(g)2(g) + + 4H4H22OO(g)(g)

(-121.2J/K) + 2(-109.6J/K) + 191.61J/K + (-121.2J/K) + 2(-109.6J/K) + 191.61J/K + 4(188.8J/K)4(188.8J/K)

NOTE: Since we know it’s Products Minus Reactants, we’ve already changed the signs of the reactants.

S = +606.4 J/K

A high positive value would be expected when solids turn to liquids or liquids turn to gas since +S means more disorder

THE EFFECT OF THE SIGNS THE EFFECT OF THE SIGNS OF OF H and H and S ON S ON

SPONTANEOUS CHANGE:SPONTANEOUS CHANGE:SITUATIONSITUATION SIGNS OF SIGNS OF H AND H AND SS COMMENTCOMMENT

11 H = - H = - S = +S = + Reaction occursReaction occurs

22 H = + H = + S = -S = - Reaction doesn’t Reaction doesn’t occuroccur

33 H = + H = + S = + S = + or or

H = - H = - S = -S = -

Consult GibbsConsult Gibbs

Gibbs Free EnergyGibbs Free Energy

GG the sign of the value for Gibbs Free the sign of the value for Gibbs Free

Energy tells us:Energy tells us: ++GG reaction reaction is not spontaneousis not spontaneous --GG reaction reaction is spontaneousis spontaneous GGrxnrxn = [= [nnGG]]productsproducts - [ - [nnGG]]reactantsreactants

G = G = H - TH - TS (temp in kelvins)S (temp in kelvins)

Notice the language “is not spontaneous” and “is spontaneous”.

This is definitive!

Gibbs Free Energy ProblemGibbs Free Energy Problem

ie: Pbie: Pb(s)(s) + ½ O + ½ O2(g)2(g) PbOPbO(s)(s) HHoo

rxnrxn = -215 kJ = -215 kJ SSoo

rxnrxn = -0.092 kJ/K = -0.092 kJ/K Is the rxn spontaneous?Is the rxn spontaneous? G = G = -215-215 - - [(298 K) (-0.092kJ/K][(298 K) (-0.092kJ/K] G = G = -188 kJ; -188 kJ; yes, the rxn yes, the rxn is spontaneousis spontaneous Again, it doesn’t just say “favors”Again, it doesn’t just say “favors”

Remember: This means at “standard”, which for temp, is 25oC or 298 K

Sign of Sign of G G

++GG Reaction Reaction Does not Does not OccurOccur

--GG Reaction Reaction OccursOccurs

Bond EnergyBond Energy

the change in enthalpy when a chemical the change in enthalpy when a chemical rxn occurs is due primarily to the rxn occurs is due primarily to the energy energy required to break the chemical bonds in required to break the chemical bonds in reactantsreactants and the and the energy produced by energy produced by forming the chemical bonds in the productsforming the chemical bonds in the products (breaking of bond = "bond dissociation") (breaking of bond = "bond dissociation")

all bond dissociations are all bond dissociations are ENDOthermicENDOthermic because energy must be supplied to break because energy must be supplied to break bonds; therefore, bonds; therefore, all BE's are + valuesall BE's are + values

Reactions and Bond EnergyReactions and Bond Energy

BEBErxnrxn = [= [nnBE ]BE ]reactants reactants - [- [nnBE]BE]productsproducts

the sign of the value for Bond the sign of the value for Bond Energy are related to enthalpy of Energy are related to enthalpy of formation, so the same formation, so the same information is giveninformation is given

Factors that Contribute Factors that Contribute to Bond Strength:to Bond Strength:

1) polarity1) polarity - a polar bond is stronger than - a polar bond is stronger than nonpolar - in general, the > the diff in e.n. nonpolar - in general, the > the diff in e.n. b/w bonded atoms, b/w bonded atoms, the more polar the bond the more polar the bond & the > the BE& the > the BE

2) # of e- pairs b/w bonded atoms2) # of e- pairs b/w bonded atoms - multiple - multiple bonds have > BE than single bondsbonds have > BE than single bonds

3) Effects related to sizes of atoms3) Effects related to sizes of atoms - If nuclei - If nuclei cannot come very close together, shared e- cannot come very close together, shared e- pair will not be effective in bonding them. pair will not be effective in bonding them. Far from nucleus bond will be weakFar from nucleus bond will be weak

Calculate Enthalpy with Calculate Enthalpy with B.E.’sB.E.’s

STEP 1 – You must know the B.C.E. 2 ClF2 ClF3(g)3(g) + 2O + 2O2(g)2(g) ClCl22OO(g)(g) + 3OF + 3OF2(g)2(g)

STEP 2 – Draw the molecules involved, with the correct numbers and STEP 2 – Draw the molecules involved, with the correct numbers and types of bonds.types of bonds.

STEP 3 – Look up Bond Energies for each. Remember, the BE STEP 3 – Look up Bond Energies for each. Remember, the BE formula has Reactants before products, not like formula has Reactants before products, not like H, H, S, S, G formulas!G formulas!

2(3*255) + 2(498) + 2 (-205) + 3(2*-184) = 1012 kJ = Hrxn