General Information "Thermodynamics", is the study of the energy changes or transfers accompanying physical and chemical processes "Thermodynamics", is the study of the energy changes or transfers accompanying physical and chemical processes helps us predict whether a rxn will occur under certain conditions helps us predict whether a rxn will occur under certain conditions DO NOT ASSUME THAT ALL REACTIONS HAPPEN AUTOMATICALLY! DO NOT ASSUME THAT ALL REACTIONS HAPPEN AUTOMATICALLY!

Spontaneity "spontaneous" - a process can occur (not necessarily fast) "spontaneous" - a process can occur (not necessarily fast) "non-spontaneous" - a process cannot occur "non-spontaneous" - a process cannot occur

General Information many spontaneous rxns release heat energy = "exothermic"; many spontaneous rxns release heat energy = "exothermic"; products have less E than reactants ("downhill") (see diagram) products have less E than reactants ("downhill") (see diagram) some rxns absorb energy = "endothermic" = products have more energy than the reactants ("uphill") (see diagram) some rxns absorb energy = "endothermic" = products have more energy than the reactants ("uphill") (see diagram)

Endothermic Exothermic E a = Activation Energy a = E a b = E a for reverse reaction c = enthalpy change

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

Four Laws of Thermodynamics Zerothlaw: If T 1 = T 2 and T 2 = T 3, then T 1 = T 3 Firstlaw: Conservation of energy Secondlaw: You can't get something from nothing; in fact, you can't even break even Third Law: When there is no motion & perfect arrangement, there is no disorder, such as at absolute zero

Enthalpy  H  H  H = heat content  H = heat content  H f = heat of formation  H f = heat of formation  H rxn = heat of reaction (which is the sum of  H f )  H rxn = heat of reaction (which is the sum of  H f )  H o = the "o" means that conditions are at standard, which for thermodynamics is at 25 o C and kPa  H o = the "o" means that conditions are at standard, which for thermodynamics is at 25 o C and kPa

Hess’ Law  H rxn = [  n  H f ] products - [  n  H f ] reactants  H rxn = [  n  H f ] products - [  n  H f ] reactants Consult a Thermodynamics chart; be careful of the units and phase (s, cr, l, aq, g) Consult a Thermodynamics chart; be careful of the units and phase (s, cr, l, aq, g) ie: 2NO (g) + O 2(g)  2NO 2(g) ie: 2NO (g) + O 2(g)  2NO 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) kJ/mol kJ/mol kJ/mol kJ/mol = kJ/mol =  H rxn = kJ/mol =  H rxn 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  H -H-H-H-HExothermic Favors spontaneity +H+H+H+HEndothermic Does not favor spontaneity

Entropy  S  S  S = a measure of the amount of disorder in a system (randomness)  S = a measure of the amount of disorder in a system (randomness) it is natural for things to become more disordered, and takes energy to make things ordered again it is natural for things to become more disordered, and takes energy to make things ordered again  S rxn = [  n  S ] products - [  n  S ] reactants  S rxn = [  n  S ] products - [  n  S ] reactants solids have low  S; gases have high  S solids have low  S; gases have high  S

Entropy

Sign of  S +S+S+S+S More disorder Favors spontaneity -S-S-S-S More order (less disorder) Does not favor spontaneity

Entropy Problem Again,  S rxn = [  n  S ] products - [  n  S ] reactants Again,  S rxn = [  n  S ] products - [  n  S ] reactants N 2 H 4(l) + 2H 2 O 2(l)  N 2(g) + 4H 2 O (g) N 2 H 4(l) + 2H 2 O 2(l)  N 2(g) + 4H 2 O (g) (-121.2J/K) + 2(-109.6J/K) J/K + 4(188.8J/K) (-121.2J/K) + 2(-109.6J/K) J/K + 4(188.8J/K) NOTE: Since we know it’s Products Minus Reactants, we’ve already changed the signs of the reactants. S = 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 OF  H and  S ON SPONTANEOUS CHANGE: SITUATION SIGNS OF  H AND  S COMMENT COMMENT 1  H = -  S = + Reaction occurs 2  H = +  S = - Reaction doesn’t occur 3  H = +  S = + or  H = -  S = - Consult Gibbs

Gibbs Free Energy  G  G the sign of the value for Gibbs Free Energy tells us: the sign of the value for Gibbs Free Energy tells us: +  Greaction is not spontaneous +  Greaction is not spontaneous -  Greaction is spontaneous -  Greaction is spontaneous  G rxn = [  n  G] products - [  n  G] reactants  G rxn = [  n  G] products - [  n  G] reactants  G =  H - T  S (temp in kelvins)  G =  H - T  S (temp in kelvins) Notice the language “is not spontaneous” and “is spontaneous”. This is definitive!

Gibbs Free Energy Problem ie: Pb (s) + ½ O 2(g)  PbO (s)  H o rxn = -215 kJ  S o rxn = kJ/K ie: Pb (s) + ½ O 2(g)  PbO (s)  H o rxn = -215 kJ  S o rxn = kJ/K Is the rxn spontaneous? Is the rxn spontaneous?  G = [(298 K) (-0.092kJ/K]  G = [(298 K) (-0.092kJ/K]  G = -188 kJ;  G = -188 kJ; yes, the rxn is spontaneous yes, the rxn is spontaneous Again, it doesn’t just say “favors” Again, it doesn’t just say “favors” Remember: This means at “standard”, which for temp, is 25 o C or 298 K

Sign of  G +G+G+G+G Reaction Does not Occur -G-G-G-G Reaction Occurs

Bond Energy the change in enthalpy when a chemical rxn occurs is due primarily to the energy required to break the chemical bonds in reactants and the energy produced by forming the chemical bonds in the products (breaking of bond = "bond dissociation") the change in enthalpy when a chemical rxn occurs is due primarily to the energy required to break the chemical bonds in reactants and the energy produced by forming the chemical bonds in the products (breaking of bond = "bond dissociation") all bond dissociations are ENDOthermic because energy must be supplied to break bonds; therefore, all BE's are + values all bond dissociations are ENDOthermic because energy must be supplied to break bonds; therefore, all BE's are + values

Reactions and Bond Energy BE rxn = [  n  BE ] reactants - [  n  BE] products BE rxn = [  n  BE ] reactants - [  n  BE] products the sign of the value for Bond Energy are related to enthalpy of formation, so the same information is given the sign of the value for Bond Energy are related to enthalpy of formation, so the same information is given

Factors that Contribute to Bond Strength: 1) polarity - a polar bond is stronger than nonpolar - in general, the > the diff in e.n. b/w bonded atoms, the more polar the bond & the > the BE 1) polarity - a polar bond is stronger than nonpolar - in general, the > the diff in e.n. b/w bonded atoms, the more polar the bond & the > the BE 2) # of e- pairs b/w bonded atoms - multiple bonds have > BE than single bonds 2) # of e- pairs b/w bonded atoms - multiple bonds have > BE than single bonds 3) Effects related to sizes of atoms - If nuclei cannot come very close together, shared e- pair will not be effective in bonding them. Far from nucleus bond will be weak 3) Effects related to sizes of atoms - If nuclei cannot come very close together, shared e- pair will not be effective in bonding them. Far from nucleus bond will be weak

Calculate Enthalpy with B.E.’s STEP 1 – You must know the B.C.E. 2 ClF 3(g) + 2O 2(g)  Cl 2 O (g) + 3OF 2(g) 2 ClF 3(g) + 2O 2(g)  Cl 2 O (g) + 3OF 2(g) STEP 2 – Draw the molecules involved, with the correct numbers and types of bonds. STEP 2 – Draw the molecules involved, with the correct numbers and types of bonds. STEP 3 – Look up Bond Energies for each. Remember, the BE formula has Reactants before products, not like  H,  S,  G formulas! STEP 3 – Look up Bond Energies for each. Remember, the BE formula has Reactants before products, not like  H,  S,  G formulas! 2(3*255) + 2(498) + 2 (-205) + 3(2*-184) = 1012 kJ =  H rxn