CHEMICAL THERMODYNAMICS The first law of thermodynamics: Energy and matter can be neither created nor destroyed; only transformed from one form to another. The energy and matter of the universe is constant. The second law of thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe. The entropy is increasing. The third law of thermodynamics: The entropy of a perfect crystal at 0 K is zero. There is no molecular motion at absolute 0 K.

STATE FUNCTIONS A property of a system which depends only on its present state and not on its pathway.  H - Enthalpy - heat of reaction - q p A measure of heat (energy) flow of a system relative to its surroundings.  H° standard enthalpy  H f ° enthalpy of formation  H° =  n  H f ° (products) -  m  H f ° (reactants)  H =  U + P  V U represents the Internal energy of the particles, both the kinetic and potential energy.  U = q + w

HEAT VS WORK energy transfer as aenergy expanded to result of a temperaturemove an object against differencea force q p w = F x d endothermic (+q)work on a system (+w) exothermic (-q)work by the system (-w) q c = -q h w = -P  V

SPONTANEOUS PROCESSES A spontaneous process occurs without outside intervention. The rate may be fast or slow. Entropy A measure of randomness or disorder in a system. Entropy is a state function with units of J/K and it can be created during a spontaneous process.  S univ =  S sys +  S surr The relationship between  S sys and  S surr  S sys  S surr  S univ Process spontaneous? + + +Yes - - -No (Rx will occur in opposite direction) + - ?Yes, if  S sys >  S surr - + ?Yes, if  S surr >  S sys

Entropy  S = S f - S i  S > q/T  S =  H/T For a reversible (at equilibrium) process  H - T  S < 0 For a spontaneous reaction at constant T & P  H - T  S If the value for  H - T  S is negative for a reaction then the reaction is spontaneous in the direction of the products. If the value for  H - T  S is positive for a reaction then the reaction is spontaneous in the direction of the reactants. (nonspontaneous for products)

S°, S°, S°, Formula J/(molK) Formula J/(molK) Formula J/(molK) Nitrogen SulfurBromine N 2 (g) S 2 (g) Br - (aq) 80.7 NH 3 (g) 193 S(rhombic) 31.9 Br 2 (l) NO(g) S(monoclinic) 32.6Iodine NO 2 (g) SO 2 (g) I - (aq) HNO 3 (aq) 146 H 2 S(g) I 2 (s) Oxygen Fluorine Silver O 2 (g) F - (aq) -9.6 Ag + (aq) 73.9 O 3 (g) F 2 (g) Ag(s) 42.7 OH - (aq) HF(g) AgF(s) 84 H 2 O(g) Chlorine AgCl(s) 96.1 H 2 O(l) 69.9 Cl - (aq) 55.1 AgBr(s) Cl 2 (g) AgI(s) 114 HCl(g) 186.8

S°, S°, S°, Formula J/(molK) Formula J/(molK) Formula J/(molK) Hydrogen CarbonCarbon (continued) H + (aq)0 C(graphite) 5.7 HCN(l) H 2 (g) C(diamond) 2.4 CCl 4 (g) Sodium CO(g) CCl 4 (l) Na + (aq) 60.2 CO 2 (g) CH 3 CHO(g) 266 Na(s) 51.4 HCO 3 - (aq) 95.0 C 2 H 5 OH(l) 161 NaCl(s) 72.1 CH 4 (g) Silicon NaHCO 3 (s) 102 C 2 H 4 (g) Si(s) 18.0 Na 2 CO 3 (s) 139 C 2 H 6 (g) SiO 2 (s) 41.5 Calcium C 6 H 6 (l) SiF 4 (g) 285 Ca 2+ (aq) HCHO(g) 219 Lead Ca(s) 41.6 CH 3 OH(l) 127 Pb(s) 64.8 CaO(s) 38.2 CS 2 (g) PbO(s) 66.3 CaCO 3 (s) 92.9 CS 2 (l) PbS(s) 91.3 HCN(g) 201.7

S° = Standard Entropy = absolute entropy S is usually positive (+) for Substances S can be negative (-) for Ions because H 3 O + is used as zero Predicting sign of  S° (+) cases l. Rx in which molecule broken 2. Rx where increase in mol of gas 3. Process where s  l or s  g or l  g  S° =  n S° (P) -  m S° (R)

APPLICATION OF THE 3RD LAW OF THERMODYNAMICS S° = standard entropy = absolute entropy Predicting the sign of  S° The sign is positive if: 1. Molecules are broken during the Rx 2. The number of moles of gas increases 3. solid  liquid liquid  gas solid  gas an increase in order occurs 1. Ba(OH) 2 8H 2 O + 2NH 4 NO 3(s)  2NH 3(g) + 10H 2 O (l) + Ba(NO 3 ) 2(aq) 2. 2SO (g) + O 2(g)  2SO 3(g) 3. HCl (g) + NH 3(g)  NH 4 Cl (s) 4. CaCO 3(s)  CaO (s) + CO 2(g)

 S° =  n S° (product) -  m S° (reactant) 1. Acetone, CH 3 COCH 3, is a volitale liquid solvent. The standard enthalpy of formation of the liquid at 25 °C is kJ/mol; the same quantity for the vapor is kJ/mol. What is  S when 1.00 mol liquid acetone vaproizes? 2.Calculate  S° at 25° for: a. 2 NiS (s) + 3 O 2(g)  2 SO 2(g) + 2 NiO 9(s) b. Al 2 O 3(s) + 3 H 2(g)  2 Al (s) + 3 H 2 O (g)

GIBBS FREE ENERGY : G G = H - TS describes the temperature dependence of spontaneity Standard conditions (1 atm, if soln=1M & 25°):  G° =  H° - T  S° A process ( at constant P & T) is spontaneous in the direction in which the free energy decreases. 1. Calculate  H°,  S° &  G° for 2 SO 2(g) + O 2(g)  2 SO 3(g) at 25°C & 1 atm

 G f °  G f °  G f ° Formula kJ/molFormula kJ/mol Formula kJ/mol NitrogenSulfur Bromine N 2(g) 0S 2(g) 80.1 Br - (aq) NH 3(g) -16 S (rhombic) 0 Br 2(l) 0 NO (g) S (monoclinic) 0.10 Iodine NO 2(g) 51 SO 2(g) I - (aq) HNO 3(aq) H 2 S (g) -33 I 2(s) 0 OxygenFluorine Silver O 2(g) 0F - (aq) Ag + (aq) 77.1 O 3(g) 163 F 2(g) 0 Ag (s) 0 OH - (aq) HF (g) -275 AgF (s) -185 H 2 O (g) Chlorine AgCl (s) H 2 O (l) Cl - (aq) AgBr (s) Cl 2(g) 0 AgI (s) HCl (g) -95.3

 G f °  G f °  G f ° Formula kJ/molFormula kJ/mol Formula kJ/mol Hydrogen CarbonCarbon (cont.) H + 0C (graphite) 0HCN (l) 121 H 2(g) 0C (diamond) 2.9CCl 4(g) SodiumCO (g) CCl 4(l) Na + (aq) CO 2(g) CH 3 CHO (g) Na (s) 0 HCO 3 - (aq) C 2 H 5 OH (l) NaCl (s) CH 4(g) Silicon NaHCO 3(s) C 2 H 4(g) 68.4Si (s) 0 Na 2 CO 3(s) C 2 H 6(g) -32.9SiO 2(s) CalciumC 6 H 6(l) 124.5SiF 4(g) Ca 2 + (aq) HCHO (g) -110Lead Ca (s) 0 CH 3 OH (l) Pb (s) 0 CaO (s) CS 2(g) 66.9PbO (s) -189 CaCO 3(s) CS 2(l) 63.6PbS (s) HCN (g) 125

STANDARD FREE ENERGY OF FORMATION  G° f The free energy change that occurs when 1 mol of substance is formed from the elements in their standard state. Calculate  G° for: 2 CH 3 OH(g) + 3 O 2 (g)  2 CO 2 (g) + 4 H 2 O(g)

INTERPRETING  G° FOR SPONTANEITY 1.When  G° is very small (less than -10 KJ) the reaction is spontaneous as written. Products dominate.  G°  G°(P) 2. When  G° is very large (greater than 10 KJ) the reaction is non spontaneous as written. Reactants dominate.  G° > 0  G°(R) <  G°(P) 3.When  G° is small (+ or -) at equilibrium then both reactants and products are present.  G° = 0 Q:Ba(OH 2 ) 8 H 2 O (g) + 2 NH 4 NO 3(g)  2 NH 3(g) + 10 H 2 O (l) + Ba(NO 3 ) 3(aq)

 G AND EQUILIBRIUM The equilibrium point occurs at the lowest free energy available to the reaction system. When a substance undergoes a chemical reaction, the reaction proceeds to give the minimum free energy at equilibrium.  G =  G° + RT 1n (Q) at equilibrium:  G = 0  G° = -RT 1n (k)  G° = 0thenK = 1  G° 1  G° > 0thenK < 1 Q: Corrosion of iron by oxygen is 4 Fe (s) + 3 O 2(g)  2 Fe 2 O 3(s) calculate K for this Rx at 25°C.

1.Calculate  Gº at 25ºc Ba SO 4 (s)  Ba 2+ (aq) + SO 4 2- (aq) What is the value for K sp at 25ºC? 2.Calculate K for Zn (s) + 2H + (aq)  Zn 2+ (aq) + H 2 (g) at 25ºc.

 Gº & Spontaneity is dependent on Temperature  G T º =  Hº - T  Sº  Hº  Sº  Gº Spontaneous at all T Non spontaneous at all T - - +/- At Low T= Spontaneous At High T= Nonspontaneous + + +/-At low T= Nonspontaneous At High T= Spontaneous Q. Predict the Spontaneity for H 2 O(s)  H 2 O(l) at -10ºc, 0ºc & 10ºc.

1. At what temperature is the following process spontaneous at 1 Atm? Br 2 (l)  Br 2 (g) What is the normal boiling point for Br 2 (l)? 2. Calculate  Gº & Kp at 35ºc N 2 O 4 (g)  2 No 2 (g) 3. Calculate  Hº,  Sº &  Gº at 25ºc and 650ºc. CS 2 (g) + 4H 2 (g)  CH 4 (g) + 2H 2 S (g) Compare the two values and briefly discuss the spontaneity of the R x at both temperature.