Chapter 17. Thermodynamics: Spontaniety, Entropy and Free Energy First Law of Thermodynamics Enthalpy -DH, DHf Second Law of Thermodynamics Third Law of Thermodynamics Entropy DS, Free Energy DG Spontaneity

Thermochemistry Heat changes during chemical reactions Thermochemical equation. eg. H2 (g) + O2 (g) ---> 2H2O(l) DH =- 256 kJ; DH is called the enthalpy of reaction. if DH is + reaction is called endothermic if DH is - reaction is called exothermic

Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

Why is it necessary to divide Universe into System and Surrounding Universe = System + Surrounding

What is the internal energy change (DU) of a system? DU is associated with changes in atoms, molecules and subatomic particles Etotal = Eke + E pe + DU DU = heat (q) + w (work) DU = q + w DU = q -P DV; w =- P DV

What forms of energy are found in the Universe? mechanical thermal electrical nuclear mass: E = mc2 others yet to discover

What is 1st Law of Thermodynamics Eenergy is conserved in the universe All forms of energy are inter-convertible and conserved Energy is neither created nor destroyed.

What exactly is DH? Heat measured at constant pressure qp Chemical reactions exposed to atmosphere and are held at a constant pressure. Volume of materials or gases produced can change. ie: work = -PDV DU = qp + w; DU = qp -PDV qp = DU + PDV; w = -PDV DH = DU + PDV; qp = DH(enthalpy )

How do you measure DU? Heat measured at constant volume qv Chemical reactions take place inside a bomb. Volume of materials or gases produced can not change. ie: work = -PDV= 0 DU = qv + w qv = DU + o; w = 0 DU = qv = DU(internal energy )

What is Hess's Law of Summation of Heat? To heat of reaction for new reactions. Two methods? 1st method: new DH is calculated by adding DHs of other reactions. 2nd method: Where DHf ( DH of formation) of reactants and products are used to calculate DH of a reaction.

Method 1: Calculate DH for the reaction: SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) DH = ? Other reactions: SO2(g) ------> S(s) + O2(g) DH = 297kJ H2SO4(l)------> H2(g) + S(s) + 2O2(g) DH = 814 kJ H2(g) +1/2O2(g) -----> H2O(g) DH = -242 kJ

SO2(g) ------> S(s) + O2(g);DH1 = 297 kJ - 1 H2(g) + S(s) + 2O2(g) ------> H2SO4(l) DH2 = -814 kJ - 2 H2O(g) ----->H2(g) + 1/2 O2(g) DH3 = +242 kJ - 3 ______________________________________ SO2(g) + 1/2 O2(g) + H2O(g) -----> H2SO4(l) ªH = DH1 + DH2 + DH3 ªH = +297 - 814 + 242 ªH = -275 kJ

Calculate Heat (enthalpy) of Combustion: 2nd method C7H16(l) + 11 O2(g) -----> 7 CO2(g) + 8 H2O(l) ; DH = ? DHf (C7H16) = -198.8 kJ/mol DHf (CO2) = -393.5 kJ/mol DHf (H2O) = -285.9 kJ/mol DHf O2(g) = 0 (zero) What method? 2nd method

DH = [Sn ( DHof) Products] - [Sn (DHof) reactants] = [-2754.5 - 2287.2] - [-198.8] = -5041.7 + 198.8 = -4842.9 kJ = -4843 kJ

Why is DHof of elements is zero? DHof, Heat formations are for compounds Note: DHof of elements is zero

What is 2nd Law of Thermodynamics Entropy (DS) of the Universe increases during spontaneous process. What is entropy Entropy (DS) : A measure of randomness or disorder of a system. Spontaneous Process: A process that occurs without outside intervention.

Definitions The Universe: The sum of all parts under consideration. System: Part of the Universe we are interested in and a change is taking place. Surrounding: Part of the Universe we are not interested in or can not observe.

Entropy DS DSuniv = entropy of the Universe DSsys = entropy of the System DSsurr = entropy of the Surrounding DSuniv = DSsys + DSsurr

DSuniv = DSsys + DSsurr DSuniv DSsys DSsurr + + + + +(DSsys>DSsurr) - + - + (DSsurr>DSsys)

Standard entropies at 25oC Substance So (J/K.mol) Substance So (J/K.mol) C (diamond) 2.37 HBr (g) 198.59 C (graphite) 5.69 HCl (g) 186.80 CaO (s) 39.75 HF (g) 193.67 CaCO3 (s) 92.9 HI (g) 206.33 C2H2 (g) 200.82 H2O (l) 69.91 C2H4 (g) 219.4 H2O (g) 188.72 C2H6 (g) 229.5 NaCl (s) 72.12 CH3OH (l) 127 O2 (g) 205.03 CH3OH (g) 238 SO2 (g) 248.12 CO (g) 197.91 SO3 (g) 256.72

Entropy Change Entropy (DS) normally increase (+) for the following changes: i) Solid ---> liquid (melting) ii) Liquid ---> gas iii) Solid ----> gas iv) Increase in temperature v) Increasing in pressure(constant volume, and temperature) vi) Increase in volume ( at constant temperature and pressure)

Free energy Gibbs free energy (G) can be used to describe the energy changes of a system. As usual, it is the changes in free energy that are of interest - DG. At constant temperature and pressure, DG is equal to: DG = DH - TDS T is the Kelvin temperature.

What is DG ? Free Energy DG = - TDSuni. DG = DH - TDS.

How do you get: DG = DH - TDS. Dsuniv= Dssys +Dssurr Dssurr = -DHsys/T) DSuniv = DSsys -DHsys/T DSuniv x T = TDSsys -DHsys : x T -DSuniv x T = -TDSsys +DHsys : x -1 -DSuniv x T = DHsys-TDSsys -DSuniv x T = DG; DG = DHsys-TDSsys or DG = DH - TDS.

What DG means If DG is - for a change it will take place spontaneously If DG is + for a change it will not take place If DG is 0 for a change it will be in equilibrium

Free energy The sign of DG indicates whether a reaction will occur spontaneously. + Not spontaneous 0 At equilibrium - Spontaneous The fact that the effect of DS will vary as a function of temperature is important. This can result in changing the sign of DG.

Predict DG at different DH, DS, T DG = DH - T DS. - - all + + + all - - /+ + high/low + -/+ - low/high -

Predict the spontaneity of the following processes from DH and DS at various temperatures. a)DH = 30 kJ, DS = 6 kJ, T = 300 K b)DH = 15 kJ,DS = -45 kJ,T = 200 K

DH = 30 kJ DS = 6 kJ T = 300 K DG = DHsys-TDSsys or DG = DH - TDS. DH = 30 kJ DS = 6 kJ T = 300 K DG = 30 kJ - (300 x 6 kJ) = 30 -1800 kJ DG = -1770 kJ b) DH = 15 kJ DS = -45 kJ T = 200 K DG = DHsys-TDSsys or DG = DH - TDS. DH = 15 kJ DS = -45 kJ T = 200 K DG = 15 kJ -[200 (-45 kJ)] = 15 kJ -(-9000) kJ DG = 15 + 9000 kJ = 9015 kJ

Calculation of DGo We can calculate DGo values from DHo and DSo values at a constant temperature and pressure. Example. Determine DGo for the following reaction at 25oC Equation N2 (g) + 3H2 (g) 2NH3 (g) DHfo, kJ/mol 0.00 0.00 -46.11 So, J/K.mol 191.50 130.68 192.3

Qualitative prediction of DS of Chemical Reactions Look for (l) or (s) --> (g) If all are gases: calculate Dn = Sn(gaseous prod.) -S n(gaseous reac.) N2 (g) + 3 H2 (g) --------> 2 NH3 (g) Dn = 2 - 4 = -2 If Dn is - DS is negative (decrease in S) If Dn is + DS is positive (increase in S)

Predict DS! 2 C2H6(g) + 7 O2(g)--> 4 CO2(g) + 6H2O(g). 2 CO(g) + O2(g)-->2 CO2(g). HCl(g) + NH3(g)-->NH4Cl(s). H2(g) + Br2(l) --> 2 HBr(g).

Calculating DS of reactions Based on Hess’s Law second method: DSo=SDSo(prod.) - SDS o(react.)

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) Calculation of standard entropy changes Example. Calculate the DSorxn at 25 oC for the following reaction. CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g) Substance So (J/K.mol) CH4 (g) 186.2 O2 (g) 205.03 CO2 (g) 213.64 H2O (g) 188.72

Calculate the DS for the following reactions using: D So = 3 D So (products) - 3 D S o(reactants) a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K mole b) 2NH 3 (g) + 3N2O (g) --------> 4N2 (g) + 3 H2O (l) D So[ NH3(g)] = 193 J/K mole ; D So [N2(g)] = 192 J/K mole; D So [N2O(g)] = 220 J/K mole; D S[ H2O(l)] = 70 J/K mole

a) 2SO2 (g) + O2 (g) ------> 2SO 3(g) D So [SO2(g)] = 248 J/K mole ; D So [O2(g)] = 205 J/K mole; D So [SO3(g)] = 257 J/K mole DSo 496 205 514 DSo = 3 DSo (products) - 3 DS o(reactants) DSo = [514] - [496 + 205] DSo = 514 - 701 DSo = -187 J/K mole

How do you calculate DG There are two ways to calculate DG for chemical reactions. i) DG = DH - TDS. ii) DGo = S DGof (products) - S DG of (reactants)

Standard free energy of formation DGfo Free energy change that results when one mole of a substance if formed from its elements will all substances in their standard states. DG values can then be calculated from: DGo = SnpDGfo products - SnpDGfo reactants

Standard free energy of formation Substance DGfo Substance DGfo C (diamond) 2.832 HBr (g) -53.43 CaO (s) -604.04 HF (g) -273.22 CaCO3 (s) -1128.84 HI (g) 1.30 C2H2 (g) 209 H2O (l) -237.18 C2H4 (g) 86.12 H2O (g) -228.59 C2H6 (g) -32.89 NaCl (s) -384.04 CH3OH (l) -166.3 O (g) 231.75 CH3OH (g) -161.9 SO2 (g) -300.19 CO (g) -137.27 SO3 (g) -371.08 All have units of kJ/mol and are for 25 oC

Calculate the DG value for the following reactions using: D Go = SD Gof (products) - S D Gof (reactants) N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ? D Gfo[ N2O5 (g) ] = 134 kJ/mole ; D Gfo [H2O(g)] = -237 kJ/mole; DGfo[ HNO3(l) ] = -81 kJ/mole N2O5 (g) + H2O(l) ------> 2 HNO3(l) ; DGo = ? DGfo 1 x 134 1 x (-237) 2 (-81) 134 -237 -162 DGo = DGof (products) - 3 DGof (reactants) DGo = [-162] - [134 + (-237)] DGo = -162 + 103 DGo = -59 kJ/mole The reaction have a negative DG and the reaction is spontaneous or will take place as written.

How do you calculate DG at different T and P DG = DGo + RT ln Q Q = reaction quotient at equilibrium DG = 0 0 = DGo + RT ln K DGo = - RT ln K If you know DGo you could calculate K

Free energy and equilibrium For gases, the equilibrium constant for a reaction can be related to DGo by: DGo = -RT lnK For our earlier example, N2 (g) + 3H2 (g) 2NH3 (g) At 25oC, DGo was -32.91 kJ so K would be: ln K = = = ln K = 13.27; K = 5.8 x 105 DGo -RT -32.91 kJ -(0.008315 kJ.K-1mol-1)(298.2K)

Calculate the D G for the following equilibrium reaction and predict the direction of the change using the equation: DG = D Go + RT ln Q ; [D Gfo[ NH3(g) ] = -17 kJ/mole N2 (g) + 3 H2 (g) W 2 NH3 (g); D G = ? at 300 K, PN2 = 300, PNH3 = 75 and PH2 = 300 N2 (g) + 3 H2 (g) W 2 NH3 (g); DG = ?

To calculate DGo Using DGo = DGof (products) - 3 DGof (reactants) To calculate DGo Using DGo = DGof (products) - 3 DGof (reactants) DGfo[ N2(g) ] = 0 kJ/mole; DGfo[ H2(g) ] = 0 kJ/mole; DGfo[ NH3(g) ] = -17 kJ/mole Notice elements have DGfo = 0.00 similar to DHfo N2 (g) + 3 H2 (g) W 2 NH3 (g); DG = ? DGfo 0 0 2 x (-17) 0 0 -34 DGo = DGof (products) - 3 DGof (reactants) DGo = [-34] - [0 +0] DGo = -34 DGo = -34 kJ/mole

To calculate Q Equilibrium expression for the reaction in terms of partial pressure: N2 (g) + 3 H2 (g) W 2 NH3 (g) p2NH3 K = _________ pN2 p3H2 p2NH3 Q = _________ ; pN2 p3H2 Q is when initial concentration is substituted into the equilibrium expression 752 Q = _________ ; p2NH3= 752; pN2 =300; p3H2=3003 300 x 3003 Q = 6.94 x 10-7

To calculate DGo DG = DGo + RT ln Q DGo= -34 kJ/mole R = 8 To calculate DGo DG = DGo + RT ln Q DGo= -34 kJ/mole R = 8.314 J/K mole or 8.314 x 10-3kJ/Kmole T = 300 K Q= 6.94 x 10-7 DG = (-34 kJ/mole) + ( 8.314 x 10-3 kJ/K mole) (300 K) ( ln 6.94 x 10-7) DG = -34 + 2.49 ln 6.94 x 10-7 DG = -34 + 2.49 x (-14.18) DG = -34 -35.37 DG = -69.37 kJ/mole