THERMODYNAMICS Chapter 19

SPONTANEOUS PROCESS  A process that occurs without ongoing outside intervention. Examples Nails rusting outdoors

Ice melting at room temperature Expansion of gas into an evacuated space Formation of water from O2(g) and H2(g): 2H2(g) + O2(g) 2H2O(g)

Why are some processes spontaneous and others not? We know that temperature has an effect on the spontaneity of a process. e.g. T>0oC  ice melts  spontaneous at this temp. H2O (s)  H2O (l) T<0oC  water freezes  spontaneous at this temp. H2O (l)  H2O (s) T=0oC  water and ice in equilibrium H2O (l) H2O (s)

Exothermic processes tend to be spontaneous. Example Rusting of nail - SPONTANEOUS! 4Fe(s) + 3O2(g)  Fe2O3(s) H = - 822.2 kJ.mol-1 Formation of water - SPONTANEOUS! 2H2(g) + O2(g) 2H2O(l) H = - 285.8 kJ.mol-1 Pt cat

However, the dissolution of ammonium nitrate is also spontaneous, but it is also endothermic. NH4NO3(s)  NH4+(aq) + NO3-(aq) H = +25.7 kJ.mol-1 So is: 2N2O5(s)  4NO2(g) + 2O2(g) H = +109.5 kJ.mol-1  a process does not have to be exothermic to be spontaneous.  something else besides sign of H must contribute to determining whether a process is spontaneous or not.

ENTROPY (S) That something else is:  extent of disorder! More disordered  larger entropy Entropy is a state function S = Sfinal - Sinitial Units: J K-1mol-1

Entropy’s effects on the mind

Examples of spontaneous processes where entropy increases: Dissolution of ammonium nitrate: NH4NO3(s)  NH4+(aq) + NO3-(aq) H = +25.7 kJ.mol-1 Decomposition of dinitrogen pentoxide: 2N2O5(s)  4NO2(g) + 2O2(g) H = +109.5 kJ.mol-1

However, entropy does not always increase for a spontaneous process At room temperature: Spontaneous Non-spontaneous

SECOND LAW OF THERMODYNAMICS The entropy of the universe increases in any spontaneous process. Suniverse = Ssystem + Ssurroundings Spontaneous process: Suniverse > 0 Process at equilibrium: Suniverse = 0 Thus Suniv is continually increasing! Suniv must increase during a spontaneous process, even if Ssyst decreases.

Q: What is the connection between sausages and the second law of thermo? A: Because of the 2nd law, you can put a pig into a machine and get sausage, but you can't put sausage into the machine and get the pig back.

Rusting nail = spontaneous process Suniv>0 For example: Rusting nail = spontaneous process 4Fe(s) + 3O2(g)  2Fe2O3(s) Ssyst<0 BUT reaction is exothermic,  entropy of surroundings increases as heat is evolved by the system thereby increasing motion of molecules in the surroundings.  Ssurr>0 +ve -ve +ve Thus for Suniv = Ssyst + Ssurr >0 Ssurr > Ssyst

Special circumstance = Isolated system: Does not exchange energy nor matter with surroundings Ssurr = 0 Spontaneous process: Ssyst > 0 Process at equilibrium: Ssyst = 0 Spontaneous  Thermodynamically favourable (Not necessarily occur at observable rate.) Thermodynamics  direction and extent of reaction, not speed.

EXAMPLE State whether the processes below are spontaneous, non-spontaneous or in equilibrium: CO2 decomposes to form diamond and O2(g) Water boiling at 100oC to produce steam in a closed container Sodium chloride dissolves in water NON-SPONTANEOUS EQUILIBRIUM SPONTANEOUS

MOLECULAR INTEPRETATION OF S Decrease in number of gaseous molecules  decrease in S e.g. 2NO(g) + O2(g)  2NO2(g) 3 moles gas 2 moles gas

Molecules have 3 types of motion: Translational motion - Entire molecule moves in a direction (gas > liquid > solid) Vibrational motion – within a molecule Rotational motion – “spinning” Greater the number of degrees of freedom  greater entropy

Decrease in temperature  decrease in thermal energy  decrease in translational, vibrational and rotational motion  decrease in entropy As the temperature keeps decreasing, these motions “shut down”  reaches a point of perfect order.

EXAMPLE Which substance has the great entropy in each pair? Explain. C2H5OH(l) or C2H5OH(g) 2 moles of NO(g) or 1.5 moles of NO(g) 1 mole O2(g) at STP or 1 mole NO2(g) at STP

THIRD LAW OF THERMODYNAMICS The entropy of a pure crystalline substance at absolute zero is zero. S(0 K) = 0  perfect order

Ginsberg's Theorem (The modern statement of the three laws of thermodynamics) 1. You can't win. 2. You can't even break even. 3. You can't get out of the game.

Entropy increases for s  l  g

EXAMPLE Predict whether the entropy change of the system in each reaction is positive or negative. CaCO3(s)  CaO(s) + CO2(g) 2SO2(g) + O2(g)  2SO3(g) N2(g) + O2(g)  2NO(g) H2O(l) at 25oC  H2O(l) at 55oC +ve -ve 3 mol gas  2 mol gas ? Can’t predict, but it is close to zero 2 mol gas  2 mol gas +ve Increase thermal energy

Standard molar entropy (So) = molar entropy for substances in their standard state NOTE So  0 for elements in their standard state So(gas) > So(liquid) > So(solid) So generally increases with increasing molar mass So generally increases with increasing number of atoms in the formula of the substance

Calculation of S for a reaction Stoichiometric coefficients (So from tabulated data)

EXAMPLE Calculate So for the synthesis of ammonia from N2(g) and H2(g): N2(g) + 3H2(g)  2NH3(g) So/J.K-1.mol-1 N2(g) 191.5 H2(g) 130.6 NH3(g) 192.5

Calculation of S for the surroundings For a process that occurs at constant temperature and pressure, the entropy change of the surroundings is: (T &P constant)

GIBB’S FREE ENERGY (G)

 - TSuniv = - TSsys + Hsys Defined as: G = H – TS - state function - extensive property Suniv = Ssys + Ssurr At constant T and P: Suniv = Ssys - Hsys T  - TSuniv = - TSsys + Hsys (at constant T & P) G = H – TS

We know: Spontaneous process: Suniv > 0 Process at equilibrium: Suniv = 0 Therefore: Spontaneous process: Process at equilibrium: < 0 -TSuniv -TSuniv = 0 Gsyst = -TSuniv

G = H – TS Spontaneity involves S H T Spontaneity is favoured by increasing S and H is large and negative.

G allows us to predict whether a process is spontaneous or not (under constant temperature and pressure conditions): G < 0  spontaneous in forward direction G > 0  non-spontaneous in forward direction/spontaneous in reverse direction G = 0  at equilibrium

But nothing about rate

Standard free energy (Go) Go = Ho – TSo Standard states: Gas - 1 atm Solid - pure substance Liquid - pure liquid Solution - Concentration = 1M Gfo = 0 kJ/mol for elements in their standard states

Tabulated data of Gfo can be used to calculate standard free energy change for a reaction as follows: Stoichiometric coefficients

Substance H (kJ mol-1) S (J K-1 mol-1) G (kJ mol-1) S (J K-1 mol-1) AgS +42.6 I2(s) +116.1 AgCl(s) 127.1 +96.2 -109.8 I2(g) +62.4 +260.6 +19.4 Al(s) 28.32 MgO(s) -601.5 +27.0 -569.2 AlCl3(s) -704.2 +110.7 -628.8 MnO2(s) -520.0 +53.1 -465.2 Al2O3(s) -1669.8 +51.0 -1576.5 N2(g) +191.5 Br2() +152.2 N2O4(g) +9.3 +304.2 +97.8 BrF3(g) -255.6 +292.4 -229.5 Na(s) +51.3 C(g) +716.7 +158.0 +671.3 NaF(s) -569.0 -546.3 C(graphite) +5.8 NaCl(s) -411.1 +72.4 -384.3 C(diamond) +1.9 +2.4 +2.9 NaBr(s) -361.1 +87.2 -349.1 CO(g) -110.5 +197.6 -137.2 NaI(s) -287.8 +98.5 -282.4 CO2(g) -393.5 +213.7 -394.4 NaOH(s) -425.6 +64.5 -379.5 CH4(g) -74.5 +186.1 -50.8 NH3(g) -46.2 +192.7 -16.4 C3H8(g) -103.8 +269.9 -23.4 N2H4() +50.6 +121.2 +149.2 Ca(s) 41.4 NO(g) +90.3 +210.6 +86.6 CaO(s) -635.1 +38.1 -603.5 NO2(g) +33.2 +240.0 CaCO3(s)(calcite) -1206.9 +92.9 -1128.8 HNO3() -174.1 +155.6 -80.8 Cl2(g) +223.0 O2(g) +205.0 Cu(s) O3(g) +142.7 +238.8 +163.2 F2(g) +202.7 P(s)(white) +41.1 Fe(s) +27.3 P4O10(s) -3010.0 +231.0 -2724.0 Fe2O3(s)(hematite) -824.2 +87.4 -742.2 PCl3(g) -287.0 +311.7 -267.8 H(g) +218.0 +114.6 +203.3 PCl5(g) -374.9 +364.5 -305.0 H2(g) +130.6 PbO2(s) -277.4 +68.6 -217.4 HCl(g) -92.3 +186.8 -95.3 S(s)(orthorhombic) +32.0 HF(g) -271.1 +173.8 -273.2 H2S(g) -20.6 +205.6 -33.4 HI(g) +26.4 +206.5 +1.6 SiO2(s)(quartz) -910.7 +41.5 -856.3 HBr(g) -36.4 +198.6 -53.5 SiCl4() -687.0 +239.7 -619.9 HCN(g) +135.1 +201.7 +124.7 SO2(g) -296.8 +248.1 -300.2 H2O(g) -241.8 +188.7 -228.6 SO3(g) -395.7 +256.6 -371.1 H2O() -285.8 +70.0 -237.2 Zn(s) +41.6 H2O2() -187.8 +109.6 -120.4 ZnO(s) -350.5 +43.6 -320.5 Hg() +75.9

EXAMPLE The combustion of propane gas occurs as follows: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Using thermodynamic data for Go, calculate the standard free energy change for the reaction at 298 K. Gfo/kJ.mol-1 C3H8(g) -23.47 CO2(g) -394.4 H2O(g) -228.57 H2O(l) -237.13

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Gfo/kJ.mol-1 C3H8(g) -23.47 CO2(g) -394.4 H2O(g) -228.57 H2O(l) -237.13 Go = [3(-394.4) + 4(-237.13)] – [(-23.47) – 5(0)] Go = -2108 kJ

Free Energy and Temperature How is change in free energy affected by change in temperature? G = H – TS H S -TS G = H - TS + - + + at all temp - + - - at all temp - at high temp + at low temp + + - + at high temp - at low temp - - - +

Note: For a spontaneous process the maximum useful work that can be done by the system: wmax = G “free energy” = energy available to do work

EXAMPLE The combustion of propane gas occurs as follows at 298K: C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho. b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Was your prediction correct?

 Ho – TSo will be less negative than Ho C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho. Go = Ho – TSo -ve -ve 6 moles gas  3 moles gas  – TSo > 0  Ho – TSo will be less negative than Ho i.e. Go will be less negative than Ho

Go = (-2220 kJ) – (298 K)(-374.46x10-3 kJ.K-1.mol-1) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Ho = -2220 kJ.mol-1 b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Was your prediction correct? Go = Ho – TSo Go = (-2220 kJ) – (298 K)(-374.46x10-3 kJ.K-1.mol-1) Go = -2108 kJ.mol-1  Prediction was correct.

EXAMPLE (TUT no. 5a) At what temperature is the reaction below spontaneous? AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s) Ho = 851.5 kJ; So = 38.5 J K-1

Assume H and S do not vary that much with temperature. At what temperature is the reaction below spontaneous? AI2O3(s) + 2Fe(s)  2AI(s) + Fe2O3(s) Ho = 851.5 kJ; So = 38.5 J K-1 Go = Ho – TSo Assume H and S do not vary that much with temperature. Set G = 0 equilibrium 0 = (851.5 kJ) – T(38.5x10-3 kJ.K-1) T = 22117 K at equilibrium For spontaneous reaction: G < 0  T > 22117 K

Free Energy and the equilibrium constant Recall: G = Change in Gibb’s free energy under standard conditions. G can be calculated from tabulated values. BUT most reactions do not occur under standard conditions. Calculate G under non-standard conditions: Q = reaction quotient R = gas constant = 8.314 J.K-1.mol-1

Under standard conditions: (1 M, 1 atm) Q = 1  ln Q = 0  G = Go At equilibrium: G = 0 and Q = Keq  If Go < 0  ln Keq > 0  Keq > 1 i.e. the more negative Go, the larger K etc. Go < 0  Keq > 1 Go > 0  Keq < 1 Go = 0  Keq = 1 Also

Calculate K for the following reaction at 25oC: 2H2O(l) 2H2(g) + O2(g) EXAMPLE Calculate K for the following reaction at 25oC: 2H2O(l) 2H2(g) + O2(g) Gfo/kJ.mol-1 H2O(g) -228.57 H2O(l) -237.13 Go = [2(0) + (0)] – [2(-237.13)] Go = 474.26 kJ.mol-1 474.26x103 J.mol-1 = -(8.314 J.K-1.mol-1)(298 K) lnK lnK = -191.4 K = 7.36x10-84