Thermodynamics Ch 19
Thermodynamics Ch 19 CHEM 1200
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This 6 page Class Notes was uploaded by Alexi Martin on Monday February 15, 2016. The Class Notes belongs to CHEM 1200 at Rensselaer Polytechnic Institute taught by Dr. Alexander Ma in Spring 2016. Since its upload, it has received 31 views. For similar materials see Chemistry II in Chemistry at Rensselaer Polytechnic Institute.
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Date Created: 02/15/16
Chapter 19Thermodynamics Defined as the study of energy and changes in the flow of energy Answers the question will the reaction absorb or release heat, will the reaction occur, will it occur spontaneously at a certain temperature? Tells nothing about the time of the reaction Balanced > an increase in randomness, enthalpy change known as delta H ( heat exchange between system and surroundings). 3 Laws 1. You cannot win you cannot create or destroy energy 2. You cannot break even delta S (entropy or the degree of randomness) always greater than 0 3. You cannot get out of the game delta s=0, you cannot get to absolute 0, and even then particles will not have no entropy First Law defined as energy cannot be created or destroyed, however it can be converted ΔEuniverse=0=ΔEsystem+ΔEsurroundings ΔEsystem= ΔEsurroundings Esystem=Kinetic Energy (system)+ Potential energy (system) in a chemical reaction form ΔE=EfEi or ΔE= EpEr endothermic reactions ΔE>0 energy energy moves into the system, exothermic reactions ΔE<0 energy moves out of the system Two methods: heat q: q+ heat absorbed Esys increases, q heat is released Esys decreases work w: w+ work done on the system Esys increase, w work done by the system Esys decreases E and ΔE are state functions and are path independent Work 1. Electrical 2.PΔV, w= PΔv where P=constant external pressure ΔE=q+ (PΔV)=qPΔV Heat at constant V Δv=0 pΔV so ΔE=qv E change heat absorbed or lose Heat at constant P open to the atmosphere enthalpy H=E+PV change in ΔH=ΔE+PΔV, VΔP=0 at constant P Convert ΔE+ΔH ΔE does not equal ΔH, unless V does not change, it only changes if a gas forms or is consumed differ ΔHΔE=PΔV assume all gases are ideal V=nRT/P ΔV=Δ(nRT/P) ot ΔV=Δn(RT/P) ΔV is caused by Δn 1 not all reactants and products are gases Δngas=(ngas)p(Δngas)> ΔH =ΔE+ P (Δn gas)(RT/P) ΔH=ΔE+ΔngasRT example 1: ΔH and ΔE difference? 2N2O5 (g)> 4NO2 (g)+ O2(g) ΔH=ΔHf,pΔHf,r ΔH=4ΔH(NO2) 2ΔHf(N2O5) Δngas=3 =4(33.8)2(11)=113 R=8.31451 J/K*mol 113=ΔE(3(298)8.314)/1000) ΔE= 106 kJ % difference= 7.43/113= 6.6% small note: Vsolid=Vliquid<<Vgas example 2: ΔE? CaCO3+(2H+)> (Ca2+)+H2O+CO2 37mL 36 18 18 22.4L ΔV=ΔVpΔVr=24.363L=24.2L (STP) no gases present ΔE=ΔH example 3: 8O2+2C6H2(NO2)3> 3N2+12CO2+6H2O ΔHf=12 (39.35)+6(241.81)2(3862.94) ΔH= 13,898.9 kJ 13989.9(15.8)298(8.314)/1000= ΔE= 13.927 kJ Enthalpy Changes and Spontaneity spontaneous occurs by itself w/o outside assistance nonspontaneous occurs w/ outside assistance Reaction rate and Spontaneity ΔH will it occur rate of reaction plays a role Direction of Change exothermic burn fuel, iron rusting, endothermic ice melting Δh and ΔE are heat is given off, energy is leaving ΔH and ΔE + heat absorbed energy absorbed ΔH influence spontaneity thermo direction of reaction, will it occur, spontaneously at T, absorb or release heat Kinetics how fast, pathway Direction of Spontaneous Change (Spontaneous) exothermic iron rusting, fuel burning Δ H and ΔE are , heat is released, energy is leaving the system endothermic ice melting, evaporation,expansion randomness and temperature ΔH and ΔE positive, heat is absorbed, energy is entering Heat Transfer release heat and cold absorbed, particles collide, transfers energy hot particles cool down, cold warms up heat flows spontaneously from hot to cold, intermolecular collisions spontaneous low probability to high probability 2 spontaneous energy is dispersed Entropy (s) [Dispersion of Energy] describes randomness, how energy is distributed probability more probable= greater S S=klnW k=1.38x10^23, w= # of energetically equivalent ways random systems require less energy W equivalent states for expansion of gas, one of the states is more probable of the 3 macro vs micro states all microstates have same macrostate (more possibilities= greater randomness Spontaneity: clear order irreversible process, increase in randomness, disorder is more common it is a state function does not matter path ΔS=SfSi or ΔS=SpSr Sproducts>Sreactants ΔS+, entropy increases, probability increases, favors spontaneity Sproducts<Sreactants ΔS, entropy decreases, probability decreases, randomness decreases, does not favor spontaneity increase entropy tend to occur spontaneously gases entropy increases as volume increases (more probability that particles are found) temperature increases entropy increases physical statecrystalline solids decrease S, liquids increase S, gases have greatest S number of particles adding particles increase # of ways more particles can be arranged +ΔS, less particles ΔS example 4: N2+3H2>2NH3 n=24= 2 Δn= so ΔS= Reactions without gas calculate mol product to reactant example 5: CaCO3 +(2H+)>(Ca2+)+H2O+CO2, + ΔS 2N2O5>4NO2+O2 3mol ΔS=+ OH+(H+)>H2O 1 mol ΔS= Both ΔS and ΔH can affect spontaneity PE decreases ΔH , ice melts ΔH+ ΔS + building collapses stores disorder + ΔS * depends on temperature* 2nd law of Thermodynamics total entropy + if process is spontaneous reversible ΔSuniverse= 0, irreversible ΔS universe> 0 ΔS universe= ΔSsystem+ ΔS surroundings entropy decreases,entropy surroundings increases ΔSsurroundings= qsurroundings/T ΔS system ΔS surroundings + Law of Conservation of E qsurroundings= q system, q system= ΔH, q surroundings= ΔH Gibbs Free Energy 3 ΔG= ΔHT ΔS ΔG<0 spontaneous ΔG>0 nonspontaneous G is a state function, units of energy, extensive property ΔH ΔS ? + spontaneous always + never spontaneous + + high T spontaneous low T spontaneous Third law of Thermodynamics abs 0, entropy 0 K=1 entropy 1 mol substance 298 K 1 atm no system can reach absolute 0, always use Kelvin + ΔS, heating increases randomness, more gases, standard state S can never equal 0 ΔS use Hess’s Law to solve ΔS= nproductsnreactants example 6: Al2O3+3H2>2Al+2H2O [2(25.3)+3(188.7)][51+3(130.6)]=+177.9 J/K Standard Free Energy Changes ΔG measured at 25 C 1 atm use ΔG equation or ΔGf use Hess’s Law Degradation of E increasing ΔS leads to an increase in E dispersion leads to an increase in degradation of E Thermodynamically Reversible process can be reversible if reaction is close to equilibrium ΔG=max possible work leads to amount of work under reversible conditions energy does not need to be lost as heat, amount of E available to perform work example 7: 246.1 kJ/mol .3371 kJ/molK ΔGT ΔS 246.1298(.3371) = 358.5 kJ/mol= ΔG System @ Equilibrium neither spontaneous nor nonspontaneous Gproducts=Greactants dynamic equilibrium ΔG=0 no work done @ equilibrium, no free energy can do work ΔG=0 Free Energy Free E mass amount of E released that can do work on the surroundings exothermic heat is released (enthalpy) increase ΔS it cannot do work ΔS is a theoretical limit 4 Phase Change=Equilibrium ΔG=0 H2O (l) reversible H2O (g) , 1st order transition only 1 T for equilibrium T= ΔH/ ΔS solid/liquid : melting/freezing point, liquid upper equilibrium= boiling example 8: Tboiling 0= 310.0929(T)= 334 K example 9: 97.6 +(.122)298 = 61.2 kJ/mol , spontaneous Affect T and ΔG run at temperatures other than 298 K equilibrium can change depending on T near 298 K ΔH and ΔS are expected to have very small values example 10: 92.38 +(.1984)298 = 33.26 kJ/mol, spontaneous 20 C 92.38 +(.1984)773= 60.98 kJ/mol, nonspontaneous 500 C example 11: CCl> C +2Cl2 95.7.1422(T)=0 T=673 greater nonspontaneous less than spontaneous example 12: 847.6 +(0.0413)T = 20,300 less than spontaneous greater than nonspontaneous ● look up free energy diagrams to see the relationship between ΔG and the reactants vs. products ΔG and the position of equilibrium K<<1 ΔG > 0 + lies close to the reactants, quick reaction, forward nonspontaneous ΔG<0 lies close to reactants, forward spontaneous ΔG G under nonstandard conditions ΔG= ΔG (nought) reactants and products are standard normal @ T, 1 atm, 1 M nonstandard ΔG=ΔG(nought) +RTlnQ at equilibrium ΔG=0 ΔG(nought)= RTlnK Effect of Change in Pressure or Molarity Q=[products]y/[reactants]x 5 example 13: ΔG =33 kJ 33.3+(8.314)298ln0.0926 Q=(0.5)^2/(3)^3=0.0926 ΔG= 44.9 kJ example 14: 71.2(8.314)298lnQ Q=(0.1)/4=0.025 ΔG= 50.7 kJ K related to ΔG ΔG= 0 Q=K 0=ΔG+RTlnK ΔG(nought)= RTlnK K=e^ΔG(nought)/RT at 298 K K<1 ΔG+ reaction spontaneous in reverse K>1 ΔG spontaneous forward K=1 ΔG=0 at equilibrium Equilibrium ΔG= RTlnK (equation above) example 15: (Ag+)+Cl > AgCl K= 1/Ksp= 1/(1.8x10^10) =5.6x10^9 ΔG(8.3145)298ln(5.6x10^9)/1000 ΔG= 56 kJ/mol example 16: ΔG= 33.3 kJ/mol k=e^ΔG/RT ΔG/RT = (33,3000)/8.314(298) =13.4 k=e^13.4= 7x10^5 example 17: ΔGpΔGr > 64.28 kJ k= e^ 64.28/8.314(298) e^25.945 1.85x10^11=k example 18: N2+3H2 > 2NH3 46.3/(8.314)700 e^7.95= 3.5x10^4=k Temperature Dependence of K ΔG(nought) = RTlnk=ΔHTΔS lnK= (ΔH/R)1/T+ΔS/R slope ΔH/RT intercept ΔS/R can determine K if you know ΔH and ΔS example 19: N2+3H2 (reversible) 2NH3 ΔH= 92.38 ΔS= 198.4 lnk= (92.38/8.314)(1/773)+198.4/8.314 k= 7.56 x10^5 <<1 nonspontaneous 6
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