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Third Law of Thermodynamics

Third Law of Thermodynamics. If increasing temperature increases entropy, then the opposite should be true also. Is it possible to decrease the temperature to the point that the entropy is zero? At what T does S = 0? If entropy is zero, what does that mean?. Third Law of Thermodynamics.

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Third Law of Thermodynamics

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  1. Third Law of Thermodynamics • If increasing temperature increases entropy, then the opposite should be true also. • Is it possible to decrease the temperature to the point that the entropy is zero? • At what T does S = 0? • If entropy is zero, what does that mean?

  2. Third Law of Thermodynamics • 3rd Law of Thermodynamics • Entropy of a perfect crystalline substance at 0 K is zero • No entropy = highest order possible • Purpose of 3rd Law • Allows S to be measured for substances • S = 0 at 0 K S = standard molar entropy The entropy of the pure substance is our zero (reference point)

  3. What is a Perfect Crystal? Perfect crystal at 0 K Crystal deforms at T > 0 K

  4. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0.

  5. Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. This allows us to calculate an absolute entropy.

  6. Measuring Heat and Enthalpies DH = qrxn Constant-Volume Calorimetry qsys = qwater + qbomb + qrxn qsys = 0 qrxn = - (qwater + qbomb) qwater = msDT qbomb = CbombDT Reaction at Constant V DH ~ qrxn No heat enters or leaves! Why only approximately the enthalpy?

  7. Constant-Pressure Calorimetry qsys = qwater + qcal + qrxn qsys = 0 qrxn = - (qwater + qcal) qwater = msDt qcal = CcalDt Reaction at Constant P DH = qrxn No heat enters or leaves! 6.5

  8. Heat of combustion C(gr) +O2(g) CO2(g) DHc = -393.5 kJ/mol 5C(gr) +5O2(g) 5CO2(g) DHc = 5(-393.5) kJ/mol CO2(g) C(gr) +O2(g) DHc = -(-393.5) kJ/mol Thermochemistry The study of heat transfer due to chemical reactions and phase changes qreaction + qcalorimeter =0 qreaction = - qcalorimeter = - CcalorimeterDT For 18 gm C(gr) or equivalently nC = 18/12=1.5 moles: DHc = 1.5(-393.5) kJ

  9. f a i b Energy is a State Function if and only if DUa = DUb • Assume the contrary and take DUa > DUb. • Go forward on a (gaining Ua) and backward along b (lossing -DUb) • During each cycle there is a net gain in energy (DUa - DUb ) > 0, available to do work and no change in the system. • We could repeat this and gain energy from doing nothing. • Nothing is Free so it’s not possible. • Not possible, therefore, DUa = DUb. Energy is conserved. • Energy is a state function • So is Enthalpy (same arguments apply), (Entropy, etc…)

  10. f a i DH1 = HC+D - HA+B b C+D A+B E+F 1 2 DH2 = HE+F - HC+D 3 DH3 = HE+F - HA+B Thermochemistry: Hess’s Law DHa = DHb = Hf - Hi DH1 + DH2 = HC+D - HA+B +HE+F - HC+D =DH3 If 2 (or more reactions) are added (or subtracted) to get a new reaction, then adding (or subtracting) enthalpy changes for the two reactions gives enthalpy change for the new reaction. Hess’s law takes advantage of the fact that Enthalpy is a state function.

  11. C(gr) + 1/2 O2(gas) CO(gas) DH1=? CO(gas) + 1/2 O2(gas) CO2(gas) DH2 =-283.0 kJ/mol CO+1/2O2 C+1/2O2 +1/2O2 CO2 C(gr) + O2(gas) CO2(gas) DH3 =-393.5 kJ kJ/mol 1 2 3 Examples of Hess’s Law When carbon burns it creates mainly carbon dioxide and water. What is the enthalpy of combining Carbon with Oxygen to make Carbon Monoxide? DH3 = DH1 + DH2 -393.5 kJ/mol= DH1 -283.0 kJ/mol DH1 = 283.0 kJ/mol + -393.5 kJ/mol = -110.5 kJ/mol

  12. C(gr) + 2 H2(gas) CH4(gas) DH=? H2(gas) + 1/2 O2(gas) H2O(liq) DH2 =-285.8 kJ/mol C(gr) + O2(gas) CO2(gas) DH1=-393.5 kJ/mol C(gr) + O2(gas) CO2(gas) DH=-393.5 kJ/mol CO2(gas) +2 H2O(liq) CH4(gas) + 2 O2(gas) DH3=890.4 kJ/mol CO2(gas) +2 H2O(liq) CH4(gas) + 2 O2(gas) DH=890.4 kJ/mol 2(H2(gas) + 1/2 O2(gas) H2O(liq) ) DH=2(-285.8) kJ/mol C(gr) + 2 H2(gas) CH4(gas) DH=-393.5+890.4 - 2(-285.8) = -74.7 kJ/mol Another Example of Hess’s Law

  13. H2O(solid) H2O(liq) DHfus = 6.007 kJ/mol C(gr) C(diamond) DH=? C(gr) C(diamond) DH=DH1-DH2=+1.883 kJ/mol C(graphite) + O2(g) CO2(g) DH1= -393.505 kJ/mol C(diamond) + O2(g) CO2(g) DH2= -395.388 kJ/mol Latent Heat in Phase Changes H2O(liq) H2O(gas) DHvap = 40.66 kJ/mol Phase changes are accompanied by latent heat. Can add heat but temp doesn’t rise until all one phase.

  14. C(gr) C(diamond) DH=? C(gr) C(diamond) DH=DH1-DH2=+1.883 kJ/mol C(graphite) + O2(g) CO2(g) DH1= -393.505 kJ/mol C(diamond) + O2(g) CO2(g) DH2= -395.388 kJ/mol What happens to diamonds in a house fire?

  15. C(s,graph) + O2(g) CO2(g) DHo298 = -393.51 kJ/mol Most stable form of C Most stable form of C Most stable form of CO2 Enthalpy of Formation (defining a zero point) DHo = standard enthalpy change for reaction when reactants and products are in their standard states Standard state of compound X is the most stable form of X at P=1 atm and T=298.15 K. DHof = standard enthalpy of formation of a compound. DHof = DH of reaction in which pure elements in standard states combine to form products in std states.

  16. C(s,graph) + 1/2O2(g) CO(g) DHof = -110.13 kJ/mol C(s,graph) + O2(g) CO2(g) DHof = -393.51 kJ/mol H2(g) + 1/2 O2(g) H2O(l) DHof = -285.5 kJ/mol H2(g) + 1/2 O2(g) + C(gr) HCOOH(l) DHof = -408.8 kJ/mol H2(g) H2(g) DHof = - 0 kJ/mol 1/2 H2(g) H(g) DHof = + 217.9 kJ/mol Examples of Enthalpy of Formation

  17. aA +bB cC + dD DH298o reactants in std states products in std states elements DH1o DH2o DH298o = DH1o +DH2o aA +bBcC + dD DH1o = - a DHfo(A) - bDHfo(B) DH2o = c DHfo(C) + dDHfo(D) Standard heats of reaction from DHof DH298o = c DHfo(C) + dDHfo(D) - a DHfo(A) - bDHfo(B)

  18. C2H5OH(l) + 3O2(g) 2CO2(g) +3H2O(l) Examples using Heats of Formation DH298o = 3 DHfo(H2O) + 2 DHfo(CO2) - DHfo(C2H2OH)

  19. CH4(g) CH3(g) + H(g) DHo= + 483 kJ/mol C2H6(g) C2H5(g) + H(g) DHo= + 410 kJ/mol CHF3(g) CF3(g) + H(g) DHo= + 429 kJ/mol CHCl3(g) CCl3(g) + H(g) DHo= + 380 kJ/mol CHBr3(g) CBr3(g) + H(g) DHo= + 377 kJ/mol Bond Enthalpies In reactions bonds are broken and bonds are formed, eg C--H Define average bond strengths: C-C DHC-C= 348 kJ C=C DHC=C= 615 kJ C-H DHC-C= 413 kJ

  20. = – Gibbs Free Energy Practical uses: surroundings & system

  21. = – Gibbs Free Energy Condition of Spontaneity Make this equation nicer:

  22. Third Law of ThemodynamicsNernst Heat Theorem The entropy of any element in equil state is defined to approach 0 as T approaches 0K. This is the reference state (sea level) for Entropy. In any thermodynamic process involving only pure phases in their equilibrium states, the entropy change S--->0 as T---> 0K for chem reaction where pure elements ---> pure compounds. Entropy of any pure substance (element or compound) in its equil state --->0 as T-->0

  23. Measuring Heat and Enthalpies q = mH2O csDT Mass of H20 measure Sp. heat -M1cs,1(Tf - 373K))= mH2O cs(Tf - 283K) Initial temps Hot metal (373K) put in cool water (283K). What is Tf? Heat absorbed by water: Hot Metal

  24. mgDh=mH2O csDT Work by weight Heating of H2O Mechanical Equivalence of Heat 1 calorie 4.184 Joules

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