Thermochemistry

1. Thermochemistry

2. Energy The capacity to do work or to produce heat.

3. The Two Types of Energy Potential:due to position or composition - can be converted to work Kinetic:due to motion of the object KE = 1/2 mv2 (m = mass, v = velocity)

4. State Function Depends only on the present state of the system - not how it arrived there. It is independent of pathway. Enthalpy is a state function It’s like going to Vegas. On the way home you know you lost $400. It doesn’t matter what casino or how many bets you placed. You still lost money.

5. First Law First Law of Thermodynamics: The energy of the universe is constant. If you lose $$ in Vegas, someone else made $$.

6. Enthalpy H = energy flow as heat (at constant pressure and volume) Thus the change of enthalpy is the change in the amount of energy of a system. A block of wood burning has a negative change in enthalpy or DH. If you lost money in Vegas, that is negative D$. If you won money, that is positive D$ .

7. Some Heat Exchange Terms specific heat capacity heat capacity per gram = J/°C g or J/K g molar heat capacity heat capacity per mole = J/°C mol or J/K mol You could calculate how much money you lost per bet or how much money you lost per casino. Two different ways of looking at it.

8. 5ways to calculate change in enthalpy • Stoichiometry- use mol ratios with heat given in a reaction • Calorimetry– use q=mc t to find heat gained or lost by water. • Hess’s Law –more to come on that • Standard heats of formation • Bond Energies

9. Stoichiometry

10. Calorimetry

11. Hess’s Law Reactants  Products The change in enthalpy is the samewhether the reaction takes place in one step or a series of steps.

13. Calculations via Hess’s Law 1. If a reaction is reversed, H is also reversed. N2(g) + O2(g)  2NO(g) H = 180 kJ 2NO(g)  N2(g) + O2(g) H = 180 kJ 2. If the coefficients of a reaction are multiplied by an integer, H is multiplied by that same integer. 6NO(g) 3N2(g) + 3O2(g) H = 540 kJ

14. Hess’s law example • Thermite is powdered aluminum plus iron III oxide creating iron and aluminum oxide. • Al(s) + Fe2O3(s)  Al2O3(s) + Fe(s) • This is extremely exothermic. • 2 Al + 3/2 O2  Al2O3∆H=-1676 kJ/mol • 2 Fe +3/2 O2  Fe2O3 ∆H=-826 kJ/mol • ??? What now???

15. But what about Standard Enthalpies of Formation? • The change in enthalpy (heat) that accompanies production of one mole of a compound from its elements in the standard state. • It’s like taking Hess’s Law and making it so much easier.

16. Change in Enthalpy Standard heat of enthalpy for any element is zero. Can be calculated from enthalpies of formation of reactants and products. Hrxn° = npHf(products) nrHf(reactants) So let’s burn something at standard conditions….. Like propane. Propane is 103.85 kJ/mole , CO2 is -393.509 kJ/mole, water vapor is -241.83 kJ/mole

17. Bond Energies Bond energy = Bonds Broken in reactants – Bonds formed in products

18. Bond Energies

19. Spontaneous Processes and Entropy

20. TWO Trends in Nature • Order  Disorder (Entropy)   • High energy  Low energy (Enthalpy) 

21. nonspontaneous spontaneous Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 0C and ice melts above 0 0C • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water forms rust

22. spontaneous nonspontaneous

23. Predict sign of DS • Sugar cube dissolves • Iodine vapor condenses • Water freezes • Liquid nitrogen evaporates

24. First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous process: DSuniv = DSsys + DSsurr > 0 Equilibrium process: DSuniv = DSsys + DSsurr = 0

25. The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C. rxn - S nS0(reactants) S nS0(products) = DS0 DS0 DS0 rxn rxn rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)] = 427.2 – [395.8 + 205.0] = -173.6 J/K•mol Entropy Changes in the System (DSsys) What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2(g)↔ 2CO2(g) S0(CO) = 197.9 J/K•mol S0(CO2) = 213.6 J/K•mol S0(O2) = 205.0 J/K•mol

26. Free Energy G = HTS(from the standpoint of the system) • G is for Gibbs Free Energy. • A process (at constant T, P) is spontaneous in the direction in which Gis negative.

27. Gibbs Free Energy For a constant-temperature process: Gibbs free energy (G) DG = DHsys -TDSsys DG < 0 The reaction is spontaneous in the forward direction. DG > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. DG = 0 The reaction is at equilibrium.

28. The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions. rxn - nDG0(reactants) S S = f Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. DG0 rxn f DG0 of any element in its stable form is zero. f nDG0 (products) f

29. - nDG0(reactants) S S = f 2C6H6(l) + 15O2(g) 12CO2(g) + 6H2O (l) DG0 DG0 DG0 - [ ] [ + ] = rxn rxn rxn [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ = Is the reaction spontaneous at 25 0C? 12DG0 (CO2) 2DG0 (C6H6) f f 6DG0 (H2O) f nDG0 (products) f What is the standard free-energy change for the following reaction at 25 0C? DG0 = -6405 kJ < 0 spontaneous

30. Effect of H and S on Spontaneity DG = DH - TDS

31. Free Energy and Pressure G = G + RT ln(Q) • Q = reaction quotient from the law of mass action. • So at equilibriumG = 0 and Q = K which means…. G = RT ln(K)

32. DG0 = -RT lnK