Presentation is loading. Please wait.

Presentation is loading. Please wait.

Review Unit 7 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Similar presentations


Presentation on theme: "Review Unit 7 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc."— Presentation transcript:

1 Review Unit 7 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2 Big Idea #5: Thermodynamics Chemical and physical processes are driven by: a decrease in enthalpy (–∆H), or an increase in entropy (+∆S), or both. Bonds break and form to lower free energy (∆G).

3 ΔS = ΔH T ΔH = q (heat) (disorder) (microstates) (dispersal of matter & energy at T) ΔE = q + w PΔV = –w (at constant P) += Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) Energy (E) (energy of ΔH and ΔS at a T) (max work done by favorable rxn) ΔG = ΔH – TΔS (–∆G sys means +∆S univ & K>1) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)

4 Enthalpy (∆H) Calculated 4 Ways 1)Bond Energies  H rxn =  (BE reactants )   (BE products ) 2)Hess’s Law  H overall =  H rxn1 +  H rxn2 +  H rxn3 … 3)Standard Heats of Formation (H f )  H =  n H f(products) –  n H f(reactants) 4)Calorimetry (lab) q = mc∆T (surroundings or thermometer) –q = ∆H ∆H/mol = kJ/mol (molar enthalpy)   (+ broken)(– formed) (NOT) (given) (NOT) (given)

5 S : dispersal of matter & energy at T (s) + ( l)  (aq) solid gas VV more microstates H 2 O (g) TT S( s ) < S( l ) < S( aq ) < S( g ) Entropy (S) (Molecular Scale)  Temperature  Volume  Particle mixing  Particle number  Particle size  S o =  n S o (products) –  m S o (reactants) +∆S (dispersal) (given)

6 Thermodynamically Favorable: (defined as) increasing entropy of the universe (∆S univ > 0) Thermodynamically Favorable ∆S univ > 0 (+Entropy of the Universe)  S univ =  S system +  S surroundings > 0 (+)(+)(+)(+) Chemical and physical processes are driven by: decrease in enthalpy (–∆H sys ) increase in entropy (+∆S sys ) causes (+∆S surr )

7  S universe =  S system + (∆S univ ) & (∆G sys )  S universe =  S system +  S surroundings > 0 For all thermodynamically favorable reactions: multiplying each term by  T:   H system T –T  S universe = –T  S system +  H system rearrange terms: –T  S universe =  H system – T  S system (Boltzmann) (Clausius)  G system =  H system – T  S system (Gibbs free energy equation)

8 –  G is thermodynamically favorable. Gibbs defined  T  S univ as the change in free energy of a system (  G sys ) or  G. Free Energy (  G) is more useful than  S univ b/c all terms focus on the system. If –  G sys, then +  S universe. Therefore… (∆S univ ) & (∆G sys ) –T  S univ =  H sys – T  S sys  G sys =  H sys – T  S sys (Gibbs free energy equation) “Bonds break & form to lower free energy (∆G).”

9 Standard Free Energy (∆G o ) and Temperature (T) The temperature dependence of free energy comes from the entropy term (–T  S  ).  G  =  H  – T  S  (on equation sheet) enthalpy term (kJ/mol) entropy term (J/mol∙K) free energy (kJ/mol) energy transferred as heat energy dispersed as disorder max energy used for work (consists of 2 terms) units must match!!! (kJ)

10 ∆G o =(∆Ho)(∆Ho)∆So∆So ( ) – T ( )  G  =  H   T  S  (high T) – (low T) + + – (high T) + (low T) – + – – + + + – – (unfav. at ALL T) (fav. at ALL T) (fav. at high T) (unfav. at low T) (unfav. at high T) (fav. at low T) – T( ) = = = = + + – – Standard Free Energy (∆G o ) and Temperature (T) Thermodynamic Favorability

11 Calculating ∆G o (4 ways) 1)Standard free energies of formation, G f  : 2)Gibbs Free Energy equation: 3)From K value (next few slides) 4)From voltage, E o (next Unit)  G  =  nG  (products) –  mG  (reactants) ff (given equation)  G  =  H  – T  S  (given equation) (may need to calc. ∆H o & ∆S o first) (given equation)

12 Free Energy (∆G) & Equilibrium (K)  G  = –RT ln K K = e^ –∆G o RT (on equation sheet) (NOT on equation sheet) R = 8.314 J∙mol –1 ∙K –1 = 0.008314 kJ∙mol –1 ∙K –1 If  G  in kJ, then R in kJ……… Solved for K : –∆G o RT = ln K

13 ∆G o = –RT(ln K)K@ Equilibrium –RT ( ) > 1 < 1 – + product favored + – reactant favored (favorable forward) (unfavorable forward) = =  G  = –RT ln K Free Energy (∆G) & Equilibrium (K)


Download ppt "Review Unit 7 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc."

Similar presentations


Ads by Google