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Slide 1 of 44 19-6 Free Energy Change and Equilibrium.

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Presentation on theme: "Slide 1 of 44 19-6 Free Energy Change and Equilibrium."— Presentation transcript:

1 Slide 1 of Free Energy Change and Equilibrium

2 Slide 2 of 44 Free Energy Change and Equilibrium CondensationEquilibriumVaporization

3 Slide 3 of 44 Relationship of ΔG° to ΔG for Non-standard Conditions 2 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) ΔG = ΔH - TΔSΔG° = ΔH° - TΔS° For ideal gases ΔH = ΔH° ΔG = ΔH° - TΔS

4 Slide 4 of 44 Relationship Between S and S° q rev = -w = RT ln VfVf ViVi ΔS = q rev T = R ln VfVf ViVi ΔS = S f – S i = R ln VfVf ViVi PiPi PfPf = -R ln PfPf PiPi S = S° - R ln P P°P° = S° - R ln P 1 = S° - R ln P

5 Slide 5 of 44 N 2 (g) + 3 H 2 (g) 2 NH 3 (g) S NH 3 = S NH 3 – Rln P NH 3 S N 2 = S N 2 – Rln P N 2 S H 2 = S H 2 – Rln P H 2 ΔS rxn = 2  (S NH 3 – Rln P NH 3 ) – (S N 2 – Rln P N 2 ) –3  (S H 2 – Rln P H 2 ) ΔS rxn = 2 S NH 3 – S N 2 –3S H 2 + Rln PN2PH2PN2PH2 P NH ΔS rxn = ΔS° rxn + Rln PN2PH2PN2PH2 P NH 3 2 3

6 Slide 6 of 44 ΔG Under Non-standard Conditions ΔG = ΔH° - TΔS ΔS rxn = ΔS° rxn + Rln PN2PH2PN2PH2 P NH ΔG = ΔH° - TΔS° rxn – TR ln PN2PH2PN2PH2 P NH ΔG = ΔG° + RT ln PN2PH2PN2PH2 2 P NH ΔG = ΔG° + RT ln Q

7 Slide 7 of 44 ΔG and the Equilibrium Constant K eq ΔG = ΔG° + RT ln Q ΔG = ΔG° + RT ln K eq = 0 If the reaction is at equilibrium then: ΔG° = -RT ln K eq

8 Slide 8 of 44 Criteria for Spontaneous Change Every chemical reaction consists of both a forward and a reverse reaction. The direction of spontaneous change is the direction in which the free energy decreases.

9 Slide 9 of 44 Significance of the Magnitude of ΔG

10 Slide 10 of ΔG° and K eq as Functions of Temperature ΔG° = ΔH° -TΔS°ΔG° = -RT ln K eq ln K eq = -ΔG° RT = -ΔH° RT TΔS° RT + ln K eq = -ΔH° RT ΔS° R +

11 Slide 11 of 44 Van’t Hoff Equation If we evaluate this equation for a change in temperature: ln = -ΔH° RT 2 ΔS° R + -ΔH° RT 1 ΔS° R + - = -ΔH° R 1 T2T2 1 T1T1 - K eq1 K eq2 ln K eq1 K eq2

12 Slide 12 of 44

13 Slide 13 of 44 Temperature Dependence of K eq ln K eq = -ΔH° RT ΔS° R + slope = -ΔH° R -ΔH°= R  slope = J mol -1 K -1  2.2  10 4 K = -1.8  10 2 kJ mol -1 Assume ΔH° and ΔS° do not vary significantly with temperature.


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