Download presentation
Presentation is loading. Please wait.
Published byJonas Gidney Modified over 10 years ago
1
Slide 1 of 44 19-6 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 3 2 3 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 3 2 2 3 ΔG = ΔH° - TΔS° rxn – TR ln PN2PH2PN2PH2 P NH 3 2 2 3 ΔG = ΔG° + RT ln PN2PH2PN2PH2 2 P NH 3 2 3 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 44 19-7 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 = -8.3145 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.
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.