Download presentation

Presentation is loading. Please wait.

Published byJonas Gidney Modified over 3 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

OK

1911209http:\\asadipour.kmu.ac.ir...46 slides. Thermodynamics 2911209http:\\asadipour.kmu.ac.ir...46 slides.

1911209http:\\asadipour.kmu.ac.ir...46 slides. Thermodynamics 2911209http:\\asadipour.kmu.ac.ir...46 slides.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To ensure the functioning of the site, we use **cookies**. We share information about your activities on the site with our partners and Google partners: social networks and companies engaged in advertising and web analytics. For more information, see the Privacy Policy and Google Privacy & Terms.
Your consent to our cookies if you continue to use this website.

Ads by Google

Ppt on generalized anxiety disorder Free ppt on brain machine interface mit Ppt on world book day dressing Ppt on business etiquettes training program Ppt on carburetor cleaner Ppt on school management system software Ppt on html tags with examples Ppt on national urban health mission Ppt on line drawing algorithms Ppt on self development tips