• Save

Loading…

Flash Player 9 (or above) is needed to view presentations.
We have detected that you do not have it on your computer. To install it, go here.

Like this presentation? Why not share!

Like this? Share it with your network

Share

Tang 05 entropy and gibb's free energy

on

  • 1,231 views

 

Statistics

Views

Total Views
1,231
Views on SlideShare
1,231
Embed Views
0

Actions

Likes
2
Downloads
0
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Microsoft PowerPoint

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Tang 05 entropy and gibb's free energy Presentation Transcript

  • 1. ENTROPY & GIBBS FREE ENERGY
  • 2. ENTROPY AND GIBBS FREE ENERGY
  • 3.
    • entropy (  S )- a measure of disorder or randomness
    • entropy =  disorder
    • The units for entropy are J/mol·K or J/K
    ENTROPY AND GIBBS FREE ENERGY
  • 4. ENTROPY AND GIBBS FREE ENERGY SECOND LAW OF THERMODYNAMICS the entropy of the universe is constantly increasing
  • 5.  S solid <  S liquid <  S gas ENTROPY AND GIBBS FREE ENERGY
  • 6. Δ S = S products – S reactants When S products >S reactants , Δ S>0 ENTROPY AND GIBBS FREE ENERGY
  • 7. Δ S = S products – S reactants When S products <S reactants , Δ S<0 ENTROPY AND GIBBS FREE ENERGY
  • 8. ENTROPY AND GIBBS FREE ENERGY Predict the sign of  S for the following reactions: H 2(g)  2 H (g)  S 2 H 2(g) + O 2(g)  2 H 2 O (l) -  S 2 NO 2(g)  N 2 O 4(g) -  S C 3 H 8(g) + 5O 2(g)  3CO 2(g) + 4H 2 O (g) +  S condensation of steam to liquid -  S
  • 9. ENTROPY AND GIBBS FREE ENERGY Predict the sign of  S for the following reactions: sublimation of dry ice +  S 2 NH 3(g)  N 2(g) + 3 H 2(g) +  S C 6 H 12 O 6(s) + 6O 2(g)  6CO 2(g) + 6H 2 O (g) +  S 2 H 2 O (l)  2 H 2(g) + O 2(g) +  S NaCl (s)  NaCl (aq) +  S
  • 10. ENTROPY AND GIBBS FREE ENERGY To solve for the entropy change of a reaction at standard conditions (SATP): Δ S° = ∑ Δ S° (products) - ∑ Δ S° (reactants) Don’t forget that the equations should be multiplied by the appropriate factor, as necessary.
  • 11. ENTROPY AND GIBBS FREE ENERGY Example 1: Calculate the standard entropy of the following reaction: C 2 H 4(g) + H 2(g)  C 2 H 6(g) Sº C 2 H 6(g) = 229.5J/mol·K Sº C 2 H 4(g) = 219.8J/mol·K Sº H 2(g) = 130.6J/mol·K Δ S° = ∑ Δ S° (products) - ∑ Δ S° (reactants) = [1 mol x 229.5J/mol·K] – [(1 mol x 219.8J/mol·K) + (1 mol x 130.6J/mol·K)] = (229.5J/K) – (350.4J/K) = -120.9J/K Therefore the standard entropy is -120.9J/K Do you think this reaction will spontaneously occur at room temperature?
  • 12. ENTROPY AND GIBBS FREE ENERGY Example 1: Calculate the standard entropy of the following reaction: C 2 H 4(g) + H 2(g)  C 2 H 6(g) Do you think this reaction will spontaneously occur at room temperature? Since molecules tend to become more and more disordered, and this reaction results in more organization (fewer moles of gas), then this reaction is most likely not spontaneous at room temperature
  • 13. ENTROPY AND GIBBS FREE ENERGY How are entropy and enthalpy related? Δ Gº = Δ Hº - T Δ Sº Gibbs free energy ( J or kJ ) enthalpy Temperature ( K ) entropy Gibbs free energy is the energy that is available to do useful work. A reaction will spontaneously occur if Δ G<0 ( exergonic reaction ) A reaction will NOT spontaneously occur if Δ G>0 ( endergonic reaction )
  • 14. ENTROPY AND GIBBS FREE ENERGY A reaction will spontaneously occur if Δ G<0 ( exergonic reaction ) A reaction will NOT spontaneously occur if Δ G>0 ( endergonic reaction )
  • 15. ENTROPY AND GIBBS FREE ENERGY Reactions with a negative Δ H and positive Δ S all have a negative Δ G Δ Gº = Δ Hº - T Δ Sº (-) - T(+) Exergonic & spontaneous at all temperatures Reactions with a positive Δ H and negative Δ S all have a positive Δ G Δ Gº = Δ Hº - T Δ Sº (+) - T(-) Endergonic & nonspontaneous at all temperatures and will only occur with the continuous input of energy
  • 16. ENTROPY AND GIBBS FREE ENERGY Reactions with a negative Δ H and negative Δ S all have a negative Δ G Δ Gº = Δ Hº - T Δ Sº (-) - T(-) Δ G will be negative in temperature conditions where the value of T Δ S is lower than the value of Δ H. Thus the reaction is spontaneous only at lower temperatures. Example: 2SO 2(g) + O 2(g)  2SO 3(g) This reaction is spontaneous at temperatures below 786ºC (1059K) and nonspontaneous above this temperature
  • 17. ENTROPY AND GIBBS FREE ENERGY Reactions with a positive Δ H and positive Δ S all have a negative Δ G Δ Gº = Δ Hº - T Δ Sº (+) - T(+) Δ G will be negative in temperature conditions where the value of T Δ S is higher than the value of Δ H. Thus the reaction is spontaneous only at higher temperatures. Example: H 2 O (s)  H 2 O (l) This reaction is spontaneous at temperatures above 0ºC (273K) and nonspontaneous below this temperature
  • 18. ENTROPY AND GIBBS FREE ENERGY Summary: Spontaneous and Non-Spontaneous Reactions Spontaneity depends on T (spontaneous at lower temperatures) Nonspontaneous (proceeds only with a continuous input of energy) Δ S<0 Spontaneous at all temperatures Spontaneity depends on T (spontaneous at higher temperatures) Δ S>0 Δ H<0 Δ H>0
  • 19. ENTROPY AND GIBBS FREE ENERGY Example 2: Is the following reaction spontaneous at SATP? CO(NH 2 ) 2(aq) + H 2 O (l)  CO 2(g) + 2NH 3(g) Δ Hº =119kJ Δ Sº = 354.8J/K = 0.3548KJ/K T = 25ºC = 298K Δ Gº = Δ Hº - T Δ Sº = (119kJ) – (298K)(0.3548kJ/K) = (119kJ) – (105.7kJ) = 13kJ Δ Gº>0, so this reaction is not spontaneous Therefore this reaction is not spontaneous at SATP
  • 20. ENTROPY AND GIBBS FREE ENERGY