Lecture12, 2-6-08 Gibbs Free Energy

Lecture12, 2-6-08 Gibbs Free Energy - SCORE DISTRIBUTION...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
SCORE DISTRIBUTION MID-TERM 1 FEB 1, 2008 0 5 10 15 20 25 30 35 40 0 10 20 30 40 50 60 70 80 90 100 Exam Score Number of Scores A, 75 - 90 B, 65-70 C, 45-60 D, 35-40 F, 10-30
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Summary – Dispersion of Energy Why this Reaction is Spontaneous? System – Energy is less dispersed in 2 moles of NH 3 than it is in 4 moles of N 2 , H 2 Therefore Δ S 0 is negative Surroundings – Energy is more dispersed in surroundings: conversion of chemical potential energy into heat released Δ H 0 = q irr is negative Δ S 0 Δ H 0 /T = -199 J/K –(-309 J/K) = +110 J/K Net change - energy is more dispersed in universe. N 2 (g) + 3 H 2 (g) 2 NH 3 (g)
Background image of page 2
System and Surroundings Undergo Changes in Chemical Reactions at Constant P and T Surroundings System Heat Exothermic Reaction ( Δ H rxn < 0) S 0 q P = H 0 Spontaneous Process S 0 - H 0 /T > 0 Dispersion of Energy in the System Dispersion of Energy due to Heat Transfer to Surroundings Reversible Pathway Irreversible Pathway Surroundings 3 rd Law
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Predicting Spontanteous Chemical Reactions, Constant P and T N 2 H 4 (l) + 2H 2 O 2 (l) N 2 (g) + 4H 2 O(g) S o rxn = 606 J/K Δ H rxn o is -642.2 kJ S 0 - H 0 /T =606 J/K –(-642200 J/298 K) S 0 - H 0 /T =606 J/K + 2155 J/K S 0 - H 0 /T = 2761 J/K > 0 Conclude Spontaneous Reaction at all T WHY
Background image of page 4
The Gibbs Free Energy A way to simplify determining whether a reaction is spontaneous or not. It accomplishes the same thing as an entropy calculation in a little different way Start with Δ S 0 Δ H 0 /T > 0, spontaneous at constant P, T T Δ S 0 Δ H 0 > 0 or Δ H 0 –T Δ S 0 < 0, spontaneous at constant P, T Define gibbs free energy: Δ G 0 = Δ H 0 –T Δ S 0
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Determining the Gibbs Free Energy Δ G 0 = Δ H 0 –T Δ S 0 Under standard thermodynamic conditions Δ H 0 = -46 kJ/mol × 2 mol = -92 kJ from table Δ S 0 = S 0 (products) – S 0 (reactants)
Background image of page 6
Image of page 7
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 24

Lecture12, 2-6-08 Gibbs Free Energy - SCORE DISTRIBUTION...

This preview shows document pages 1 - 7. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online