CHEM /21/11 IV. 2nd Law of THERMO

Slides:



Advertisements
Similar presentations
CHAPTERCHAPTERCHAPTERCHAPTER 19. Under standard conditions — ∆G o sys = ∆H o sys - T∆S o sys free energy = total energy change for system - energy change.
Advertisements

CAUSES OF CHANGE Order and Spontaneity. Enthalpy and Reactions Some reactions happen easily, but some others do not. Sodium and chlorine readily react.
Lecure 8: The Second and Third Laws of Thermodynamics Reading: Zumdahl 10.5, 10.6 Outline –Definition of the Second Law –Determining  S –Definition of.
Heat Capacity Amount of energy required to raise the temperature of a substance by 1C (extensive property) For 1 mol of substance: molar heat capacity.
Thermodynamics Chapter st Law of Thermodynamics Energy is conserved.  E = q + w.
Thermodynamics Chapter 19 Brown-LeMay. I. Review of Concepts Thermodynamics – area dealing with energy and relationships First Law of Thermo – law of.
Atkins’ Physical Chemistry Eighth Edition Chapter 3 – Lecture 2 The Second Law Copyright © 2006 by Peter Atkins and Julio de Paula Peter Atkins Julio de.
In general, the more atoms in its molecules, the greater is the entropy of a substance Entropy is a function of temperature.
Causes of Change Order and Spontaneity Gibbs Energy and Predicting Spontaneity.
Chapter 18: Thermodynamics Renee Y. Becker Valencia Community College.
Chapter 20 Thermodynamics and Equilibrium. Overview First Law of Thermodynamics Spontaneous Processes and Entropy –Entropy and the Second Law of Thermodynamics.
CHAPTER 16: SPONTANEITY, ENTROPY, & FREE ENERGY Dr. Aimée Tomlinson Chem 1212.
Thermodynamics Mr. Leavings. Objectives Use the laws of thermodynamics to solve problems, identify energy flow within a system, determine the classification.
 S univ =  S sys +  S surr Must be ____ for spontaneity as dictated by… + …the Second Law of Thermodynamics  S surr =  H / T (at constant P) So that:
ENTROPY Measure of: Level of disorder in a system or Number of Microscopic Energy Levels Available to a Molecule (i.e. microstates)
V. Determining Spontaneity It would be easier if we could determine spontaneity by just considering changes in the system. We derive an equation from the.
Chapter 19 Chemical Thermodynamics. Spontaneity of Physical and Chemical Changes Spontaneous changes happen without any continuing outside influences.
Spontaneous Processes and Entropy First Law “Energy can neither be created nor destroyed“. The energy of the universe is constant. Spontaneous Processes.
Entropy. 1 st Law of Thermodynamics Energy is neither created nor destroyed –The energy of the universe is constant –Energy just changes from one form.
Thermodynamics (the other part) Topic 19 in Red Book Chapter 16 in textbook.
Energy Changes in Chemical Reactions -- Chapter First Law of Thermodynamics (Conservation of energy)  E = q + w where, q = heat absorbed by system.
Entropy and Free Energy Thermodynamics: the science of energy transfer – Objective: To learn how chemists predict when reactions will be product-favored.
Chapter 19 Spontaneity, entropy and free energy (rev. 11/09/08)
1 Vanessa N. Prasad-Permaul Valencia College CHM 1046.
Thermodynamics the study of energy transformations and transfer THREE QUESTIONS to be addressed: 1. Will a reaction occur when two substances are mixed.
Find the change in entropy when 87.3 g of water vapor condense, given that water’s heat of vaporization is 5.99 kJ/mol K = – = –77.9 For.
The 3 rd Law of Thermodynamics Valentim M. B. Nunes ESTT-IPT May 2015.
SPONTANEOUS REACTIONS. Spontaneity 1 st Law of Thermodynamics- energy of the universe is ________. Spontaneous Rxns occur without any outside intervention.
CHAPTER TWO (18) Entropy, Free Energy, and Equilibrium
Heating / Cooling Curve Calculations
Thermodynamics Why Chemical Reactions Happen
Lecture 3 Entropy and Euler’s homogenous theorem
Aim # 18: What is the Gibbs Free Energy?
Entropy = S Entropy is disorder or randomness.
Third Law of Thermodynamics
The Third Law of Thermodynamics
Chapter Thermodynamics and Electrochemistry
Unit 10: Thermodynamics. Unit 10: Thermodynamics.
Molar Ratios By Elise Rippelmeyer Elise Rippelmeyer
Unit 5: Thermochemistry
Spontaneity & Entropy
Unit 10: Thermodynamics.
CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics Chapter 19 November 19, 2018 Chapter 19.
Spontaneity, entropy and free energy
Calculating ΔH using molar heats of formation
Dr. Namphol Sinkaset Chem 201: General Chemistry II
Entropy changes in irreversible Processes
Entropy and the 2nd Law of Thermodynamics
Catalyst.
Thermo Grudgeball.
Assign.# 6.5 – 2nd and 3rd Law of Thermodynamics
"Sometimes the best helping hand you can get is a good, firm push."
Entropy and Free Energy
Entropy = S Entropy is disorder randomness dispersal of energy.
Fun With Molar Ratios By: Andrea Moody.
Spontaneity, entropy and free energy
Entropy & Chemical Reactions
Entropy, Free Energy and Equilibrium
CHEM 433 – 10/4/11 III. 1st Law: Adiabatic changes (2.6)
CHEM /13/11 III. 2nd Law of THERMO
THE 2ND LAW OF THERMODYNAMICS:
CHEM /26/11 IV. 2nd Law of THERMO
III. 1st Law: CHEM 433 – 10/6/11 Thermochemistry ( )
Spontaneity, entropy and free energy
Topics 5 & 15 Chemical Thermodynamics
Lecture 3 Entropy and result of Euler’s homogenous theorem
- “Heats” of formation fH° - DrU° ?
Reaction Energy.
Thermodynamics tutorhour 6
Spontaneity, entropy and free energy
Presentation transcript:

CHEM 433 - 10/21/11 IV. 2nd Law of THERMO • T dep. (heating/cooling) -> Measuring Sm° • Third Law Entropies - Standard Reaction Entropy Changes - T dependence of S READ: CHAPTER 3… HW #5 : Out when I get that far ….

How to Measure molar S: Measure Cp at many T’s Fit to equaiton and integrate: Cp/T dT) Measure trsS (as needed) Add up all the parts - this is eqn on board…

(s) vs. (l) vs. (g) ? Size ? (Svib) Why are most gas values lower than CO2? (this is subtle…)

Nernst Heat Theorem: For all processes: S —> 0 as T—> 0 (provided all substances are perfectly crystalline…) 3rd LAW of THERMO: The entropy of a perfect crystalline substance is zero at T=0. 3rd Law Entropies: Molar entropies obtained via assumption that S(0) = 0 (for all substances - crystalline or otherwise…)

For: H2 (g) + 1/2 O2 (g) —> H2O (l) rS° = ? Sm° = 130.7 205.0 69.9 (all values in J/K mol at 298K…) BTW: we calculated Ssurr for this last week - —> ~1000 J/K mol (or 1 kJ/K mol)