Chem 163 chapter 20
This presentation is the property of its rightful owner.
Sponsored Links
1 / 24

CHEM 163 Chapter 20 PowerPoint PPT Presentation


  • 104 Views
  • Uploaded on
  • Presentation posted in: General

CHEM 163 Chapter 20. Spring 2009. 3-minute exercise. Is each of the following a spontaneous change? Water evaporates from a puddle A small amount of sugar dissolves in hot tea Methane burns in air A hamburger becomes uncooked. Thermodynamics. First Law: Law of Conservation of Energy.

Download Presentation

CHEM 163 Chapter 20

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Chem 163 chapter 20

CHEM 163Chapter 20

Spring 2009


3 minute exercise

3-minute exercise

Is each of the following a spontaneous

change?

  • Water evaporates from a puddle

  • A small amount of sugar dissolves in hot tea

  • Methane burns in air

  • A hamburger becomes uncooked


Thermodynamics

Thermodynamics

First Law:

Law of Conservation of Energy

Limitation:

Internal E of a system

Explains change, but not direction

heat

work

Second Law:

Systems change towards more disorder


Spontaneous change

Spontaneous Change

  • Change that occurs without continuous E input

  • Change can only be spontaneous in one direction, under a given set of conditions

  • Enthalpy (∆H):

    • heat gained or lost at constant P

    • Sign of ∆H

      • Exothermic or endothermic

      • No information about spontaneity

  • Entropy (∆S):

    • Freedom of particle motion (dispersed E of motion)


Energy levels

Energy Levels

  • Each atom or molecule has quantized E levels

    • Electronic

    • Kinetic

      • Vibrational, rotational, translational

  • Microstate: combined E at any given point

    • each microstate is equally possible (equal E)

      for a given set of conditions

Number of microstates

entropy

Boltzmann constant

(J/K)

= R/NA

= 1.38 x 10-23 J/K


Entropy change microstates

Entropy Change: Microstates

for 1 mol


Entropy change heat changes

Entropy Change: Heat Changes

  • Remove 1 grain of sand

  • Gas does work on piston

    • absorbs heat to maintain E

  • Works for tiny changes (totally reversible)


Always increasing entropy

Always Increasing Entropy

All real processes occur spontaneously in direction that increases the entropy of the universe.

  • Perfect crystal at T = 0 K has S = 0

    • 1 microstate

  • Standard Molar Entropy (S°)

    • S increase from 0 to standard state

      • 1 atm (gases)

      • 1 M (solutions)

      • Pure substance, most stable form (liquids/solids)


What affects s

What affects S°?

As # of microstates (or kinetic E) increases, S increases

  • Temperature change

    ↑ T ↑ S

  • Phase change

    absorb heat ↑ S


Chem 163 chapter 20

  • Dissolution

Ions: increased S, except small, highly charged ions

Molecules (solid or liquid): ∆S ≈ 0

Gases: decreased S

  • Atomic Size

    • heavier atoms

    • allotropes

 closer E levels

 more microstates

Molecular Complexity

more complex

 more types of movement

 more microstates

Only applies to molecules in same physical state


3 minute practice

3-minute practice

What is the sign of ∆Ssys?

  • A pond freezes in winter

  • Atmospheric CO2 dissolves in the ocean

  • 2 K (s) + F2 (g)  2KF (s)


Chem 163 chapter 20

Standard Entropy of Reaction

Increasing disorder:

Decreasing disorder:

Predict ∆Srxn:

Change in # moles of gas

Calculate ∆Srxn:

N2 (g) + 3H2 (g)  2 NH3 (g)


S universe

∆Suniverse

Decrease in ∆Ssys only if greater increase ∆Ssurr

System acts as heat sink or drain

  • Exothermic

  • Endothermic

at constant P

Measure ∆Hsys to determine ∆Ssurr


Spontaneous at 298 k

Spontaneous at 298 K?

3

2

Balance equation!

Calculate ∆Ssys

Calculate ∆Hsys

Calculate ∆Ssurr

Calculate ∆Suniv

N2 (g) + H2 (g)  NH3 (g)


Entropy at equilibrium

Entropy at Equilibrium

Approaching equilibrium:

At equilibrium:

No net change


5 minute practice

5-minute Practice

Calculate ∆Ssys for the combustion reaction of ammonia (producing nitrogen dioxide and water vapor).


Gibbs free energy

Gibbs Free Energy

  • Measure of spontaneity

  • Combines enthalpy and entropy

Spontaneous if…

∆Suniv > 0

∆Gsys < 0

T∆Suniv > 0

-T∆Suniv < 0


Chem 163 chapter 20

∆G

Spontaneous process

Nonspontaneous process

Process at equilibrium

Standard Free Energy Change

Standard Free Energy of Formation :

E change when 1 mol of compound is made from its elements in standard states

Element in standard state:


G and work constant t p

∆G and work (constant T & P)

  • Nonspontaneous process:

    • Process may occur if work is done to the system

    • How much work is needed?

w = ∆G

  • Spontaneous process:

    • ∆G = maximum useful work done by the system

  • wmax only if process is totally reversible

  • Actually does less w <wmax

    • Extra E lost as heat


Useful work

Useful Work

  • Excludes work done by or on atmosphere

  • Some free energy is always lost to heat

∆Hsys?

< 0

∆Ssys?

> 0

∆Gsys = wmax that can be done by system

∆Gsys?

< 0

∆Gsys > w actually done by system

In some multistep reactions, ∆G from one reaction can cause an otherwise nonspontaneous reaction to occur.

“coupling of reactions”


What about t

What about T?

  • Typically ∆H > T∆S

  • For ∆G to be negative, need ∆H to be…

  • What about at high T?

negative

  • T∆S term can dominate

  • Sign of ∆S becomes important

4 situations:

∆H < 0 & ∆S > 0

∆H > 0 & ∆S < 0

∆H > 0 & ∆S > 0

∆H < 0 & ∆S < 0

∆G < 0

∆G > 0

∆G > 0 at low T;

∆G < 0 at high T

∆G < 0 at low T;

∆G > 0 at high T


High t v low t

High T v. Low T?

Spontaneous “limit” at what temperature?


Equilibria and g

Equilibria and ∆G

  • If Q < K, reaction…

  • If Q > K, reaction…

  • If Q = K, at equilibrium

Q/K

< 1

∆G < 0

Q/K

> 1

∆G > 0

Q/K

= 1

∆G = 0

proportional

0

Make Q standard state (all values = 1)


Homework due monday may 11 th

Homework due MONDAY, May 11th

Chap 20: #19, 26, 30, 41, 50, 52, 58, 70, 78, 106


  • Login