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### CHEM 163Chapter 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

Limitation:

Internal E of a system

Explains change, but not direction

heat

work

Second Law:

Systems change towards more disorder

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

- 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

for 1 mol

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

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°?

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

- Temperature change

↑ T ↑ S

- Phase change

absorb heat ↑ S

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

What is the sign of ∆Ssys?

- A pond freezes in winter
- Atmospheric CO2 dissolves in the ocean
- 2 K (s) + F2 (g) 2KF (s)

Increasing disorder:

Decreasing disorder:

Predict ∆Srxn:

Change in # moles of gas

Calculate ∆Srxn:

N2 (g) + 3H2 (g) 2 NH3 (g)

∆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?

3

2

Balance equation!

Calculate ∆Ssys

Calculate ∆Hsys

Calculate ∆Ssurr

Calculate ∆Suniv

N2 (g) + H2 (g) NH3 (g)

5-minute Practice

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

Gibbs Free Energy

- Measure of spontaneity
- Combines enthalpy and entropy

Spontaneous if…

∆Suniv > 0

∆Gsys < 0

T∆Suniv > 0

-T∆Suniv < 0

∆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)

- 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

- 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?

- 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?

Spontaneous “limit” at what temperature?

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 11th

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

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