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ENGR 145: Chemistry of Materials

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Additional Information for Exam 3. Covers lectures 19 25 and the associated reading assignments ... Hypoeutectic Pb-Sn alloy at still higher magnification. ... – PowerPoint PPT presentation

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Title: ENGR 145: Chemistry of Materials


1
Additional Information for Exam 3
  • Covers lectures 19 25 and the associated
    reading assignments
  • Arrive 5 minutes before the start of the exam
  • Closed-book, closed notes
  • Bring
  • Calculators (memories cleared)
  • Notes on a 3 x 5 card
  • Periodic table (handed out in recitation)
  • Provided
  • Fundamental constants

2
Reading Assignments for Exam 3
  • Lecture 19 Enthalpy Changes in Nonreacting
    Systems
  • OGN 5.4-5.6 7.2, 7.3, 7.5
  • Lecture 20 Spontaneous Processes Thermodynamic
    Equilibrium
  • OGN 7.5, 8.1, 8.3, 8.5-8.7
  • Lecture 21 Chemical Equilibrium
  • OGN 9.1-9.2 eq. 9.4 9.6
  • Lecture 22 Two-Component Systems Continuous
    Solid Solutions
  • Callister 9.1-9.6
  • Lecture 23 Two-Component Systems Eutectics
  • Callister 9.7
  • Lecture 24 Equilibrium vs. Nonequilibrium
    Solidification
  • Callister 9.6-9.7 lecture slides
  • Lecture 25 Role of Bonding in Binary Systems
    Practice Problems
  • Lecture slides

3
First Law of Thermodynamics OGN 7.3
  • Heat and work are forms in which energy is
    transferred into or out of a system
  • The first law of thermodynamics describes the net
    change in energy ?E transferred between a system
    and its surroundings
  • q heat into system w work done by
    surroundings
  • One type of work change in volume ?Vsurr (
    ?Vsys) at constant pressure P
  • Net change in energy of system at constant
    pressure

4
Calorimetry Two Approaches OGN 7.3
OGN Fig. 7.6
?HP M ? cS ? ?T
5
One-Component Phase Diagrams OGN Ch. 5
  • Similar one-component diagrams for many different
    substances

OGN Fig. 5.21
6
Zeroes of Enthalpy and Entropy
  • Elements in standard states at 298.15 K have H
    0
  • Pure substances in equilibrium form as T ? 0 K
    have S ? 0

7
Enthalpy vs. T for a Phase Transition
  • H vs. T for melting of a solid or freezing of a
    liquid
  • Slopes heat capacities (Which is larger? Why?)
  • ?H _at_ Tm ?Hf

Cliq
?Hf
enthalpy
Csolid
8
Computing Entropy Changes OGN 8.5
  • How to compute ?S for a process
  • Examples for several types of processes OGN 8.5
  • ?S in n moles of a single nonreacting phase
    during a ?T (P const)
  • ?S for a phase transition

9
Gibbs Free Energy for a Phase Transition OGN
8.7
At constant T and P
spontaneous(may happen)
reversible (may go either way)
non-spontaneous (can happen only with energy
input from surroundings)
10
?G for a Phase Transition OGN 8.7
  • Example melting freezing of H2O

?Gfreezinggt0
?Gfreezinglt0
At 0C (273 K)
OGN Fig. 8.10
11
?G for a Chemical Reaction OGN 8.7
  • Competition between
  • the change in enthalpy (?H) and the change in
    entropy (?S)
  • At eqm, ?G 0 ? Teqm ?H /?S
  • At other temperatures, sign of ?G depends on
    signs of ?H and ?S

12
?G, K, T (OGN eq. 9.4)
  • The temperature dependence of K is linked to ?G
  • for reaction aA bB ? cC dD with Gibbs f.e. ?G
  • K exp?G/RT or
  • ?G lt 0 ? K gtgt 1 ? C and D are predominant
  • ?G gt 0 ? K is small A and B are predominant

for solutions
for gases
13
Examples of Le Châteliers Principle
  • Ti(s) O2(g) ? TiO2(s)
  • Reaction consumes gas from left to right
  • P? ? reaction shifts to right (i.e., to relieve
    pressure)
  • pO2? ? reaction shifts to right (i.e., to consume
    oxygen)
  • ?Hf,298 944.7 kJ mol1 (exothermic) ?Sf,298
    185.33 J mol1 K1 (lt0, as expected
    why?)?Gf,298 889.4 kJ mol1 K pO21
    exp?G/RT K' exp?H/RT
  • T? ? K? ? reaction shifts to left (to less heat
    given off) (toward reduction of the oxide to
    metal)

14
Lever Rule Callister 9.6
  • Tie line tells
  • phases present (L, ?)
  • their compositions (CL, Ca)
  • how much of each phase is present for a given
    overall composition C0 the lever rule

liquidus curve
solidus curve
Callister Fig. 9.2b
15
Review Equilibrium Solidification of a Solid
Solution
  • Assume equilibrium is reached at each temperature
  • Above liquidus T 100 liquid
  • Initial formation of solid
  • Continued growth of solid
  • Solidification of last liquid
  • Below solidus T 100 solid
  • Note
  • Composition of liquid changes from 35 wt Ni to
    23 wt Ni
  • Composition of solid changes from 49 wt Ni to
    35 wt Ni
  • These processes take time

16
Nonequilibrium Solidification of a Solid Solution
  • Start from 100 liquid
  • Initial solid forms eqm compn
  • Additional solid eqm compn at surface only
  • Average solid compn is Ni-rich (dotted line)
  • ? remaining liquid is Cu-rich
  • Last liquid has a lower liquidus temperature than
    at eqm
  • Grains have compositional gradients from interior
    to grain boundary
  • On reheating, grain boundaries may melt
    prematurely

17
The Pb-Sn System Callister 9.7
  • Solidification of 15 wt Sn, 85 wt Pb
  • 100 liquid
  • L??
  • 100 ?
  • So far, process is same as solidification of a
    continuous solid solution
  • ? ? ??
  • Solvus reaction
  • ? precipitates inside ? grains

Callister Fig. 9.10
18
The Pb-Sn System Callister 9.7
  • Solidification of eutectic liquid (CE)
  • L ? ? ? at TE
  • Characteristic lamellar (layered) microstructure
  • Individual eutectic colonies nucleate and grow
    until they impinge on each other

Callister Fig. 9.11
Callister Fig. 9.12
19
The Pb-Sn System Callister 9.7
  • Solidification of 40 wtSn, 60 wt Pb
  • A hypoeutectic composition (C lt CE)
  • 100 liquid
  • Primary ? forms
  • Last primary forms
  • Eutectic reaction

Callister Fig. 9.14
Callister Fig. 9.15
20
Real Data EMSE 270, Spring 06
  • Hypoeutectic Pb-Sn alloy at still higher
    magnification. Light phase ? (Pb-rich) dark
    phase ? (Sn-rich)

? precip-itates in primary ? why?
eutectic
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