Chem 1310: Introduction to physical chemistry Part 0: Some preliminaries

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Chem 1310 Introduction to physical chemistry
Part 0 Some preliminaries

• Peter H.M. Budzelaar

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Course Outline

• This course is about thermodynamics and
  kinetics.
• Both help you to make quantitative statements
  about chemical reactions.
• Thermodynamics is concerned with initial and
  final states (I and F).
• Kinetics studies how you get from I to F.

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Course Outline

• Chapter 6 is about energy.
• Chapter 7 handles kinetics.
• Chapters 14-17 discuss equilibria, in general and
  more specific cases.
• Chapter 18 links energy with equilibria in a
  quantitative fashion.
• For details, see your printed outline.

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Things you should be able to do

• There are a few techniques you should have
  mastered in high school or in CHEM1300. If you
  still have problems with them, you will have a
  serious problem completing this course.
• To test this, do the exercises on the next slide,
  in class, on paper, anonymously.
• Then grade them yourself, using the answers
  provided, and hand in the corrected answers.

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Things you should be able to do

1. Balance the following equation ?C3H8 ?O2
   ?CO2 ?H2O
2. Calculate the molecular weight of C6H5Cl.
3. 21 g of glucose (C6H12O6) is dissolved in 320 mL
   of water. What is the final concentration of
   glucose (in mol/L)?
4. 200 mL of this solution is diluted to 730 mL.
   What is the new concentration?
5. How much chlorine (in g) is present in 1 L of
   chlorine gas at 1 atm, room temperature?

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Things you should be able to do

1. C3H8 5 O2 3 CO2 4 H2O
2. C6H5Cl 612.01 51.01 35.45 112.56
3. (21/180.18) mol / 0.32 L 0.36 mol/L
4. (0.20.36) mol / 0.73 L 0.099 mol/L
5. Ideal gas 1 mol is 22.414 L at 0C, ca 24 L at
   20C. 1 L Cl2 (1/24) mol (1/24) 235.45 g
   2.95 gIf you assumed 1 mol is 22.414 L, don't
   worry but remember to correct for the temperature
   next time!

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Linear equation in 1 unknown

• a x b c
• a x c-b
• x (c-b)/a

30.3 x - 23.7 7.1 x 5.0 (30.3-7.1) x - 23.7
5.0 23.2 x - 23.7 5.0 23.2 x 28.7 x 1.24
(not 1.23707...)
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Quadratic equation

• a x2 b x c 0

2.0 x2 -7.0 x - 0.3 0
Notes a) often one solution is "unphysical" b)
if b2 4ac, you might loose precision
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Logarithms

• x log y Û y 10x
• log (ab) log a log b
• log (a/b) log a - log b
• log (an) n log a
• x ln y Û y ex

log 3.0 0.48 log 0.3 -0.52 log 30000
4.48 log x 0.7Þ x 100.7 5.0
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Logarithms (2)

• log 0 does not exist (would be -)
• c.f. 1/x
• log x does not exist for x lt 0
• c.f. Öx