Chem 1310: Introduction to physical chemistry Part 0: Some preliminaries Peter H.M. Budzelaar.

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Presentation transcript:

Chem 1310: Introduction to physical chemistry Part 0: Some preliminaries Peter H.M. Budzelaar

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.

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

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.

Things you should be able to do 1.Balance the following equation: □C 3 H 8 + □O 2  □CO 2 + □H 2 O 2.Calculate the molecular weight of C 6 H 5 Cl g of glucose (C 6 H 12 O 6 ) is dissolved in 320 mL of water. What is the final concentration of glucose (in mol/L)? 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?

Things you should be able to do 1.C 3 H O 2  3 CO H 2 O 2.C 6 H 5 Cl: 6* * = (21/180.18) mol / 0.32 L = 0.36 mol/L 4.(0.2*0.36) mol / 0.73 L = mol/L 5.Ideal gas: 1 mol is L at 0°C, ca 24 L at 20°C. 1 L Cl 2 : (1/24) mol = (1/24) * 2*35.45 g = 2.95 g If you assumed 1 mol is L, don't worry but remember to correct for the temperature next time!

Linear equation in 1 unknown a x + b = c a x = c-b x = (c-b)/a 30.3 x = 7.1 x +5.0 ( ) x = x = x = 28.7 x = 1.24 (not )

Quadratic equation a x 2 + b x + c = x x = 0 Notes: a) often one solution is "unphysical" b) if b 2 » 4ac, you might loose precision

Logarithms x = log y  y = 10 x log (a  b) = log a + log b log (a  b) = log a - log b log (a n ) = n log a x = ln y  y = e x log 3.0 = 0.48 log 0.3 = log = 4.48 log x = 0.7  x = = 5.0

Logarithms (2) log 0 does not exist (would be -  ) c.f. 1/x log x does not exist for x < 0 c.f.  x