Week 4 Key Practice Problems

Week 4 Key Practice Problems - CHEMISTRY 201X Practice...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHEMISTRY 201X September 10-14, 2007 Practice Problems (35) Chlorine, bromine, and iodine form pentafluorides. At 273 K, one of these is a solid, one is a liquid and one is a gas. Specify which is which, and explain your reasonings. Without knowing anything about the structures, we can assume that London dispersion forces will dominate. Then, at 273 K, IF 5 should be a solid , BrF 5 should be a liquid , and ClF 5 should be a gas. Melting Point (K) Boiling Point (K) IF 5 283 371 BrF 5 212 313 ClF 5 170 260 (36) How do expect the boiling point of hydrogen peroxide, H 2 O 2 , to compare with those of fluorine, F 2 , and hydrogen sulfide, H 2 S, which are two substances with molar masses comparable with that of hydrogen peroxide? H 2 O 2 (MW = 34 g/mol) can have hydrogen-bonding, so we can expect a higher boiling point (150.2 ° C) than for F 2 (MW = 38 g/mol) ( 188 ° C) and H 2 S (MW = 34 g/mol) ( 60.2 ° C). (37) The van der Waals constant, b (L/mol), is related to the volume per molecule in the condensed (liquid) state, and thus to the size of a molecule. The following combination of van der Waals constants, a / b , has units of L atm/mol. 1 L atm (= 101.325 J) is a unit of energy, so a / b is proportional to the strength of the attractive forces between molecules. Use the van der Waals constants below to rank the attractive forces from strongest to weakest. Gas a (L 2 atm/mol 2 ) b (L/mol) a / b (L atm/mol) H 2 0.2444 .02661 9.185 N 2 1.390 .03913 35.52 HCl 3.667 .04081 89.86 SO 2 6.714 .05636 119.13 So, the attractive forces increase in strength from H 2 , N 2 , HCl, SO 2 . This can be explained by both London dispersion forces and the effect of dipole-dipole forces for HCl and SO 2 over the nonpolar molecules H 2 and N 2 . (38)* Since the van der Waals constant b is related to the volume excluded per mole of molecules, it should be proportional to N A σ 3 , where N A is Avogadro’s number (molecules/mole) and σ is the distance of closest approach in the Lennard-Jones potential. (a) Use the following table of information and plot b vs. N A σ 3 . Do you see any correlation between the two values? Can you write a mathematical equation based on the graph?
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
CHEMISTRY 201X September 10-14, 2007 Practice Problems Gas a (L 2 atm/mol 2 ) b (L/mol) σ (10 10 m) ε (10 21 J) H 2 0.2444 0.02661 2.93 0.511 N 2 1.390 0.03913 3.70 1.312 O 2 1.360 0.03183 3.58 1.622 Ar 1.345 0.03219 3.40 1.654 CH 4 2.253 0.04278 3.82 2.045 Gas a (L 2 atm/mol 2 ) b (L/mol) N A σ 3 (10 2 L/mol) N A 2 εσ 3 (L 2 atm/mol 2 ) H 2 0.2444 0.02661 1.5147 0.0460 N 2 1.390 0.03913 3.0503 0.238 O 2 1.360 0.03183 2.7631 0.267 Ar 1.345 0.03219 2.3669 0.233 CH 4 2.253 0.04278 3.3568 0.408 b (10 2 L/mol) 2.6 2.8 3.0 3.2 3.4 3.6 3.8 4.0 4.2 4.4 N A σ 3 (10 2 L/mol) 1.0 1.5 2.0 2.5 3.0 3.5 4.0 N A σ 3 = 1.03 b 0.9547 (Nearly linear – O 2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 8

Week 4 Key Practice Problems - CHEMISTRY 201X Practice...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online