Dec. 2-6, 1999 NAME____________________________

 

Chemistry 103 Section 1

HOUR EXAM #3

IMPORTANT NOTE--For partial credit be sure to show your reasoning. Please attempt all parts of all problems--no partial credit can be given for blank spaces. If you need more paper, write on the back of the page and indicate clearly that you have done so.

 

There will be no partial credit on Problem 1.

 

(Possibly Useful )Physical Constants and Equations

NA = 6.022 x 1023 R = 3.29 x 10 15 Hz

Mass of an electron 9.1 x 10-28 g RH = 2.179 x 10-18 J

Mass of a proton 1.67 x 10-24 g h = 6.63 x 10 -34 J.s/photon

Mass of a neutron 1.67 x 10-24 g c = 3.0 x 10 8 m/s

E = hn E= hc/l E = -RH/n2

H = E + PV Q = m Cp (DT)

Density of water 1.00 g/cm3 Specific heat of water = 4.18 J/g °C

R= 8.3145 J/(K.mol) R= 0.082 L.atm/(K.mol)

0°C = 273.15 K PV = nRT v = (3RT/M)1/2

Volume of a sphere = 4/3 p r3 Pythagorean Theorem c2 = a2 + b2 p = 3.14

 

Problem 1 ________

Problem 2 ________

Problem 3 ________

Problem 4________

Problem 5 ________

Total _________

 

 

 

NAME____________________________

Problem 1 (20 points)

Multiple choice problems. (Correct answers are in red)

1) Label each picture as solid, liquid, or gas.

____liquid_____ __gas________ ___solid____

 

2) Mercury is a liquid at room temperature because

a) its atoms are strongly hydrogen bonded to each other

b) it is very dense.

c) there are strong London Dispersion Forces between mercury atoms.

d) It is used to make barometers and manometers.

3) In Molecular Orbital theory, the antibonding orbitals

  1. have maximal electron density in between the two nuclei
  2. have lower energy than the corresponding bonding orbitals
  3. never have electrons in real, stable molecules
  4. are of higher energy than the corresponding bonding orbitals.

4) HI has a higher boiling point than HBr because

  1. HBr is more polar.
  2. HI is more polar.
  3. HI makes stronger hydrogen bonds.
  4. HI has stronger London Dispersion forces due to its larger molar mass.

 

5) If a substance is above its critical temperature then

  1. it must be a solid
  2. it has very high surface tension
  3. it must be a supercritical liquid
  4. it cannot be a liquid
    5.

     

     

    NAME____________________________

    Problem 2 (30 points)

    In class we calculated that the unit cell of a face centered cubic structure which is a closest packed structure is 26% empty space. Calculate the percentage of empty space in the unit cell of a simple cubic lattice.

  5. Describe your problem-solving strategy in three steps using words and pictures.
  1. Each cube has 8 x 1/8 = 1 spherical atoms per cell
  2. The atoms touch along the edge whose length is 2r
  3. The fraction of space occupied is volume of 1 atom/volume of the cube

 

 

b) Solve the problem using equations and numbers.

Volume of 1 atom = 4/3pr3

Volume of 1 cube = (2r)3= 8r3

% empty space = 100x (1-4/3pr3/8r3)= 48 %

_________48___ % empty space.

 

  1. Is your answer reasonable? What upper or lower limits do you expect?

The answer seems reasonable. We expect an answer greater than 26% and less than 100%.

 

Problem 3 (20 points)

Suppose that a student who is surrepticiously eating her lunch during Dr. Lukacs' demonstration of frozen Memory Metals accidentally swallows a drop of spilled liquid nitrogen, N2 (l) . Assume that one drop of liquid nitrogen has a density of 1.149 g/mL and a volume of 0.050 mL. What volume of gas will be produced in the student's stomach at a temperature of 37°C and 1 atm.?

Strategy

  1. Use the density and volume to find the mass and number of moles of nitrogen
  2. Use the Ideal Gas Law to find the volume.

Mass = 1.149 g/mL x 0.050 mL = .05745 g

Moles nitrogen = 0.05745 g x 1mole/ 28.02 g = .00205 moles

PV = nRT so V = nRT/P = .00205 moles x .082 Latm/K mol x (273 + 37) / 1 atm

= .052 L or 52 mL

 

Problem 4 (20 points)

 

 

 

 

 

 

 

 

For all O2 species For all C2 species

The molecular orbitals for O2 and C2 are arranged in energetic order.

  1. The bond order for O2+ is (8-3)/2 = 2.5 11 e
  2. s2s2 s2s*2 s2p2 p2p2 , p2p2 p2p*1
  1. The bond order for O2 is ___(8-4)/2 = 2_______12 e
  2. s2s2 s2s*2 s2p2 p2p2 , p2p2 p2p*2
  3. The bond order for O2 - is ____(8-5)/2 = 1.5_____13 e
  4. s2s2 s2s*2 s2p2 p2p2 , p2p2 p2p*2, p2p*1
  1. The bond order for C2+ is __(5-2)/2= 1.5________ 7 e s2s2 s2s*2 p2p2 , p2p1
  2. e) The bond order for C2 is ___(6-2)/2 = 2_____ 8e s2s2 s2s*2 p2p2 , p2p2

f) The bond order for C2 - is ___(7-2)/2= 2.5_____ 9e s2s2 s2s*2 p2p2 , p2p2 s2p1

Does loss of an electron result in a stronger (for O) or weaker (for C) bond?

Does gaining an electron result in a stronger (for C) or weaker (for O) bond?

Can you generalize? If not, explain why.

The answers depend on whether the highest occupied molecular orbital (HOMO) and the lowest unoccupied molecular orbital (LUMO) are bonding or antibonding. For O2 the highest HOMO is antibonding so gaining an electron weakens the bond whereas losing an electron strengthens the bonds. For C2 the HOMO and LUMOs are bonding orbitals so gaining electrons strengthens the bond and losing electrons weakens it.

Problem 5 (10 points)

Explain why a balloon filled with Helium gas deflates more quickly than an identical balloon filled with air. Assume that air is composed of O2 and N2 and that both balloons are at the same temperature, volume, and pressure.

Balloons deflate as gases leak out via effusion.

Because Helium has a much lower molar mass (4 g/mol) than does N2 (28 g/mol) it atoms are travelling much faster than are the nitrogen molecules. Therefore the rate of effusion through the balloon material will be much faster.