CHEM 35

General Chemistry

EXAM #2

October 18, 2000

 

Name:         Anne Serkey               

 

SSN:                                                 

 

Lab T.A.:                                          

 

INSTRUCTIONS:  Read through the entire exam before you begin. Answer all of the questions. For questions involving calculations, show all of your work -- HOW you arrived at a particular answer is MORE important than the answer itself! Circle your final answer to numerical questions.

 

The entire exam is worth a total of 150 points. Attached are a periodic table and a formula sheet jam-packed with useful stuff. Good Luck!

 

Page

Possible Points

Points Earned

2

20

20

3

30

30

4

15

15

5

20

20

6

15

15

7

20

20

8

30

30

TOTAL:

150

150

 

1.     (10 pts) Pure acetic acid (HC2H3O2; MW = 60.05 g/mol) is a liquid with a density of 1.049 g/mL at 25 oC. Calculate the molarity (mol/L) of a solution of acetic acid made by dissolving 15.00 mL of pure acetic acid at 25 oC in enough water to make 100.0 mL of solution.

 

 

 

 

 

15.00 mL HAc x 1.049 g HAc x 1 mol HAc x 1000 mL = 2.620316

100.0 mL Soln      mL HAc    60.05 g HAc     L       mol HAc/L

 

                                           = 2.630 M HAc

 

 

 

 

 

 

 

 

2.     (10 pts) How many mL of 0.155 M HCl are needed to react completely with 35.0 mL of a 0.101 M Ba(OH)2 solution? The unbalanced reaction equation is:

 

HCl(aq) + Ba(OH)2(aq) ® H2O(l) + BaCl2(aq)

 

 

 

 

Balanced:     2HCl(aq) + Ba(OH)2(aq) ® 2H2O(l) + BaCl2(aq)

 

35.0 mL Ba(OH)2   L   x 0.101 mol Ba(OH)2 x 2 mol HCl x     L        x 1000 mL =

               1000 mL         L          1 mol Ba(OH)2 0.155 mol HCl      L

 

                                                      = 45.6129 mL HCl

                                           = 45.6 mL HCl

 

3.      (10 pts) The Hindenburg was a famous hydrogen-filled dirigible that exploded in 1937. If the Hindenburg held 2.0 x 105 m3 of hydrogen gas (H2) at 25. oC and 1.0 atm, what mass (kg) of hydrogen was present?

 

V = 2.0 x 105 m3 x 1000 L = 2.0 x 108 L

                   1 m3

 

n = PV =      (1.0 atm)(2.0 x 108 L)    =  8.174547 x 106 mol H2

    RT   (0.08206 L-atm/mol-K)(298.15 K)

 

8.1745 x 106 mol H2 x 2.01588 g H2 x    1 kg  = 1.64789 x 104 kg

                        mol H2       1000 g

                                            = 1.6 x 104 kg H2

 

4.     (10 pts) Calculate the molar mass (g/mol) of a gas if 3.75 g occupies 0.935 L at 430. torr at 35.0 oC.

 

430 torr x 1 atm   = 0.56579 atm

          760 torr

 

n = PV =    (0.56579 atm)(0.935 L)      = 2.09205 x 10-2 mol

    RT   (0.08206 L-atm/mol-K)(308.15 K)

 

Molar mass =     3.75 grams    = 179.25 g/mol = 179. g/mol

             2.09205 x 10-2 mol

 

5.     A mixture containing 0.538 mol He(g), 0.315 mol Ne(g), and 0.103 mol Ar(g) is confined in a 5.00-L vessel at 25.0 oC.

 

a.      (5 pts) Calculate the total pressure (atm) of the mixture.

 

Ntotal = 0.538 + 0.315 + 0.103 = 0.956 mol

 

Ptotal = nRT = (0.956 mol)(0.08206 L-atm/mol-K)(298.15 K)

        V                 5.00 L

 

           =  4.6779 atm = 4.68 atm

 

b.     (5 pts) Calculate the partial pressure (atm) of He in the mixture.

 

PHe = XHePtotal     XHe = nHe/ntotal = 0.538 mol = 0.562762

                                0.956 mol

PHe = (0.562762)(4.6779 atm)

   = 2.63256 atm = 2.63 atm

6.     (5 pts) A balloon made of rubber that is permeable to small molecules is filled with helium to a pressure of 1 atm. This balloon is then placed in a box that contains pure hydrogen (H2), also at a pressure of 1 atm. Will the balloon expand or contract? Explain.

 

The balloon with EXPAND. The rate of effusion of the H2 is greater than that of He, due its lower molar mass (recall: r1/r2 = [(M2/M1)]½), so H2 will enter the balloon faster than He will leave the balloon and it will expand.

 

7.     Vessel A contains CO gas at 0 oC and 1 atm. Vessel B contains CO2 gas at 20 oC and 0.5 atm. The two vessels have the same volume.

 

a.      (3 pts) Which vessel contains more molecules? Briefly explain.

 

VESSEL A – Since the absolute temperatures are almost the same ( 273 versus 293 K), we can say that the number of moles of gas is proportional to the pressure. Since PA = 2 x PB, there are more moles of gas in vessel A.

 

b.     (3 pts) In which vessel is the average kinetic energy of the molecules greater? Briefly explain.

 

VESSEL B – Kinetic energy is proportional to temperature, so the gas at the greater temperature will have the greater average kinetic energy. TB > TA, so the molecules in vessel B have a greater K.E.

 

c.      (4 pts) In which vessel is the rms speed of the molecules greater? Briefly explain.

 

VESSEL A – We need to take into account both the temperature and the  molar mass, as vrms is proportional to (T/M)½. The molar mass difference is greater than the temp difference, so the vessel with the lighter gas molecules (vessel A) will contain the faster molecules (even though they are at a slightly lower temp).

 

8.     The ideal gas law is often used under conditions in which gases are unlikely to behave in an ideal fashion.

 

a.      (5 pts) Calculate, using the ideal gas law, the pressure (atm) of 40.0 mol of argon gas contained in a volume of 2.00 L at 300 K.

 

P = nRT = (40.0 mol)(0.08206 L-atm/mol-K)(300 K) = 492.36 atm

     V                  2.00 L

                                               = 492. atm

 

b.     (8 pts) Now, using the van der Waals equation, calculate the pressure (atm) of the same argon gas sample.

 

P =  nRT  - an2

    V-nb     V2

 

  = (40.0 mol)(0.08206 L-atm/mol-K)(300 K) – (1.36 atm-L2/mol2)(40.0 mol)2

      (2.00 L – (40.0 mol)(0.0320 L/mol)              (2.00 L)2

 

  =  1367.67 atm – 544.00 atm  =  823.67 atm  = 824. atm

 

c.      (7 pts) If there is a greater than 10% difference in the two pressure values, explain why the ideal gas law fails. If there is less than a 10% difference in the two pressure values, indicate under what conditions you would expect to find a more significant difference between the two calculated pressures.

 

The pressure difference between the IGL and the VDW equation is much more than 10% under these conditions (it’s > 40%). The IGL fails under conditions of high pressure and low temperature and we have very high pressure here. The VDW equation corrects for the IGL’s ignorance of both the volume occupied by the gas molecules themselves as well as their intermolecular attraction. The former would tend to elevate the pressure by increasing intermolecular repulsion, while the latter would tend to decrease the pressure by decreasing the volume occupied by the gas species clustering closer together (due to attractive forces). Clearly, the repulsive forces dominate here as the VDW equation pressure is nearly double that predicted by the IGL.

 

 

9.     Consider two gas samples: He and Ar at 300 K.

 

a.      (10 pts) Sketch the Maxwell-Boltzmann speed distribution profile for both gases; use the same set of axes for both gases. Calculate the rms speed for both gases and indicate the location of these values on the plot. Please note: the ONLY quantitative aspect of this plot should be these rms speeds – the rest should be strictly a qualitative sketch showing clearly any differences between the two gases.

 

vrm (Ar) = (3RT/M)½ = [3(8.3145 J/mol)(300K)/(39.948 x 10-3 kg/mol)]½

vrm (Ar) = 433 m/s

 

vrm (He) = (3RT/M)½ = [3(8.3145 J/mol)(300K)/(4.0026 x 10-3 kg/mol)]½

vrm (He) = 1370 m/s

 

F(v)
 
 

 

 

 

 

 

 

 

 

 

 

 


b.     (5 pts) How would the speed distribution for Ar be affected by an increase in temperature to 1000 K? Sketch speed distribution profiles for Ar (on a new set of axes, please) at the initial temperature (300 K) and at 1000 K. As before, use the calculated rms speed as the sole quantitative calibration point for these qualitative speed distribution plots.

 

vrm (Ar) = (3RT/M)½ = [3(8.3145 J/mol)(1000K)/(39.948 x 10-3 kg/mol)]½

vrm (Ar) = 790 m/s

F(v)
 
 

 

 

 

 

 

 

 

 



10. Each of the following reactions produces a gas as one of its products. For each reaction: write a balanced reaction equation, classify the reaction as being a precipitation, acid/base, or oxidation/reduction reaction, and give evidence to support your classification by either identifying the precipitate compound, the acid and the base, or by indicating which of the reactants are oxidized and which are reduced.

 

a.     CaCO3(s) + HNO3(aq) ®  products

 

Balanced reaction equation (3 pts):

 

          CaCO3(s) + 2 HNO3(aq) ® Ca(NO3)2(aq) + H2O(l) + CO2(g)

 


Reaction type (1 pt):    precipitation        or       acid/base   or       redox  (circle one)

 

Evidence (1 pt):

Acid: HNO3  Base: CaCO3

 

b.     Cu(s) + HNO3(aq) ® products

 

Balanced reaction equation (3 pts):

 

          Cu(s) + 2 HNO3(aq) ® Cu(NO3)2(aq) + H2(g)

 


Reaction type (1 pt):    precipitation        or       acid/base   or       redox  (circle one)

 

Evidence (1 pt):

       Oxidized: Cu     Reduced: H+

 

11. (2 pts each) Determine the oxidation state of sulfur in each of the following compounds:

 

a.      H2S  -2

 

b.     S8   0

 

c.  SCl2    +2

 

d.  Na2SO3  +4

 

e.      SO42-   +6

 

12.    A sample of 70.5 mg of K3PO4 (MW = 212.266 g/mol) is added to 15.0 mL of 0.050 M AgNO3, resulting in the formation of a precipitate.

 

a.      (5 pts) Write a balanced net ionic equation for this reaction.

 

     3 Ag+(aq) + K3PO4(s) ® Ag3PO4(s) + 3 K+(aq)

 

b.     (5 pts) What is the limiting reactant in this reaction.

 

0.0705 g K3PO4 x    mol K3PO4       x 3 mol AgNO3  x      L       x 1000 mL =

                 212.266 g K3PO4     1 mol K3PO4   0.050 mol AgNO3    L

 

                                                            19.9 mL AgNO3

                                                                                (needed)

            Only 15 mL AgNO3 available, so

                AgNO3 is LIMITING REACTANT

 

c.      (5 pts) Calculate the mass (g) of precipitate formed in this reaction

 

0.0150 L AgNO3 x  0.050 mol AgNO3 x 1 mol Ag3PO4  x 418.576 g Ag3PO4 =

                        L          3 mol AgNO3          mol Ag3PO4

 

                                                      = 0.10464 g Ag3PO4

                                           = 0.10 g Ag3PO4

 

13.  (10 pts) Determine the Lewis structures (yes, there is more than one) for SO2, and compute oxidation numbers and formal charges for the sulfur and oxygen atoms.

                   _        _

16 e- ->    |O = S – O|  «  |O - S = O|

 

Oxidation Numbers: O ® -2     S ® +4

 

Formal Charges: fc = #valence electrons – (#bonds + # nonbonded e-)

 

          S = 6 – (3 + 2) = +1

          O = 6 – (2 + 4) = 0 (double-bonded oxygen)

          O = 6 – (1 + 6) = -1 (single-bonded oxygen)

 

   b.(5 pts) Do the oxidation numbers or the formal charges more realistically indicate the charge distribution within the molecule? Explain.

 

Oxidation numbers assume bonds are totally ionic, while formal charges assume bonds are mostly covalent. In SO2, bonding is between two non-metals and, so, is mostly covalent. Thus, the formal charges are better indicators of the charge distribution.

Betcha thought that I forgot all about the

EXTRA CREDIT! – 10 pts

 

A newspaper article about the danger of global warming from the accumulation of greenhouse gases such as carbon dioxide states that "reducing driving your car by 20 miles a week would prevent release of over 1000 pounds of CO2 per year into the atmosphere."  Is this a reasonable statement? Assume that gasoline is octane (C8H18 - MW = 114.1502 g/mol) and that it is burned completely to CO2 and H2O in the engine of your car. Note that you will need to make some reasonable guesses as to the gas mileage of your car, the density of octane, etc.

 

      2 C8H18 + 25 O2 ® 16 CO2 + 18 H2O

 

Assuming: 20 miles/gallon

          4 quarts/gallon

          1 L = 1 quart

          density of gas = density of water = 1 g/mL

 

20 miles x  gallon  x 52 weeks = 52 gallons/year

 week      20 miles    year

 

52 gallons x  4 quarts  x  1 L  x  1000 mL  x 1 g C8H8 =

               gallon      1 qt       L         mL

 

                                  = 2.08 x 105 g C8H18

 

2.08 x 105 g C8H18  x    1 mol C8H18 x 16 mol CO2 x 44 g CO2 =

                     114.15 g C8H18   2 mol C8H18   mol CO2

 

                                  = 6.414 x 105 g CO2

 

6.414 x 105 g CO2 x       kg       x 2.2046 lb =

                       1000 g          kg

 

                             = 1.414 x 103 lb CO2

 

                             = 1,400 lb CO2/year

 

                        Yes! The statement is

                             Reasonable.