This in turn increases the entropy of the Universe by 76A3OY

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									CHEM 1000 A and V                               Mid-Year Examination                                 December, 2009



                                       Mid Year Exam December 12, 2009


Part A. Answer all questions (5 marks each).



A1. Why is light of only certain wavelengths observed in an atomic spectrum?



The light is a result of transitions of electrons from one quantized energy to another. Since the energy levels
are quantized, so are the differences between them, and so are the photon energies.




A2. Why is a molecule with a triple bond in it likely to be very reactive?



A triple bond consists of one sigma-bond plus two pi-bonds. The pi-electrons are delocalized (far from the
nuclei) and so are not tightly bound. Since they are easily removed, this makes the molecule very reactive.




A3. What type of hybrid orbitals is the As atom using to bond with the Cl atoms in AsCl 3?



5 + (3 x 7) = 26 electrons. Thus it is AX3E, i.e. the As atoms has four electron groups around it. Thus, it is using
sp3 hybrid orbitals.


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CHEM 1000 A and V                               Mid-Year Examination                                   December, 2009




A4. Can Ca(s) be used as a reducing agent for Ni+2(aq)? Why or why not?



Yes. Ca is above Ni in the activity series. Thus, Ca is more easily oxidized that Ni. Thus, Ca gets oxidized and
Ni+2 gets reduced.




A5. Explain why a reaction having ΔH < 0 tends to make the entropy of the universe increase.



The heat from the system moves into the surroundings. This increases the entropy of the surroundings. This
in turn increases the entropy of the Universe, since the surroundings are part of the Universe.




A6. Suppose the reaction Zn(s) + Sn+2(aq) Ý Zn+2(aq) + Sn(s) is at equilibrium. Will diluting the solution with water
have an effect on the position of the equilibrium? Why or why not?



Dilution will have no effect since Q = [Zn+2] / [Sn=2], and dilution will not change the value of Q.




A7. Dissolving CuCl2(s) in H2O(l) to make Cu+2(aq) and Cl(aq) ions has ΔSo < 0. Why?



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CHEM 1000 A and V                            Mid-Year Examination                               December, 2009

Although the ions in CuCl2(s) have low entropy, making the aqueous ions reduces the entropy of the
associated water molecules even more.




A8. CuCl2(s) spontaneously dissolves in water even though ΔSo < 0. Why?



Because ΔHo < 0




A9. Why are the ionization potentials of phosphorus (P) and sulphur (S) approximately equal?



P has three unpaired p-electrons. Although S has four, and the fourth should be more easily removed since it
is paired with one other, thus decreasing the ionization potential, the S atom also has one more proton,
making the effective nuclear charge greater, increasing the ionization potential. The net effect is no change.




A10. Give two reasons why the van der Waals equation more accurately predicts gas pressure than the ideal
gas law, especially at high pressures.



The van der Waals equation takes account of molecular volumes and intermolecular attractive forces, neither
of which the ideal gas law does.




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CHEM 1000 A and V                           Mid-Year Examination                             December, 2009



A11. Write the four reactions collectively known as the Chapman mechanism for ozone in the stratosphere.



O2 + h → O + O

O + O2 → O3

O3 + h → O2 + O

O + O3 → 2 O2




A12. Give two reasons why 1 kg of CH4(g) causes more warming of the atmosphere than 1 kg of CO2(g).



   1. CH4 has a lower molecular weight, so 1 kg of it is more moles than 1 kg of CO 2

   2. CH4 has more modes of vibration, hence absorbs more wavelengths of IR




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CHEM 1000 A and V                            Mid-Year Examination                             December, 2009



Part B. Answer all questions on this page (20 marks total).



B1. Ammonium nitrate can be used as an explosive. It decomposes according to:



                                     NH4NO3(s) → N2(g) + 2 H2O(g) + ½ O2(g).



Assuming the gases behave ideally, calculate the total volume of the products at 300 oC and 1 atm when 12.0 g
of ammonium nitrate decomposes.



 12 g 
              0.15 mol NH 4 NO3
 80 g / mol 



x 3.5 mol gas / mol NH4NO3 = 0.525 mol gas



     nRT 0.525 mol(0.082 L atm mol1K 1)(300  273)K
V                                                    24.7 L
      p                     1 atm



B2. A sample of an unknown gas diffuses in 11.1 minutes. An equal volume of H2(g) in the same apparatus
under the same conditions effuses in 2.42 minutes. Calculate the molecular weight of the unknown gas.




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CHEM 1000 A and V                              Mid-Year Examination                            December, 2009

time?? rate H2   MW??
              
timeH2 rate??    MWH2
                      2
MW??  time??     
                
MWH2  timeH2
     
                  
                  
                           2
             time??                   1  11.1 
                                                    2
                                                                 1
MW??  MWH2                2.02 g mol           42.5 g mol
             timeH                        2.42 
                   2     



B3. Calculate the pressure exerted by 1.00 mol of CH4(g) in a 500 mL vessel at 25.0oC using:

   (a) The ideal gas law



            nRT 1.00 mol(0.082 L atm K 1mol1)(25  273)K
       p                                                  48.9 atm
             V                   0.500 L

   (b) The van der Waals equation. For CH4(g), a = 2.25 L2 atm mol-2 and b = 0.0428 L mol-1



                      2
    nRT     n
p        a 
   V  nb   V
                                                                          2
  1.00 mol(0.082 L atm K 1mol1)(25  273)K        L2atm  1.00 mol 
                                             2.25                 
      0.500 L  1.00 mol(0.0428 L mol1 )           mol2  0.500 L 
 53.45 atm  9.00 atm
 44.5 atm




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CHEM 1000 A and V                             Mid-Year Examination                              December, 2009



Part C. Answer any five of questions C1 – C7 (20 marks each). If you answer more than five, the best five
will be used to calculate your total mark.



C1. Nitrogen dioxide decomposes according to 2 NO2(g) Ý 2 NO(g) + O2(g). At a certain temperature, Kp is
extremely small (4.4810-13). If 0.75 atm of NO2(g) is placed in a vessel at this temperature, find the
equilibrium partial pressures of all three gases.



                                   p(NO2), atm                  p(NO), atm                   p(O2), atm

Initial, atm                           0.75                          0                            0

Change, atm                             -2x                          +2x                         +x

Equilibrium, atm                     0.75 – 2x                       2x                           x




At equilibrium,

 p NO 2 pO2
               Kp
  p NO2 2
   (2x)2 x
               4.48  1013
 (0.75  x) 2




Here, Kp is so small, we can make the approximation that x << 0.75, so the expression simplifies to




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CHEM 1000 A and V                                Mid-Year Examination                            December, 2009

(2x) 2 x
          4.48  10 13
(0.75) 2
4 x 3  (0.75) 2 (4.48  10 13 )
x 3  (0.75)2 (4.48  10 13 ) / 4
x 3  6.3  10 14
x  3.98  10 5



Thus, pNO2 = 0.75 – 2x = 0.75 atm

pNO = 2x = 7.96 x 10-5 atm

pO2 = x = 3.98 x 10-5 atm



Check: (7.96 x 10-5)2(3.98 x 10-5) / (0.75 – 2(3.98 x 10-6))2 = 4.48 x 10-13 = Kp




C2. Given the following information, calculate the ionization energy of lithium (in kJ mol-1):




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CHEM 1000 A and V                                     Mid-Year Examination   December, 2009

Lattice energy of LiF(s)                      1050 kJ mol-1

Electron affinity of fluorine                         328 kJ mol-1

ΔHfo (LiF(s))                                         617 kJ mol-1

Enthalpy of sublimation of lithium metal             161 kJ mol-1

Bond dissociation energy of fluorine                 159 kJ mol-1




Li(s) → Li(g)                     +161 kJ mol-1



Li(g) → Li+(g) + e-               I1 (Li)



½ F2(g) → F(g)                    ½ (+159 kJ mol-1)


        -       -                        -1
F(g) + e → F (g)           -328 kJ mol



Li+(g) + F-(g) → LiF(s)           -(1050) kJ mol-1



_______________________________________



Li(s) + ½ F2(g) → LiF(s)          -617 kJ mol-1




Thus, the ionization potential of lithium is found by difference:



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CHEM 1000 A and V                             Mid-Year Examination   December, 2009

I1(Li) = -617 – (+161 + ½ (159) – 328 – 1050) = 520.5 kJ mol-1




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CHEM 1000 A and V                             Mid-Year Examination                               December, 2009



C3. (a) Two possible structures for nitrosyl chloride are Cl=NO and ClN=O.



       (i)      Draw complete Lewis diagrams for each of these structures.



                ::Cl=N:-O:::                                :::Cl-N:=O::




       (ii)     Calculate formal charges and decide which structure is more likely:




                                     Cl=NO                                           ClN=O



                        Cl             N               O                   Cl           N              O

Enter the
formal charge           +1
                                       0               -1                  0            0               0
of each atom:

More likely
structure
                                                                                  More likely!
(check one!)


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CHEM 1000 A and V                             Mid-Year Examination                            December, 2009



(b) What hybridization is each of the four indicated atoms using in the molecule of Aspirin shown below?




                                                                  3
                                          1     H         H
                                                    C C       O
                                              H C     C C O H
                                                  C C    H
                                                            H
                                                H     O C C
                                                              O H
                                                     4
                                                                       2




   Atom 1:__sp2 ___ Atom 2:__sp3_____          Atom 3:___sp2____      Atom 4:___sp3____



(Don’t forget that there are lone pairs on atoms 3 and 4!)




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CHEM 1000 A and V                             Mid-Year Examination                                 December, 2009



C4. Use the data in the table to answer the following questions about the reaction 2 NO2(g) Ý N2O4(g).



                                                      NO2(g)                              N2O4(g)

           ΔHfo, kJ mol-1                              33.2                                 9.16

           So, J K-1 mol-1                            240.0                                304.2



   (a) Calculate ΔHo (kJ mol-1)

       ΔHo = ΔHfo(N2O4(g)) - 2 ΔHfo(NO2(g))

       = 9.16 – 2(33.2)

       = -57.24 kJ mol-1




   (b) Calculate ΔSo (J K-1 mol-1)

       ΔHo = So(N2O4(g)) - 2 So(NO2(g))

       = 304.2 – 2(240.0)

       = -175.8 J K-1 mol-1



   (c) Calculate ΔGo (kJ mol-1)



       ΔGo = ΔHo – TΔSo

       = -57,240 J mol-1 – (25 + 273)K(-175.8 J K-1 mol-1)

       = -4851.6 J mol-1

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CHEM 1000 A and V                                                        Mid-Year Examination   December, 2009

      = -4.85 kJ mol-1



   (d) Calculate Kp at 50oC



                 G o                   4851 J mol1            
                
                 RT            
                                   8.314 J K 1mol1 ( 50  273 )K 
                                                                    
       Kp  e          
                             e                                    
                                                                         6.09



                                                   o
   (e) If the total pressure at 50 C is 1 bar, calculate the partial pressure of each gas.



      Since the total pressure is 1 bar, then pNO2 + pN2O4 = 1



                                     p N2 O4
      But, K p  6.09                       2
                                                       Thus, pN2O4 = 6.09(pNO2)2
                                     p NO2

      pNO2 + 6.09(pNO2)2 = 1

      6.09(pNO2)2 + pNO2 – 1 = 0



      Solving this quadratic, pNO2 = 0.331 atm



      pN2O4 = 1 – pNO2 = 1bar - 0.331 bar = 0.669 bar



                 p N2 O4         0.669
      Check:             2
                                        6.09  K p
                 p NO2           0.3312




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CHEM 1000 A and V                               Mid-Year Examination                                  December, 2009



C5. (a) Ozone (O3) is destroyed in the upper atmosphere according to the reaction O 3(g)  O2(g) + O(g),
                                                                                          UV light


which requires breaking one of the bonds in the ozone molecule. If the bond energy is 445 kJ mol-1, what is
the maximum wavelength of photon capable of doing this (in nm)?




445 kJ mol-1 / 6.02 x 1023 mol-1 = 7.39 x 10-22 kJ per bond = 7.39 x 10-19 J per bond



             c
  E  h  h
             
            hc 6.63  10 34 J s (3.00  108 m s 1)
  thus,                                            2.69  10 7 m  269 nm
            E            7.39  10 19 J




(b) We know about s, p, d and f-orbitals. Electrons will begin to fill g-orbitals after the f-orbitals are filled. Use
Aufbau and your knowledge of quantum numbers to determine the atomic number of the lightest element
(as yet undiscovered) that will contain an electron in a g-orbital.



Orbitals are filled according to the Aufbau principle. We can determine when the first g-orbital will accept an
electron accordingly. The first g-orbital will fill in the n=5 principle shell, and energy-wise, just after the 8s
orbital fills. Thus, count the electrons up to that point and add one:




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CHEM 1000 A and V                              Mid-Year Examination                              December, 2009

1s

2s 2p

3s 3p 3d

4s 4p 4d 4f

5s 5p 5d 5f 5g

6s 6p 6d 6f 6g

7s 7p 7d 7f 7g

8s 8p 8d 8f 8g




1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f145d10 6p6 7s2 5f14 6d10 7p6 8s2 5g1



And we see that the 5g electron is number 121, i.e. element 121 will be the lightest element containing this
electron.




C6. (a) A sample of a gas is at 50oC in a 50.0 L container at 1.1 atm. The gas is expanded to 75 L and cooled to
0oC. Calculate the new pressure.



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CHEM 1000 A and V                             Mid-Year Examination                               December, 2009

p1V1 p2 V2
    
 T1   T2
             p1V1T2 1.1 atm(50 L)(273 K)
Thus, p2                                0.62 atm
              T1V2   (50  273) K (75 L)




   (b) 1.00 mol of He(g) under certain conditions exerts a pressure of 5.00 atm. 1.00 mol of Cl2(g) under the
   same conditions exerts a lower pressure. Why?



   Chlorine is a larger molecule than the very small He atoms. Larger molecules have larger attractive forces
   for one another. Hence the Cl2 molecules are pulled together, reducing the pressure they exert.




   (c) Calculate the density of Cl2(g) at 10.0 atm and 100oC (in g/L).



             p(MW)        10 atm(0.070 kg mol1)
                                                     2.29 102 kg L1  22.9 g L1
               RT    0.082 L atm K 1mol1(100  273)K




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CHEM 1000 A and V                            Mid-Year Examination                               December, 2009



C7. Predict the shapes of the following molecules. Wrong name = zero marks.

   (a) AsCl52



       Electrons = 5 + (5 x 7) + 2 = 42. Making the five bonds uses 10 of these. Completing the octets on the Cl
       atoms uses another 30. This leaves two, which are placed as a lone pair on the As atom. Thus the
       shorthand notation is AX5E. This is a square pyramid.




   (b) PCl5



       Electrons = 5 + (5 x 7) = 40. Making the five bonds uses 10 of these. Completing the octets on the Cl
       atoms uses the other 30. Thus the shorthand notation is AX 5. This is a trigonal bipyramid.




   (c) O3



       Electrons = 3 x 6 = 18. Making the two single bonds uses 4 of these. Completing the octets on the
       terminal atoms uses another 12, leaving two as a lone pair on the central O atom. Thus the shorthand
       notation is AX2E, which is bent. (note that moving lone pairs inwards to make double bonds does not
       change the shape).




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CHEM 1000 A and V                           Mid-Year Examination                              December, 2009




   (d) CF3



      Electrons = 4 + (3 x 7) + 1 = 26. Making the three bonds uses 6 of these. Completing the octets on the F
      atoms uses another 18. This leaves two, which are placed as a lone pair on the C atom. Thus the
      shorthand notation is AX3E. This is a trigonal pyramid.




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