Midterm, Spring 2000, Chemistry 82/92

General Directions:

  1. Please answer all the questions in the Blue Book (which should have your name and social security number on the front).

b. A Periodic Table, a Table of Standard Reduction Potentials and a Table of various simple forms of integrated rate equations is attached.

  1. Please show your work (so that minimal credit will be lost for silly errors)
  2. Your papers will be returned after the Spring Break.
  3. You may keep the examination paper.
  4. I do not discuss grades over the telephone or on the internet. I will be pleased to meet with you if you have questions about the examination, grading, etc. after your examination is returned to you.

 

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  1. [10 points] Calculate the equilibrium constant (Keq) at 25 oC for the redox reaction:

    2. O2 (g) + 4 H + (aq) + 4 Fe 2+ (aq) ¾ ¾ ¾ ¾ ® 4 Fe 3+ (aq) + 2 H2O (l)

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    O2 (g) + 4 H + (aq) + 4e - ® H2O (l) Eo = 1.23 V

    4 Fe 2+ (aq) ® 4 Fe 3+ (aq) + 4e - Eo = - 0.77 V

    Eocell = 0.46 V and thus

    log K = n Eo /0.0591 V = (4)(0.46) / 0.0591 = 31.1

    So K = 1 x 1031

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  2. [15 points] Make a sketch of a Galvanic cell made up of some specified anode and cathode. Identify the components, the sites of oxidation and reduction and any other features you understand to be important.

    3. ----------------------------------------------------------------

    See your textbook or even your laboratory reports for complete details. Any cell would do but, having chosen the anode and cathode, it is necessary to specify the direction of electron flow (as given in the Table of Reduction Potentials provided) the solutions that made up the cells, the presence of a bridge for ion mobility, etc. Many answers are possible.

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  3. [10 points] For the reaction:

    4. Cr2O7 = (aq) + 14 H +(aq) + 6 I - (aq) ¾ ¾ ® 2 Cr 3+ (aq) + 3 I2 (s) + 7 H2O (l)

    the Standard emf (Eo) is 0.79 V.

    Calculate the emf (E) generated when Cr2O7 = (aq) = 2.0 M, H +(aq) = 1.0 M, I - (aq) = 1.0 M and Cr 3+ (aq) = 1.0 x 10-5 M.

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    Q = [Cr 3+] 2 / ([Cr2O7 - 2] [H +]14 [I - ] 6) = 5 x 10 - 11

    E = 0.79 V - (0.059 / 6) log (5 x 10 - 11)

    E = 0.89 V

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  4. [10 points] Given the observations which resulted in the Rydberg Constant (RH =

    5. -2.18 x 10-18 J), calculate how much energy is required to ionize the one electron of a hydrogen atom (i.e. what is the ionization potential [IP] of hydrogen)?

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    E = 2.18 x 10-18 J

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  5. [10 points] Account for the experimental observation that SnCl3 - is found to be tetrahedral. Sketch a representation of the anion.

    6. Use VSEPR. Filled orbitals push (repel) other filled orbitals. Minimum energy is found at tetrahedral geometry.

     

     

     

     

  6. [10 points] Account for the experimental observation that while CO2 is linear, SO2 is bent.

    7. -------------------------------------------------------------------------------

    The extra electron pair on sulfur causes electron - electron repulsion

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  7. [10 points] Account for the experimental observation that while the both neutral O2 and the O2 + cation are paramagnetic, the

O2 = dianion is diamagnetic. NOTE: You are expected to draw some sort of diagram(s).

 

 

 

 

8.[10 points] Account for the experimental observation that while CF4 is tetrahedral and the four fluorines are identical, SF4 is not tetrahedral and the four fluorines are not identical. Predict a structure for SF4

 

 

9. [15 points] The first order rate constant (k) for hydrolysis (reaction with water) of an insecticide sprayed to control malaria mosquitos at 12 oC is 1.45 /year. A quantity of this insecticide was washed into a lake in June of 1998 and the overall concentration of the insecticide in samples averaged 5.0 x 10-7 g/cm3. If the average temperature of the lake is 12 oC what was the concentration a year later (June 1999)?

 

 

Some necessary items…

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Planck's constant = 6.626 x 10-34 J s

Á = Faraday = 96,500 J / V - mol

R = 8.314 J / K - mol

RH = -2.18 x 10-18 J

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E = hn = hc/ l

E = Eo - (2.303RT / nF) log Q

D E = RH [1/(n1)2 - 1/(n2)2 ]

D Go = - nFEo

D Go = -2.303 RT log K

D Go = D Ho - TD So