CHM 1382
Answers to Examination 2 (Spring 1999)

 

Short answer questions

 

1.      (a) p = 3 [FeO(s), Fe(s), and gas]
  c = s
- ra = 4 - 1 - 0  = 3 [2nd reaction does not occur, a = 0 for first rxn because each of its products are in different phases]
f = c
- p + 2 = 3 - 3 + 2 = 2

(b) p = 3 [s in part (a)]
c = s
- r - a = 6 - 2 - 0 = 4 [2nd reaction occurs, but no new constraints since each of its products are in different phases for both rxns]
f = 4
- 3 + 2 = 3

 

 

2.      (a) F (compare with answer to b)
(b) T
(c ) F
(d) F (usually T, but the counterexample is a eutectic mixture)
(e) T (even if solution is not ideal, the partial pressures vs. x should not exhibit a maximum)
(f) T

 

 

3.      Compare experiments 2 and 3 first:

Then from experiments 1 and 2

Any of the experiments can be used to determine k. From experiment 1:

 

 

4.      (a) T
(b) T
(c) T (Arrhenius law)
(d) T
(e) F (the stated condition is necessary but not sufficient to conclude an elementary rxn)
(f) T [see answer to (e)]

  

5.     

(a)
     
(b) (low-boiling azeotrope is shown)

6.      The tie line at 400 K intersects the boundary of the two-phase region at approximately 15 wt% and 85 wt%. From where a=H2O and b=aniline, we obtain

 

 

 

Problems

1.      Rewriting the first-order expression for each nuclide,

If we define t=0 to be the time when [238U] = [235U], then ln[238U]0 = ln[235U]0 and we can write

The k values are available from the half-lives:

Solving for t gives

 

 

2.

(a) Regions of phase diagram are labeled below

(b) Cooling curve is shown below. There is only 1 halt when solid A freezes out: all other features are breaks.(in DSC, breaks will appear as steps and halts will appear as peaks)

(c ) X and Y correspond to points where three phases are present. At X, the phases are liquid A, liquid B, and solid A; at Y, the phases are solid A, solid B, and liquid B. According to the phase rule f = c + p + 2  = 2-3+2 = 1 at these points. When pressure is held constant, there are no remaining degrees of freedom.


 

 

 

 


3. The only trick to this problem is recognizing the P*benzene = 760 torr at the normal boiling point.  Then

(a)    

(b)     The solution will boil when the total pressure is 500 torr. Expressing total pressure in terms of xbenz,