1.
(a)
p = 3 [FeO(s), Fe(s), and gas]
c
= s - r – a = 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)]
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5.

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

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,
