Chem 3411 Solution Set 5 - Mindanao State University - Iligan Institute Of Technology - 2010 Page 13
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14CHEM 3411, Fall 2010
Solution Set 5
6
Exercise 5.27b pg 203
Exercise
Sketch the phase diagram of the system NH
/N
H
given that the two substances do not form a compound with
3
2
4
◦
◦
each other, that NH
freezes at T
(NH
) =
78
C and N
H
freezes at T
(N
H
) = +2
C and that a eutectic is
3
m
3
2
4
m
2
4
◦
formed when the mole fraction of N
H
is χ
= 0.07 and that the eutectic melts at T
(eutectic) =
80
C.
2
4
N
H
m
2
4
In terms of given variables, this is written:
◦
T
(NH
) =
78
C
m
3
◦
T
(N
H
) = +2
C
m
2
4
χ
(eutectic) = 0.07
N
H
2
4
◦
T
(eutectic) =
80
C
m
Strategy
For each component, we can use its pure melting/freezing temperature to determine one point on the coexistence
curve between the phase where this component partially precipitates out of liquid (Liquid + X phase) and the phase
where both liquids are completely miscible (Liquid phase). Additionally, the eutectic point falls on each coexistence
curve and the rest of each coexistence curve can be estimated. The eutectic melting temperature also serves as the
coexistence line between the solid phase and liquid phases. As an example of such a phase diagram showing similar
information see Figure 5.51 on page 186 of the text book. The solution for this problem is shown on the next page.
It also worth mentioning the significance of the eutectic point. This is the moles fraction composition where both
compounds are always well mixed; neither will ever precipitate out of solution regardless of temperature.
Although not requested, we have also included an example of determing the phase composition of a sample with
◦
χ
= 0.60 and T =
40
C. This point is shown as the blue open circle on the plot. At this temperature, tie lines
N
H
2
4
are constructued between the sample point and the phase boundaries; the homogenous liquid mixture on the left
and the pure solid N
H
on the right. The endpoints of each tie line correspond to χ
= 0.34 and χ
= 1.00.
2
4
N
H
N
H
2
4
2
4
We can now use the lever rule (equation 5.46 on pg 178) to determine the phase composition of this sample.
n
l
= n
l
α
α
β
β
where n
denotes the moles of the i-th phase and l
is the length of the tie line between the i-th phase and the sample
i
i
point. For our sample we’ll use this equation in the following form.
n
l
= n
l
liquid
liquid
N
H
(solid)
N
H
(solid)
2
4
2
4
where n
and n
are the moles of the homogenous liquid mixture and the moles of the percipitated solid
liquid
N
H
(solid)
2
4
N
H
respectively. The l
and l
similarly correspond to the tie line lengths of each phase.
2
4
liquid
N
H
(solid)
2
4
The length of the each tie line is calculated as
l
= 0.60
0.34 = 0.26
liquid
l
= 1.00
0.60 = 0.40
N
H
(solid)
2
4
Now we’ll use the lever rule equation to solve for the mole fraction of the liquid phase χ
.
liquid
13
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