Page 168 - Modern physical chemistry

160 Relationships among Reactants

So fonnula (7.64) yields

Kc = Kp(RTr~n = (1.130 x 10 40 bar- 1I2 1(0_0831451 bar K- i mol- i )(298-15 K)J'2 = 5.63 x 104OM- 1I2 _

ExampleZl0

How does the fraction of dissociation a vary with pressure P for (a) reaction (7.69),
(b) reaction (7.73)?
(a) Begin with 1 mole PC1 5 and 0 mole PCl 3 and Clz• Then
no =a =1, b=c=O, x=a,

and equation (7.72) reduces to

(b) Begin with 1 mole N 20 4 and 0 mole N0 2 • Then
no =a=l, b=O, x=a,
and equation (7.75) reduces to
4a 2 p 4a 2 p
K =( I-a l+a l-a
p
X )=- 2 '

ExampleZll

For N Z0 4 (g) the log K f is -17.132, while for N0 2 (g) it is -8.981. Calculate Kp for
reaction (7.73).
Proceed as in example 7.8 to get

10gKp =2(logKr)N02 -(logKr)Nz 4 =2(-8.981)-(-17.132)=-0.830
0
whence
Kp = 0.148 bar.

ExampleZ12

What is the density of an equilibrium mixture of N Z0 4 and NOz at 25 C and 1.000 bar?
0
With the results in examples 7.10 (b) and 7.11, we have
2
4a ( 1.000 bar)
_-'--__ -L = 0.148 bar
l-a 2
whence
a =0.189.
With equation (7.77), we find that
W PM (1.000 bar )(92.011 g mOl-I)
p=_= I = =312g1-1
I
V RT[I+{V-l)a] (0.0831451 bar K- mol-I X298.15 KX1.189)· .