Page 207 - Physical Chemistry

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Chapter 6 is attained at 25°C and 2.00 atm. Find the equilibrium composition.
Reaction Equilibrium in Ideal Gas
Mixtures Carrying out step 1 of the preceding scheme, we use Appendix data to get

¢G° >1kJ>mol2 2151.312 97.89 4.73
298
For step 2, we have G° RT ln K° and
P
4730 J>mol 18.314 J>mol-K21298.1 K2 ln K°
P
ln K° 1.908 and K° 0.148
P
P
For step 3, let x moles of N O react to reach equilibrium. By the stoi-
2
4
chiometry, 2x moles of NO will be formed and the equilibrium mole numbers
2
will be
n 10.300 x2 mol and n 10.500 2x2 mol (6.42)
N 2 O 4 NO 2
[Note that the equilibrium extent of reaction is j x mol and the equations in
(6.42) satisfy n n n j.]
i i,0 i
Since T and P are fixed, we use Eq. (6.41) in step 4(a) to write
0.500 2x 0.300 x
P x P P, P x P P
0.800 x 0.800 x
NO 2 NO 2 N 2 O 4 N 2 O 4
since n (0.300 x) mol (0.500 2x) mol (0.800 x) mol.
i i
Performing step 5, we have
3P >P°4 2
K° NO 2
P
P >P°
N 2 O 4
2
10.500 2x2 1P>P°2 2 0.800 x 0.250 2x 4x 2 P
0.148
2
10.800 x2 2 10.300 x21P>P°2 0.240 0.500x x P°
The reaction occurs at P 2.00 atm 1520 torr, and P° 1 bar 750 torr.
Thus 0.148(P°/P) 0.0730. Clearing of fractions, we find
2
4.0730x 2.0365x 0.2325 0
The quadratic formula x [ b (b 4ac) ]/2a for the solutions of ax
2
2
1/2
bx c 0 gives
x 0.324 and x 0.176
The number of moles of each substance present at equilibrium must be positive.
Thus, n(N O ) (0.300 x) mol 0, and x must be less than 0.300. Also,
2 4
n(NO ) (0.500 2x) mol 0, and x must be greater than 0.250. We have
2
0.250 x 0.300. The root x 0.324 must therefore be discarded. Thus x
0.176, and step 6 gives
n1N O 2 10.300 x2 mol 0.476 mol
4
2
n1NO 2 10.500 2x2 mol 0.148 mol
2
Exercise
For O(g) at 4200 K, G° 26.81 kJ/mol. For a system whose initial com-
f
position is 1.000 mol of O (g), find the equilibrium composition at 4200 K and
2
3.00 bar. (Answer: 0.472 mol of O , 1.056 mol of O.)
2