2000 Chemistry Spring 2001 Section B Assignment 2
Solutions
1. Since this is a liquid, the mole fractions are the appropriate measures of activity. Fortunately,
we are given the data in exactly the right form: If 6.42% of the molecules are in the chair form
Xchair 0 0642. Moreover, the rest (93.58%) must be in the boat form, i.e. Xboat 0 9358.
The equilibrium constant is
a2
NOBr
aNO 2 aBr2
Q
PNOBr P 2
PNO P 2 PBr2 P
0 969 2
0 105 2 0 400
213
2. We need to determine the reaction ratio Q:
14 6
0 9358
0 0642
Xboat
Xchair
aboat
achair
K
Since Q is larger than K , there is excess product. Accordingly, NOBr will react to form NO
and bromine.
(a) We can solve this problem by working out how the initial and nal pressures are related:
0
PCOCl2
PCOCl2
0 20
aCOCl2
aCO aCl2
K
03
PCOCl2 (atm)
0
PCOCl2
For this reaction,
PCl2 (atm)
0.3
0 3 PCOCl2
PCOCl2 P
PCO PCl2
PCOCl2
PCOCl2 0 3
PCOCl2
2
0 20 0 PCOCl2
0 3 018 0 12PCOCl2
2
0 20PCOCl2
2
0 20PCOCl2 1 12PCOCl2
5 58 or 0 016 atm
0 018
Only the second solution makes any physical sense. Thus, PCOCl2
1
PCO (atm)
initial
0.3
nal equilibrium 0 3 PCOCl2
3.
0 016 atm.
(b)
3
mol
aH2 CO3
aCO2 aH2O
K
4.
n
T
V
P
PV
RT
0 016 atm 101 325 Pa atm
1633 Pa
3
1m
10 L
0 01 m3
1000 L
600 273 15 K 873 K
1633 Pa 0 01 m3
2 2 10
8 314 510 J K 1mol 1 873 K
n
We are given K and the pressure of carbon dioxide (from which we can easily compute
its activity). We want to solve for the activity of H2 CO3 (from which we can obtain the
concentration). We still need to know the activity of the solvent, i.e. its mole fraction.
aH2 CO3
H2 CO3
K aCO2 aH2O
10
10
5
3 36
mol L
2
8 33 mol
0 31 0 31 mol
33 Pa
10 2
101 325 Pa
XH2 O
nNaCl
150 g
8 33 mol
18 0152 g mol
18 g
0 31 mol
58 443 g mol
nH2 O
nH2 O nNa
nCl
8 33
nH2 O
0 931
0 931
10
10
5