Fullerton College Office of Special Programs
SI Review 111B
Handout #2 Solutions
Problem 1:
a.
SbCl5(g)
b.
2BrNO(g)
c.
CH4(g) + 2H2S(g)
d.
2CO(g) + O2(g)
SbCl3(g) + Cl2(g)
Keq 
[SbC 3l ][C l2 ]
[SbC 5l ]
2NO(g) + Br2(g)
Keq 
[N O ]2[Br2 ]
[BrN O 2]
CS2(g) + 4H2(g)
2CO2(g)
Keq 
Keq 
[C S2 ][H2 ]4
[C H4 ][H2S]2
[C O2 ]2
[C O 2][O2 ]
Problem 2:
KC 
0 .1 8 5
[C H3O H ]

 136
[C O ][H2 ]2 0 .1 0 50.1 1 42
Problem 3:
Since this is KP, the units of pressure will be in atm (atmospheres). Before solving the problem,
the pressures should be converted from torr into atm.
1 1 7torr SO2 x
2 5 5torr C l2 x
1 atm
 0 .1 539 atm SO2
7 6 0 .0torr
1 atm
 0 .3 355 atm C l2
7 6 0 .0torr
So,
KP 
[SO2 ][C l2 ]
[SO2 ][C l2 ]
 [SO2C l2 ] 
[SO2C l2 ]
KP
[SO2C l2 ] 
(0 .1 539 )(0 .3 355 )
 1 .7 7x 1 0-5 atm (or 0 .0 1 3 5torr)
2 .9 1x 1 03
Problem 4:
2COF2(g)
CO2(g) + CF4(g)
KP = 2.2 x 106 at 298 K
Products are favored if the equilibrium constant is large and reactants are favored if the
equilibrium constant is small.
a.
COF2(g)
1
/2CO2(g) + 1/2CF4(g)
KP '  KP 
1/2

 2 .2 x 1 06

1/2
 1 .5 x 1 03
Big K, so products are favored!
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b.
6COF2(g)
3CO2(g) + 3CF4(g)

KP '  KP   2 .2 x 1 06
3

3
 1 .1 x 1 019
Big K, so products are favored!
2
c.
2CO2(g) + 2CF4(g)
4COF2(g)
2
 1 
1


  
KP '  
  2 .1 x 1 0-13
6 

 2 .2 x 1 0 
 KP 
Small K, so reactants are favored!
Problem 5:
Note: Because equilibrium constants are usually not written with units, it’s best to exclude them
from these types of calculations.
a.
KP  KC(RT )n where n  2 - 1  1 , so KP  5 .9 x 1 0-3 x (0 .0 8 2 0 6x 2 9 8 1)  0 .1 4
b.
KP  KC(RT )n where n  2 - 4  - 2 , so KP  3 .7 x 1 08 x (0 .0 8 2 0 6x 2 9 8 -2
)  6 .2 x 1 05
c.
KP  KC(RT )n where n  2 - 2  0 , so KP  KC  4 .1 0x 1 0-31
Problem 6:
CaCO3(s)
CaO(s) + CO2(g)
KP  pCO2  1 .0 4
The equilibrium constant expression has been written to include gases, but not solids. The
reaction quotient expression will have the same form as the equilibrium constant expression and
will, therefore be dependent on the pressure of the carbon dioxide.
a.
Q  pCO2  2 .5 5 1 .0 4 K
The system will shift to the left, towards the reactants, and the mass of CaO will decrease.
b.
Q  pCO2  1 .0 4 1 .0 4 K
The system is at equilibrium, so the mass of CaO will not change.
c.
Q  pCO2  1 .0 4 1 .0 4 K
The system is at equilibrium, so the mass of CaO will not change.
d.
Q  pCO2  0 .2 1 1 1 .0 4 K
The system will shift to the right, towards the products, and the mass of CaO will increase.