1
A sample of PCl
5
gas is introduced into a flask at
250
o
C at a pressure of exactly 1 atm, but some
of this gas decompose to PCl
3
and Cl
2
to
produce an equilibrium total pressure of 1.98
atm.
What is the value of the equilibrium
constant, K
p
, for this reaction?
PCl
5
(g)
PCl
3
(g) + Cl
2
(g)
A) 50 atm
B) 1.98 atm
C) 48 atm
D) 2.0 atm
Partial pressure is proportional to moles
and concentration, so you can work with
partial pressures just like using moles or
concentration.
Construct the RICE table.
Let x = partial
pressure of PCl
5
that decomposes.
Solve for x using the equilibrium total
pressure.
Substitute into the K
p
expression to find K
p
.
2
PCl
5
(g)
PCl
3
(g) + Cl
2
(g)
1 atm
0
0
-x
+x
+x
1 atm – x
x
x
2
2
1
1
2
1.98
1.98
1
0.98
(.98)
48
.02
p
p
x
K
x
x
x
x
K
atm
One equation in two
unknowns, need another
equation.
The total final pressure was
1.98 atm.

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