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Calculation of Gas Phase Equilibrium
From K we can determine quantitatively the extent of reaction, i.e.
concentration of products and reactants at equilibrium
At T = 250
o
C (thermodynamic) K = 2.15 for
PCl
5
(g)
PCl
3
(g) + Cl
2
(g)
If
0.1000 atm of PCl
5
is place in a closed container at 250
o
C,
a.
What are the pressures of all gases at equilibrium?
b.
What is the extent of reaction? (what % PCl
5
unreacted.)
All K problems solved by the same basic technique:
1.
Tabulate initial concentrations/pressures
2.
Use stoichiometric coefficients to indicate changes due to reaction
3.
Use
I C E
table (initial, change, equilibrium)
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View Full Document Make a table including all chemical species involved
PCl
5
PCl
3
Cl
2
I
nitial
0.1000 atm
0
0
C
hange
–x
+ x
+ x
E
quil.
0.1000 – x
x
x
x = partial pressure
(# mols) of PCl
5
that reacts
Since P = nRT/V for
T, V constant P
∝
n
K
=
P
PCl
3
P
Cl
2
P
PCl
5
=
x
2
0.1
x
= 2.15
PCl
5
(g)
PCl
3
(g)
+ Cl
2
(g)
x
2
+ 2.15
x
 0.215 = 0
Solve by quadratic formula:
ax
2
+ bx + c = 0
x
=

b
b
2
 4
ac
2
a
Here:
a = 1, b = 2.15, c = –0.215
Two roots for x:
x = 0.0957 atm or x = –2.25 atm.
Which correct?
x = amt. of PCl
5
that reacts, must be between 0 and 0.100
X
At equil., P
PCl5
= 0.1000–.0957 = .0043 atm;
P
PCl3
= 0.0957 atm = P
Cl2
% PCl
5
unreacted
0.0043
atm
0.1000
atm
100 = 4.3%
Rxn nearly goes to
completion
Q
:
the reaction quotient and the direction of spontaneous change
Remember for the gas phase reaction:
aA(g) +
bB(g)
→
cC(g) +
dD(g)
∆
G
= D
G
o
+
RT
ln
P
C
/
P
ref
(
)
c
P
D
/
P
ref
(
)
d
P
A
/
P
ref
(
)
a
P
B
/
P
ref
(
)
b
At equilibrium,
∆
G = 0 & [
] = K
Away from equilibrium call [
] = Q
≡
reaction quotient, and thus:
∆
G
= D
G
o
+
RT
ln
Q
∆
G
o
= 
RT
ln
K
Substituting for
∆
G
o
, we have:
∆
G
= 
RT
ln
K
+
RT
ln
Q
=
RT
ln
Q
K
This expression will tell us about the direction of spontaneous
change for
any
set of reactant & product concentrations
(Remember:
∆
G
<
0 for spontaneous direction)
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If
Q < K
, ln(Q/K) < 0 and
∆
G < 0; and reaction is spontaneous
towards products as written.
Numerator of Q must increase
relative to denominator to approach K value.
Reaction
proceeds to right (products)
.
2.
If Q > K, Q/K > 0, ln(Q/K) > 0 and
∆
G > 0; and reaction is
spontaneous towards reactants. Numerator of Q must decrease
relative to denominator to approach K value.
Reaction
proceeds to left (reactants)
.
3.
If Q = K, ln(Q/K) = ln(1) = 0 and
∆
G = 0; system is already at
equilibrium; no changes in concentrations observed.
∆
G
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This note was uploaded on 04/17/2008 for the course CAS CH131 taught by Professor Zigler during the Spring '08 term at BU.
 Spring '08
 Zigler

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