(b) The Equilibrium Constant and the Reaction Quotient
The reaction quotient and the equilibrium constant for a reaction have the same expression, but
the numerical values can be different. The reaction quotient is, in fact,
more general
, because
Q
takes values for any conditions.
K
takes values specified by the thermodynamics of the process.
Now, we can relate
∆
G
°
to the equilibrium constant
K
and
∆
G
to the reaction quotient
Q
by
recognizing that when
∆
G
< 0;
Q
<
K
and when
∆
G
= 0,
Q
=
K
. Therefore, we have the
following equation:
∆
G
(
T
) =
∆
G
°
(
T
) +
RT
ln
Q
=
−
RT
ln
K
+
RT
ln
Q
=
ln
Q
RT
K
.
General
Standard
Conditions
Conditions
where
R
= 8.314 J/K mol and
T
is in degrees Kelvin.
To summarize equilibrium then:
Sign of
∆
G
Size of
Q
vs.
K
Spontaneous?
−
Q < K
Yes (left to right)
0
Q = K
Equilibrium (quasistatic)
+
Q > K
No (left to right)
Example
: How can we use
Q
and
K
?
Consider the following
homogeneous equilibrium
:
2 NO(g) + O
2
(g)
ZZX
YZZ
2 NO
2
(g) at
T =
750 K
∆
H
0
=
−
113.1 kJ
∆
S
0
=
−
145.3 J/K
∆
G
0
(750 K) =
−
4.13 kJ
2
2
2
NO
2
NO
O
(750 K)
1.94
p
p
K
p
p
=
=
2 NO
2
(g)
ZZX
YZZ
N
2
O
4
(g) is another, simultaneous equilibrium
that also occurs.
2
4
2
N O
6
2
NO
' (750 K)
6.55
10
p
p
K
p
−
=
=
×
Consider the following conditions:
NO
p
2
O
p
2
NO
p
Q
vs.
K
p
∆
G
Comments / Impact on
2
4
N O
p
1.00 atm
1.00 atm
0 atm
Q
= 0 <
K
p
< 0
NO
2
forms; also forms N
2
O
4
2.00 atm
1.00 atm
2.00 atm
Q
= 1 <
K
p
< 0
NO
2
forms; also forms N
2
O
4
1.00 atm
1.00 atm
4.00 atm
Q
= 16 >
K
p
> 0
NO
2
consumed; also reduces N
2
O
4

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