Page 214 - Physical Chemistry

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To help figure out the direction of the equilibrium shift, we define the reaction Section 6.6
quotient Q for the reaction 0 → n m at some instant of time as Shifts in Ideal-Gas Reaction Equilibria
i
i
i
P
Q q 1P 2 n i (6.49)
P
i
i
where P is the partial pressure of gas i in the system at a particular time, and the sys-
i
tem is not necessarily in equilibrium. We imagine that the change to the equilibrium
system is made instantaneously, so the system has no time to react while the change
is being made. We then compare the value of Q the instant after the change is made
P
with the value of K . If Q K , the equilibrium position will shift to the right so as
P P P
to produce more products (which appear in the numerator of Q ) and increase Q until
P P
it equals K at the new equilibrium position. If we find Q K , then the system is in
P P P
equilibrium after the change and the change produces no shift in the equilibrium po-
sition. If Q K , the equilibrium shifts to the left. Alternatively, we can compare K°
P P P
n
with Q° (P /P°) i .
P i i
Isobaric Temperature Change
Suppose we change T, keeping P constant. Since d ln y (1/y) dy, Eq. (6.36) gives
2
2
dK°/dT K° H°/RT . Since K° and RT are positive, the sign of dK°/dT is the same
P
P
P
P
as the sign of H°.
If H° is positive, then dK°/dT is positive; for a temperature increase (dT 0),
P
dK° is then positive, and K° increases. Since P x P, and P is held fixed during the
i
i
P
P
change in T, the instant after T increases but before any shift in composition occurs,
all partial pressures are unchanged and Q° is unchanged. Therefore the instant after
P
the temperature increase, we have K° Q° and the equilibrium must shift to the right
P
P
to increase Q°. Thus for an endothermic reaction ( H° 0), an increase in tempera-
P
ture at constant pressure will shift the equilibrium to the right.
If H° is negative (an exothermic reaction), then dK°/dT is negative and a posi-
P
tive dT gives a negative dK°. An isobaric temperature increase shifts the equilibrium
P
to the left for an exothermic reaction.
These results can be summarized in the rule that an increase in T at constant P in
a closed system shifts the equilibrium in the direction in which the system absorbs heat
from the surroundings. Thus, for an endothermic reaction, the equilibrium shifts to the
right as T increases.
Isothermal Pressure Change
Consider the ideal-gas reaction A ∆ 2B. Let equilibrium be established, and suppose
we then double the pressure at constant T by isothermally compressing the mixture to
half its original volume; thereafter, P is held constant at its new value. The equilibrium
constant K is unchanged since T is unchanged. Since P x P, this doubling of P
P i i
doubles P and doubles P (before any shift in equilibrium occurs). This quadruples
A B
2
the numerator of Q P /P and doubles its denominator; thus Q is doubled. Before
P B A P
the pressure increase, Q was equal to K , but after the pressure increase, Q has been
P P P
increased and is greater than K . The system is no longer in equilibrium, and Q will
P P
2
have to decrease to restore equilibrium. Q P /P decreases when the equilibrium
P B A
shifts to the left, thereby decreasing P and increasing P . Thus an isothermal pres-
B A
sure increase shifts the gas-phase equilibrium A ∆ 2B to the left, the side with fewer
moles in the balanced reaction.
Generalizing to the ideal-gas reaction aA bB

∆ eE f F

, we
see that if the total moles e f

on the right is larger than the total moles