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Chapter 6 a b

on the left, an isothermal increase in pressure will increase the numera-
Reaction Equilibrium in Ideal Gas tor of the reaction quotient
Mixtures
e
e
P P f p x x f p P e f p
E F
Q E F
a b p
a
N 2 (g) 3H 2 (g) ∆ 2NH 3 (g) P P P B b p x x P a b p
A B
A
more than the denominator and will therefore shift the equilibrium to the left (the side
with fewer moles) to reduce Q to K . If e f

is less than a b

, a pres-
P P
sure increase shifts the equilibrium to the right. If e f

equals a b

,
a pressure increase has no effect on Q and does not shift the equilibrium. Since the
P
system’s volume is proportional to the total number of moles of gas present, we have
the rule that an increase in P at constant T in a closed system shifts the equilibrium in
the direction in which the system’s volume decreases.
Although K depends on T only, the equilibrium composition of an ideal-gas re-
P
action mixture depends on both T and P, except for reactions with n 0. Figure 6.12
plots the equilibrium extent of reaction versus T at three pressures for N (g) 3H (g)
2 2
∆ 2NH (g), assuming ideal-gas behavior.
3
The italicized rules for shifts produced by an isothermal pressure change and by
Figure 6.12 an isobaric temperature change constitute Le Châtelier’s principle. These two rules
can be proved valid for any reaction, not just ideal-gas reactions (see Kirkwood and
Equilibrium extent of reaction
Oppenheim, pp. 108–109).
versus T at several pressures for
the ammonia-synthesis reaction
with an initial composition of Isochoric Addition of Inert Gas
1 mol N and 3 mol H . A pressure
2
2
increase at fixed T increases the Suppose we add some inert gas to an equilibrium mixture, holding V and T constant.
yield of NH . Since P n RT/V, the partial pressure of each gas taking part in the reaction is unaf-
i
3
i
fected by such an addition of an inert gas. Hence the reaction quotient Q P is
n i
P i i
unaffected and remains equal to K . Thus, there is no shift in ideal-gas equilibrium for
P
isochoric, isothermal addition of an inert gas. This makes sense because in the absence
of intermolecular interactions, the reacting ideal gases have no way of knowing
whether there is any inert gas present.
Addition of a Reactant Gas
Suppose that for the reaction A B ∆ 2C D we add some A to an equilibrium
mixture of A, B, C, and D while holding T and V constant. Since P n RT/V, this ad-
i i
dition increases P and does not change the other partial pressures. Since P appears
A A
in the denominator of the reaction quotient (6.49) (n is negative), addition of A at
A
constant T and V makes Q less than K . The equilibrium must then shift to the right
P P
in order to increase the numerator of Q and make Q equal to K again. Thus, addi-
P P P
tion of A at constant T and V shifts the equilibrium to the right, thereby consuming
some of the added A. Similarly, addition of a reaction product at constant T and V
shifts the equilibrium to the left, thereby consuming some of the added substance.
Removal of some of a reaction product from a mixture held at constant T and V shifts
the equilibrium to the right, producing more product.
It might be thought that the same conclusions apply to addition of a reactant while
holding T and P constant. Surprisingly, however, there are circumstances where
constant-T-and-P addition of a reactant will shift the equilibrium so as to produce
more of the added species. For example, consider the ideal-gas equilibrium N 3H
2 2
∆ 2NH . Suppose equilibrium is established at a T and P for which K [Eq. (6.26)] is
3 x
3
2
8.33; K 8.33 [x(NH )] /x(N )[x(H )] . Let the amounts n(N ) 3.00 mol, n(H )
2
2
3
x
2
2
1.00 mol, and n(NH ) 1.00 mol be present at this T and P. Defining Q as Q
3 x x
2
3
(x ) , we find that, for these amounts, Q (0.2) /0.6(0.2) 8.33. Since Q K ,
n i
i i x x x
the system is in equilibrium. Now, holding T and P constant, we add 0.1 mol of N .
2