The Foundations of Chemistry

Both [N 2 O 4 ] and [NO 2 ] increasebecause of the large decrease in volume. However, the
number of molesof N 2 O 4 increases, while the number of molesof NO 2 decreases. We predict this
from LeChatelier’s Principle.

You should now work Exercise 66.

V

n

is 

no.

L

mol



KPhas no units for the same reasons

that Kchas no units.

17-9 Partial Pressures and the Equilibrium Constant 733

Problem-Solving Tip:There Are Several Ways to Solve

Equilibrium Problems

When a stress is applied to a system originally at equilibrium, it is no longer at equilib-

rium. As we did in Example 17-12(b), we can apply the stress to the old equilibrium values

and then treat these as the new “initial values.” Alternatively, we could adjust the original

concentrations valuesto reflect the stress, and then treat these as the new “initial values.”

We could consider [N 2 O 4 ]initialin Example 17-12(b) as theoriginal0.800 mol of N 2 O 4

from part (a) in the newvolume, 0.500 L. That is, [N 2 O 4 ]initial0.800 mol/0.500 L

1.60 M,with no NO 2 having yet been formed. [NO 2 ]initial 0 M.From that starting

point, the reaction would proceed to the right.

PARTIAL PRESSURES AND THE EQUILIBRIUM

CONSTANT

It is often more convenient to measure pressures rather than concentrations of gases.
Solving the ideal gas equation, PVnRT,for pressure gives

P (RT)orPM(RT)

The pressure of a gas is directly proportional to its concentration (n/V). For reactions in
which all substances that appear in the equilibrium constant expression are gases, we some-
times prefer to express the equilibrium constant in terms of partial pressures in atmospheres
(KP) rather than in terms of concentrations (Kc).
In general for a reaction involving gases,

aA(g)bB(g) 34 cC(g)dD(g) KP

For instance, for the following reversible reaction,

N 2 (g)3H 2 (g) 34 2NH 3 (g) KP

EXAMPLE 17-13 Calculation of KP

In an equilibrium mixture at 500°C, we find PNH 3 0.147 atm, PN 2 6.00 atm, and PH 2 
3.70 atm. Evaluate KPat 500°C for the following reaction.

N 2 (g)3H 2 (g) 34 2NH 3 (g)

(PNH 3 )^2



(PN 2 )(PH 2 )^3

(PC)c(PD)d



(PA)a(PB)b

n



V

17-9