The Foundations of Chemistry

Plan

We are given equilibrium partial pressures of all reactants and products. So we write the expres-

sion for KPand substitute partial pressures in atmospheres into it.

Solution

KP7.11 10 ^5

You should now work Exercises 70 and 73.

(0.147)^2



(6.00)(3.70)^3

(PNH 3 )^2



(PN 2 )(PH 2 )^3

nrefers to the numbers of moles of
gaseous substances in the balanced
equation, notin the reaction vessel.

734 CHAPTER 17: Chemical Equilibrium

Problem-Solving Tip:In KPCalculations, Gas Pressures Must Be

Expressed in Atmospheres

One error that students sometimes make when solving KPproblems is to express pres-

sures in torr. Remember that these pressures must be expressed in atmospheres.

RELATIONSHIP BETWEEN KPAND Kc

If the ideal gas equation is rearranged, the molar concentration of a gas is

 or M

Substituting P/RTfor n/Vin the Kcexpression for the N 2 –H 2 –NH 3 equilibrium gives

the relationship between Kcand KPfor thisreaction.

Kc

[N

[N

2 ]

H

[H

3 ]

2

2

]^3



KP(RT)^2 and KPKc(RT)^2

In general the relationship between Kcand KPis

KPKc(RT)
n or KcKP(RT)
n 
n(ngas prod)(ngas react)

For reactions in which equal numbers of moles of gases appear on both sides of the equa-

tion, 
n0 and KPKc.

In Example 17-1, we saw that for the ammonia reaction at 500°C (or 773 K),

Kc0.286. We can describe this equilibrium in terms of partial pressures using KP.

NH 3 (g)3H 2 (g) 34 2NH 3 (g) 
n 2  4  2

KPKc(RT)
n(0.286)[(0.0821)(773)]^2 7.10 10 ^5

This is essentially the same value we obtained in Example 17-13.

R

1

T



2



R

1

T



4

(PNH 3 )^2



(PN 2 )(PH 2 )^3



P

R

N

T

H 3



2





P

R

N

T

^2



P

R

H

T

^2



3

P



RT

P



RT

n



V

17-10