The Foundations of Chemistry

Solving by the quadratic formula gives

x5.49 10 ^3 and 3.06 10 ^2 (extraneous root)

[SbCl 5 ](0.00669x) M(0.006690.00549) M1.20 10 ^3 M

__? g SbCl 5 5.00 L

1.20 1

L

0 ^3 mol



2

m

99

ol

g

 1.79 g SbCl 5

Let us now solve the same problem using KPand partial pressures.

(b) Plan (using KP)

Calculate the initial partial pressure of SbCl 5 , and write the reaction summary. Substitution of

the representations of the equilibrium partial pressures into KPgives their values.

(b) Solution (using KP)

We calculate the initial pressureof SbCl 5 in atmospheres, using PVnRT.

PSbCl 5 0.396 atm

Clearly, PSbCl 3 0 and PCl 2 0 because only SbCl 5 is present initially. We write the reaction

summary in terms of partial pressures in atmospheres, because KPrefers to pressures in atmo-

spheres.

Let ydecrease in pressure (atm) of SbCl 5 due to reaction. In terms of partial pressures,

the reaction summary is

SbCl 5 34 SbCl 3  Cl 2

initial 0.396 atm 0 0

change due to rxn yatm yatm yatm

equilibrium (0.396y) atm yatm yatm

KP1.48

0.5861.48yy^2 y^2 1.48y0.586 0

Solving by the quadratic formula gives

y0.325 and 1.80 (extraneous root)

PSbCl 5 (0.396y)(0.3960.325)0.071 atm

We use the ideal gas law, PVnRT,to calculate the number of moles of SbCl 5.

n0.0060 mol SbCl 5

__? g SbCl 5 0.0060 mol

2

m

99

ol

g

 1.8 g SbCl 5

You should now work Exercises 75 and 78.

(0.071 atm)(5.00 L)



0.0821 m

L

o

a

l

t

m

K

(721 K)

PV



RT

(y)(y)



0.396y

(PSbCl 3 )(PCl 2 )



PSbCl 5

(10.0 g)

1

29

m

9

o

g

l

0.0821 

m

L

o

a

l

t

m

K

(721 K)



5.00 L

nRT



V

The partial pressureof each substance is
proportional to the number of molesof
that substance.

If we did not know the value of KP,
we could calculate it from the known
value of Kc, using the relationship
KPKc(RT)
n.

We see that within roundoff range
the same result is obtained by both
methods.

736 CHAPTER 17: Chemical Equilibrium