The Foundations of Chemistry

For gas-phase reactions,we can calculate the amounts of substances present at equilib-
rium using either KPor Kc. The results are the same by either method (when they are
both expressed in the same terms). To illustrate, let’s solve the following problem by both
methods.

EXAMPLE 17-14 Using Kcand KP

We place 10.0 grams of SbCl 5 in a 5.00-liter container at 448°C and allow the reaction to
attain equilibrium. How many grams of SbCl 5 are present at equilibrium? Solve this problem
(a) using Kcand molar concentrations and (b) using KPand partial pressures.

SbCl 5 (g) 34 SbCl 3 (g)Cl 2 (g) at 448°C, Kc2.51 10 ^2 and KP1.48

(a) Plan (using Kc)

We calculate the initial concentration of SbCl 5 , write the reaction summary and represent the
equilibrium concentrations; then we substitute into the Kcexpression to obtain the equilibrium
concentrations.

(a) Solution (using Kc)

Because we are given Kc, we use concentrations. The initial concentration of SbCl 5 is

[SbCl 5 ]

10.0

5.0

g

0

S

L

bCl 5



1

29

m

9

o

g

l

0.00669 MSbCl 5

Let xmol/L of SbCl 5 that react. In terms of molar concentrations, the reaction summary is

SbCl 5 34 SbCl 3  Cl 2

initial 0.00669 M 00

change due to rxn x M x M x M

equilibrium (0.00669x) Mx Mx M

Kc

2.51 10 ^2 

x^2 1.68 10 ^4 2.51 10 ^2 x

x^2 2.51 10 ^2 x1.68 10 ^4  0

(x)(x)



0.00669x

[SbCl 3 ][Cl 2 ]



[SbCl 5 ]

17-10 Relationship Between Kpand Kc 735

Problem-Solving Tip:Be Careful About the Value of R

To decide which value of Rto use when you convert between Kcand KP, you can reason

as follows. Kcinvolves molar concentrations, for which the units are mol/L; KPinvolves

pressures expressed in atm. Thus the appropriate value of Rto use for these conversions

must include these units. We use 0.08206 , rounded to the number of places

appropriate to the problem.

LTatm



molTK