Physical Chemistry , 1st ed.

These equilibrium amounts are in terms ofmoles,not in terms of pressure.

We are given the total pressure at equilibrium as well as the temperature. We

can use the ideal gas law to convert the moles of each species into pressures

of each species, then sum the pressures and require that this sum equals 0.750

atm. Thus, at equilibrium, we have

Pressure (atm) I 2 2I

Equilibrium

where we have left the units off the variables for clarity. You should be able

to recognize the units that go with each value. These pressures represent the

partial pressures of the species at equilibrium for this reaction. We use them

in the expression for the equilibrium constant:

K

(p

p

I

I

/

2 /

p

p

°

°

)^2



We can substitute the partial pressures into the above expression and get

K

K

K

2

which is subject to the condition that

0.750

It is this second equation, where it is understood that the units are atm, that

is most immediately solvable. By evaluating each fractional expression, we

find that

0.4923 82.05x164.1x0.750

82.05x0.258

x3.14  10 ^3

where in the last step we have limited our final answer to three significant fig-

ures. If we want the equilibrium amount of I 2 and I atoms, we need to solve

the appropriate expressions. For the number of moles of reactants and prod-

ucts, we have

mol I 2 6.00  10 ^3 x6.00  10 ^3 3.14  10 ^3

2.86  10 ^3 mol I2`

mol I  2 x2(3.14  10 ^3 ) 6.28  10 ^3 mol I

To get the equilibrium partial pressures, in terms of which the equilibrium

constant is written, we need to use the following expressions:

pI 2 0.235 atm

pI(6.28 ^10 0.515 atm

 (^3) )(0.08205)(1000)

1.00

(2.86  10 ^3 )(0.08205)(1000)



1.00

(2x)(0.08205)(1000)



1.00

(6.00  10 ^3 x)(0.08205)(1000)



1.00

(6.00  10 ^3 x)(0.08205)(1000)



1.00

(2x)(0.08205)(1000)



1.00

(2x)(0.08205)(1000)

1.00

(6.00 ^10 ^3 x)(0.08205)(1000)

1.00

QP

128 CHAPTER 5 Introduction to Chemical Equilibrium