Physical Chemistry Third Edition

7.6 The Temperature Dependence of Chemical Equilibrium. The Principle of Le Châtelier 339

volume (the stress) will increase the pressure by a smaller amount than if the reaction

were somehow prevented from shifting.

EXAMPLE7.19

For the reaction of Example 7.3, calculate the equilibrium degree of dissociation if the volume

is reduced to 10.00 L at 298.15 K. Interpret the results in terms of the principle of Le Châtelier.

Solution

Letαequal the degree of dissociation. Assume that the gases are ideal.

K 0. 148 

(2α)^2

1 −α

(1.000 mol)RT

P◦V



(2α)^2

1 −α

(1.000 mol)(8.3145 J K−^1 mol−^1 )(298.15 K)

(100000 Pa)(0.01000 m^3 )



(2α)^2

1 −α

(2.479)

9. 916 α^2

1 −α

9. 916 α^2 + 0. 148 α− 0. 148  0

α

− 0. 148 +

√

(0.148)^2 +4(9.916)(0.148)

2(9.916)

 0. 115

We have disregarded a negative root to the quadratic equation sinceαcannot be negative.

The decrease in volume causes the reaction to shift to the left, so that fewer moles of gas are

present in the smaller volume than in the larger volume of Example 7.3.

The principle of Le Châtelier also applies to the addition of a reactant or a product

to an equilibrium system.

EXAMPLE7.20

For the system of Example 7.3, find the effect of adding an additional 0.500 mol of NO 2.

Solution

Letαbe defined by

n(N 2 O 4 )( 1 .000 mol)( 1. 000 −α)

n(NO 2 )( 1 .000 mol)( 2. 000 α+ 0. 500 )

0. 148 

( 2. 000 α+ 0. 500 )^2

1. 000 −α

( 1 .000 mol)

RT

P◦V

RT

P◦V



(

8 .3145 J K mol−^1

)

( 298 .15 K)

(100000 Pa)

(

0 .02446 m^3

)

 1 .0135 mol−^1

0. 148

1. 0135

 0. 146 

( 2. 000 α+ 0. 5000 )^2

1. 000 −α