Physical Chemistry Third Edition

312 7 Chemical Equilibrium

which is the same as

4. 054 α^2 + 0. 148 α− 0. 148  0

Use of the quadratic formula gives

α

− 0. 148 ±

√

(0.148)^2 +(4)(4.054)(0.148)

2(4.054)

 0. 174

We have disregarded the negative root to the quadratic equation, since a negative value

ofαis not physically possible if no NO 2 is initially present. The total amount of gas is

(1.000 mol)(1−α+ 2 α) 1 .174 mol, so the equilibrium total pressure is

Peq

(1.174 mol)

(

8 .3145 J K−^1 mol−^1

)

(298.15 K)

0 .02446 m^3

 1. 190 × 105 Pa 1 .174 atm

It is always the case that when a quadratic equation is solved in a chemistry problem,

only one of the two roots corresponds to a physically possible situation.

We now consider reactions involving pure liquids and solids as well as gases. If the

pressure of the system does not differ very much fromP◦, then Eq. (6.3-14) gives for

the activity of a pure liquid or solid

ai≈1 (pure liquid or solid nearP◦) (7.2-3)

The activities of pure liquids or solids can be omitted from an activity quotient unless

the pressure is very large.

EXAMPLE 7.4

a.Write the equilibrium constant quotient for the reaction CaCO 3 (s)
CaO(s)+CO 2 (g).

b.Find the value of the equilibrium constant at 298.15 K andPeq(CO 2 ) at 298.15 K.

Solution

a.Since the activities of the solids are nearly equal to unity,

K

aeq(CaO)aeq(CO 2 )

aeq(CaCO 3 )

≈aeq(CO 2 )≈

Peq(CO 2 )

P◦

b.From the Gibbs energy changes of formation

∆G◦(1)(− 603 .501 kJ mol−^1 )− 394 .389 kJ mol−^1

+(−1)(− 1128 .79 kJ mol−^1 )

 130 .90 kJ mol−^1