**Practice Problem 11**

Use values of *H*^{o} and *S*^{o} for the following reaction at 25°C to estimate
the equilibrium constant for this reaction at the temperature of boiling water (100°C),
ice(0°C), a dry ice-acetone bath (-78°C), and liquid nitrogen (-196°C):

2 NO_{2}(*g*) N_{2}O_{4}(*g*)

**Solution**

Using a standard-state enthalpy of formation and absolute entropy data table, we find the following information:

**Compound
** *H*_{f}^{o}(kJ/mol)
S°(J/mol-K)

NO_{2}(*g*)
33.18 240.06

N_{2}O_{4} (*g*)
9.16 304.29

According to these data, the reaction is favored by enthalpy:

*H*^{o
}= (1 mol N_{2}0_{4} x 9.16 kJ/mol) - (2 mol NO_{2} x
33.18 kJ/mol)

= -57.20 kJ

But it is not favored by entropy:

*S*^{o
}= (1 mol N_{2}0_{4} x 304.29 J/mol-K) - (2 mol NO_{2}
x 240.06 J/mol-K)

= -175.83 J/K

If we assume that these values of *H*^{o} and *S*^{o} are still
valid at 100°C, the value of *G*^{o} at this temperature is 8400 J:

*G*^{o}_{373
}= *H*^{o}_{298}
- T *S*^{o}_{298}

= -57,200 J - (373 K)(-175.83 J/K) = 8400 J

Repeating this calculation at the other temperatures gives the following results:

100°C: *G*^{o} = 8.4 kJ

0°C: *G*^{o }= -9.2 kJ

-78°C: *G*^{o} = -22.9kJ

-196°C: *G*^{o} = -43.7
kJ

We now write the equation for the relationship between *G*^{o }and the
equilibrium constant for the reaction:

*G*^{o}
= - *RT* ln *K*

and solve for the natural log of the equilibrium constant:

Let's start by calculating the value of ln *K _{p} *when the
reaction is at 100°C:

The equilibrium constant at this temperature is therefore 0.067:

*K* = *e*^{-2.71} = 0.067

Repeating this calculation at the other temperatures gives the following results:

100°C: *K _{p }= *0.067

0°C: *K _{p}*
= 58

-78°C: *K _{p}*
= 1.4 x 10

-196°C: *K _{p}*
= 3.8 x 10

At 100°C, the unfavorable entropy term is realtively important, and the
equilibrium lies on the side of NO_{2}. As the reaction is cooled, the
entropy term becomes less important, and the equilibrium shifts toward N_{2}O_{4}.
At the temperature of liquid nitrogen all of the NO_{2} condenses to form N_{2}O_{4}.