Practice Problem 11

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 NO₂(g) N₂O₄(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)

NO₂(g) 33.18 240.06

N₂O₄ (g) 9.16 304.29

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

H^o= (1 mol N₂0₄ x 9.16 kJ/mol) - (2 mol NO₂ x 33.18 kJ/mol)

= -57.20 kJ

But it is not favored by entropy:

S^o= (1 mol N₂0₄ x 304.29 J/mol-K) - (2 mol NO₂ 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^o₃₇₃= H^o₂₉₈ - T S^o₂₉₈

= -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^oand 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:

Ex21_11.gif (1628 bytes)

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 10⁶

-196°C: K_p = 3.8 x 10²⁹

At 100°C, the unfavorable entropy term is realtively important, and the equilibrium lies on the side of NO₂. As the reaction is cooled, the entropy term becomes less important, and the equilibrium shifts toward N₂O₄. At the temperature of liquid nitrogen all of the NO₂ condenses to form N₂O₄.