The Foundations of Chemistry

This very small number indicates that at equilibrium almost no N 2 and O 2 are converted to

NO at 25°C. For all practical purposes, the reaction does not occur at 25°C.

You should now work Exercise 86.

A very important application of the relationships in this section is the use of measured

Kvalues to calculate 
G^0 rxn.

EXAMPLE 17-19 K versus 
G^0 rxn

The equilibrium constant, KP, for the following reaction is 5.04 1017 at 25°C. Calculate


G^0298 for the hydrogenation of ethylene to form ethane.

C 2 H 4 (g)H 2 (g) 34 C 2 H 6 (g)

Plan

We use the relationship between 
G^0 and KP.

Solution

G^0298 RTln KP

(8.314 J/molTK)(298 K) ln (5.04 1017 )

1.01 105 J/mol

101 kJ/mol

You should now work Exercise 88(a).

EVALUATION OF EQUILIBRIUM CONSTANTS AT

DIFFERENT TEMPERATURES

Chemists have determined equilibrium constants for thousands of reactions. It would be

an impossibly huge task to catalog such constants at every temperature of interest for each

reaction. Fortunately, there is no need to do this. If we determine the equilibrium constant,

KT 1 , for a reaction at one temperature, T 1 , and also its 
H^0 , we can then estimatethe equi-

librium constant at a second temperature, T 2 , using the van’t Hoff equation.

ln    

Thus, if we know 
H^0 for a reaction and Kat a given temperature (say 298 K), we can

use the van’t Hoff equation to calculate the value of Kat any other temperature.

EXAMPLE 17-20 Evaluation of KPat Different Temperatures

We found in Example 17-18 that KP4.4 10 ^31 at 25°C (298 K) for the following reac-

tion. 
H^0 180.5 kJ/mol for this reaction. Evaluate KPat 2400. K.

1



T 2

1



T 1

H^0



R

KT 2



KT 1

17-13

742 CHAPTER 17: Chemical Equilibrium

Compare the form of this equation to
the Arrhenius equation (Section 16-8)
and to the Clausius–Clapeyron
equation (Section 13-9).