The Foundations of Chemistry

value of Kand the more favorable is the formation of products. We think of some reac-
tions as going to “completion.” These generally have very negative
G^0 rxnvalues. The
more positive the
G^0 rxnvalue, the smaller is the value of Kand the less favorable is the
formation of products.

EXAMPLE 17-17 K versus
G^0 rxn

Use the data in Appendix K to calculate KPfor the following reaction at 25°C.

2N 2 O(g) 34 2N 2 (g)O 2 (g)

Plan

The temperature is 25°C, so we evaluate
G^0 rxnfor the reaction from
Gf^0 values in Appendix
K. The reaction involves only gases, so Kis KP. This means that
G^0 rxnRTln KP. We solve
for KP.

Solution

G^0 rxn[2 
Gf^0 N 2 (g)
Gf^0 O 2 (g)][2 
Gf^0 N 2 O(g)]

[2(0)0][2(104.2)]208.4 kJ/mol, or 2.084 105 J/mol

This is a gas-phase reaction, so
G^0 rxnis related to KPby

G^0 rxnRTln KP

ln KP84.1

KPe84.1 3.3 1036

The very large value of KPtells us that the equilibrium lies veryfar to the right. This reac-
tion is so slow at 25°C, however, that very little N 2 O decomposes to N 2 and O 2 at that
temperature.

You should now work Exercise 84.

EXAMPLE 17-18 K versus
G^0 rxn

In Examples 15-16 and 15-17 we evaluated
G^0 rxnfor the following reaction at 25°C and found
it to be 173.1 kJ/mol. Calculate KPat 25°C for this reaction.

N 2 (g)O 2 (g) 34 2NO(g)

Plan

In this example we use
G^0 rxnRTln KP.

Solution

G^0 rxnRTln KP

ln KP

69.9

KPe69.9 4.4 10 ^31

1.731 105 J/mol



(8.314 J/molTK)(298 K)

G^0 rxn



RT

2.084 105 J/mol



(8.314 J/molTK)(298K)

G^0 rxn



RT

Units cancel when we express 
G^0 in

joules per mole. We interpret this as

meaning “per mole of reaction”—that

is, for the number of moles of each

substance shown in the balanced

equation.

On some calculators, we evaluate exas

follows: Enter the value of x,then

press INV followed by ln x.

17-12 Relationship Between 
G^0 rxnand the Equilibrium Constant 741