The Foundations of Chemistry

17-13 Evaluation of Equilibrium Constants at Different Temperatures 743

Problem-Solving Tip:Use the Correct K

The Kvalues that appear in the van’t Hoff equation represent the thermodynamic equi-

librium constant(see Section 17-12). For a gas-phase reaction (such as that in Example

17-20), Krepresents KP; if the value of Kcwere given, we would have to convert it to

KP(see Section 17-10) before using the van’t Hoff equation.

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

Plan

We are given KPat one temperature, 25°C, and the value of
H^0. We are given the second
temperature, 2400. K. These data allow us to evaluate the right side of the van’t Hoff
equation, which gives us ln (KT 2 /KT 1 ). Because we know KT 1 , we can find the value for
KT 2.

Solution

Let T 1 298 K and T 2 2400. K. Then

ln    

Let us first evaluate the right side of the equation.

ln    63.8

Now take the inverse logarithm of both sides.

e63.85.1 1027

Solving for KT 2 and substituting the known value of KT 1 , we obtain

KT 2 (5.1 1027 )(KT 1 )(5.1 1027 )(4.4 10 ^31 ) 2.2 10 ^3 at 2400. K

You should now work Exercise 88(b–f ).

KT 2



KT 1

1



2400. K

1



298 K

1.805 105 J/mol



8.314 J/molTK

KT 2



KT 1

1



T 2

1



T 1

H^0



R

KT 2



KT 1

In Example 17-20 we see that KT 2 (KPat 2400. K) is quite small, which tells us that the
equilibrium favors N 2 and O 2 rather than NO. Nevertheless, KT 2 is very much larger than
KT 1 , which is 4.4 10 ^31. At 2400. K, significantly more NO is present at equilibrium,
relative to N 2 and O 2 , than at 298 K. So automobiles emit small amounts of NO into
the atmosphere, sufficient to cause severe air pollution problems. Catalytic converters
(Section 16-9) are designed to catalyze the breakdown of NO into N 2 and O 2.

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

This reaction is spontaneous. Catalysts do not shift the position of equilibrium. They
favor neither consumption nor production of NO. They merely allow the system to reach
equilibrium more rapidly. The time factor is very important because the NO stays in the
automobile exhaust system for only a very short time.

The KPvalue could be converted to Kc

using the relationship KcKP(RT)
n

(Section 17-10).

2400 K is a typical temperature

inside the combustion chambers of

automobile engines. Large quantities

of N 2 and O 2 are present during

gasoline combustion, because the

gasoline is mixed with air.