The Foundations of Chemistry

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Thus, for the reaction of N 2 and H 2 to form NH 3 at 500°C, we can write

Kc0.286

The small value of Kcindicates that the equilibrium lies to the left.

The value of Kcfrom Example 17-1 is much different from the value given earlier for
the same reaction at 25°C. For this reaction, products are favored at the lower tempera-
ture (Kc3.6 108 at 25°C), whereas reactants are favored at the higher temperature
(Kc0.286 at 500°C). The dependence of Kcon temperature will be discussed later in
this chapter; for now, we note that it can depend strongly on temperature.

EXAMPLE 17-2 Calculation of Kc
We put 10.0 moles of N 2 O into a 2.00-L container at some temperature, where it decomposes
according to

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

At equilibrium, 2.20 moles of N 2 O remain. Calculate the value of Kcfor the reaction.
Plan
We express all concentrations in moles per liter. The mole ratio from the balanced chemical
equation allows us to find the changes in concentrations of the other substances in the reac-
tion. We use the reaction summary to find the equilibrium concentrations to use in the Kc
expression.
Solution
At equilibrium, 2.20 mol N 2 O remain, so

__? mol N 2 O reacting10.00 mol N 2 O initial2.20 mol N 2 O remaining
7.80 mol N 2 O reacting

The initial [N 2 O](10.00 mol)/(2.00 L)5.00 M;the concentration of N 2 O that reacts is
(7.80 mol)/(2.00 L)3.90 M.From the balanced chemical equation, each 2 mol N 2 O that
react produces 2 mol N 2 and 1 mol O 2 , or a reaction ratio of

1 mol N 2 O reactingS1 mol N 2 formedS^12 mol O 2 formed

We can now write the reaction summary.

2N 2 O(g) 34 2N 2 (g)  O 2 (g)
initial 5.00 M 00
change due to rxn 3.90 M 3.90 M ^12 (3.90)1.95 M
equilibrium 1.10 M 3.90 M 1.95 M

We put these equilibrium concentrations into the equilibrium constant expression and eval-
uate Kc.

Kc24.5

You should now work Exercises 24 and 28.

(3.90)^2 (1.95)

(1.10)^2

[N 2 ]^2 [O 2 ]

[N 2 O]^2

[NH 3 ]^2

[N 2 ][H 2 ]^3

714 CHAPTER 17: Chemical Equilibrium

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