1.1 What is Chemistry?

http://www.ck12.org Chapter 19. Chemical Equilibrium

0 .2000 mol− 2 x= 0 .0204 mol

x= 0 .0898 mol

At equilibrium, our reaction mixture contains 0.0204 moles of NO 2 and 0.0898 moles of N 2 O 4. Because the reaction
is taking place in a 1-L container, the concentrations of these two components will be 0.0204 M and 0.0898 M,
respectively. Plugging these values into the equilibrium constant expression, we get the following:

Kc=

[N 2 O 4 ]

[NO 2 ]^2

Kc=

[ 0. 0898 ]

[ 0. 0204 ]^2

Kc= 2. 16 × 103

The equilibrium constant for this reaction has a value of 2.16× 103 for the temperature at which these data were
collected.

Now we will consider a different experiment in which we start with the product of this reaction (as written above),
but no reactant:

Example 19.5

0.8450 moles of pure N 2 O 4 is placed in a rigid 1-liter container at a fixed temperature (the same temperature as in
the previous example problem). The system is allowed to reach equilibrium, at which point the concentration of
NO 2 is found to be 0.01966 M. Calculate the equilibrium constant for the following reaction:

2NO 2 (g)N 2 O 4 (g)

Answer:

Again, we are given the equilibrium concentration of NO 2 , so now we need to find the concentration of N 2 O 4. Start
by setting up an ICE table, including any known information:

TABLE19.4:Example 19.5 ICE Table Initial Setup

2NO 2 (g)  N 2 O 4 (g)

Initial amount 0 mol 0.8450 mol

Change

Equilibrium amount 0.01966 mol

Because we are starting with no reactant, we know that the amount of reactant will increase and the amount of
product will decrease before an equilibrium is reached. Therefore, we can add the following information:

TABLE19.5:Example 19.5 ICE Table I

2NO 2 (g)  N 2 O 4 (g)

Initial amount 0 mol 0.8450 mol

Change +2x mol -x mol

Equilibrium amount 0.01966 mol 0.8450-x mol

Using the information given for the final amount of NO 2 , solve forx: