**Practice Problem 5**

Ammonia is made from nitrogen and hydrogen by the following reversible reaction.

N_{2}(*g*) + 3 H_{2}(*g*) 2 NH_{3}(*g*)

Assume that the initial concentration of N_{2} is
0.050 moles per liter and the initial concentration of H_{2}
is 0.100 moles per liter. Calculate the equilibrium
concentrations of the three components of this reaction at 500^{o}C
if the equilibrium constant for the reaction at this temperature
is 0.040.

**Solution**

We start, as always, by building a representation for the problem based on the following general format.

We then calculate the initial reaction quotient and compare it with the equilibrium constant for the reaction.

The reaction quotient (*Q*_{c} = 0) is
smaller than the equilibrium constant (*K*_{c}
= 0.040), so the reaction has to shift to the right to reach
equilibrium. This will result in a decrease in the concentrations
of N_{2} and H_{2} and an increase in the NH_{3}
concentration. The relationship between the changes in the
concentrations of N_{2}, H_{2}, and NH_{3}
as this reaction comes to equilibrium is determined by the
stoichiometry of the reaction, as shown in Figure 16.7.

Substituting this information into the equilibrium constant expression for the reaction gives the following equation.

Since *Q*_{c} for the initial
concentrations and *K*_{c} for the
reaction are both smaller than 1, let's try the assumption that
is small enough that subtracting it from 0.100 -- or even
subtracting 3 from 0.100 -- doesn't make a significant change.
This assumption gives the following approximate equation.

Solving this equation for gives the following result.

** ~ 1.0 x 10**^{-3}**
****M**** **

Now we have to check our assumptions. Is significantly smaller
than 0.100? Yes, is about 1% of the initial concentration of N_{2}.

0.001 |
x 100% = 1% |

0.100 |

Is 3 significantly smaller than 0.100? Once again, the answer
is yes, 3 is only about 2% of the initial concentration of H_{2}.

3(0.001) |
x 100% = 3% |

0.100 |

We can therefore use this approximate value of to determine
the equilibrium concen- trations of N_{2}, H_{2},
and NH_{3}.

[NH_{3}] = 2 ~ **0.0020 ****M****
**

[N_{2}] = 0.100- ~ **0.099 ****M****
**

[H_{2}] = 0.100 **- 3**** **~**0.097
****M**** **

Note that only 1% of the nitrogen is converted into ammonia under these conditions.

We can check the validity of our calculation by substituting this information back into the equilibrium constant expression.

Once again, we have reason to accept the assumption that is
small compared with the initial concentrations because the
equilibrium constant calculated from these data agrees with the
value of *K*_{c} given in the problem
within experimental error.