# In the reaction N_2(g) + 3H_2(g) -> 2NH_3(g), N_2 = 10.00 M, H_2 = 8.00 M, and NH_3 = 2.00 M. What is the equilibrium constant K, for this reaction?

Dec 23, 2015

#### Answer:

$7.81 \cdot {10}^{- 4}$

#### Explanation:

Before doing any calculation, take a look at the values of the equilibrium concentrations for the three chemical species that take part in the reaction.

Notice that the concentrations of the two reactants, nitrogen gas, ${\text{N}}_{2}$, and hydrogen gas, ${\text{H}}_{2}$, are hogher than the concentration of the product, ammonia, ${\text{NH}}_{3}$.

This tells you that, at this given temperature, the reaction favors the reactants over the formation of the product. Simply put, the equilibrium lies to the left, so you can expect the equilibrium constant, ${K}_{c}$, to be smaller than one.

For your given reaction

${\text{N"_text(2(g]) + color(red)(3)"H"_text(2(g]) rightleftharpoons color(blue)(2)"NH}}_{\textrm{3 \left(g\right]}}$

the equilibrium constant takes the form

${K}_{c} = \left({\left[{\text{NH"_3]^color(blue)(2))/(["N"_2] * ["H}}_{2}\right]}^{\textcolor{red}{3}}\right)$

Plug in your values to get

${K}_{c} = {\left(2.00\right)}^{\textcolor{b l u e}{2}} / \left(10.00 \cdot {8.00}^{\textcolor{red}{3}}\right) = \frac{4.00}{10.00 \cdot 512}$

${K}_{c} = \textcolor{g r e e n}{7.81 \cdot {10}^{- 4}}$

The answer is rounded to two sig figs.