# What is K_(eq) for the reaction N_2+3H_2 -> 2NH_3 if the equilibrium concentrations are [NH_3] = 3.0 M, [N_2] = 2.0M, and [H_2] = 1.0M?

Jan 5, 2016

${K}_{c} = 4.5$

#### Explanation:

As you know, the equilibrium constant for a given chemical equilibrium depends on

• the equilibrium concentrations of the chemical species that take part in the reaction
• the stoichiometric coefficients of these chemical species

More specifically, the equilibrium constant is defined as the ratio between the product of the equilibrium concentrations of the products and the product of the equilibrium concentrations of the reactants, each raised to the power of their stoichiometric coefficients.

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

By definition, the equilibrium constant for this reaction, ${K}_{c}$, will be equal to

K_c = (["NH"_3]^color(blue)(2))/(["N"_2] * ["H"_2]^color(red)(3))" ", where

$\left[{\text{NH}}_{3}\right]$, $\left[{\text{N}}_{2}\right]$, $\left[{\text{H}}_{2}\right]$ represent the equilibrium concentrations of the three species.

In your case, you know that these concentrations are equal to

["NH"_3] = "3.0 M"

["H"_2] = "1.0 M"

["N"_2] = "2.0 M"

This means that the equilibrium constant will be equal to - keep in mind that equilibrium constant are usually used without corresponding units

${K}_{c} = {3.0}^{\textcolor{b l u e}{2}} / \left(2.0 \cdot {1.0}^{\textcolor{red}{3}}\right) = \frac{9}{2} = \textcolor{g r e e n}{4.5}$

If you want, you can include the units in the calculation to get

K_c = (3.0^color(blue)(2) "M"^color(blue)(2))/("2.0 M" * 1.0^color(red)(3)"M"^color(red)(3)) = "4.5 M"^(-2)