# Question #82dc7

Feb 27, 2017

Here's what I got.

#### Explanation:

The equilibrium reaction given to you looks like this

$\textcolor{b l u e}{2} {\text{NH"_ (3(g)) rightleftharpoons "N"_ (2(g)) + color(red)(3)"H}}_{2 \left(g\right)}$

Now, the equilibrium constant for this reaction can be expressed using equilibrium concentrations, in which case you have ${K}_{c}$, or equilibrium partial pressures, in which case you have ${K}_{p}$.

For ${K}_{c}$, you have the ratio between the equilibrium concentrations of the two products raised to the power of their respective stoichiometric coefficients and the equilibrium concentration of the reactant raised to the power of its stoichiometric coefficient.

${K}_{c} = \left({\left[{\text{N"_ 2] * ["H"_2]^color(red)(3))/(["NH}}_{3}\right]}^{\textcolor{b l u e}{2}}\right)$
For ${K}_{p}$, you have the ratio between the equilibrium partial pressures of the two products raised to the power of their respective stoichiometric coefficients and the equilibrium partial pressure of the reactant raised to the power of its stoichiometric coefficient.
${K}_{p} = \left({\left({\text{N"_2) * ("H"_2)^color(red)(3))/(("NH}}_{3}\right)}^{\textcolor{b l u e}{2}}\right)$