# At a certain temperature, 4.0 mol NH_3 is introduced into a 2.0 L container, and the NH_3 partially dissociates to 2NH_3(g)\rightleftharpoonsN_2(g)+3H_2(g). At equilibrium, 2.0 mol NH_3 remains. What is the value of K_c?

## A clearer version: At a certain temperature, 4.0 mol $N {H}_{3}$ is introduced into a 2.0 L container, and the $N {H}_{3}$ partially dissociates to $2 N {H}_{3} \left(g\right) \setminus r i g h t \le f t h a r p \infty n s {N}_{2} \left(g\right) + 3 {H}_{2} \left(g\right)$. At equilibrium, 2.0 mol $N {H}_{3}$ remains. What is the value of ${K}_{c}$? Note: I started an ICE table but I don't know what to put in the ${N}_{2}$ and $3 {H}_{2}$ columns.

Jun 27, 2018

Remember to include the coefficients in the change in concentratio, as well as the exponents. I get ${K}_{c} = 1.69$.

If ${K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta {n}_{\text{gas}}}$, what would ${K}_{p}$ be if this reaction has been occurring at $\text{298 K}$? $R = \text{0.08206 L"cdot"atm/mol"cdot"K}$. Highlight below to see.

$\textcolor{w h i t e}{\text{From the reaction, "Deltan_"gas} = \left(1 + 3\right) - \left(2\right) = 2 ,}$
$\textcolor{w h i t e}{{\text{since there are 3 mols H"_2, "1 mol N"_2, "and 2 mols NH}}_{3.}}$

$\textcolor{w h i t e}{\text{So,}}$

$\textcolor{w h i t e}{{K}_{p} = 1.69 \cdot {\left(\text{0.08206 L"cdot"atm/mol"cdot"K" cdot "298 K}\right)}^{\left(1 + 3\right) - \left(2\right)}}$

$\textcolor{w h i t e}{\approx 1010 \text{ in implied units of atm}}$

Let's first find the concentrations, because ${K}_{c}$ is in terms of concentrations.

$\text{4.0 mols NH"_3/("2.0 L") = "2.0 M}$

$\text{2.0 mols NH"_3/("2.0 L") = "1.0 M}$

The ICE table uses ${\text{NH}}_{3}$ having the initial concentration of $\text{2.0 M}$ and final of $\text{1.0 M}$. Therefore, we should be able to find $x$ already, but let's assume we don't know $x$ yet.

$\textcolor{red}{2} {\text{NH"_3(g) rightleftharpoons "N"_2(g) + color(red)(3)"H}}_{2} \left(g\right)$

$\text{I"" "2.0" "" "" "" "0" "" "" } 0$
$\text{C"" "-color(red)(2)x" "" "+x" "" } + \textcolor{red}{3} x$
$\text{E"" "2.0-color(red)(2)x" "" "x" "" } \textcolor{red}{3} x$

Remember that the coefficients go into the change in concentration and exponents.

Now, the ${K}_{c}$ would be:

${K}_{c} = \frac{x {\left(\textcolor{red}{3} x\right)}^{\textcolor{red}{3}}}{2.0 - \textcolor{red}{2} x} ^ \textcolor{red}{2}$

$= \frac{27 {x}^{4}}{2.0 - 2 x} ^ 2$

But like mentioned, we know $x$. This is because

$2.0 - 2 x = \text{1.0 M}$.

Therefore:

$x = \text{0.5 M}$.

As a result,

$\textcolor{b l u e}{{K}_{c}} = \frac{27 {\left(0.5\right)}^{4}}{2.0 - 2 \left(0.5\right)} ^ 2$

$= \textcolor{b l u e}{1.69}$