# For the reaction, K_c = 0.513 at 500 K. N_2O_4(g) rightleftharpoons 2NO_2(g). If a reaction vessel initially contains an N_2O_4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N_2O_4 and NO_2 at 500 K?

Oct 10, 2016

The equilibrium concentrations of ${\text{N"_2"O}}_{4}$ and ${\text{NO}}_{2}$ are 0.0115 mol/L and 0.0769 mol/L, respectively.

#### Explanation:

Let's set up an ICE table to solve the problem.

$\textcolor{w h i t e}{m m m m m m m l l} {\text{N"_2"O"_4 ⇌ "2NO}}_{2}$
$\text{I/mol·L"^"-1} : \textcolor{w h i t e}{m m} 0.0500 \textcolor{w h i t e}{m m l l} 0$
$\text{C/mol·L"^"-1":color(white)(mml)"-2"xcolor(white)(mml)"+2} x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{l} 0.0500 - x \textcolor{w h i t e}{m l l} 2 x$

K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = (2x)^2/("0.0500-"x) = 0.513

$\frac{4 {x}^{2}}{\text{0.0500-} x} = 0.513$

4x^2 = 0.513("0.0500-"x) = "0.025 65" - 0.513x

$4 {x}^{2} + 0.513 x - \text{0.025 65} = 0$

$x = \text{0.038 46}$

["N"_2"O"_4] = ("0.0500-"x) color(white)(l)"mol/L" = "(0.0500 - 0.038 46)"color(white)(l) "mol/L" = "0.0115 mol/L"

["NO"_2] = 2x color(white)(l)"mol/L" = "2 × 0.038 46 mol/L" = "0.0769 mol/L"

Check:

${K}_{c} = \left(\left[{\text{NO"_2]^2)/(["N"_2"O}}_{4}\right]\right) = {0.0769}^{2} / 0.0115 = 0.514$

Close enough! It checks!