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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?

Md Ariful Islam

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Md Ariful Islam

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

$color(white)(mmmmmmmll)"N"_2"O"_4 ⇌ "2NO"_2$
$"I/mol·L"^"-1":color(white)(mm)0.0500color(white)(mmll)0$
$"C/mol·L"^"-1":color(white)(mml)"-2"xcolor(white)(mml)"+2"x $
$"E/mol·L"^"-1":color(white)(l)0.0500 - xcolor(white)(mll)2x $

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

$(4x^2)/("0.0500-"x) = 0.513$

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

$4x^2 + 0.513x - "0.025 65" = 0$

$x = "0.038 46"$

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

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

Check:

$K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = 0.0769^2/0.0115 = 0.514$

Close enough! It checks!

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