What is the value of Kc here?

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What is the value of Kc here?

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Answer 1 · 2016-09-11T18:23:16

$K_{c} = 0.006 94$ $K_c = "0.006 94"$

Explanation:

Your chemical equation is

$N_{2} O_{4} ⇌ {2NO}_{2}$ $"N"_2"O"_4 ⇌ "2NO"_2$

$K_{c} = \frac{{[{NO}_{2}]}_{2}^{2}}{[N_{2} O_{4}]}$ $K_c = (["NO"_2]^2)/(["N"_2"O"_4])$

We can setup an ICE table to solve this problem.

$m m m m m m m l N_{2} O_{4} ⇌ {2NO}_{2}$ $color(white)(mmmmmmml)"N"_2"O"_4 ⇌ "2NO"_2$
${I/mol\cdotL}^{-1} : m l l 0.0300 m m l l 0$ $"I/mol·L"^"-1": color(white)(mll)0.0300color(white)(mmll)0$
${C/mol\cdotL}^{-1} : m m l - x m m m +2 x$ $"C/mol·L"^"-1":color(white)(mml) "-"xcolor(white)(mmm)"+2"x$
${E/mol\cdotL}^{- 1} : m 0.0236 m m l 2 x$ $"E/mol·L"^-1": color(white)(m)0.0236color(white)(mml)2x$

From the $N_{2} O_{4}$ $"N"_2"O"_4$ concentrations, we see that

$0.0300 - x = 0.0236$ $0.0300 - x = 0.0236$

∴ $x = 0.0300 - 0.0236 = 0.0064$ $x = "0.0300 - 0.0236 = 0.0064"$

So, ${[{NO}_{2}]}_{eq} = 2 x l mol/L = (2 \times 0.0064) mol/L = 0.0128 mol/L$ $["NO"_2]_text(eq) = 2xcolor(white)(l) "mol/L" = "(2 × 0.0064) mol/L" = "0.0128 mol/L"$

$K_{c} = \frac{{0.0128}^{2}}{0.0236} = 0.006 94$ $K_"c" = 0.0128^2/0.0236 = "0.006 94"$

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What is the value of Kc here?

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