# Question 909ba

Apr 1, 2017

${K}_{\textrm{c}} = 0.064$

#### Explanation:

The chemical equation is

${\text{2SO"_3 ⇌ "2SO"_2 + "O}}_{2}$

The initial amount of ${\text{SO}}_{3}$ is

0.80 color(red)(cancel(color(black)("g SO"_3))) × ("1 mol SO"_3)/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "0.009 99 mol SO"_3

The volume of the container is 1 L, so the initial concentration of ${\text{SO}}_{3}$ is

["SO"_3] = "0.009 99 mol"/"1 L" = "0.09 99 mol/L"

We can setup an ICE table to solve this problem.

$\textcolor{w h i t e}{m m m m m m m m} {\text{2SO"_3 color(white)(l)⇌ color(white)(l)"2SO"_2 + "O}}_{2}$
$\text{I/mol·L"^"-1": color(white)(mm)"0.009 99} \textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m m} 0$
$\text{C/mol·L"^"-1":color(white)(mmll) "-2"xcolor(white)(mmmll)"+2"xcolor(white)(mll)"+} x$
$\text{E/mol·L"^"-1": color(white)(m)"0.009 99 - 2} x \textcolor{w h i t e}{m l} 2 x \textcolor{w h i t e}{m m l l} x$

We know that 80 % of the ${\text{SO}}_{3}$ has decomposed.

∴ The concentration remaining at equilibrium is

$\text{0.20 × 0.009 99 = 0.002 00}$

$\text{0.009 99 -"color(white)(l)2x = "0.002 00}$

$2 x = \text{0.009 99 - 0.002 00" = "0.007 99}$

$x = \text{0.007 99"/2 = "0.004 00}$

So,

["SO"_3]_text(eq) = "0.002 00 mol/L"

["SO"_2]_text(eq) = 2x color(white)(l)"mol/L" = "0.007 99 mol/L"

["O"_2]_text(eq) = x color(white)(l)"mol/L" = "0.004 00 mol/L"

K_c = (["SO"_2]^2["O"_2])/(["SO"_3]^2) = (("0.007 99")^2 × "0.004 00")/("0.002 00")^2 = 0.064#