# Question 7cd04

Apr 24, 2016

${K}_{\text{eq}} = 0.0019$

#### Explanation:

$\text{A" ⇌ "3B" + "C}$

Remember that the numbers in an ICE table must be molarities, not moles.

We must therefore convert our numbers to molarities by using the formula

color(blue)(|bar(ul(color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "

I: ["A"] = "3 mol"/"2 L" = "1.5 mol/L"

E: ["B"] = "0.6 mol"/"2 L" = "0.3 mol/L"

Now, let's set up an ICE table in the usual way.

$\textcolor{w h i t e}{m m m m m m m m} \text{A"color(white)(l) ⇌color(white)(l) "3B" +color(white)(l) "C}$
$\text{I/mol·"^"-1} : \textcolor{w h i t e}{m m m} 1.5 \textcolor{w h i t e}{m m m} 0 \textcolor{w h i t e}{m m l} 0$
$\text{C/mol·L"^"-1":color(white)(mml)"-} x \textcolor{w h i t e}{m m} + 3 x \textcolor{w h i t e}{l} + x$
$\text{E/mol·L"^"-1} : \textcolor{w h i t e}{m} 1.5 - x \textcolor{w h i t e}{m l l} 3 x \textcolor{w h i t e}{m m} x$

We know that at equilibrium,

["B"] = "0.3 mol/L" = 3xcolor(white)(l) "mol/L"

Therefore, $x = 0.1$.

The other equilibrium concentrations must be

["C"] = xcolor(white)(l) "mol/L" = "0.1 mol/L"

["A"] = (1.5 – x) color(white)(l)"mol/L" = "(1.5 – 0.1)"color(white)(l)"mol/L" = "1.4 mol/L"

K_"eq" = (["B"]^3["C"])/(["A"]) = (0.3^3×0.1)/1.4 = 0.0019#