# Question fa88a

Apr 26, 2015

The equilibrium constant for your reaction will be 0.0667.

The first thing you need to do is determine the initial concentration of sulfur trioxide placed in the flask. To do that, use the compound's molar mass to determine how many moles you're working with

120.0cancel("g") * ("1 mole "SO_3)/(80.066cancel("g")) ~= "1.500 moles "# $S {O}_{3}$

The initial concentration will then be

${\left[S {O}_{3}\right]}_{0} = \text{1.500 moles"/"2.00 L" = "0.750 M}$

Use an ICE table to determine what the expression of ${K}_{c}$ will look like once the equilibrium is established

$\text{ } \textcolor{red}{2} S {O}_{3 \left(g\right)} r i g h t \le f t h a r p \infty n s \textcolor{red}{2} S {O}_{2} + {O}_{2 \left(g\right)}$
I......0.75................0...............0
C....(-$\textcolor{red}{2} x$).............(+$\textcolor{red}{2}$x)..........(+x)
E...0.75-2x............2x...............x

By definition, the equilibrium constant will be

${K}_{c} = \frac{\left[{O}_{2}\right] \cdot {\left[S {O}_{2}\right]}^{\textcolor{red}{2}}}{{\left[S {O}_{3}\right]}^{\textcolor{red}{2}}} = \frac{x \cdot {\left(2 x\right)}^{2}}{0.75 - 2 x} ^ 2 = \frac{4 {x}^{3}}{0.75 - 2 x} ^ 2$

You know that the equilibrium concentration of oxygen is 0.150 M, so you can replace $x$ in the above equation with that value to get

${K}_{c} = \frac{4 \cdot {0.150}^{3}}{0.750 - 0.150} ^ 2 = \textcolor{g r e e n}{0.0667}$

Apr 26, 2015

${K}_{c} = 0.0675 \text{mol/l}$

$2 S {O}_{3 \left(g\right)} r i g h t \le f t h a r p \infty n s S {O}_{2 \left(g\right)} + {O}_{2 \left(g\right)}$

Mass $S {O}_{3}$ = 120g

${M}_{r} \left[S {O}_{3}\right] = 80.0$

So no. moles $S {O}_{3} = \frac{120}{80} = 1.5$

$2 S {O}_{3 \left(g\right)} r i g h t \le f t h a r p \infty n s S {O}_{2 \left(g\right)} + {O}_{2 \left(g\right)}$

Initial moles:

$S {O}_{3} = 1.5$

$S {O}_{2} = 0$

${O}_{2} = 0$

If $x$ moles of $S {O}_{3}$ are used up then the number of moles at equilibrium $\Rightarrow$

$S {O}_{3} = \left(1.5 - x\right)$

$S {O}_{2} = x$

${O}_{2} = \frac{x}{2}$

We are told $\left[{O}_{2}\right] = 0.15 \text{M}$

$c = \frac{n}{V}$ so $n = c V = 0.15 \times 2 = 0.3$

So $\frac{x}{2} = 0.3$

$x = 0.3 \times 2 = 0.6$

So moles at equilibrium$\Rightarrow$

$S {O}_{3} = \left(1.5 - 0.6\right) = 0.9$

$S {O}_{2} = 0.6$

${O}_{2} = 0.3$

So to get equilibrium concentrations we divide each by the total volume of 2L.

So equilibrium conc.s $\Rightarrow$

$\left[S {O}_{3}\right] = \frac{0.9}{2} = 0.45$

$\left[S {O}_{2}\right] = \frac{0.6}{2} = 0.3$

$\left[{O}_{2}\right] = \frac{0.3}{2} = 0.15$

${K}_{c} = \frac{{\left[S {O}_{3}\right]}^{2}}{{\left[S {O}_{2}\right]}^{2} \left[{O}_{2}\right]}$

$= \frac{{0.3}^{2} \times 0.15}{{0.45}^{2}}$

${K}_{c} = 0.0675 \text{mol/l}$