# 3 moles of "SO"_3 gas are taken in an "8-L" container at 800^@"C". At equilibrium, 0.6 moles of "O"_2 are formed. What is the value of K_c for 2"SO"_2(g) + "O"_2(g) rightleftharpoons 2"SO"_3(g) ?

Dec 30, 2017

${K}_{c} = 50$

#### Explanation:

The first thing to notice here is that the problem wants you to figure out the equilibrium constant, ${K}_{c}$, for this equilibrium reaction

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

but that you start with sulfur trioxide as the only gas present in the reaction vessel. This implies that the reverse reaction, i.e. the reaction that consumes sulfur trioxide and produces sulfur dioxide and oxygen gas, is favored here.

$2 {\text{SO"_ (3(g)) rightleftharpoons 2"SO"_ (2(g)) + "O}}_{2 \left(g\right)}$

The equilibrium constant for this reaction is

${K}_{c}^{'} = \frac{1}{K} _ c$

Now, you know that you start with $3$ moles of sulfur trioxide in an $\text{8-L}$ container, so you can say that the initial concentration of sulfur trioxide is

["SO"_ 3]_ 0 = "3 moles"/"8 L" = 3/8color(white)(.)"M"

Now, the balanced chemical equation tells you that in order for the reaction to produce $2$ moles of sulfur dioxide and $1$ mole of oxygen gas, it must consume $2$ moles of sulfur trioxide.

At equilibrium, you know that you have

["O"_ 2] = "0.6 moles"/"8 L" = 3/40color(white)(.)"M"

You can say that in order for the reaction to produce $\frac{3}{40}$ $\text{M}$ of oxygen gas, it must also produce

["SO"_ 2]_ "produced" = 2 xx ["O"_ 2]_ "consumed"

["SO"_ 2]_ "produced" = 2 xx 3/40color(white)(.)"M"

["SO"_ 2 ]_ "produced" = 3/20color(white)(.)"M"

and consume

["SO"_ 3]_ "consumed" = ["O"_ 2]_ "produced"

["SO"_ 3]_ "consumed" = 3/40color(white)(.)"M"

This means that the equilibrium concentration of sulfur trioxide will be

["SO"_ 3] = ["SO"_ 3]_ 0 - ["SO"_ 3]_ "consumed"

["SO"_ 3] = 3/8color(white)(.)"M" - 3/40color(white)(.)"M"

["SO"_ 3] = 3/10color(white)(.)"M"

By definition, the equilibrium constant for the initial equilibrium when the reverse reaction is favored is equal to

${K}_{c}^{'} = \left({\left[{\text{SO"_2]^2 * ["O"_2])/(["SO}}_{3}\right]}^{2}\right)$

This implies that the equilibrium constant for the initial equilibrium when the forward reaction is favored will be

${K}_{c} = \frac{1}{K} _ {c}^{'}$

${K}_{c} = \left(\left[{\text{SO"_3]^2)/(["SO"_2]^2 * ["O}}_{2}\right]\right)$

Plug in the values you have for the equilibrium concentrations of the three gases to find--I'll leave the expression without added units!

${K}_{c} = \frac{{\left(\frac{3}{10}\right)}^{2}}{{\left(\frac{3}{20}\right)}^{2} \cdot \frac{3}{40}} = \frac{160}{3} = 53.3$

Rounded to one significant figure, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{K}_{c} = 50}}}$

So, this tells you that at ${800}^{\circ} \text{C}$, the equilibrium

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

has

${K}_{c} = 50$

which implies that, at this temperature, the forward reaction is favored, i.e. the equilibrium will lie to the right.