# At a certain temperature, 0.660 mol of SO3 is placed in a 4.50-L container, according to the reaction: 2SO3(g) --> 2SO2(g) + O2(g), at equilibrium, 0.130 mol of O2 is present, how would you calculate Kc?

Dec 6, 2015

${K}_{c} = 0.0122$

#### Explanation:

For a given equilibrium reaction, the equilibrium constant is calculated by taking the ratio between the equilibrium concentrations of the products and the equilibrium concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.

For your equilibrium reaction

$\textcolor{red}{2} {\text{SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_text(2(g]) + "O}}_{\textrm{2 \left(g\right]}}$

the equilibrium constant, ${K}_{c}$, will take the form

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

So, you know that the starting concentration of sulfur trioxide is equal to

$\textcolor{b l u e}{c = \frac{n}{V}}$

["SO"_3]_0 = "0.660 moles"/"4.50 L" = "0.1467 M"

The starting concentrations of the two products will be equal to zero, since initially, the reactant is the only one present in the reaction container.

This means that you can use an ICE table to help you determine how the concentrations of the species will change once the reaction starts

${\text{ " color(red)(2)"SO"_text(3(g]) " "rightleftharpoons" " color(blue)(2)"SO"_text(2(g]) " "+" " "O}}_{\textrm{2 \left(g\right]}}$

color(purple)("I")" " " "0.1467" " " " " " " " " "0" " " " " " " " " "0
color(purple)("C")" " " "(-color(red)(2)x)" " " " " "(+color(blue)(2)x)" " " " " "(+x)
color(purple)("E")" "0.1467-color(red)(2)x" " " " " "color(blue)(2)x" " " " " " " " " " "x

Now, you know that the reaction container will contain $0.130$ moles of oxygen gas at equilibrium. This means that the equilibrium concentration of oxygen gas will be equal to

["O"_2] = "0.130 moles"/"4.50 L" = "0.02889 M"

Now, take a look at the ICE table. The equilibrium concentration of oxygen gas is said to be equal to $x$. This means that the equilibrium concentrations of the other two chemical species involved in the reaction will be

["SO"_3] = 0.1467 - 2 * 0.02889 = "0.08892 M"

["SO"_2] = 2 * 0.02889 = "0.05778 M"

This means that the equilibrium constant for this reaction at this certain temperature will be

${K}_{c} = \frac{{0.05778}^{2} \cdot 0.02889}{0.08892} ^ 2 = \textcolor{g r e e n}{0.0122}$

The answer is rounded to three sig figs.