# Question 9c357

Dec 6, 2015

["O"_2] = 6.76 * 10^(-2)"M"

#### Explanation:

As you know, 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 this reaction

$\textcolor{red}{2} {\text{SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_2 + "O}}_{2}$

the equilibrium constant has the following expression

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

Rearrange the equation to solve for $\left[{\text{O}}_{2}\right]$ and plug in your values to get

["O"_2] = (["SO"_3]^color(red)(2))/(["SO"_2]^color(blue)(2)) * K_c

["O"_2] = ( (4.46 * 10^(-2))^2 color(red)(cancel(color(black)("M"^2))))/((2.92 * 10^(-2))^2 color(red)(cancel(color(black)("M"^2)))) * 2.90 * 10^(-2)#

$\left[\text{O"_2] = 2.333 * 2.90 * 10^(-2) = color(green)(6.76 * 10^(-2)"M}\right)$

Now, does this result make sense?

Notice that in the expression for ${K}_{c}$, equal equilibrium concentrations for ${\text{SO}}_{3}$ and ${\text{SO}}_{2}$ will result in

$\left[{\text{O}}_{2}\right] = {K}_{c}$

Any different in magnitude between the equilibrium concentrations of sulfur trioxide and sulfur dioxide will be amplified by their respective stoichiometric coefficients.

This tells you that anytime the equilibrium mixture contains more ${\text{SO}}_{3}$ than ${\text{SO}}_{2}$, the concentration of ${\text{O}}_{2}$ will have to be greater than the value of ${K}_{c}$.