Question #9c357

1 Answer
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

color(red)(2)"SO"_text(3(g]) rightleftharpoons color(blue)(2)"SO"_2 + "O"_2

the equilibrium constant has the following expression

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

Rearrange the equation to solve for ["O"_2] 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)

["O"_2] = 2.333 * 2.90 * 10^(-2) = color(green)(6.76 * 10^(-2)"M")

Now, does this result make sense?

Notice that in the expression for K_c, equal equilibrium concentrations for "SO"_3 and "SO"_2 will result in

["O"_2] = 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 "SO"_3 than "SO"_2, the concentration of "O"_2 will have to be greater than the value of K_c.