# How would you calculate K for the following equilibrium when [SO_3] = 0.0160 mol/L, [SO_2] = 0.00560 mol/L. and [O_2] = 0.00210 mol/L?

Jul 16, 2016

How would you calculate K for the following equilibrium when [SO3] = 0.0160 mol/L, [SO2] = 0.00560 mol/L. and [O2] = 0.00210 mol/L?

K = 3.89 × 10^3

You don’t identify "the following equilibrium", so I assume that it is

${\text{2SO"_2 + "O"_2 ⇌ "2SO}}_{3}$

Then

K = ["SO"_3]^2/(["SO"_2]^2["O"_2])

and

K = 0.0160^2/(("0.005 60")^2 × "0.002 10") = (2.56×10^"-4")/(3.136 × 10^"-5" × 2.10 ×10^"-3") = 3.89 × 10^3

If the equilibrium is

${\text{2SO"_3 ⇌ 2"SO"_2 + "O}}_{2}$

K = (["SO"_2]^2["O"_2])/ ["SO"_3]^2 = 1/3890 = 2.57 × 10^"-4"

Jul 16, 2016

Here's what I got.

#### Explanation:

Start by writing out the equilibrium reaction given to you

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

Now, the equilibrium constant for this equilibrium, ${K}_{c}$, is calculated by using the equilibrium concentrations of the chemical species that take part in the reaction -- keep in mind that solids and liquids are excluded from the expression of the equilibrium constant!

More specifically, the equilibrium constant is equal to the equilibrium concentration of the product raised to the power of its stoichiometric coefficient divided by the product of the equilibrium concentrations of the reactants, also raised to the power of their respective stoichiometric coefficients.

In your case, you have sulfur trioxide, ${\text{SO}}_{3}$, as the product and sulfur dioxide, ${\text{SO}}_{2}$ and oxygen gas, ${\text{O}}_{2}$, as the reactants.

Notice that the equilibrium concentration of the product is higher than the equilibrium concentrations of the two reactants. This tells you that the above equilibrium lies to the right, which implies that ${K}_{c} > 1$.

You will have

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

All you have to do now is use the date given to you by the problem

K_c = ( 0.0160^color(blue)(2) color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(blue)(2)))))/(0.00560^color(red)(2)color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(red)(2)))) * "0.00210 mol L"^(-1))

${K}_{c} = \text{3887.3 mol"^(-1)"L}$

Now, the equilibrium constant is usually given without added units, which means that your answer will be

${K}_{c} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{3890} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs

Mind you, you can also have

$\textcolor{b l u e}{2} {\text{SO"_ (3(g)) rightleftharpoons color(red)(2)"SO"_ (2(g)) + "O}}_{2 \left(g\right)}$

In this case, the equilibrium constant is

K_"c rev" = (["SO"_2]^color(red)(2) * ["O"_2])/(["SO"_3]^color(blue)(3))

Plug in your values to find

K_"c rev" = (0.00560^color(red)(2)color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(red)(2)))) * "0.00210 mol L"^(-1))/( 0.0160^color(blue)(2) color(darkgrey)(cancel(color(black)(("mol L"^(-1))^color(blue)(2)))))

${K}_{\text{c rev" = "0.000257 mol L}}^{- 1}$

${K}_{\text{c rev}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{0.000257} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs
$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{\text{c rev}} = \frac{1}{K} _ c} \textcolor{w h i t e}{\frac{a}{a}} |}}}$