# 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?

##### 2 Answers

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?

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

Then

and

If the equilibrium is

#### Answer:

Here's what I got.

#### Explanation:

Start by writing out the equilibrium reaction given to you

#color(red)(2)"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons color(blue)(2)"SO"_ (3(g)) #

Now, the *equilibrium constant* for this equilibrium, **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,

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

You will have

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

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 = "3887.3 mol"^(-1)"L"#

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

#K_c = color(green)(|bar(ul(color(white)(a/a)color(black)(3890)color(white)(a/a)|))) -># rounded to threesig figs

Mind you, you can also have

#color(blue)(2)"SO"_ (3(g)) rightleftharpoons color(red)(2)"SO"_ (2(g)) + "O"_ (2(g))#

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_"c rev" = "0.000257 mol L"^(-1)#

The answer would thus be

#K_"c rev" = color(green)(|bar(ul(color(white)(a/a)color(black)(0.000257)color(white)(a/a)|))) -># rounded to threesig figs

As a final note, notice that you have

#color(purple)(|bar(ul(color(white)(a/a)color(black)(K_"c rev" = 1/K_c)color(white)(a/a)|)))#