9.2g of dinitrogen tetroxide is taken in a closed 1L vessel and heated till the following equilibrium is reached. At equilibrium, #50%# of the #"N"_2"O"_4(g)# is dissociated. What is the equilibrium constant (in mol/liter) ?
#"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#
1 Answer
Explanation:
The equilibrium reaction given to you looks like this
#"N"_ 2"O"_ (4(g)) rightleftharpoons color(blue)(2)"NO"_ (2(g))#
Your first goal here is to figure out the initial concentration of the dinitrogen tetroxide. You know that because the vessel has a volume of
Use the molar mass of dinitrogen tetroxide to find the number of moles present in the sample
#9.2 color(red)(cancel(color(black)("g"))) * ("1 mole N"_2"O"_4)/(92.011color(red)(cancel(color(black)("g")))) ~~ "0.100 moles N"_2"O"_4#
This means that the initial concentration of dinitrogen tetroxide is equal to
Now, the problem tells you that at equilibrium,
So, if half of the initial concentration gets converted to nitrogen dioxide, you can say that the equilibrium concentration of dinitrogen tetroxide will be
#["N"_ 2"O"_ 4]_ "eq" = 1/2 * "0.100 mol L"^(-1) = "0.0500 mol L"^(-1)#
Notice that for every molecule of dinitrogen tetroxide that dissociates,
This means that the equilibrium concentration of nitrogen dioxide will be twice the concentration of dinitrogen tetroxide that dissociated.
#["NO"_ 2]_ "eq" = color(blue)(2) * "0.0500 mol L"^(-1) = "0.100 mol L"^(-1)#
By definition, the equilibrium constant for this reaction will be
#K_c = (["NO"_ 2]_ "eq"^color(blue)(2))/(["N"_ 2"O"_ 4]_ "eq")#
Plug in your values to find
#K_c = ("0.100 mol L"^(-1))^color(blue)(2)/("0.0500 mol L"^(-1)) = color(green)(bar(ul(|color(white)(a/a)color(black)("0.20 mol L"^(-1))color(white)(a/a)|)))#
The answer is rounded to two sig figs.