At #25^@ "C"#, for the reaction #"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#, the equilibrium constant is #K_c = 5.85xx10^(-3)#. #"20.0 g"# of #"N"_2"O"_4# were added to a #"5.00-L"# container. What is the molar concentration of #"NO"_2# at equilibrium?
#N_2O_4(g) ⇌ 2NO_2(g)#
1 Answer
I got
Since you know the chemical formula for dinitrogen tetroxide, you can calculate its molar mass, which allows you to get the initial molar concentration.
If we define the molar mass of substance
#M_(m,N_2O_4) = 2M_(m,N) + 4M_(m,O)#
#= 2("14.007 g/mol") + 4("15.999 g/mol")#
#=# #"92.01 g/mol"#
Therefore, the
#color(green)(n_(N_2O_4)) = 20.0 cancel("g N"_2"O"_4(g)) xx "1 mol"/(92.01 cancel("g N"_2"O"_4(g)))#
#=# #color(green)("0.217 mols")#
and so, you can find the concentration that it was initially at, by assuming that as a gas, it fills the entire container (i.e. use the volume of the container as your volume):
#color(green)(["N"_2"O"_4(g)]_0) = ("0.217 mols")/("5.00 L") = color(green)("0.0435 M")#
At this point, you have enough information to do the remainder of the problem.
The balanced reaction was:
#"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#
so if
#["N"_2"O"_4(g)]_(eq) = 0.0435 - x# #"M"# ,
#["NO"_2(g)]_(eq) = 2x# #"M"# ,
and:
#K_c = 5.85xx10^(-3) = (["NO"_2(g)]^2)/(["N"_2"O"_4(g)])#
#= (2x)^2/(0.0435 - x) = (4x^2)/(0.0435 - x)#
Since
WITH THE SMALL X APPROXIMATION
#5.85xx10^(-3)*0.0435 ~~ 4x^2#
#=> color(green)(x) ~~ |sqrt(1/4 (5.85xx10^(-3)*0.0435))|#
#=# #color(red)("0.00797 M")#
WITHOUT THE SMALL X APPROXIMATION
#5.85xx10^(-3)*0.0435 - 5.85xx10^(-3)x - 4x^2 = 0# which we solve using the quadratic formula to get:
#color(green)(x) = (-(-5.85xx10^(-3)) pm sqrt((-5.85xx10^(-3))^2 - 4(-4)(5.85xx10^(-3)*0.0435)))/(2(-4))#
#= [...]# which, if you take the physically reasonable answer, gives you
#=# #color(green)("0.00728 M")#
which is
So, we take this more accurate answer, and plug it back into the original expression for
#color(blue)(["NO"_2(g)]_(eq)) = 2x#
#= 2("0.00728 M")#
#=# #color(blue)("0.0146 M")#