# 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?

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#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")#