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?

Question

#N_2O_4(g) ⇌ 2NO_2(g)#

Peyton Adams · Accepted Answer

I got #"0.0146 M"#.

You can determine the initial molar concentration of dinitrogen tetroxide by calculating its molar mass, which can be done using the chemical formula.

If we define the molar mass of substance #bb(i)# as #M_(m,i)#, then:

#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 #"mol"#s of this substance added to the container are:

#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")#

Therefore, by assuming that it fills the entire container as a gas (i.e., using the volume of the container as your volume), you can determine the concentration that it was at initially:

#color(green)(["N"_2"O"_4(g)]_0) = ("0.217 mols")/("5.00 L") = color(green)("0.0435 M")#

You now have enough knowledge to solve the remaining steps of the problem.

The response that was in balance was:

#"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#

so if #x# concentration of #"N"_2"O"_4(g)# was stoichiometrically transformed into #2x# concentration of #"NO"_2(g)# (not #x#!), then

#["N"_2"O"_4(g)]_(eq) = 0.0435 - x# #"M"#,

#["NO"_2(g)]_(eq) = 2x# #"M"#,

additionally:

#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 #K_c > 10^(-5)#, it is probably not a good idea to make a small #x# approximation, but we will give that a shot and compare to the true answer.

In the case of 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 APPROXIMATION OF THE SMALL X

#5.85xx10^(-3)*0.0435 - 5.85xx10^(-3)x - 4x^2 = 0#

which the quadratic formula is used to solve, yielding:

#color(green)(x) = (-(-5.85xx10^(-3)) pm sqrt((-5.85xx10^(-3))^2 - 4(-4)(5.85xx10^(-3)*0.0435)))/(2(-4))#

#= [...]#

which, if you accept the response that makes sense physically, gives you

#=# #color(green)("0.00728 M")#

which is #9.59%# error (more than an acceptable #5%# error), and so, that is why you should not make a small #x# approximation when #K_c > 10^(-5)# (same with #K_p# and #K_(sp)#).

So, we take this more accurate answer, and plug it back into the original expression for #["NO"_2(g)]_(eq)# to get:

#color(blue)(["NO"_2(g)]_(eq)) = 2x#

#= 2("0.00728 M")#

#=# #color(blue)("0.0146 M")#

Olivia Anton · Answer

undefined To solve this problem, we need to start by calculating the initial concentration of N2O4 in the container. We do this by converting the given mass of N2O4 to moles and then dividing by the volume of the container.

Given:
Mass of N2O4 = 20.0 g
Volume of container = 5.00 L

Next, we use the stoichiometry of the reaction to determine the change in concentration of N2O4 and NO2 at equilibrium. Since 1 mole of N2O4 produces 2 moles of NO2, the change in concentration of N2O4 is -2x and the change in concentration of NO2 is +4x, where x is the change in concentration.

We then use the equilibrium constant expression, Kc, to set up an equation:

Kc = [NO2]^2 / [N2O4]

Substituting the equilibrium concentrations into the equation and solving for x gives us the change in concentration of N2O4 and NO2 at equilibrium.

Finally, we use the initial concentration of N2O4 and the calculated change in concentration to find the equilibrium concentration of NO2.

Calculations:

1. Calculate initial concentration of N2O4:
   moles of N2O4 = mass / molar mass = 20.0 g / 92.02 g/mol = 0.217 mol
   initial concentration of N2O4 = moles / volume = 0.217 mol / 5.00 L = 0.0434 M

2. Set up equilibrium expression:
   Kc = (4x)^2 / (0.0434 - 2x)

3. Substitute Kc and initial concentration into the equation:
   5.85x10^(-3) = (4x)^2 / (0.0434 - 2x)

4. Solve for x:

5.85x10^(-3) = (16x^2) / (0.0434 - 2x)
   5.85x10^(-3)(0.0434 - 2x) = 16x^2
   0.0002539 - 0.0117x = 16x^2
   16x^2 + 0.0117x - 0.0002539 = 0

Using the quadratic formula, we find:
   x ≈ 0.012 M

5. Calculate equilibrium concentration of NO2:
   [NO2] = initial concentration + change in concentration
   [NO2] = 0 + (4 * 0.012) = 0.048 M

Therefore, the molar concentration of NO2 at equilibrium is approximately 0.048 M.