For the reaction, #K_c = 0.513# at 500 K. #N_2O_4(g) rightleftharpoons 2NO_2(g)#. If a reaction vessel initially contains an #N_2O_4# concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of #N_2O_4# and #NO_2# at 500 K?

1 Answer
Ernest Z.
Oct 10, 2016

The equilibrium concentrations of #"N"_2"O"_4# and #"NO"_2# are 0.0115 mol/L and 0.0769 mol/L, respectively.

Explanation:

Let's set up an ICE table to solve the problem.

#color(white)(mmmmmmmll)"N"_2"O"_4 ⇌ "2NO"_2#
#"I/mol·L"^"-1":color(white)(mm)0.0500color(white)(mmll)0#
#"C/mol·L"^"-1":color(white)(mml)"-2"xcolor(white)(mml)"+2"x #
#"E/mol·L"^"-1":color(white)(l)0.0500 - xcolor(white)(mll)2x #

#K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = (2x)^2/("0.0500-"x) = 0.513#

#(4x^2)/("0.0500-"x) = 0.513#

#4x^2 = 0.513("0.0500-"x) = "0.025 65" - 0.513x#

#4x^2 + 0.513x - "0.025 65" = 0#

#x = "0.038 46"#

∴ #["N"_2"O"_4] = ("0.0500-"x) color(white)(l)"mol/L" = "(0.0500 - 0.038 46)"color(white)(l) "mol/L" = "0.0115 mol/L"#

#["NO"_2] = 2x color(white)(l)"mol/L" = "2 × 0.038 46 mol/L" = "0.0769 mol/L"#

Check:

#K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = 0.0769^2/0.0115 = 0.514#

Close enough! It checks!

Related questions
See all questions in Dynamic Equilibrium
Impact of this question
18955 views around the world
You can reuse this answer
Creative Commons License