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?

Question

Mateo Aguilar · Accepted Answer

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

To solve the issue, let's set up an ICE table. #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"# Verify: #K_c = (["NO"_2]^2)/(["N"_2"O"_4]) = 0.0769^2/0.0115 = 0.514# It checks and is sufficiently close!

Luna Chen · Answer

undefined To solve this, we use the equilibrium expression and initial concentration to set up an ICE table and then solve for the equilibrium concentrations.

Given:
Kc = 0.513
Initial concentration of N2O4 = 0.0500 M

Using the equilibrium expression:
Kc = [NO2]^2 / [N2O4]

Setting up the ICE table:
N2O4(g) ↔ 2NO2(g)
Initial (M): 0.0500     0
Change (x): -x          +2x
Equilibrium (M): 0.0500 - x      2x

Substitute equilibrium concentrations into the equilibrium expression:
0.513 = (2x)^2 / (0.0500 - x)

Solve for x:
x = 0.0204

Substitute x back into the equilibrium concentrations:
[N2O4]eq = 0.0500 - 0.0204 = 0.0296 M
[NO2]eq = 2(0.0204) = 0.0408 M

Therefore, at equilibrium:
[N2O4] = 0.0296 M
[NO2] = 0.0408 M