N2O4(g) ⇌ 2 NO2(g). For the reaction, Kc = 0.513 at 500 K. If a reaction vessel initially contains an N2O4 concentration of 0.0500 M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?

The equilibrium concentrations are [N₂O₄] = 0.0115 mol/L; [NO₂] = 0.0769 mol/L.

First, write the balanced chemical equation for the equilibrium and set up an ICE table.

N₂O₄ ⇌ 2NO₂ I/mol·L⁻¹: ##0.0500##; ##0## C/ mol·L⁻¹: ##-x##; ##+2x## E/ mol·L⁻¹: ##0.0500 - x##; ##2x##

##K_"eq" = (["NO"_2]^2)/["N"_2"O"_4]= (2x)^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 = 0.0385##

##["N"_2"O"_4] = (0.0500 - x) " mol/L" = (0.0500 – 0.0385) " mol/L" = "0.0115 mol/L"##

##["NO"_2] = 2x " mol/L" = "2 × 0.0385 mol/L" = "0.0769 mol/L"##

Check: ##K_"eq" = (["NO"_2]^2)/["N"_2"O"_4]= 0.0769^2/0.0115 = 0.514##. Close enough!