Question

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?

Answer 1

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!