Question

What is the value of Kc here?

Answer 1

`K_c = "0.006 94"`

Explanation:

Your chemical equation is

`"N"_2"O"_4 ⇌ "2NO"_2`

`K_c = (["NO"_2]^2)/(["N"_2"O"_4])`

We can setup an ICE table to solve this problem.

`color(white)(mmmmmmml)"N"_2"O"_4 ⇌ "2NO"_2`
`"I/mol·L"^"-1": color(white)(mll)0.0300color(white)(mmll)0`
`"C/mol·L"^"-1":color(white)(mml) "-"xcolor(white)(mmm)"+2"x`
`"E/mol·L"^-1": color(white)(m)0.0236color(white)(mml)2x`

From the `"N"_2"O"_4` concentrations, we see that

`0.0300 - x = 0.0236`

∴ `x = "0.0300 - 0.0236 = 0.0064"`

So, `["NO"_2]_text(eq) = 2xcolor(white)(l) "mol/L" = "(2 × 0.0064) mol/L" = "0.0128 mol/L"`

`K_"c" = 0.0128^2/0.0236 = "0.006 94"`