What is the value of Kc here?

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1 Answer
Sep 11, 2016

Answer:

#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"#