# Question 562d1

Jun 3, 2017

$\text{20 dm"^3"/mol}$

${K}_{c}$ has the units given by the right-hand side. You are normally not asked to report the units... but you will need to know how to figure them out when solving for concentrations.

Recall that stoichiometric coefficients become the exponents of each concentration (because each equilibrium reaction direction is an elementary reaction from kinetics).

${K}_{c} = \left({\left[{\text{N"_2"O"_4])/(["NO}}_{2}\right]}^{2}\right)$

Always make sure you wrote down the problem correctly. You are actually given that ["NO"_color(red)(bb2)] = "0.1 mol/dm"^3, and ["N"_2"O"_4] = "0.2 mol/dm"^3. That can be rewritten into $\text{mol/L}$ or $\text{M}$ if you wish. ${\text{1 L" = "1 dm}}^{3}$.

So:

color(blue)(K_c) = ("0.2 M")/(("0.1 M")^2)#

$= {\text{20 M"^1"/M}}^{2}$

$= \textcolor{b l u e}{{\text{20 M}}^{- 1}}$

or $\textcolor{b l u e}{\text{dm"^3"/mol}}$, since $1 - 2 = - 1$ for the exponents, and $\text{M"^(-1) = 1/"M" = 1/("mol"/"dm"^3) = "dm"^3/"mol}$.