Suppose a 500. mL flask is filled with 1.6 mol of N, and 0.10 mol of O,. The following reaction becomes possible: N,(2) +0,(g) -2NO(2) The equilibrium constant K for this reaction is 2.74 at the temperature of the flask. Calculate the equilibrium molarity of N,. Round your answer to two decimal places.

Equilibrium constant $\left({\mathrm{K}}_{\mathrm{C}}\right)$ for a reaction is a constant and it is the ratio between equilibrium concentration of products and reactants.

The given balanced equation is as follows.

${\mathrm{N}}_{2\left(\mathrm{g}\right)}+{\mathrm{O}}_{2\left(\mathrm{g}\right)}\iff 2{\mathrm{NO}}_{\left(\mathrm{g}\right)}$

${\mathrm{K}}_{\mathrm{C}}=\frac{{\left[\mathrm{NO}\right]}^2}{\left[{\mathrm{N}}_2\right]\left[{\mathrm{O}}_2\right]}$

Equilibrium constant for this reaction is $2.74$.

Initial concentration of ${\mathrm{N}}_2 \mathrm{and} {\mathrm{O}}_2$ can be calculated as follows.

$\begin{array}{rcl}\mathrm{Concentration}& =& \frac{\mathrm{no} \mathrm{of} \mathrm{moles}}{\mathrm{Volume} \mathrm{in} \mathrm{liters}}\\ {\mathrm{C}}_{{\mathrm{N}}_2}& =& \frac{1.6 \mathrm{mol}}{0.5 \mathrm{L}}=3.2 \mathrm{mol}/\mathrm{L}=3.2 \mathrm{M}\\ {\mathrm{C}}_{{\mathrm{O}}_2}& =& \frac{0.10 \mathrm{mol}}{0.5 \mathrm{L}}=0.2 \mathrm{mol}/\mathrm{L}=0.2 \mathrm{M}\end{array}$

Now we can construct the ICE table.

$$\begin{gathered} {\mathrm{N}}_{2\left(\mathrm{g}\right)} + {\mathrm{O}}_{2\left(\mathrm{g}\right)} \iff 2{\mathrm{NO}}_{\left(\mathrm{g}\right)} \\ \mathrm{Initial} \mathrm{conc}. 3.2 \mathrm{M} 0.2 \mathrm{M} 0 \\ \mathrm{Change} \mathrm{in} \mathrm{conc}. -\mathrm{x} -\mathrm{x} +2\mathrm{x} \\ \mathrm{Equilibrium} \mathrm{conc}. \left(3.2-\mathrm{x}\right) \mathrm{M} \left(0.2-\mathrm{x}\right) \mathrm{M} 2\mathrm{x} \end{gathered}$$