# Question #17305

Sep 16, 2015

${K}_{c} = 6.94 \times {10}^{- 3}$

#### Explanation:

${N}_{2} {O}_{4} \left(g\right) \to 2 N {O}_{2} \left(g\right)$

${K}_{c} = \frac{{\left[N {O}_{2} \left(g\right)\right]}^{2}}{\left[{N}_{2} {O}_{4} \left(g\right)\right]}$

We need to find the equilibrium concentrations for ${N}_{2} {O}_{4} \left(g\right) \text{ and } 2 N {O}_{2} \left(g\right)$; therefore we can use $\textcolor{b l u e}{\text{ICE table}}$ ( I don't know how to format it into a table!):

Remember, the volume is equal to 1.000 L and therefore,
${\left[{N}_{2} {O}_{4} \left(g\right)\right]}_{0} = \frac{n}{V} = \frac{0.0300 m o l}{1.000 L} = 0.0300 M$

at equilibrium: $\left[{N}_{2} {O}_{4} \left(g\right)\right] = \frac{n}{V} = \frac{0.0236 m o l}{1.000 L} = 0.0236 M$

${N}_{2} {O}_{4} \left(g\right) \text{ "->" } 2 N {O}_{2} \left(g\right)$

Initial: $\text{ " " " " ""0.0300M" " " " " " " " " " ""0M}$

Change: $\text{ " " " " "color(blue)(- x)"M" " " " " " " " " " "color(blue)(+2x)"M}$

Equilibrium: $\text{ " "0.0236M" " "" " " " " " " "+2x"M}$

Therefore, at equilibrium, $\left[{N}_{2} {O}_{4} \left(g\right)\right] = \left(0.0300 - x\right) M = 0.0236 M$

$\implies x = 0.0300 - 0.0236 = 0.0064 M$

$\left[N {O}_{2} \left(g\right)\right] = + 2 x M = 2 \times 0.0064 M = 0.0128 M$

${K}_{c} = \frac{{\left[N {O}_{2} \left(g\right)\right]}^{2}}{\left[{N}_{2} {O}_{4} \left(g\right)\right]} = \frac{{\left(0.0128\right)}^{2}}{0.0236} = 0.00694 = 6.94 \times {10}^{- 3}$