# At 27°C ,1atm, 20% of N2O4 dissociated into NO2. Determine the partial pressure of each gas at equilibrium and the equilibrium constant Kp at this temperature(assume that the number of mole of N2O4 before dissociation is 1 mole)?

Nov 23, 2015

${P}_{N {O}_{2}} = 0.33 a t m$

${P}_{{N}_{2} {O}_{4}} = 0.67 a t m$

${K}_{P} = 0.163$

#### Explanation:

The dissociation reaction of ${N}_{2} {O}_{4}$ is:

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

For this question, the temperature and pressure are constant. If 20% dissociates of $1 m o l$ of ${N}_{2} {O}_{4}$, therefore, $0.2 m o l$ will be consumed. Using the $I C E$ table:

$\text{ " " " " " " " " } {N}_{2} {O}_{4} \left(g\right) \to 2 N {O}_{2} \left(g\right)$
$I n i t i a l \text{ " " " " "1mol" " " } 0 m o l$
$\text{Change" " " " "-0.2mol" " " } + 0.4 m o l$
$\text{Equilibrium" " " "0.8mol" " " " " " } 0.4 m o l$

From the equilibrium number of moles, we can find the mole fraction of each gas:

${\chi}_{N {O}_{2}} = {n}_{N {O}_{2}} / {n}_{\text{total}} = \frac{0.4 \cancel{m o l}}{1.2 \cancel{m o l}} = 0.33$

${\chi}_{{N}_{2} {O}_{4}} = {n}_{{N}_{2} {O}_{4}} / {n}_{\text{total}} = \frac{0.8 \cancel{m o l}}{1.2 \cancel{m o l}} = 0.67$

or ${\chi}_{{N}_{2} {O}_{4}} = 1 - {\chi}_{N {O}_{2}} = 0.67$

Since the pressure is constant ${P}_{\text{total}} = 1 a t m$

${\chi}_{N {O}_{2}} = \frac{{P}_{N {O}_{2}}}{P} _ \left(\text{total")=>P_(NO_2)=chi_(NO_2)xxP_("total}\right) = 0.33 \times 1 a t m = 0.33 a t m$

${P}_{{N}_{2} {O}_{4}} = {\chi}_{{N}_{2} {O}_{4}} \times {P}_{\text{total}} = 0.67 \times 1 a t m = 0.67 a t m$

The equilibrium constant ${K}_{P}$ expression can be written as:

${K}_{P} = \frac{{P}_{N {O}_{2}}^{2}}{P} _ \left({N}_{2} {O}_{4}\right) = {\left(0.33\right)}^{2} / 0.67 = 0.163$