Answer to Question #123438 in General Chemistry for elif

Question

in a vessel at 70 C and 1 atm pressure, 65.6% of dinitrogen tetroxide gas dissociate into nitrogen dioxide gas. what is the partial pressure of each gas?

Answer 1

Ans:

The dissociation reaction of N₂O₄ is:

N₂O₄(g)→2NO₂(g)

For this question, the temperature and pressure are constant. If 65.6% dissociates of 1mol of N2O4, therefore, 0.656 mole will be consumed. Using the ICE table:

         N₂O₄(g)         →            2NO₂(g)

Initial     1 mole                        0 mole

Change   −0.656 mole          1.312 mole

Equilibrium 0.344 mole            1.312 mole

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

χ_NO2 = n_NO2/n_total = 0.344 mole/1.656 mole = 0.208

χ_N2O4 = n_N2O4/n_total = 1.312 mole/1.656 mole = 0.792

Since the pressure is constant P_total=1atm

χ_NO2 = P_NO2/P_total ⇒ P_NO2 = χ_NO2 × P_total = 0.208 × 1atm=0.208 atm (Ans.)

χ_N2O4 = P_N2O4/P_total ⇒ P_N2O4 = χ_N2O4 × P_total = 0.792 × 1atm=0.792 atm (Ans.)