Question

In a 1.5 liter flask in which the vacuum has been made, 0.08 moles of N2O4 are introduced and heated to 35 ° C. Part of N2O4 dissociates into NO2 according to the chemical balance:
N2O4 (g) ↔ 2NO2 (g)
When equilibrium is reached, the total pressure is 2.27 atm.

Calculate the partial pressure of NO2 at equilibrium

Answer 1

Date : Soutien ܠܦܓܕܨܠ 210, 49) Initial: Dina mol +22 changes Final -N 10:08-W) 2x Total sumber of miles at equilikodum 0:00- +22 n. 2 0.08 +1 V = 1.5L equilibmium pressure, pa 2.22.alom volume of of flase Temperature Te 35°C erom moles al equilibonium is Ideal alas gas equation, the total numbers of pu RT N = 3 0.13459 mol from from @ + (2.27 atm) (1.52) 10.0821 Latm ma'si) (35+273.15)K we get 2 = (0.13459 -0.08)mol 10:13459 -0.08)mol = 0.05459 Therefore, the partial pressure of No, at equilibrium is Paro Px 21 ช) = 2.27 atm (2x0-05459 mo 0.13459 mol 1.041434 atom