Question

Dinitrogen tetroxide decomposes to nitrogen dioxide: N2O4(g)→2NO2(g) ΔrH∘=55.3kJmol−1 At 298 K, a reaction vessel initially contains 0.100 bar of N2O4. When equilibrium is reached, 58% of the N2O4 has decomposed to NO2.

What percentage of N2O4 decomposes at 389 K? Assume that the initial pressure of N2O4 is the same (0.100 bar).

Please use bar and not atm.

Answer 1

We are given that are given that initial po essure of Noy = 0.100 bar At equillibrium N, Ou decomposes 58%.. Firetty we will calculate the prevure of No Oy at equilibonim. PNoy = equilibrium posure of Ny C XN - mole fraction of Non Ny – initial pressure of Na .: Prusa = 11-0.58 ) (0.10 bar) = 0.042 bar 2 No (9) Ng Oylg) 0.10 - u 0.042 u mo 202 At equilli borum , 0.848 is left with No Oy, So, Pos = a (alo -0.042) = 0.116 bar Equillibrium prenw constant * k = (PNoz ² Lolculated a (0.116)2 0.042 - 0.32 ryou (k, = 0.30

Given that, Mn = 55.34 Ime at T = 298k This value is for prenure constant K . Now we cwill colulate pressure constant kg, at specified temperature T = 3894 Using Max 2.303 R - 7 log (ky) - 5513 X 1000 71001 8.31486 m x 2.303 .32 298 38 55.34 1000 & mol x 91 8.314 x 2.303 mx 1159224 do ) = 9.267 0:32 (2.267) kg = (0.32) x 10" [ty = 59017] Equrlloboium constant expeewion for ke, Ng Oy /9) - 2NO, (8) I Oslo с- +2x 22 0.10-u

)! k DR ('No = (a (0.1 -x) NOY (93) = 59.17 59.17 (0.1 -0 40 + 59.17 - 5.917 = 0 Solving for u, where a=4, b= 59.17., c= -5.917 u= – 6 + 62.40 - - 59.17 + 3501.08 + 94,672 a = -59.17 € 59.96 = 0.09875 we will not take eve value of ne. D = 2.1 = 2x 0.09875 = 0.1975 bar Puson = 0.10 x = 0.00125 bar & pos = 0.10 har 9 k = of age of Pausen No Oy remains at 389 k = No Cu x 100 PO4 % of Ny Cavemains = 0.00195 x 100 0.10 = 1.25% Nou = 1.951.) remains; then (100-1.25)% decomposes. Decomposition of No Oy = (100-1.25)ol. = 98.75% of so was comumed.