13.8g of N2O4 was placed in a 1L reaction vessel at 400K

Question

13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium ?

mehul gupta · Accepted Answer

Given, Total volume (V) = 1L Mass of N2O4 = 13.8g Molar mass of N2O4 = 14(2) + 16(4) = 92g Number of moles N2O4(n) = 13.8g92g13.8g92g = 0.15 Gas constant (R) = 0.083 bar L mol−1 K−1 Temperature (T) = 400 K According to ideal gas equation, pV = nRT p × 1 L = 0.15 mol × 0.083 bar L mol−1 K−1x 400 K ∴∴ p = 4.98 bar For this reaction, N2O4 (g) ⇌ 2NO2 (g) Initial pressure 4.98 bar 0 At equilibrium (4.98 – x) bar 2x bar Hence, Ptotal at equilibrium = PN2O4PN2O4 + PNO2PNO2 9.15 = (4.98 - x) + 2x 9.15 = 4.98 + x ∴∴ x = 9.15 - 4.98 = 4.17 bar Partial pressures at equilibrium are, PN2O4PN2O4 = 4.98 - 4.17 = 0.81 bar PNO2PNO2 = 2x = 2 x 4.17 = 8.34 bar Kp = (PNO2)2PN2O4(PNO2)2PN2O4 = (8.34)20.81(8.34)20.81 = 85.87 Kp = Kc (0.083 x 4.00)1 ∴∴ Kc = 85.870.083×40085.870.083×400 = 2.586

Physical Chemistry> 13.8g of N2O4 was placed in a 1L reaction...

13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium ?

sudhanshu , 10 Years ago

mehul gupta

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Physical Chemistry> 13.8g of N2O4 was placed in a 1L reaction...

13.8g of N2O4 was placed in a 1L reaction vessel at 400K and allowed to attain equilibrium ?

sudhanshu , 10 Years ago

mehul gupta

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