1.

y At 340 K one atmospheric pressure N2O4 is 66% dissociated into NOp what volume of 10 g No.occupy under these conditions.Na

Answer»

N2O4(g) <==> 2NO2(g)10.0gT = 340KP = 1.00atm

Assuming we started with 10.0g of N2O4 and, 66% of it decomposed, that will leave 3.4g of N2O4, and make 6.6g of NO2. Convert those to moles, and then plug them into the ideal gas equation.

At equilibrium....3.4 g N2O4 .... 0.03696 mol N2O4 ...... carry extra sig figs for intermediate answersand6.6g NO2 ....... 0.1435 mol NO2

Total moles = 0.1805 moles of N2O4 and NO2, at a total pressure of 1.00 atm. Find the volume.

PV = nRTV = nRT / P = 0.1805 mol x 0.0821Latm/molK x 340K / 1.00 atm = 5.04L

You will get 5.02L if you use 0.18 moles of gas.



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