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5.) (14 points) Nitrogen tetroxide decomposes according to the reaction6N 2 O 4( g)2 NO 2 ( g) .At 55˚C and 1 atm, the average molecular weight of N 2 O 4 and NO 2 present at equilibrium is 61.2 g/mol.Determine the equilibrium € constant. [Note: atomic € mass N = 14.007 g/mol; atomic mass O = 15.999g/mol.]We can use the average molecular weight to determine the equilibrium mole fractions. The equation for theaverage molecular weight M isM = x NO2 M NO2 + x N2 O 4M N2 O 4,€where x is the gas phase mole fraction and M is the molecular weight. Since the sum of the mole fractions mustequal 1, we have€x NO2 + x N2 O 4= 1 ,or€x NO2 = 1 − x N2 O 4.Substituting into the expression for the average molecular weight yieldsThen,€M = ( 1− x N2 O 4)M NO2 + x N2 O 4M N2 O 461.2 g/mol = 1− x N2 O 4( ) 46.005g/mol61.2 = 46.005 x N2 O 4+ 46.00515.195 = 46.005 x N2 O 4x N2 O 4= 0.330.( ) + x N2 O 4( 92.010 g/mol)€x NO2 = 1 − x N2 O 4= 1 − 0.330x NO2 = 0.670.Knowing the mole fractions, we can now determine the equilibrium constant. Treating the gases as ideal gases,the equilibrium constant can be determined as€K eq ="$#"$#P NO2P !P N2 O 4P !2%'&.%'&€
5.) Continued7In terms of mole fractions, the equilibrium constant becomes€K eq ===2" x NO2 P %$# P ! '&" x N2 O 4P %$# P ! '&2x NO2 Px N2 O 4P !K eq = 1.38.( 0.670) 2 1.013bar0.330( )( )( 1bar)Here, we have used the equilibrium pressure, P =1 atm = 1.013 bar.
Page 2: 1 b.) Continued2The equilibrium con
Page 5: 4.) (15 points) True/false, short a
Page 9: 7.) (15 points) True/false, short a

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