Solution N2O4 → 2NO2initial moles; 1 0at equilibrium; 1-x 2xtotal moles at equilibrium=1-x +2x=1+xM wt of equilibrium mixture=2×34 5 (V D )=69M wt at initial=92so x can be calculated as followsinitial molesmoles at equilibrium=M wt of equilibrium mixtureInitial M wt 11+x=6992x=0 33total moles at equilibrium=1 33moles of NO2 at equilibrium=0 66% abundance of NO2 at equilibrium mixture=0 661 33×100=49 62% or approx 50%

Poll ddd ittt

Solution

N_{2} O_{4} \to 2 N O_{2} i n i t i a l m o l e s; 1 0 a t e q u i l i b r i u m; 1 - x 2 x t o t a l m o l e s a t e q u i l i b r i u m = 1 - x + 2 x = 1 + x M . w t . o f e q u i l i b r i u m m i x t u r e = 2 \times 34.5 (V . D .) = 69 M . w t . a t i n i t i a l = 92 s o x c a n b e c a l c u l a t e d a s f o l l o w s \frac{i n i t i a l m o l e s}{m o l e s a t e q u i l i b r i u m} = \frac{M . w t . o f e q u i l i b r i u m m i x t u r e}{I n i t i a l M . w t .} \frac{1}{1 + x} = \frac{69}{92} x = 0.33 t o t a l m o l e s a t e q u i l i b r i u m = 1.33 m o l e s o f N O_{2} a t e q u i l i b r i u m = 0.66 % a b u n d a n c e o f N O_{2} a t e q u i l i b r i u m m i x t u r e = \frac{0.66}{1.33} \times 100 = 49.62 % o r a p p r o x 50 %

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