138 gm of `N_2O_4(g)` is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm `"mol"^

Question

138 gm of `N_2O_4(g)` is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm `"mol"^

138 gm of `N_2O_4(g)` is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm `"mol"^(-1) K^(-1)`)
A. `alpha`=degree of dissociation of `N_2O_4=0.25`
B. `K_p` of `N_2O_4 hArr 2NO_2` (g) will be 9 atm
C. Total pressure at equilibrium =6.75 atm
D. The density of equilibrium mixture will be 16.83 gm/litre.

Monjur Alahi

6 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - B,C,D
`{:(,N_2O_4" "hArr,2NO_2),(t=0,a,0),(t,a(1-alpha),2aalpha):}`
vapour density`=46/(1+alpha)=30.67`
so `1+alpha=1.5=0.5=50%`
Total pressure`=(1.5xx1.5xx0.082xx300)/8.2=6.75` atm So, `K_p=(4alpha^2)/(1-alpha^2)P=9` atm
and for density of mixture `=138/8.2` gm/L=16.83 gm/L

6 views Answered 2023-02-05 15:29