InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Density of equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at `1 atm` and `384 K` is `1.84 g dm^(-3)`. Calculate the equilibrium constant of the reaction. `N_(2)O_(4)hArr2NO_(2)` |
|
Answer» We know `Pm=dRT` `1xxm=1.84xx0.0821xx384` `m=29xx2` Vapour density (d) at equilibrium `=29` Initial vapour density`=M//2=92//2=46` therefore, degree of dissociation is: `x=(D-d)/((n-1)d)=(46-29)/(29)=0.586` For reaction `N_(2)O_(4)hArr2NO_(2)` `{:(t=0,1,,0,),(t_(eq),1-x,,2x,"Total moles"=1+x):}` `p_(N_(2)O_(4))=(1-x)/(1+x)xxP, p_(NO_(2))=(2x)/(1+x)xxP` `K_(p)=(4x^(2)P)/(1-x^(2))=(4xx(0.586)^(2)xx1)/(1-(0.586)^(2))=2.09 "atm"` |
|