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When `N_(2)O_(5)(g)` is heated it dissociates to give `N_(2)O_(3)` and `O_(2) . K_(c)` for `N_(2)O_(5)rarr N_(2)O_(3)+O_(2)` is `7.75` and `K_(c)` for `N_(2)O_(3)rarr N_(2)O+O_(2)` is `4.0 mol L^(-1)`. (both `K_(c)` are at same temperature) `4` mol `N_(2)O_(5)` in `1.0L` vessel is kept at a certain temperature. the concentration of `O_(2)` was `4.5 mol L^(-1)`. Find the concentration of `N_(2)O_(5), N_(2)O_(3)`, and `N_(2)O` at equilibrium. |
Answer» `I[{:(N_(2)O_(5),hArr,N_(2)O_(3),+,O_(2),,K_(c )=7.75),(4-x,,x,,x,):}` `II[{:(N_(2)O_(3),hArr,N_(2)O,+,O_(2),,K_(c )=4),(x,,0,,x,),(x-y,,y,,x+y,):}` `:. x+y =[O_(2)]=4.5` From `I, K_(c )=([N_(2)O_(3)][O_(2)])/([N_(2)O_(5)])` `7.75=([N_(2)O_(3)][O_(2)])/([N_(2)O_(5)])=((x-y)(4.5))/(4-x)` From `II, K_(c )=([N_(2)O][O_(2)])/([N_(2)O_(3)]) 4=(yxx(4.5))/(x-y)` Solve to find `x` and `y [x=3.06, y=1.44]` |
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