Saved Bookmarks
| 1. |
The dissociation of ammonium hydrogen sulphide in a closed container produces a pressure of `10 atm` at `200^@C`. The value of `K_p` isA. `25`B. `50`C. `100`D. `75` |
|
Answer» Correct Answer - A `{:(,NH_(4)^(+)HS^(-)(s)hArrNH_(3)(g)+H_(2)S(g)),("Initial moles (assume)"," a 0 0"),("Change"," -x +x +x"),("Equilibrium moles",bar(" a-x x x")):}` Total moles of gases `=x+x=2x` (`NH_4HS` is a solid. Thus, its moles are not counted. ) `P_(NH_3)=`(Mole fraction `_NH_3`) (Total pressure) `=(("Moles"_(NH_3))/("Total moles"))("Total pressure")` `=((x)/(2x))*P` Similarly, `P_(H_2S)=((x)/(2x))*P` `K_P=P_(NH_3)P_(H_2S)=((x)/(2x)*P)((x)/(2x)*P)` `=(P/2)^2` Since `P=10atm`, we have `K_P=((10)/(2))^2=(5)^2=25` Note that `NH_4HS` being a solid does not appear in `K_p` expression. |
|