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Solid Ammonium carbamate dissociates as: `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g).` In a closed vessel, solid ammonium carbonate is in equilibrium with its dissociation products. At equilibrium, ammonia is added such that the partial pressure of `NH_(3)` at new equilibrium now equals the original total pressure. Calculate the ratio of total pressure at new equilibrium to that of original total pressure. Also find the partial pressure of ammonia gas added. |
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Answer» `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)` Let `P`= original equilibrium pressure. From the mole ratio of `NH_(3)` and `CO_(2)` at equilibrium, we have `p_(NH_(3))=2/3 P` and `p_(CO_(2))=P/3` `rArr K_(p)=(p_(NH_(3)))^(2).p_(CO_(2))=(2/3P)^(2) (P/3)=4/27 P^(3)` Now `NH_(3)` is added such that `p_(NH_(3))=P` Find the pressure of `CO_(2)` `rArr 4/27 P^(3)=P^(2) p_(CO_(2)) rArr p_(CO_(2))=4/27 P` Total new pressure `=P_("new")=p_(NH_(3))+p_(CO_(2))` `rArr P_("new")=P+4/27 P=31/27 P` `rArr` Ratio =`P_("new")/P_("original")=(31/27P)/P=31/27` Let `x` be the partial pressure of `NH_(3)` added at original equilibrium. `NH_(2)COONH_(4)(s)hArr2NH_(3)(g)+CO_(2)(g)` `{:("At equilibrium",,2/3 P,,1/3 P),("When" NH_(2) " is added",,2/3P+x,,1/3 P),("At new equilibrium",,2/3P+x-2y,,1/3P-y):}` `rArr 2/3 P+x-2y=p_(NH_(3))` and `1/3 P-y=P_(CO_(2))=4/27P` `rArr` Solve to get: `x=19/27 P` |
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