Consider the following two equilibria simultaneously established in a rigid vessel at a particular temperature NH_(2)COOMH_(4)(s)hArr2NH_(3)+CO_(2)(g) CaCO_(3)(s)hArrCaO(s)+CO_(2)(g) Now, on adding some amout of NH_(3) to the reaction vessel, the orignal equilibrium is disturbed and a new equilibrium state is obtained. On comparing the following at the initial & final equilibrium states, select the INCORRECT statement (s):

Consider the following two equilibria simultaneously established in a

1.	Consider the following two equilibria simultaneously established in a rigid vessel at a particular temperature NH_(2)COOMH_(4)(s)hArr2NH_(3)+CO_(2)(g) CaCO_(3)(s)hArrCaO(s)+CO_(2)(g) Now, on adding some amout of NH_(3) to the reaction vessel, the orignal equilibrium is disturbed and a new equilibrium state is obtained. On comparing the following at the initial & final equilibrium states, select the INCORRECT statement (s):
Answer» Nothing can be SAID about the number of moles of `CO_(2)` gas in reaction vessel. Nothing can be said about the number of moles of `NH_(3)` gas in reaction vessel. Number of moles of `NH_(3)` gas would have definitely increased. Number of moles of `CaCO_(3)` solid gas would have definitely DECREASED. Solution :Upon addition of `NH_(2),1^(st)` equilibrium will shift backward leading to decrease in amount of `CO_(2)`. As a result,`2^(nd)` equilibrium will shift forward compensating for the amount of decreased `CO_(2)`. So, `n_(caco_(3)(s)` will definitely decrease Now, `K_(P_(1))=P_(NH_(3))^(2)xxP_(CO_(2)) "&" K_(P_(2))=P_(CO_(2))` Since no change in temperature would have taken place, `K_(P_(1)=K_(P_(2)` will have same values at both equilibrium states. So, amount of `NH_(2) "&" CO_(2)` at the two equilibrium states will remain the same.