Following equilibrium is established to decomposing of Ammonium carbonate NH_4COONH_2 in closed vessel at 700 K temperature. NH_4COONH_(2(s)) hArr 2NH_(3(g)) + CO_2 At initial if there is vaccum and at equilibrium total pressure is P bar than derive the value of K_p with respect to P.

Following equilibrium is established to decomposing of Ammonium carbon

1.	Following equilibrium is established to decomposing of Ammonium carbonate NH_4COONH_2 in closed vessel at 700 K temperature. NH_4COONH_(2(s)) hArr 2NH_(3(g)) + CO_2 At initial if there is vaccum and at equilibrium total pressure is P bar than derive the value of K_p with respect to P.
Answer» Solution :At EQUILIBRIUM the RATIO of `NH_(3(g))` and `CO_(2(g))` is 2:1. If total pressure = P, So, `p_(NH_3)=2/3P` and `p_(CO_2)=1/3 P` `{:("Reaction:",NH_4COONH_(2(s)) hArr 2NH_(3(g)) , + CO_(2(g))),("Pressure at equilibrium :", p_(NH_3)=2/3P , p_(CO_2)=1/3P):}` `therefore K_p=(p_(NH_3))^2(p_(CO_2))=(2/3P)^2 (1/3P)` `therefore K_p=4/27P^3` `P^3=27/4K_p` So, `P=3(K_p/4)^(1/3)`