Chemistry - Chemical Equilibrium Question with Solution | TestHub

In case of heterogeneous equilibrium. The achive mass of solid is unit so not consider in the expression of${\text{K}}_{\text{p}}\text{and}{\text{K}}_{\text{c}}$

${\text{NH}}_2{\text{COONH}}_4\left(\text{s}\right)\rightleftharpoons 2{\text{NH}}_3\left(\text{g}\right)+{\text{CO}}_2\left(\text{g}\right)$

At equilibrium$\Rightarrow$2 p and   p  will be partial pressures of ammonia and carbon dioxide gas respectively.

${\text{K}}_{\text{P}}\text{=}{\left[{\text{P}}_{{\text{NH}}_{\text{3}}}\right]}^{\text{2}}\left[{\text{P}}_{{\text{CO}}_{\text{2}}}\right]\text{=}{\left[\text{2P}\right]}^{\text{2}}\left[\text{P}\right]\text{=}{\text{4P}}^{\text{3}}$

$4{\text{P}}^3=2·9\times 10^{-5}{\text{atm}}^3$

${\text{P}}^3=0·725\times 10^{-5}{\text{atm}}^3=7·25\times 10^{-6}{\text{atm}}^3$

$\text{P}=\sqrt[3]{7·25\times 10^{-6}}=1·94\times 10^{-2}\text{atm}$

$\Sigma \text{P}={\left({\text{P}}_{\text{Total}}\right)}_{\text{equilibrium}}=2\text{P}+\text{P}=3\text{P}$

=$3\times 1·94\times 10^{-2}\text{atm}$

=$5·82\times 10^{-2}\text{atm}$