question_answer

\[N{{H}_{2}}COON{{H}_{4}}(s)\,\,\,2N{{H}_{3}}(g)+C{{O}_{2}}(g)\] equilibrium constant of given reaction is\[2.9\,\,\times \,\,1{{0}^{-}}^{5}\,at{{m}^{3}}\]. The total pressure of gases in equilibrium with \[{{\operatorname{NH}}_{3}}COON{{H}_{4}}(s)\,\,at\,\,30{}^\circ C\] is

A)  0.0194 atm                  

B)  0.194 atm

C)  5.82 atm                     

D)  0.0582 atm \[\underset{1-x}{\mathop{{{\operatorname{NH}}_{2}}COON{{H}_{4}}(s)}}\,\,\,\,\,2N{{H}_{\underset{2x}{\mathop{3}}\,}}\,(g)\,+\,\,C{{O}_{\underset{x}{\mathop{2}}\,}} (g)\] \[{{K}_{p}}={{({{\rho }_{N{{H}_{3}}}})}^{2}}\times {{\rho }_{C{{O}_{2}}}}\] \[2.9\,\,\times \,\,1{{0}^{-}}^{5}={{\left( 2x \right)}^{2}}\times x\] \[2.9\,\,\times \,\,1{{0}^{-}}^{5}=4{{x}^{3}}\] \[\operatorname{x}= 0.0194 atm\] \[\operatorname{Total} pressure = 2x +x=3x = 3\,\,\times \,\, 0.0194\] \[= 0.0582 atm\]