A) \[S=\{{{K}_{sp}}/{{(6912)}^{1/7}}\}\]
B) \[S={{\{{{K}_{sp}}/144\}}^{1/7}}\]
C) \[S={{\{{{K}_{sp}}/6912\}}^{1/7}}\]
D) \[S={{\{{{K}_{sp}}/6912\}}^{7}}\]
Correct Answer: B
Solution :
\[N{{H}_{2}}COON{{H}_{4}}(s)2N{{H}_{3}}(g)+C{{O}_{2}}(g)\] \[{{K}_{P}}=\frac{{{\left( {{P}_{N{{H}_{3}}}} \right)}^{2}}\times \left( {{P}_{C{{O}_{2}}}} \right)}{{{P}_{N{{H}_{2}}COON{{H}_{4}}}}(s)}={{\left( {{P}_{N{{H}_{3}}}} \right)}^{2}}\times \left( {{P}_{C{{O}_{2}}}} \right)\] As evident by the reaction, \[N{{H}_{3}}\] and \[C{{O}_{2}}\] are formed in molar ratio of 2 : 1. Thus if P is the total pressure of the system at equilibrium, then \[{{P}_{N{{H}_{3}}}}=\frac{2\times P}{3}\] \[{{P}_{C{{O}_{2}}}}=\frac{1\times P}{3}\] \[{{K}_{P}}={{\left( \frac{2P}{3} \right)}^{2}}\times \frac{P}{3}=\frac{4{{P}^{3}}}{27}\] Given \[{{K}_{P}}=2.9\times {{10}^{-5}}\] \[\therefore \]\[2.9\times {{10}^{-5}}=\frac{4{{P}^{3}}}{27}\] \[{{P}^{3}}=\frac{2.9\times {{10}^{-5}}\times 27}{4}\] \[P={{\left( \frac{2.9\times {{10}^{-5}}\times 27}{4} \right)}^{{}^{1}/{}_{3}}}=5.82\times {{10}^{-2}}atm\]
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