A) \[0.242\times {{10}^{-4}}\text{ }at{{m}^{2}}\]
B) \[1\times {{10}^{-4}}\text{ }at{{m}^{2}}\]
C) \[4.9\times {{10}^{-3}}\text{ }at{{m}^{2}}\]
D) \[0.242\text{ }at{{m}^{2}}\]
Correct Answer: D
Solution :
\[N{{H}_{4}}S{{H}_{\left( s \right)}}\rightleftharpoons N{{H}_{3}}_{\left( g \right)}+{{H}_{2}}{{C}_{s}}_{(g)}\,\,{{H}_{2}}{{S}_{\left( g \right)}}\] \[t=0\text{ }\,\,\frac{5.1}{5.1}\,\,\,=\,\,0.1\,mole\] \[t=t~~\,\,\,0.1-0.1\times 0.3~\,\,\,\,~\,\,0.03~~~0.03\] \[{{\eta }_{N{{H}_{3}}}}+{{\eta }_{{{H}_{2}}S}}\,=\,\,0.06\] \[PV=nRT\] \[{{P}_{T}}\times 3=0.06\times 0.082\times 600\] \[{{P}_{T}}=0.984\] \[\therefore \text{ }\,\,{{P}_{N{{H}_{3}}}}={{P}_{{{H}_{2}}S}}=0.492\] \[\therefore \,\,\,Kp={{(0.492)}^{2}}=0.242\,\,at{{m}^{\text{2}}}\]You need to login to perform this action.
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