For the reaction, PCl 5(g) leftharpoons PCl 3(g)+ Cl 2(g) in a 3 litre vessel at 250° C , Kc is 0.04 Calculate the initial number of moles of PCl 5 if equilibrium concentration of Cl 2 is 0.15 M

Question

For the reaction, PCl 5(g) leftharpoons PCl 3(g)+ Cl 2(g) in a 3 litre vessel at 250° C , Kc is 0.04 Calculate the initial number of moles of PCl 5 if equilibrium concentration of Cl 2 is 0.15 M (A) 2.1 (B) 0.56 (C) 0.71 (D) 0.24

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a431bf139f1fcac0fefcbb800bf4055a-.png" /> Kc=([ PCl 3][ Cl 2]/[ PCl 5]) ⇒ 0.04=(0.15 × 0.15/(C-x)) ⇒ C-x=0.5625 M Initial concentration of PCl 5(C)=0.5625+0.15=0.7125 M Initial number of moles of PCl 5= Concentration × Volume (L) =0.7125 × 3=2.1375

Q. For the reaction, PCl5(g)​⇌PCl3(g)​+Cl2(g)​ in a 3 litre vessel at 250∘C,Kc​ is 0.04 Calculate the initial number of moles of PCl5​ if equilibrium concentration of Cl2​ is 0.15M

2.1

0.56

0.71

0.24

Kc​=[PCl5​][PCl3​][Cl2​]​ ⇒0.04=(C−x)0.15×0.15​ ⇒C−x=0.5625M Initial concentration of PCl5​(C)=0.5625+0.15=0.7125M Initial number of moles of PCl5​= Concentration × Volume (L) =0.7125×3=2.1375

Q. For the reaction, $PC l_{5 (g)} ⇌ PC l_{3 (g)} + C l_{2 (g)}$ in a 3 litre vessel at $25 0^{\circ} C, K_{c}$ is $0.04$ Calculate the initial number of moles of $PC l_{5}$ if equilibrium concentration of $C l_{2}$ is $0.15 M$