PCl 5 leftharpoons PCl 3+ Cl 3 Kc=1.844 3.0 moles of PCl 5 is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl 5 at equilibrium is × 10-3 (Round off to the Nearest Integer)

Question

PCl 5 leftharpoons PCl 3+ Cl 3 Kc=1.844 3.0 moles of PCl 5 is introduced in a 1 L closed reaction vessel at 380 K. The number of moles of PCl 5 at equilibrium is × 10-3 (Round off to the Nearest Integer) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

PCl 5(g) leftharpoons PCl 3(g)+ Cl 2(g) Kc=1.844 ⇒ ([ PCl 3][ Cl 2]/[ PCl 5])=(x2/3-x)=1.844 ⇒ x2+1.844-5.532=0 ⇒ x=(-1.844+√(1.844)2+4 × 5.532/2) ≅ 1.604 ⇒ Moles of PCl 5=3-1.604 ≅ 1.396

Q. PCl5​⇌PCl3​+Cl3​Kc​=1.844 3.0 moles of PCl5​ is introduced in a 1L closed reaction vessel at 380K. The number of moles of PCl5​ at equilibrium is_______ ×10−3 (Round off to the Nearest Integer)

PCl5(g)​⇌PCl3(g)​+Cl2(g)​Kc​=1.844 ⇒[PCl5​][PCl3​][Cl2​]​=3−xx2​=1.844 ⇒x2+1.844−5.532=0 ⇒x=2−1.844+(1.844)2+4×5.532​​ ≅1.604 ⇒ Moles of PCl5​=3−1.604≅1.396

Q. $PC l_{5} ⇌ PC l_{3} + C l_{3} K_{c} = 1.844$ $3.0$ moles of $PC l_{5}$ is introduced in a $1 L$ closed reaction vessel at $380 K$ . The number of moles of $PC l_{5}$ at equilibrium is_______ $\times 1 0^{- 3}$
(Round off to the Nearest Integer)