MCQ
$4.0\,moles$ of argon and $5.0\, moles$ of $PCI _{5}$ are introduced into an evacuated flask of $100\, litre$ capacity at $610\, K$. The system is allowed to equilibrate. At equilibrium, the total pressure of mixture was found to be $6.0 \,atm$. The $K _{ p }$ for the reaction is ...... [Given : $R=0.082 \,L \,atm\, K ^{-1}\, mol ^{-1}$ ]
  • $2.25$
  • B
    $6.24$
  • C
    $12.13$
  • D
    $15.24$

Answer

Correct option: A.
$2.25$
a
$PCl _{5}=5$ $mole$

$Ar =4$ $mole$

$P _{\text {Total }}=\frac{9 \times 0.82 \times 610}{100}=4.5 \,atm$

$P _{ PCI _{5}}=\frac{5 \times 4.5}{9}=2.5 ; P _{ Ar }=\frac{4 \times 4.5}{9}=2$

$PCl _{5} \rightleftharpoons PCl _{3}+ Cl _{2}$

$2.5- P \quad P \quad P$

$P_{\text {total }}=2.5-P+P+P+P_{A r}=6$

$P =1.5$

$K_{p}=\frac{1.5 \times 1.5}{1}=2.25$

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