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  3. Chemistry
  4. 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)

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Q. $PCl _{5} \rightleftharpoons PCl _{3}+ Cl _{3}\,\, K_{c}=1.844$ $3.0$ 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_______ $\times 10^{-3}$
(Round off to the Nearest Integer)

JEE MainJEE Main 2021Equilibrium

Solution:

$PCl _{5(g)} \rightleftharpoons PCl _{3(g)}+ Cl _{2(g)} \quad K_{c}=1.844$
$\Rightarrow \frac{\left[ PCl _{3}\right]\left[ Cl _{2}\right]}{\left[ PCl _{5}\right]}=\frac{x^{2}}{3-x}=1.844$
$\Rightarrow x^{2}+1.844-5.532=0$
$\Rightarrow x=\frac{-1.844+\sqrt{(1.844)^{2}+4 \times 5.532}}{2}$
$\cong 1.604$
$\Rightarrow$ Moles of $PCl _{5}=3-1.604 \cong 1.396$