$PCl_{5} \rightleftharpoons PCl_{3} + Cl_{2} \quad K_{c} = 1.844$
$3.0 \ \text{moles}$ of $PCl_{5}$ is introduced in a $1 \ \text{L}$ closed reaction vessel at $380 \ \text{K}$. The number of moles of $PCl_{5}$ at equilibrium is $..... \times 10^{-3}$. (Round off to the Nearest Integer)

  • A
    $1500$
  • B
    $1292$
  • C
    $1400$
  • D
    $5123$

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Similar Questions

In the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,initially $1 \text{ mole}$ each of $PCl_5$ and $PCl_3$ are present. At equilibrium,$x \text{ moles}$ of $PCl_5$ remain. What is the total number of moles at equilibrium (in $- x$)?

For the ideal gas reaction,$X + Y \rightleftharpoons Z$,a mixture with $n_{X} = 1 \, mol$,$n_{Y} = 3 \, mol$ and $n_{Z} = 2 \, mol$ is at equilibrium at $300 \, K$ and $1 \, bar$. If the pressure is isothermally increased to $2 \, bar$,the number of moles of $X$ in the new equilibrium is closest to $......$

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