PCl 5 ⇌ PCl 3 + Cl 2 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 — × 10 - 3 (Round off to the Nearest Int

Question

PCl 5 ⇌ PCl 3 + Cl 2 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 — × 10 - 3 (Round off to the Nearest Integer)

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The dissociation of PCl₅ is given by the equilibrium:

PCl₅(g) $⇌$ PCl₃(g) + Cl₂(g)

Initial moles:

[PCl₅] = 3.0 mol / 1 L = 3.0 M

[PCl₃] = 0 M

[Cl₂] = 0 M

Let $x$ be the change in concentration of PCl₅ at equilibrium.

Equilibrium concentrations:

[PCl₅] = $3.0 - x$ M

[PCl₃] = $x$ M

[Cl₂] = $x$ M

The equilibrium constant $K_{c}$ is given by:

$K_{c} = \frac{[ PC l _{3} ] [ C l _{2} ]}{[ PC l _{5} ]}$

$1.844 = \frac{( x ) ( x )}{( 3.0 - x )}$

$1.844 (3.0 - x) = x^{2}$

$5.532 - 1.844 x = x^{2}$

$x^{2} + 1.844 x - 5.532 = 0$

Using the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ :

$x = \frac{- 1.844 \pm ( 1.844 ) ^{2} - 4 ( 1 ) ( - 5.532 )}{2 ( 1 )}$

$x = \frac{- 1.844 \pm 3.399 + 22.128}{2}$

$x = \frac{- 1.844 \pm 25.527}{2}$

$x = \frac{- 1.844 \pm 5.052}{2}$

Since concentration cannot be negative, we take the positive root:

$x = \frac{- 1.844 + 5.052}{2} = \frac{3.208}{2} = 1.604$

The number of moles of PCl₅ at equilibrium is $3.0 - x$ :

Moles of PCl₅ = $3.0 - 1.604 = 1.396$ mol

Rounding off to the nearest integer and expressing in the format $_\times 1 0^{- 3}$ :

$1.396 m o l = 1396 \times 1 0^{- 3} m o l$

Chemistry - Chemical Equilibrium Question with Solution | TestHub

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