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Effective Equilibrium Constant

For the given...

Question

For the given reaction
$P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$
If 2.0 mol of $P C l_{5} (g)$ is placed in a 2 L vessel at 300 K and 0.8 mole of $P C l_{5} (g)$ dissociated, what will be the equilibrium constant $(K_{c})$ of the reaction?

A

0.266

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B

0.366

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C

0.466

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D

3.75

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Solution

The correct option is D 3.75
Given reaction is
$P C l_{5} (g) ⇌ P C l_{3} (g) + C l_{2} (g)$

$\begin{matrix} P C l_{5} & P C l_{3} & C l_{2} Initial & 2 & 0 & 0 Change & - 0.8 & 0.8 & 0.8 Equilibrium & 1.2 & 0.8 & 0.8 Molar concentration & 0.6 & 0.4 & 0.4 \end{matrix}$

The expression for the equilibrium constant is $K_{c} = \frac{[P C l_{3}] [C l_{2}]}{[P C l_{5}]}$
$K_{c} = \frac{0.4 \times 0.4}{0.6} = 0.266$

$P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)$
$K_{c}$ for this reaction = 3.75

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Similar questions

Q. For the given reaction

P C l_{3} (g) + C l_{2} (g) ⇌ P C l_{5} (g)

P C l_{5} (g)

is placed in a 2 L vessel at 300 K and 0.8 mole of

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dissociated, what will be the equilibrium constant

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Effective Equilibrium Constant

Standard XII Chemistry

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