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Law of Mass Action

In the follow...

Question

In the following equilibrium taking place at $298 K$ in $5 L$ flask, moles of $S O_{2}$ and $S O_{3}$ are equal at equilibrium.
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{C} = 5$
Thus number of moles of $O_{2} (g)$ at equilibrium are_________.

A

1

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B

2

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C

5

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D

2.5

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Solution

The correct option is B $2$
Given
Kc=5
Volume of flask=5L
Temperature=298K
$M o l e s o f S O_{2} = M o l e s o f S O_{3}$ =x
Solution
$2 S O_{2} + O_{2} = 2 S O_{3}$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} * [O_{2}]}$
5=10/y
y=2
Moles of $O_{2}$ is 2
The correct option is B

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