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Law of Mass Action

K c for 2S...

Question

$K_{c}$ for $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ in a $10 l i t$ flask at certain $T$ is $100 l i t . m o l^{- 1}$ . Now , if equilibrium pressures of $S O_{2}$ and $S O_{3}$ are equal , then mass of $O_{2}$ present at equilibrium is:

A

6.4 g

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B

12.8 g

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C

3.2 g

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D

16 g

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Solution

The correct option is C $3.2 g$
It is given that the equilibrium pressures of $S O_{2}$ and $S O_{3}$ are equal. Therefore, the number of moles of $S O_{2}$ and $S O_{3}$ are also equal at equilibrium.
$a - 2 x = 2 x$
$a = 4 x$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$100 = \frac{1}{[O_{2}]}$ (Since concentration of $S O_{3}$ and $S O_{2}$ are equal)

$100 = \frac{10}{a - x} = \frac{10}{3 x}$
$3 x = 0.1$
Number of moles of $O_{2}$ at equilibrium are $a - x = 3 x = 0.1$
Mass of $O_{2} = 0.1 \times 32 = 3.2 g$

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Similar questions

Q. The equilibrium constant

k_{p}

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a t m^{- 1}

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the partial pressure of

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