Q. For a reaction : $2\left(SO\right)_{2}\left(\right.g\left.\right)+O_{2}\left(\right.g\left.\right)\rightleftharpoons2\left(SO\right)_{3}\left(\right.g\left.\right)$ , $1.5$ moles of $SO_{2}$ and $1$ mole of $O_{2}$ are taken in a $2L$ vessel. At equilibrium the concentration of $SO_{3}$ was found to be $0.35molL^{- 1}$ . The $K_{C}$ for the reaction would be :
Solution:
$\frac{X}{2}=0.35$
$x=0.70$
$\left[SO_{2}\right]\Rightarrow \frac{1 . 5 - 0 . 7}{2}=0.4$
$\left[O_{2}\right]\Rightarrow \frac{1 - 0 . 7 / 2}{2}=0.325$
$K_{C}=\frac{\left(\right. 0 . 35 \left(\left.\right)^{2}}{\left(\right. 0 . 4 \left(\left.\right)^{2} \times 0 . 325}$
$K_{C}=2.35$
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