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Standard XII

Chemistry

Effect on Kc

A reaction sy...

Question

A reaction system in equilibrium according to reaction $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ in one litre vessel at a given temperature was found to be $0.12 mol$ each $S O_{2}$ and $S O_{3}$ and $5$ mole of $O_{2}$ .In another vessel of one litre contains $32 g$ of $S O_{2}$ at tempreture.What mass in grams of $O_{2}$ must be added to this vessel to order that at equilibrium $20 %$ of $S O_{2}$ in oxidized to $S O_{2}$ ?

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Solution

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} = \frac{(0.12)^{2}}{(0.12)^{2} (5)} = 0.2$
$Moles of S O_{2} = \frac{32}{64} = 0.5 mol$
Assume $y moles$ of $O_{2}$ is added,
$\begin{matrix} 2 S O_{2} (g) & + & O_{2} (g) & ⇌ 2 S O_{3} (g) 0.5 & y & 0 2 S O_{2} (g) & + & O_{2} (g) ⇌ & 2 S O_{3} (g) Moles at eqm & 0.5 - 2 x & y - x & 2 x \end{matrix}$
According to the question,
$2 x = 0.2 \times 0.5$
$\Rightarrow x = 0.05 moles$
$K_{c} = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]} = \frac{(0.12)^{2}}{(0.12)^{2} (5)} = \frac{(2 x)^{2}}{(0.5 - 2 x)^{2} \times (y - x)} = 0.2$
Solving the equation, by putting $x = 0.05$ , we get,
$y = 0.3625 mol, 0.3625 \times 32 g = 11.6 g$

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