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Byju's Answer

Standard XII

Chemistry

Characteristics of Chemical Equilibrium

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 $0.12$ mole each of ${S O}_{2}$ and ${S O}_{3}$ and $5$ mole of $O_{2}$ . In another vessel of one litre contains $32$ g of ${S O}_{2}$ at the same temperature. What mass of $O_{2}$ must be added to this vessel in order that at equilibrium $20$ % of ${S O}_{2}$ is oxidised to ${S O}_{3}$ :

0.4125

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13.2

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1.6

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11.6

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Solution

The correct option is D $11.6$ g
$K_{c} = \frac{{[S O_{3}]}^{2}}{{[S O_{2}]}^{2} [O_{2}]} = \frac{{0.12}^{2}}{{0.12}^{2} \times 5} = 0.2$

At constant temperature, $K_{c}$ will remain constant.
Note : 32g of $S O_{2}$ = 0.5 mol
Assume $x m o l e s$ of $O_{2}$ is added

$2 S O_{2} 0.5 + O_{2} x \to 2 S O_{3} 0$

$2 S O_{2} 0.5 - 2 y + O_{2} x - y \to 2 S O_{3} 2 y$

According to question,
$2 y = 0.2 \times 0.5$
$y = 0.05 m o l e s$

Now,
$K_{c} = \frac{{[S O_{3}]}^{2}}{{[S O_{2}]}^{2} [O_{2}]} = \frac{{(2 y)}^{2}}{{(0.5 - 2 y)}^{2} \times (x - y)} = 0.2$

Solving the equation by putting $y = 0.05$ , we get

$x = 0.3625 m o l e s = 0.3625 \times 32 g = 11.6 g$

Hence, option D is correct

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Similar questions

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