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Standard XII

Chemistry

Law of Mass Action

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 oxidized to $S O_{3}$ ?

0.4125

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11.6

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1.6

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Solution

The correct option is B $11.6$ g
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
$K_{c} = \frac{{(0.12)}^{2}}{{(0.12)}^{2} \times 5} = 0.2$
Another vessel
$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$
Moles of equilibrium $0.5 - 2 x$ $y - x$ $2 x$
As per given,
$2 x = \frac{20}{100} \times 0.5 = 0.1$
$K_{c} = \frac{{(0.1)}^{2}}{{(0.4)}^{2} (y - 0.1)} = 0.20 y = 0.4125 M o l e$
Therefore $,$ Mass of $O_{2}$ added $= 13.2 g$
Hence, option $(B)$ is correct.

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Similar questions

Q. 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

S O_{2}

at tempreture.What mass in grams of

O_{2}

must be added to this vessel to order that at equilibrium

20 %

S O_{2}

in oxidized to

S O_{2}

Q. A reaction system in equilibrium according to the equation

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

in 1 litre reaction vessel at a given temperature was found to contain 0.11 mol of

S O_{2}

, 0.12 of

S O_{3}

and 0.05 mol of

O_{2.}

Another 1 litre reaction vessel contains 64 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 half of

S O_{2}

is oxidised to

S O_{3}

Q. At 800 K a reaction mixture contained 0.5 mole of

S O_{2}

. 0.12 mole of

O_{2}

and 5 mole of

S O_{3}

at equilibrium.

K_{c}

for the equilibrium

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

is 833 lit/ mole. If the voiume of the container is 1 litre, calculate how much

O_{2}

is to be added at this equilibrium in order to get 5.2 moles of

S O_{3}

at the same temperature.

Q. One mole of

S O_{3}

was placed in a litre reaction flask at a given temperature when the reaction equilibrium was established in the reaction.

2 S O_{3} ⇌ 2 S O_{2} + O_{2}

the vessel was found to contain

0.6

mol of

S O_{2}

. The value of the equilibrium constant is :

Q. One mole of

S O_{3}

was placed in a litre reaction vessel at a certain temperature , the following equilibrium was established,

2 S O_{3} ⇌ 2 S O_{2} + O_{2}

. At equilibrium 0.6 moles of

S O_{2}

were formed. The equilibrium constant of the reaction will be:

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The correct option is B 11.6 g2SO2(g)+O2(g)⇌2SO3(g)Kc=(0.12)2(0.12)2×5=0.2Another vessel 2SO2(g)+O2(g)⇌2SO3(g)Moles of equilibrium 0.5−2x y−x 2xAs per given, 2x=20100×0.5=0.1Kc=(0.1)2(0.4)2(y−0.1)=0.20y=0.4125MoleTherefore, Mass of O2 added =13.2gHence, option (B) is correct.