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Standard XII

Chemistry

Arrhenius Acid

At constant t...

Question

At constant temperature in one liter vessel when the reaction $2 S O_{3} (g) ⇌ 2 S O_{2} (g)$ is at equilibrium, the $S O_{2}$ concentration is 0.6 M, initial concentration of $S O_{3}$ is 1M. The equilibrium constant is ?

2.7

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1.36

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0.34

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0.675

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Solution

The correct option is D 0.675
solution:
$2 S O_{3} ⇌ 2 S O_{2} + O_{2}$
(1-0.6) 0.6 0.3

$K_{c} = \frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}} = \frac{[0.6]^{2} [0.3]}{[0.4]^{2}} = 0.675$

hence the correct option:D

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Similar questions

Q. One mole of

S O_{3}

was placed in a vessel of

1

L capacity at a certain temperature when the following equilibrium was established.

2 S O_{3} (g) ⇌ 2 S O_{2} (g) + O_{2} (g)

At equilibrium,

0.6

moles of

S O_{2}

were formed. The equilibrium constant

K_{c}

for the reaction will be:

Q. For the reaction,

2 S O_{2 (g)} + O_{2 (g)} ⇌ 2 S O_{3 (g)}

What is

K_{c}

when the equilibrium concentration of

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

and

[S O_{3}] = 1.90 M

Q. In the following gaseous phase equilibrium at constant temperature.

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

The concentration of

[S O_{2}] = 3.0 \times 10^{- 3} M, [O_{2}] = 3.5 \times 10^{- 3} M, [S O_{3}] = 5.0 \times 10^{- 2} M

. Calculate equilibrium constant.

Q. A vessel of one litre capacity containing

1

mole of

{S O}_{3}

is heated till a state of equilibrium is attained.

2 {S O}_{3 (g)} ⇌ 2 {S O}_{2 (g)} + O_{2 (g)}

At equilibrium,

0.6

moles of

{S O}_{2}

had formed. The value of equilibrium constant is:

Q. For the equilibrium

2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)

the concentration of each substance at equilibrium is:

[S O_{2}] = 0.60 M, [O_{2}] = 0.82 M

and

[S O_{3}] = 1.90 M

, the K

_{c}

is:

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At constant temperature in one liter vessel when the reaction 2SO3(g)⇌2SO2(g) is at equilibrium, the SO2 concentration is 0.6 M, initial concentration of SO3 is 1M. The equilibrium constant is ?

The correct option is D 0.675solution:2SO3⇌2SO2+O2(1-0.6) 0.6 0.3Kc=[SO2]2[O2][SO3]2=[0.6]2[0.3][0.4]2=0.675hence the correct option:D

At constant temperature in one liter vessel when the reaction $2 S O_{3} (g) ⇌ 2 S O_{2} (g)$ is at equilibrium, the $S O_{2}$ concentration is 0.6 M, initial concentration of $S O_{3}$ is 1M. The equilibrium constant is ?

The correct option is D 0.675
solution:
$2 S O_{3} ⇌ 2 S O_{2} + O_{2}$
(1-0.6) 0.6 0.3

$K_{c} = \frac{[S O_{2}]^{2} [O_{2}]}{[S O_{3}]^{2}} = \frac{[0.6]^{2} [0.3]}{[0.4]^{2}} = 0.675$

hence the correct option:D