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Standard XII

Chemistry

Characteristics of Equilibrium Constant

2SO2+O2⇌ 2SO3...

Question

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ . Starting with $2$ moles $S O_{2}$ and $1$ mole $O_{2}$ in $1$ L flask, mixture required $0.4$ mole $M n O_{4}^{-}$ in acidic medium. Hence, equilibirum constant $K$ is:

2

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Solution

The correct option is A $2$
$0.4 m$ of $K M n O_{4} = 1$ mole of $S O_{2} = 1$ mole of $S O_{3}$
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$
$2$ $1$ $0$
$1$ $0.5$ $1$
$K = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$= \frac{1^{2}}{1^{2} \times 0.5} = 2$

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Similar questions

2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}

starting with

2

mol of

{S O}_{2}

and

1

mol of

O_{2}

1

litre flask, mixture required

0.4

mol of

M n O_{4}^{-}

in acid medium. Hence

K_{c}

is:

2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}

. Starting with

2

mol

{S O}_{2}

and

1 m o l

O_{2}

1 L

flask, mixture required

0.4 m o l

M n O_{4}^{-}

in acidic medium. Hence

K_{c}

is :

Q. In a study of equilibrium

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

.
Starting with 2 mole

S O_{2}

and 1.5 mole

O_{2}

in 5 litre flask.Equilibrium mixture required 0.4 mole

K M n O_{4}

in acidic medium.Hence

K_{C}

is:

Q. A mixture of

S O_{3}

S O_{2}

and

O_{2}

gases is maintained in a

10.0 L

flask at a temperature at which equilibrium constant

K_{c}

fo the reaction

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

100

Then choose the correct statements :

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.

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2SO2+O2⇌2SO3. Starting with 2 moles SO2 and 1 mole O2 in 1L flask, mixture required 0.4 mole MnO4− in acidic medium. Hence, equilibirum constant K is:

The correct option is A 20.4m of KMnO4=1 mole of SO2=1 mole of SO32SO2+O2⇌2SO32 1 01 0.5 1K=[SO3]2[SO2]2[O2]=1212×0.5=2

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ . Starting with $2$ moles $S O_{2}$ and $1$ mole $O_{2}$ in $1$ L flask, mixture required $0.4$ mole $M n O_{4}^{-}$ in acidic medium. Hence, equilibirum constant $K$ is:

The correct option is A $2$
$0.4 m$ of $K M n O_{4} = 1$ mole of $S O_{2} = 1$ mole of $S O_{3}$
$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$
$2$ $1$ $0$
$1$ $0.5$ $1$
$K = \frac{[S O_{3}]^{2}}{[S O_{2}]^{2} [O_{2}]}$
$= \frac{1^{2}}{1^{2} \times 0.5} = 2$