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Standard VIII

Chemistry

Characteristics of Chemical Reaction

2SO2+O2⇌ 2SO3...

Question

$2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}$ starting with $2$ mol of ${S O}_{2}$ and $1$ mol of $O_{2}$ in $1$ litre flask, mixture required $0.4$ mol of $M n O_{4}^{-}$ in acid medium. Hence $K_{c}$ is:

80

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Solution

The correct option is A $80$
0.4 Mole of $M n O_{4} -$ =1 mole of SO2=1 Mole of SO3
$2 S O_{2} + O_{2} ⟺ 2 S O_{3} 210 1 0.5 1 K_{C} = \frac{[S O {3]}^{2}}{[S 02]^{2} [02]} = \frac{1}{1 \times 0.5} = 2$

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Similar questions

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 :

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

. Starting with

2

moles

S O_{2}

and

1

mole

O_{2}

1

L flask, mixture required

0.4

mole

M n O_{4}^{-}

in acidic medium. Hence, equilibirum constant

K

is:

Q. In a 2 litre flask, if the equilibrium

2 S O_{3} ⇌ 2 S O_{2} + O_{2}

is established, the equilibrium concentration of

S O_{2}

is 1.2 mol L

^{- 1}

and the initial concentration of

S O_{3}

is 2 mol L

^{- 1}

. Find

K_{c}

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:

{S O}_{2}

react with

O_{2}

as follows :

{2 S O}_{2} + O_{2} \to {2 S O}_{3}

The rate of disappearance of

{S O}_{2}

2.4 \times 10^{- 4}

mol

{l i t}^{- 1} {m i n}^{- 1}

, then :

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2SO2+O2⇌2SO3 starting with 2 mol of SO2 and 1 mol of O2 in 1 litre flask, mixture required 0.4 mol of MnO−4 in acid medium. Hence Kc is:

The correct option is A 800.4 Mole of MnO4−=1 mole of SO2=1 Mole of SO32SO2+O2⟺2SO321010.51KC=[SO3]2[S02]2[02]=11×0.5=2

$2 {S O}_{2} + O_{2} ⇌ 2 {S O}_{3}$ starting with $2$ mol of ${S O}_{2}$ and $1$ mol of $O_{2}$ in $1$ litre flask, mixture required $0.4$ mol of $M n O_{4}^{-}$ in acid medium. Hence $K_{c}$ is:

The correct option is A $80$
0.4 Mole of $M n O_{4} -$ =1 mole of SO2=1 Mole of SO3
$2 S O_{2} + O_{2} ⟺ 2 S O_{3} 210 1 0.5 1 K_{C} = \frac{[S O {3]}^{2}}{[S 02]^{2} [02]} = \frac{1}{1 \times 0.5} = 2$