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Ways to Define Concentration

The volume ra...

Question

The volume ratio fo $S O_{2}, O_{2}$ and $S O_{3}$ in the reaction for the formation of $S O_{3}$ is :

1 : 2 : 2

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2 : 2 : 1

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2 : 1 : 2

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1 : 1 : 2

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Solution

The correct option is C 2 : 1 : 2
The balanced chemical equation can be written as:
$2 S O_{2} + O_{2} \to 2 S O_{3}$
Thus, the volume ratio is $2 : 1 : 2$ .

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Similar questions

O_{2} \to S O_{3}

S O_{2} + 1 / 2 O_{2} \to S O_{3}

S O_{2}

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

{S O}_{2} + \frac{1}{2} O_{2} \to {S O}_{3} + y

{S O}_{2}

K_{c}

S O_{3}

S O_{2}

O_{2}

g m o l^{- 1}

In the reaction

$S + \frac{3}{2} O_{2} \to S O_{3} + 2 x k J$ and $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} + y k J$

Heat of formation of $S O_{2}$ is

S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J

{S O}_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J

H_{2} {S O}_{4}

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