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Sulphuric Acid

Identify the ...

Question

Identify the equations involved in bleaching action of $S O_{2}$ :

A

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

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B

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

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C

S O_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 [H]

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D

S O_{3} + H_{2} O \to H_{2} S O_{4}

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Solution

The correct option is B $S O_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 [H]$
The bleaching action of sulphur dioxide is because of the formation of nascent hydrogen in the presence of moisture. This nascent hydrogen reduces coloured substances, thus bleaching them.

$S O_{2} + 2 H_{2} O \to H_{2} S O_{4} + 2 [H]$

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Similar questions

S + O_{2} \to S O_{2}, △ H = - 298.2 k J m o l e^{- 1}

S O_{2} + 1 / 2 O_{2} \to S O_{3}, △ H = - 98.7 k J m o l e^{- 1}

S O_{3} + H_{2} O \to H_{2} S O_{4}, △ H = - 130.2 k J m o l e^{- 1}

H_{2} + 1 / 2 O_{2} \to H_{2} O, △ H = - 287.3 k J m o l e^{- 1}

the enthalpy of formation of

H_{2} S O_{4}

298 K

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

the enthalpy of formation of

H_{2} S O_{4}

298

S + O_{2} \to S O_{2}

Δ H = - 298.2

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

Δ H = - 98.7

S O_{3} + H_{2} O \to H_{2} S O_{4}

Δ H = - 130.2

H_{2} + \frac{1}{2} O_{2} \to H_{2} O

Δ H = - 287.3

Then the enthalpy of formation of

H_{2} S O_{4}

S + O_{2} ⟶ S O_{2}; Δ H = - 298.2 k J

S O_{2} + 1 / 2 O_{2} ⟶ S O_{3}; Δ H = - 98.7 k J

S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J

H_{2} + 1 / 2 O_{2} ⟶ H_{2} O; Δ H = - 287.3 k J

Then the enthalpy of formation of

H_{2} S O_{4} a t 298 K

S + O_{2} ⟶ S O_{2}

Δ H = - 398.2 k J

S O_{2} + \frac{1}{2} O_{2} ⟶ S O_{3}

Δ H = - 98.7 k J

S O_{3} + H_{2} O ⟶ H_{2} S O_{4}

Δ H = - 130.2 k J

H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O

Δ H = - 227.3 k J

The enthalpy of formation of sulphuric acid at

298 K

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