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Reactivity Trends (Group 2)

The crystals ...

Question

The crystals of ferrous sulphate on heating gives:

A

F e O + S O_{2} + H_{2} O

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B

F e_{2} O_{3} + H_{2} S O_{4} + H_{2} O

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C

F e_{2} O_{3} + S O_{2} + S O_{3}

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D

F e O + S O_{3} + H_{2} S O_{4} + H_{2} O

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Solution

The correct option is C $F e_{2} O_{3} + S O_{2} + S O_{3}$
On heating, ferrous sulphate crystals lose water and anhydrous ferrous sulphate ( $F e S O_{4}$ ) is formed. So their colour changes from light green to white. On further heating, anhydrous ferrous sulphate decomposes to form ferric oxide ( $F e_{2} O_{3}$ ), sulphur dioxide ( $S O_{2}$ ) and sulphur trioxide ( $S O_{3}$ ).

$2 F e S O_{4} Δ \to F e_{2} O_{3} + S O_{2} + S O_{3}$
The reaction for the decomposition of ferrous sulphate is as above.

So, the correct option is $C$

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Similar questions

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} \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} ⟶ 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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