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Hess' Law

S+O2→SO2; Δ H...

Question

$S + O_{2} \to {S O}_{2}; Δ H = - 298.2 k J$
${S O}_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J$
${S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; Δ H = - 130.2 k J$
$H_{2} + \frac{1}{2} O_{2} \to H_{2} O; Δ H = - 227.3 k J$
The heat of formation of $H_{2} {S O}_{4}$ will be:

A

- 754.4 k J

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B

+ 320.5 k J

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C

- 650.3 k J

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D

- 433.7 k J

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Solution

The correct option is A $- 754.4 k J$
Since, on adding all the 4 reaction given in question we will get a final reaction,
$S + 2 O_{2} + H_{2} \to H_{2} S O_{4}$
Which is basically the formation reaction of $H_{2} S O_{4}$
Hence, the heat of formation of $H_{2} S O_{4}$ = sum of heat of reactions of all four reaction, i.e,
$Δ H_{f (H_{2} S O_{4})} = Δ H_{1} + Δ H_{2} + Δ H_{3} + Δ H_{4} = - 298.2 - 98.7 - 130.2 - 227.3 = - 754.4 K J .$
hence, answer is option $A$ .

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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} ⟶ 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

S + O_{2} \to S O_{2}; Δ H = - 398.2 k J; S O_{2} + \frac{1}{2} O_{2} \to S O_{3}; Δ H = - 98.7 k J

S O_{2} + 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 sulphuric acid at

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}

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