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Heat of Formation

If S+O2⟶ SO...

Question

If $S + O_{2} ⟶ S O_{2}; Δ H = - 398.2 k J : S O_{2} + \frac{1}{2} O_{2} ⟶ S O_{3}; Δ H = - 98.7 k J$
$S O_{3} + H_{2} O ⟶ H_{2} S O_{4}; Δ H = - 130.2 k J : H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O_{3}; Δ H = - 227.3 k J$
The enthalpy of formation of sulphuric acid at $298 K$ will be:

A

- 854.4 k J

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B

- 754.4 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 $- 854.4 k J$
Solution:- (A) $- 854.4 k J$
Given:-
$S + O_{2} ⟶ S O_{2} Δ H = - 398.2 k J . . . . . (1)$
$S O_{2} + \frac{1}{2} O_{2} ⟶ S O_{3} Δ H = - 98.7 k J . . . . . (2)$
$S O_{3} + H_{2} O ⟶ H_{2} S O_{4} Δ H = - 130.2 k J . . . . . (3)$
$H_{2} + \frac{1}{2} O_{2} ⟶ H_{2} O Δ H = - 227.3 k J . . . . . (4)$
Adding ${e q}^{n} (1), (2), (3) & (4)$ , we have
$H_{2} + S + 2 O_{2} ⟶ H_{2} S O_{4} Δ H = - 854.4 k J$
Hence the enthalpy of formation of sulphuric acid is $- 854.4 k J$ .

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Similar questions

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 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} \to {S O}_{2}; △ H = - 398.2 k J : {S O}_{2} + \frac{1}{2} O_{2} \to {S O}_{3}; △ H = - 98.7 k J

{S O}_{3} + H_{2} O \to H_{2} {S O}_{4}; △ H = - 130.2 k J : H_{2} + \frac{1}{2} O_{2} \to H_{2} O, △ H = - 227.3 k J

The enthalpy of formation of sulphuric acid at 298 K will be:

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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